---
title: "Au B Sc Iii Chemistry Iiok"
book: "test"
category: "MA"
publisher: "Ratan Prakashan Mandir Pvt. Ltd."
type: "Educational Material"
---
According to Latest Syllabus
For Dr. Bhimrao Ambedkar University (A.U.) Examination
# VIKAS
## GUIDE BOOK
B.Sc. III
### CHEMISTRY-II
Km. Pankaj
Published by
Ratan Prakashan Mandir Pvt. Ltd.
2nd Floor, Centre Plaza, Parinay Kunj, Lajpat Kunj Marg, Agra-282002
Copyright Authors & Publishers
Published by
#### Ratan Prakashan Mandir Pvt. Ltd.
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ISBN: 978-93-7940-938-6
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B. Sc. III Chemistry II (Organic Chemistry)
Syllabus
Unit I
- (a) Spectroscopy : U. V. Spectroscopy and its application in organic chemistry. I. R. Spectroscopy and its applications in organic chemistry. N. M. R. Spectroscopy and its applications in organic chemistry.
- (b) Synthetic Dyes : Colour and constitution (electronic concept). Classification of dyes. Chemistry and synthesis of Methyl orange, Congo red, Malachite green, Crystal violet, Phenolphthalein, Flourescein, Alizarin and Indigo.
Unit II
- (a) Organometallic Compounds : Organomagnesium compounds : the Grignard reagent-formation, structure and chemical reactions, Organozinc compounds : formation and chemical reactions. Organolithium compounds : formation and chemical reactions.
- (b) Organosulphur Compounds : Nomenclature, structural features, Methods of formation and chemical reaction of thiols, thioethers, sulphonic acids, sulphonamides and sulphaguanidine.
Unit III
- (a) Haterocyclic Compound : Introduction : Molecular orbital picture and aromatic characteristics of pyrrole, furan thiophene and pyridine. Methods of synthesis and chemical reactions with particular emphasis on the mechanism of electrophilic substitution. Mechanism of Nucleophilic substitution reactions in pyridine derivatives. Comparison of basicity of pyridine, piperidine and pyrrole.
Introduction to condensed five and six membered heterocycles. Preparation and reactions of indole, quinoline and isoquinoline with special reference to Fisher-indole synthesis, Skraup synthesis and Bischler-Napieralski synthesis. Mechanism of electrophilic substitution reaction of indole, quinoline and isoquinoline.
- (b) Fats, Oils and Detergents : Natural fats, edible and industrial oils of vegetable origin, common fatty acids, glycerides, hydrogenation of unsaturated
oils. Saponification value, iodine value, acid value. Soaps, synthetic detergents, alkyl and aryl sulphonates.
Unit IV
Carbohydrates : Classification and nomenclature. Monosaccharides, mechanism of osazone formation, interconversion of glucose and fructose, chain lengthening and chain shortening of aldoses. Configuration of monosaccharides. Erythro and threo diastereomers. Conversion of glucose into mannose. Formation of Glycosides, ethers and esters. Determination of ring size of monosaccharides. Cyclic structure of D (+) glucose. Mechanism of mutarotation.
Structure of ribose and deoxyribose.
An introduction to diasaccharides (maltose, sucrose and lactose) and polysaccharides (starch and cellulose without involving structure determination).
Unit V
- (a) Amino Acids, Peptides, Proteins and Nucleic Acids : Classification, structure and stereochemistry of amino acids. Acid-base behaviour, isoelectric point and electrophoresis. Preparation and reactions of α-amino acids.
Structure and nomenclature of peptides and proteins. Classification of proteins. Peptide structure determination, end group analysis, selective hydrolysis of peptides. Classical peptide synthesis, solid-phase peptide synthesis. Structures of peptides and proteins. Levels of protein structure. Protein denaturation/renaturation.
Nucleic acids : Introduction. Constituents of nucleic acids. Ribonucleosides and ribonucleotides. The double helical structure of DNA.
- (b) Synthetic Polymers : Addition or chain-growth polymerisation. Free radical vinyl polymerization, ionic vinyl polymerisation, Ziegler-Natta polymerisation and vinyl polymers.
Condensation or step growth polymerisation. Polyesters, Polyamides, phenol-formaldehyde resins, urea-formaldehyde resins, epoxy resigns and polyurethanes.
Natural and synthetic rubbers.
Question Index
Q. No.
P. No.
Section ‘A’
Multiple Choice Type Questions
1 to 207 11-31
Very Short Answer Type Questions
1 to 50 32-42
Section ‘B’
Short Answer Type Questions
1\.
Draw the structure of a compound with each of the following molecular formulae that will show only one peak in its NMR spectrum :
(a) C3H6Cl2 (b) C5H12 (c) C2H6O (d) C4H6 42
2\.
Give a structure consistent with the following NMR data :
Molecular formula = C10H14
(a) Singlet at δ 1.30, 9H (b) Singlet at δ 7.28, 5H 42
3\.
4\.
Write a short note on saponification value. 42
How many signals (ignoring the splitting patterns) would you see in the NMR spectra of the following compounds ?
- (a) Butanone (b) Cyclobutane
- (c) p-xylene (d) 2-Propanol 43
5\.
6\.
What do you mean by a dye ? 43
Give the chemistry of crystal voilet. Also give their methods of preparation and uses. 43
7\.
8\.
What are organometallic compounds ? Give examples. 45
What happens when
- (i) Grignard reagent reacts with acetone.
- (ii) Grignard reagent reacts with ethyl orthoformate. 46
9\.
A chemist prepared the two isomeric ketones (A) and (B), placed them in separate flasks, but forgot to label them. How could you differentiate the two by UV spectroscopy 46
10\.
11\.
12\.
13\.
14\.
What are polysaccharides ? 47
Write a short note on glycosidic linkage. 47
What are amino acids ? 47
What is the effect of heat on α-amino acids ? 48
Suggest the structure of a compound each with the following NMR structural features ?
- (a) An alcohol with two NMR peaks
- (b) A compound C5H10 with a single NMR peak
- (c) A compound C4H6 with a single NMR peak. 49
- 15\. Differentiate between primary and secondary structures of proteins. 49
- 16\. Discuss the principle of IR spectroscopy. How will you distinguish between the following pairs of compounds on the basis of IR spectroscopy :
- (a) Ethyl alcohol and diethyl ether,
- (b) Acetic acid and ethyl acetate ? 49
- 17\. How would you distinguish between the following pairs of compounds by IR spectroscopy ?
- (a) CH3CH2CH2N(CH3)2 and CH3CH2CH2NH2
- (b) CH3CH2CH2COOH and CH3CH2COOCH3
- (c) CH3CH2COCH3 and CH3CH2COOCH3 49
- 18\. Two compounds, (A) and (B), have the same molecular formula C2H6O. They have different IR spectra. Compound (A) shows a strong/broad absorption at 3400 cm–1, while compound (B) does not. Suggest formulae for (A) and (B) which account for the difference. 50
- [19. The UV spectrum for a compound with formula C3H6O shows a weak absorption band at 280 nm. The NMR spectrum shows only one signal, a singlet. What is the structure of the compound ?50](#bookmark16)
- [20. What do you mean by spectroscopy ? Give principle.50](#bookmark17)
- [21. What are soaps ? Give their types.51](#bookmark18)
- [22. Write a short note on cleanising action of soap.52](#bookmark19)
- [23. Give the constitution of isoquinoline.52](#bookmark20)
- [24. What do you mean by mutarotation ? Give its mechanism.53](#bookmark21)
- [25. Write a short note on iso-electric point of amino acids.54](#bookmark22)
- 26\. Give the various colour tests used for the identification of protein. 55
- [27. What is the relation between amino acids and proteins ?55](#bookmark23)
- [28. Write a short note on denaturation of proteins.56](#bookmark24)
- [29. What are polymers ?57](#bookmark25)
- 30\. How the following polymers are synthesised ? Give their important uses.
- [(i) Nylon (ii) Cellulloid.58](#bookmark26)
- 31\. Predict the signal pattern of the CH3 protons in the NMR spectra of the following compounds :
O
(a) CH3CHBr2
II
- (b) CH –C–OH
OH
II
- [(c) CH3CH2Br (d) CH3–CH–CH358](#bookmark27)
- 32\. How will you distinguish between 1, 3-pentadiene and 1, 4-pentadiene
[by UV spectroscopy ?59](#bookmark28)
- [33. How will you distinguish between benzene and anthracene by UV spectroscopy ?59](#bookmark29)
- [34. Two isomeric dienes (A) and (B) having the molecular formula C5H8 absorb at lmax 223 nm and lmax 178 nm respectively. Write the structures of the two isomers.59](#bookmark30)
- 35\. Arrange the following compounds in the increasing order of their UV absorption maxima :
- (a) Ethylene (b) Naphthalene
[(c) Anthracene (d) 1, 3-Butadiene59](#bookmark31)
- 36\. Give a structure consistent with the following NMR data :
Molecular formula = C3H5Cl3
- (a) Singlet at δ 2.20, 3H (b) Singlet at δ 4.02, 2H 59
- [37. The NMR spectrum of compound (A), C5H12, gives only one signal, a singlet. What is the structure of (A) ?59](#bookmark32)
- 38\. Suggest a structure consistent with the following NMR data : Molecular formula : C9H12
- (a) Singlet at δ 6.78, 3H (b) Singlet at δ 2.25, 9H 60
- 39\. How would you distinguish between the following pair of compounds
by NMR spectroscopy ?
O
,.
(a) CH3–C–CH3 and
O
II
CH3CH2–C–H
O
(b) II
(b) CH3–C–CH3
and
O
II
CH –C–OCH
60
- 40\. The NMR spectrum of compound (A) C2H6O, shows one signal only, a singlet. Deduce the structure of (A). 60
- 41\. (a) Mixture of furan and ammonia is treated with steam in presence of
Al2O3.
- (b) Pyroll is heated with chloroform and caustic potash.
- (c) Pyroll reacts with diazonium salts. 60
- 42\. What do you mean by oils and fats ? Give the difference between them.
61
- 43\. Write short notes on the following :
- (a) Hydrogenation and Hydrogenolysis
- (b) Hydrolysis of oils and fats
- [(c) Rancidification.62](#bookmark33)
- [44. What do you mean by invert sugar ?63](#bookmark34)
- [45. Write a short note on epimerisation.64](#bookmark35)
- [46. Give the mechanism of osazone formation.65](#bookmark36)
- [47. Write short note on reformat sky reaction.66](#bookmark37)
- [48. What do you mean by thioethers ? How they are named?67](#bookmark38)
- [49. Give the oxidation reactions of thioalcohols.67](#bookmark39)
- [50. What are thioalcohols ? Give two methods for their preparation.68](#bookmark40)
- [51. Give the reaction of aldehyde and ketones over thiols.68](#bookmark41)
- [52. Write a short note on Sulphaguanidine.69](#bookmark42)
- [53. What do you mean by sulphonation ? Give its mechanism.69](#bookmark43)
- 54\. What are heterocyclic compounds? Explain with suitable example. 70
- 55\. (a) Pyroll reacts with hydroxyl amine.
- (b) Pyridine reacts with mercuric and platinum salts.
- [(c) Pyridine is treated with halogens.71](#bookmark44)
- 56\. What happens when ?
- (a) Thiophene reacts with acetyl chloride in presence of SnCl4.
- (b) Thiophene is shaked with conc. H2SO4 and a crystal of its atom. 72
- [57. What is Grignard’s reagent. How it is prepared ?72](#bookmark45)
- [58. How organozinc compounds are prepared ?73](#bookmark46)
- 59\. How many NMR signals do you expect from each of the following compounds ? Indicate also the splitting pattern of the various signals.
- [(a) CH3OCH3 (b) CH3OCH2CH3 (c) CH3CH2OH73](#bookmark47)
- [60. How will you distinguish between the three dibromobenzenes by their NMR spectra ?74](#bookmark48)
- [61. Write a note on auxochrome.74](#bookmark49)
- [62. What do you mean by chromophore and chromogen ?74](#bookmark50)
- 63\. Select chromophore, chromogen and auxochrome from the following :
- (i) (oy n=n xcx nCh
(ii) NaSO
_files/AU20B.20Sc.20III20Chemistry20II(OK)-1.png)CH3
CH,
Section ‘C’
Essay Type Questions
- 1\. Give the α-helix structure and β-flat sheet and β-pleated sheet structure for proteins. 75
- 2\. Give the constituent of nucleic acids.
Or
What do you mean by nucleosides and nucleotides.
Or
[Write short note on nucleosides and nucleotides.77](#bookmark51)
- [3. What are proteins ? How they are classified ? Discuss in brief.80](#bookmark52)
- [4. What are condensation polymers? Give their mechanism.81](#bookmark53)
- [5. What do you mean by formaldehyde resins.82](#bookmark54)
- [6. Explain the mechanism of addition polymerisation.85](#bookmark55)
- 7\. How following polymers are synthesised ? Give their important
application.
- (i) Polythene
- (ii) Polystyrene
- [(iii) Terylene86](#bookmark56)
- [8. What do you mean by ultra-violet spectroscopy ? Give its applications in organic chemistry.87](#bookmark57)
- 9\. (a) Give the Skrup’s synthesis of quinoline. Also discuss its following properties :
- (i) Oxidation (ii) Reduction (iii) Nitration
- [(b) Describe the structure of quinolene.89](#bookmark58)
- [10. What do you mean by IR spectroscopy ? Give the applications to organic compounds.91](#bookmark59)
- [11. Give the modern theories of colour and constitution.94](#bookmark60)
- [12. Classify the dyes on the basis of their application.96](#bookmark61)
- 13\. How the following dyes are synthesized ? Explain in brief.
- (a) Malachite Green (b) Methyl Orange
- [(c) Congo Red (d) Phenolphthalein97](#bookmark62)
- [14. What do you mean by synthetic detergents ? Explain with suitable examples.99](#bookmark63)
- 15\. Give the chemistry of following dyes. Also give their methods of preparation and uses.
- (a) Fluorescene
- (b) Alizarin
- (c) Indigo 101
- 16\. Give preparation and properties of benzene sulphonic acid. 104
- 17\. How followings are prepared from Grignard’s reagents :
- (a) Primary, Secondary and Tertiary alcohols
- (b) Alkanes
- (c) Esters
- (d) Ethyl, methyl ketone. 107
- 18\. How pyridine is synthesized from acrolien ? Give its properties and structure. 109
- 19\. How isoquinoline is synthesised from Bischler-Naiperalski method ? Discuss the following properties of iso-quinoline :
- (i) Basic nature of nitrogen
- (ii) Reduction
- (iii) Oxidation
- (iv) Electrophilic substitution
- (v) Nucleophilic substitution 112
- 20\. Give the preparation and important properties of thiophene. 114
- 21\. How would you convert the following :
- (a) Aldose into ketose or glucose into fructose
- (b) Ketone into aldose or fructose into glucose
- (c) Aldosein to next higher aldose
- [(d) Aldose into the next lower aldose.116](#bookmark64)
- [22. Give the Cyclic structure of D-glucose.118](#bookmark65)
- [23. Discuss the method of preparation and important properties of α-amino acid.120](#bookmark66)
Section ‘A’
Multiple Choice Type Questions
- 1\. Which among the following is the structure of sulphaguanidine ?
NH
(a)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-2.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-3.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-4.jpg)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-5.png)
(d)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-6.png)
NH
Il
SO2NH–C–NH2
- 2\. Which among the following is not a heterocyclic compound :
CH – CH
|| ||
- (a) CH CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-7.png)
HC
- (b) |
HC
C – CHO
||
CH
O
CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-8.png)
CH – CH
HC
(c) |
HC
\\
CH || CH
N
||
(d) CH
||
CH
S
- 3\. Pyroll with KOH forms potassium salt. By this reaction the......nature of
pyroll is confirmed :
- (a) acidic
(c) amphoteric
(b) basic
(d) allotrope.
12
4.
Vikas, 2009 (A. U. )
The molecular formula of oxazole is :
CH – CH
|| ||
(a) CH CH
(b)
CH – N
|| ||
CH CH
NH
O
CH
HC
(c) |
HC
\\
CH || CH
(d)
N
CH
HC
|
HC
\\
N
N || CH
5\.
Which of the following compounds does not follow Huckel rule ?
O
(d)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-9.png)
N
6\.
Which of the following compounds is sulphanilamide ?
(a) O2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-10.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-11.png)SO2H
(b) NH
SO2NH2
7\.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-12.png)
Above reaction is called : (a) Friedal-Craft's reaction (c) Coupling reactions
- (b) Gattermam Koch reaction (d) Alkylation.
O
CH3COCl
AlCl3
O COCH3
above reaction is called :
(a) Friedal-Craft’s reaction
- (c) Wurtz reaction
(b) Sandmayer reaction (d) Fitting reaction.
- B. Sc. III Chemistry II
- 9. Which among the following compounds is heterocyclic ?
(a) N j (b) 0O
(c) s (d) All of these.
- 10. Teflon is prepared by the polymerization of :
- (a) Butadiene (b) Vinyl cyanide
- (c) Vinyl chloride (d) Tetrafluoroethylene.
- 11. Bakelite is obtained from :
- (a) Phenol and formaldehyde
- (b) Adipic acid and hexamethylene diamine
- (c) Dimethyl terephthalate and ethylene glycol
- (d) Neoprene.
- 12. Nylon-6, 6 is obtained from :
- (a) Adipic acid and hexamethylene diamine
- (b) Tetrafluoroethylene
- (c) Vinyl cyanide
- (d) Vinyl benzene.
- 13. Natural rubber is a polymer of :
- (a) Propene (b) Isoprene
- (c) Formaldehyde (d) Phenol.
- 14. Neoprene is a polymer of the following monomer :
- (a) Chloroprene (b) Isoprene
- (c) Isobutane (d) Isopentene.
- 15. Which of the following is a thermosetting polymer ?
- (a) Bakelite (b) Nylon-6, 6
- (c) Polyethylene (d) Teflon.
- 16. Which of the following is an example of a condensation polymer ?
- (a) Nylon-6, 6 (b) Teflon
- (c) Polypropylene (d) Orlon.
- 17. Orion is prepared by the polymerization of :
- (a) Vinyl cyanide (b) Allyl alcohol
- (c) Vinyl chloride (d) Allyl chloride.
- 18. The mutarotation of glucose is characterised by :
- (a) a change from an aldehyde to ketone sturucture
- (b) a change of specific rotation from a (+) to a (–) value
- (c) the presence of an intramolecular bridge strucutre
- (d) the irreversible change from α-D to the β-D form.
- 19. Starch :
- (a) is a trisaccharide
- (b) is also called amylose
- (c) is also called amylopectin
- (d) is a mixture of amylose + amylopectin.
- 20. α-D-glucopyranose is a (n) :
- (a) hemiacetal (b) hemiketal
- (c) acetal (d) ketal.
- 21. Which among the following is monosaccharide ?
- (a) sucrose (b) maltose
- (c) galactose (d) cellulose.
- 22. Which of the following statement is false about cellulose ?
- (a) it is a polymer of glucose molecules joined in β-1, 4 linkages
- (b) it is a major component of cotton
- (c) it is used in the manufacture of dacron fibres
- (d) it is used in the manufacture of rayon fibres.
- 23. Which of the following compounds reduces Tollen’s reagent ?
- (a) glucose (b) sucrose
- (c) methanol (d) acetic acid.
- 24. The monosaccharide obtained by hydrolysis of strach is :
- (a) D-glucose (b) maltose
- (c) D-galactose (d) D-ribose.
- 25. Glycine is a unique amino acid because it :
- (a) has no chiral carbon
- (b) has a sulphur containing R group
- (c) cannot form a peptide bond
- (d) is an essential amino acid.
- 26. Which of the following organic ions results when glycine is treated with concentrated HCl ?
- (a) NH3CH2COOH (b) NH2CH2COO–
- (c) NH3CH2COO– (d) HOCH2COO–.
- 27. The name of the compound is :
- (a) Furon (b) Hydrofuron
- (c) Tetrahydrofuron (d) Dihydrofuron.
CH = CH
- 28. Structural formula | O is of :
- (a) thiophene
(b) furan
(d) pyridine.
(c) pyroll
- 29. Which of the following compounds reduces Tollen’s reagent ?
- (a) Glucose (b) Sucrose
- (c) Methanol (d) Acetic acid.
- 30. The reagent that can be used to differentiate and aldose and a ketose is :
- (a) Bromine water (b) Fehling’s solution
- (c) Tollens’ reagent (d) None of these.
- 31. Which of the following products is not derived from cellulose ?
- (a) Rayon (b) Insulin
- (c) Gun cotton (d) Paper.
- 32. Which is a disaccharide ?
- (a) Ribose (b) Fructose
- (c) Sucrose (d) Glucose.
- 33. Glucose can’t be classified as :
- (a) a hexose (b) an oligosaccharide
- (c) an aldose (d) a monosaccharide
- 34. Which of the following statements is false about glucose ?
- (a) It is a reducing sugar (b) It is a disaccharide
- (c) It has a pyranose form (d) It is a poly alcohol.
- 35. Monosaccharides are cassified according to :
- (a) the number of carbon in the molecule
- (b) whether they contain an aldehyde or a ketone group
- (c) their configuration relationship to glyceraldehyde
- (d) all of the above.
- 36. All of the following monosaccharides give the same osazone except :
- (a) Galactose (b) Glucose
- (c) Fructose (d) Mannose.
- 37. Mutarotation is a term related to :
- (a) Interconversion of anomers
- (b) Relationship of D-and L-families
- (c) Hydrolysis of sucrose
- (d) Number of monosaccharides in a carbohydrate.
- 38. Which of the following is the main structural feature of proteins ?
- (a) Peptide linkage (b) Ester linkage
- (c) Ether linkage (d) α, β-linkage.
- 39. The linear arrangement of amino acid units in proteins is called :
- (a) primary structure (b) secondary structure
- (c) tertiary structure (d) quaternary structure.
- 40. When glycine is heated, it forms :
- (a) Diketopiperazine (b) Acrylic acid
- (c) Butyric acid (d) Butyrlactam.
16
Vikas, 2009 (A. U. )
41\.
Simple lipids are :
- (a) Triglycerides (b) Phosphotriglycerides
- (c) Ester of lower acids (d) None of these.
42\.
Antibodies are :
- (a) Carbohydrates (b) Proteins
- (c) Lipids (d) Enzymes.
43\.
The energy stored in the cells of a living body is in the form of :
- (a) Fats (b) Glucose
- (c) ATP (d) Proteins.
44\.
A ‘base-sugar-phosphate’ unit in nucleic acid is called :
- (a) Phosphotide (b) Base phosphate
- (c) Nucleoside (d) Nucleotide.
45\.
The five elements present in most naturally occurring proteins are :
- (a) C, H, O, P and S (b) N, C, H, O and I
- (c) N, S, C, H, and O (d) C, H, O, S, and I
46\.
The nitrogen content of proteins can be quantitatively determined by :
- (a) Carius method (b) Keldahl’s method
- (c) Victor Meyer’s method (d) Rast method.
47\.
An aqueous solution of glycine is neutral because of the formation of :
- (a) Carbanion (b) Zwitter ion
- (c) Carbonium ions (d) Free radicals.
48\.
Which one of the following compounds form Zwitter ions ?
- (a) carbonyl compounds (b) amino acids
- (c) phenol (d) heterocyclic compounds.
49\.
Vapours of acetylene and HCN on passing in red-hot tube gives :
- (a) Pyridine (b) Quinolene
- (c) Pyroll (d) Indole.
50\.
The nature of pyridine is :
- (a) Acidic (b) Basic
- (c) Neutral (d) None of these.
51\.
The nitrozen atom present in pyridine is :
- (a) Primary (b) Secondary
- (c) Tertiary (d) Quarternary.
52\.
The reaction of pyridine and sodamide is called :
- (a) Arylation (b) Chichibabine reaction
- (c) Fischer reaction (d) William’s reaction.
53\.
Soap is :
- (a) a mitxture of slats of fatty acids
- (b) a salt of glycerol
- (c) a mixture of ethers
- (d) a mixture of aromatic ethers.
- 54. Fats differs from waves in that fats have :
- (a) more unsaturation (b) higher melting points
- (c) a glycerol backbone (d) longer fatty acids.
- 55. Fatty acids are :
- (a) unsaturated dicarboxylic acids
- (b) long chains alkanoic acid
- (c) aromatic carboxylic acids
- (d) aromatic dicarboxylic acids.
- 56. Fast and oils are :
- (a) monoesters of glycerol (b) diesters of glycerol
- (c) triesters of glycerol (d) diesters of glycol.
- 57. The degree of unsaturation of a fat can be determined by means of its :
- (a) Iodine number (b) Octane number
- (c) Saponification number (d) Melting point.
- 58. Oleic acid is a fatty acid contianing :
- (a) 12 carbons (b) 14 carbons
- (c) 16 carbons (d) 18 carbons.
- 59. Sodium or potassium salts of fatty acids are called :
- (a) Proteins (b) Terpenes
- (c) Carbohydrates (d) Soaps.
- 60. Alkaline hydrolysis of oils (or fats) is called :
- (a) Saponification (b) Fermentation
- (c) Diazotisation (d) Rancidification.
- 61\. Which of the following gives a tertiary alcohol when treated with Grignard reagent ?
O
Il
- (a) H–C–H
II
(b) CH3–C–H
(c)
CH
CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-13.png)C = O
(d) None of these.
- 62. Ketones react with Grignard reagents to form an addition product which on hydrolysis gives a :
- (a) Primar alcohol (b) Tertiary alcohol
- (c) Secondary alcohol (d) Ketone.
- 63. Phenyl magnesium bromide reacts with acetaldehyde to form an addition product which undergoes acid hydrolysis to give :
- (a) Diphenyl carbinol (b) Benzyl alcohol
- (c) Methyl phenyl carbinol (d) Benzoic acid.
- 64\. Which among the following is not an organo metallic compound ?
- (a) C2H5MgBr (b) (C2H5)3Zn
- (c) C2H5ONa (d) C2H5Li.
- 65\. Which of the following is least soluble in water ?
- (a) CH3OH (b) CH3CH2OH
- (c) CH3SH (d) HOCH2CH2OH.
- 66\. Which of the following compounds will react with methyl magnesium bromide to give tert-butyl alcohol ?
- (a) Acetyl chloride (b) Acetone
- (c) Isopropyl alcohol (d) Acetaldehyde.
- 67\. Which is the best reagent to accomplish the following conversion ?
CH3CH2Br----—^ CH3CH3
- (a) conc. H2SO4 (b) Na
- (c) conc. HCl (d) Mg, then H2O.
- 68\. What is the major product of the following reaction ?
O
|| H O/H+
CH3CH2 - C - H + CH3MgBr------> X —-----> Y.
- (a) 1-Butanol (b) Butanal
- (c) 2-Butanol (d) Butanone.
- 69\. Which of the following compounds will react with methyl magnesium iodide followed by acid hydrolysis to give ethyl alcohol ?
- (a) Ethylene (b) Acetaldehyde
- (c) Formaldehyde (d) Acetone.
- 70\. Ethyl magnesium iodide reacts with formaldehyde to give a product which on acid hydrolysis forms :
- (a) an aldehyde (b) a primary alcohol
- (c) a ketone (d) a secondary alcohol.
- 71\. The structural formula of DMSO is :
O
II
- (a) CH –S–CH
33
O
(c) CH3–S–CH3
O
OH
I
- (b) CH3–S–CH3
OH
(d) None of these.
- 72\. The electronic spectra of molecule requires energy for electronic transition corresponding to wavelengths :
(a) 100 nm-800 nm (b) 400 nm-500 nm
- (c) below 20 nm (d) None of these.
73\.
A molecule can absorb IR radiation only when :
- (a) Its natural frequency coincides with incident IR radiation.
- (b) There is change in its dipole moment.
- (c) When the molecules are symmetrical.
- (d) All of the above.
74\.
The shif of absorption band towards higher wavelength is called :
(a) Red shift (b) Blue shift
(c) Hypochromic (d) Hyperchromic.
75\.
O2 and N2 molecules are IR inactive because :
- (a) They have zero dipole (b) They show no vibrations
- (c) They are linear molecules (d) None of these.
76\.
Which of the following electromagnetic radiation has higher frequency ?
- (a) Radio wave (b) Infrared
- (c) Ultraviolet (d) X-rays.
77\.
The infrared region lies between wavelengths :
(a) 400 nm-800 nm (b) Below 400 m
(c) 350 nm -800 nm (c) 100 nm-200 nm.
78\.
Which of the factor lowers down the frequency in the IR absorption spectrum ?
- (a) Low bond length (b) Hydrogen bonding
- (c) Low electronegativity (d) All of these.
79\.
In the IR spectrum of conjugated molecule, the stretching frequency decreases due to :
- (a) Decrease in bond length (b) Disappearance of conjugation
- (c) Increase in bond length (d) All of these.
80\.
Which of the following relationship is correct regarding molecular energy levels :
- (a) electronic vibrational rotational
( ) rotational vibrational electronic
- (c) electronic rotational vibrational
() vibrational electronic rotational.
81\.
The designation D or L before the name of a monosaccharide :
- (a) indicate the direction of rotation of polarised light
- (b) indicate the length of the carbon chain in the carbon hydrates
- (c) indicate the position of the OH group on the carbon next to the primary alcoholic group
- (d) indicate the position of the asymmetric carbon atoms in the carbohydrate.
82\.
Which of the following carbohydrates is not a reducing sugar :
- (a) glucose (b) sucrose
- (c) fructose (d) lactose.
20
Vikas, 2009 (A. U. )
83\.
Which of the following product is not derived from cellulose ?
- (a) rayon (b) insulin
- (c) gun cotton (d) paper.
84\.
Which of the following carbohydrates will not give a precipitate of Cu2O when heated with Benedict solution ?
- (a) maltose (b) glucose
- (c) sucrose (d) fructose.
85\.
α-D-Glucopyranose is a(n) :
(a) hemiacetal (b) hemiketal
(c) acetal (d) ketal.
86\.
Which of the following statements is false about glyceraldehyde ?
- (a) Its IUPAC name is 1, 2-dihydroxypropanal
- (b) It is isomeric with 1, 3-dihydroxypropanone
- (c) It is optically active
- (d) It shows mutarotation.
87\.
Common table sugar is :
- (a) Glucose (b) Sucrose
- (c) Fructose (d) Maltose.
88\.
A zwitterion is :
- (a) an ion that is positively charged in solution
- (b) an ion that is negatively charged in solution
- (c) a compound that can ionise both as a base and an acid.
- (d) a carbohydrate with an electrical charge.
89\.
A zwitter ion has which of the following properties :
- (a) no net charge (b) a high melting point
- (c) soluble in water (d) all of these.
90\.
The pH at which the amino acid shows no tendency to migrate when placed in an electrical field is known as its :
- (a) Isoelectric point (b) Dipole moment
- (c) Iodine number (d) Wavelength.
91\.
Ninhydrin test is given by :
- (a) Carbohydrates (b) Proteins
- (c) Alkanes (d) Alkenes.
92\.
Which of the following tests is not used for testing proteins ?
- (a) Ninhydrin test (b) Biuret test
- (c) Xanthoproteic test (d) Tollen’s test.
93\.
A protein solution on warming with concentrated HNO3 may turn yellow. This test is called :
- (a) Xanthoproteic test (b) Ninhydrin test
- (c) Biuret test (d) Million’s test.
- 94. Glycine is :
- (a) NH2CH2COOH (b) NH2CH2CH2CH2CH2NH2
- (c) NO2CH2CH2COOH (d) BrCH2COOH.
- 95. Glycine reacts with nitrous acid to form :
- (a) Glycolic acid (b) Diketopiperazine
- (c) Methylamine (d) Ethyl alcohol.
- 96. The double helical structure of DNA is held together by :
- (a) sulphur-sulphur linkages (b) peptide bonding
- (c) hydrogen bonding (d) glycosidic bonds.
- 97. Which of the following statements is false about an aldohexose ?
- (a) It is a monosaccharide
- (b) It contains a potential aldehyde group
- (c) α-D-Glucopyranose is an aldohexose
- (d) Fructose is an aldohexose.
- 98. Which among the following is a disaccharide ?
- (a) glucose (b) maltose
- (c) fructose (d) starch.
- 99. Which of the following carbohydrates is sweeter than sucrose ?
- (a) Glucose (b) Fructose
- (c) Lactose (d) None of these.
- 100\. By approximately what factor is the sweetness of saccharin greater
than that of sugar ?
- (a) 5
- (c) 500
- 101\. A reducing sugar will :
- (a) react with Fehling’s solution
- (c) have fewer calories
- 102\. The principal sugar in blood is :
- (a) glucose
- (c) sucrose
(b) 50
(d) 1000.
(b) not react with Fehling’s solution
- (d) always be a ketone.
(b) fructose
(d) glucose.
- 103\. Which among the following is not a monosaccharide ?
- (a) ribose (b) fructose
- (c) sucrose (d) glucose.
- 104\. The number of asymmetric carbon atome in the α-D-glucopyransoe molecule is :
- (a) 2 (b) 3
- (c) 4 (d) 5
- 105\. All of the following monosaccharides give the same osazone except :
- (a) galactose (b) glucose
- (c) fructose (d) mannose.
- 106\. Starch :
- (a) is a trisaccharide
- (b) is also called amylose
- (c) is also called amylopectin
- (d) is a mixture of amylose + amylopectin.
107\.
108\.
109\.
110\.
111\.
112\.
113\.
114\.
115\.
116\.
117\.
Which among the following statements is wrong (a) All the dyes are coloured.
- (b) All coloured compounds are not dyes
- (c) Dye remain ineffected by atmospheric changes
- (d) Dyes are insoluble in water.
Which among the following is not an azo dye
(a) Bismark brown
(c) Methyl red
- (b) Methyl orange
(d) Malachite green.
Which among the following is a vet dye
- (a) Indigo
(c) Alizarin
Mordant dye is a :
(a) Phenolphthalein
- (c) Alizarin
(b) Malachite green
(d) Aro-dye.
(b) Malachite green (d) Anthraquinone.
Which among the following is not an auxochrome ?
- (a) – NH2 (b) – NH2
- (c) – OH (d) – NO2.
Which is a chromophere ?
- (a) – NO (b) – SO3H
- (c) – COOH (d) – NH2.
Which dye is used as an pergetive ?
- (a) Alizarin (b) Pararosanline
- (c) Picric acid (d) Florescence.
Alizarin with which cation will give red colour to the cloth ?
- (a) Fe3+ (b) Cr3+
- (c) Al3+ (d) Ba2+.
Which is not used in the use of mordant dye ?
- (a) Acid (b) Metal ion
- (c) Phenol (d) Benzene.
The reason for the colour of an organic compound :
- (a) chromophase (b) oxochrome
- (c) resonance (d) a and b both.
Liquid oils can be converted to solid fats by :
- (a) Hydrogenation (b) Saponification
- (c) Hydrolysis (d) Oxidation or double bonds.
B. Sc. III Chemistry II 23
- 118\. Both stearic and linoleic acid have 18 carbons. Linoleic acids is
unsaturated while stearic is saturated. The melting point of stearic acid :
- (a) is higher than linoleic acid
- (b) is lower than linoleic acid
- (c) is same as linoleic acid
- (d) cannot predict, insufficient information.
- 119\. Saponification of a fat :
- (a) always results in the formation of insoluble soaps
- (b) produces glycerol and soaps
- (c) is used in the production of detergents
- (d) is used in the production of lactic acid.
- 120\. α-D-glucose is different from β-D-glucose :
- (a) in the configuration at C-1
- (b) because they are mirror images of each other
- (c) because they are enantiomers
- (d) because they are geometrical isomers.
- 121\. Upon hydrolysis, proteins give :
- (a) Amino acids (b) Hydroxy acids
- (c) Fatty acids (d) Alcohols.
- 122\. Complete hydrolysis of proteins produces :
- (a) Ammonia and carbon dioxide (b) Urea and uric acid
- (c) A mixture of amino acids (d) Glycogen and a fatty acid.
- 123\. Proteins are :
- (a) polyamides (b) polymers of ethylene
- (c) α-Aminocarboxylic acids (d) Polymers of propylene.
- 124\. Which of the following polymers contain nitrogen :
- (a) PVC (b) Teflon
- (c) Nylon (d) Terylene.
- 125\. Adipic acid reacts with hexamethylene diammine to form :
- (a) Bakelite (b) Nylon-6, 6
- (c) Terylene (d) Nylon-6, 8.
- 126\. Ethylene glycol reacts with dimethyl terephthalate to form :
- (a) Nylon-6,6 (b) Teflon
- (c) Dacron (d) Orlon.
- 127\. The monomers for Buna-S are 1, 3-butadiene and :
- (a) Ethylene glycol (b) Adipic acid
- (c) Styrene (d) Caprolactum.
- 128\. Which of the following statements is not true ?
- (a) Natural rubber is a hydrocarbon
- (b) Natural rubber is made of isoprene units
(c) Natural rubber is a polymer of 1, 3-Butadiene
(d) Natural rubber can be vulcanized.
129\. Which of the following is an example of condensation polymer
(a) Polythene
(c) Orlon
(b) PVC
(d) Terylene.
130\. Rotational spectra are observed in the :
(a) Near infrared region
(c) Visible region
(b) Far infrared region (d) Ultraviolet region.
- 131. Frank-Condon principle is related to :
- (a) Time required for electronic transition to occur
- (b) Absorption of light
- (c) Time of electronic transition and change in inter-nuclear distance
- (d) None of these.
- 132\. Which of the transitions need larger energy ?
- (a) o ^ o\*
- (c) n ^ o\*
(b) n ^ n\*
(d) o ^ n\*.
133\. Which among the following is a natural dye
(a) Phenolphthalein
(c) Alizarin
(b) Indigo
(d) Orange-I.
134\. Which among the following is a direct dye
(a) Malachite green
(c) Congo-red
(b) Alizarin
(d) Martines yellow.
135\. Which of the following is an azodye
(a) Orange-I
(c) Indigo
(b) Malachite green (d) Alizerin.
136\. Triphenyl methane dye is :
- (a) Alizarin
(c) Martius
- 137\. Phthaline dye is :
(b) Methyl orange
(d) Malachite green.
(a) Phenolphthalein
(c) Martius yellow
(b) Methyl orange
(d) Malachite green.
138\. A dye absorbs wavelength of blue coloured light. The colour of the dye will be :
(a) Blue
(c) Green
(b) Red
(d) Yellow.
139\. Where malachite green is not used
- (a) in dyeing wool
- (c) In the identification of cerium
(b) In dyeing tissues (d) as a pergative.
140\. Synthesis of indigo starts with :
(a) Aniline
(c) Phthalic anhydride
(b) Anthraquinone
(d) Alizarine.
- 141. Pyroll on reaction with benzene diazonium chloride gives 2-phenyl azopyroll. This reaction is called :
- (a) coupling (b) addition
- (c) displacement (d) elimination.
- 142\. Sodium succinate on heating with phosphorus tri-sulphide gives :
- (a) Pyroll (b) Furan
- (c) Thiophene (d) Pyragallol.
- 143\. When a mixture of acetylene and H2S is passed over alumina at 400°C, we get :
- (a) Pyridine (b) Oxazole
- (c) Thiophene (d) Furan.
- 144\. Thiophene on bromination gives :
- (a) 2-bromothiophene (b) 5-bromothiophene
- (c) 2, 5-dibromothiophene (d) None of these.
- 145\. Which of the following carbohydrates will not give a red precipitate of
Cu2O when heated with Benedicts’ solution ?
- (a) Maltose (b) Glucose
- (c) Sucrose (d) Fructose.
- 146\. Common table sugar is :
- (a) glucose (b) sucrose
- (c) fructose (d) maltose.
- 147\. The reagent that can be used to differentiate an aldose and a ketose is :
- (a) Bromine water (b) Fehling’s solution
- (c) Tollen’s reagent (d) None of these.
- 148\. Which of the following statement is false about glyceraldehyde ?
- (a) its IUPAC name is 1, 2-dihydroxypropanal
- (b) it is isomeric with 1, 3-dihydroxy propanone
- (c) its is opically active
- (d) it shows mutarotation.
- 149\. Which of the following statements is false about sucrose ?
- (a) It is also called table sugar
- (b) It may be fermented by yeast to produce alcohol
- (c) It reduces Fehling’s solution
- (d) It does not reduce Tollen’s reagent.
- 150\. The sugar that yields only glucose on hydrolysis is :
(a) Lactose
(c) Maltose
(b) Sucrose
(d) Fructose.
- 151\. A reducing sugar will :
- (a) react with Fehling’s test
(b) not react with Fehling’s test
(c) have fewer calories
(d) always be a ketone.
- 152. Which of the following statements is false about α-D-glucose ?
- (a) it has a pyranose ring
- (b) it is a hemiacetal
- (c) it shows muta-rotation
- (d) it is the purest form of table sugar.
- 153\. Which of the following dyes is used in the manufacture of boot polish ?
- (a) Bismark brown (b) Melachite green
- (c) Pareosaniline (d) Eosin.
- 154\. Magenta (or Rosaniline) is used in the preparation of :
- (a) Tollen’s reagent (b) Sariff’s reagent
- (c) Phenolphthaline (d) Indigo.
- 155\. Grignard reagent do not show any reaction with :
- (a) alkoxy alkanes (b) alkanones
- (c) alkyl alkanoate (d) acyl halides.
- 156\. Grignard reagent CH3CH2.MgBr can be used to prepare :
- (a) Ethane (b) 3-ethyl-3-pentanol
- (c) Propanoic acid (d) all of the these.
- 157\. n-propyl magnesium bromide on treatment with CO2 and further hydrolysis gives :
- (a) Acetic acid (b) Propanoic acid
- (c) Butanoic acid (d) Formic acid.
- 158\. A lipid which contains glycerol, fatty acids, phosphoric acid and an alcoholic base is called :
- (a) Phosphoglyceride (b) Sphingo lipid
- (c) Glycolipid (d) None of these.
- 159\. Digestion of proteins involves :
- (a) changes in secondary structure only
- (b) cleavage of peptide linkages
- (c) removal of all carboxyl groups in the form of CO2
- (d) removal of all NH2 groups in the form of NH3.
- 160\. Irreversible precipitation of proteins caused by heating is called :
- (a) Polymerisation (b) Denaturation
- (c) Electrophoresis (d) Inversion.
- 161\. Precipitation or coagulation of proteins may be caused by :
- (a) Heat (b) Changes in pH
- (c) Heavy metal salts (d) All of these.
- 162\. Which of the following is responsible for hereditary character of the cell ?
(a) DNA
(c) Proteins
(b) RNA
(d) Hormones.
- 163. Which one of the following is not present in RNA ?
- (a) Uracil (b) Thymine
- (c) Ribose (d) Phosphate.
- 164\. The primary structure of a protein refers to :
- (a) Whether the protein is firbous or globular
- (b) The amino acid sequence in the polypeptide chain
- (c) The orientation of the amino acid side chains in space
- (d) The presence or absence of an α-helix.
- 165\. The α-helix is held in a coiled conformation partially because of :
- (a) Optical activity (b) Hydrogen bonding
- (c) Resonance (d) Delocalisation.
- 166\. The backbone of each strand of DNA is formed by :
- (a) Nucleotides (b) Sugar phosphate linkage
- (c) Nucleosides (d) None of these.
- 167\. Mutarotation is a term related to :
- (a) interconversion of anomers
- (b) relationship of D- and L-families
- (c) hydrolysis of sucrose
- (d) number of monosaccharides in carbohydrates.
- 168\. By approximately what factor is the sweetness of saccharin greater than that of sugar ?
- (a) 5 (b) 50
- (c) 500 (d) 1000.
- 169\. Which of the following carbohydrate is sweeter than sucrose ?
- (a) glucose (b) fructose
- (c) lactose (d) none of these.
- 170\. The sugar that yields only glucose on hydrolysis is :
- (a) lactose (b) sucrose
- (c) maltose (d) fructose.
- 171\. Which of the following statement is flase about sucrose ?
- (a) it is also called table sugar
- (b) it may be fermented by yeast to produce alcohol
- (c) it reduces Fehling’s solution
- (d) it does not reduce Tollen’s reagent.
- 172\. Which of the following statements is false about an aldohexose ?
- (a) it is monosaccharide
- (b) it contains a potential aldehyde group
- (c) α-D-glucopyranose is an aldohexose
- (d) fructose is an aldohexose.
- 173. The monosaccharide obtained by hydrolysis of starch is :
- (a) D-Glucose (b) Maltose
- (c) D-Galactose (d) D-Ribose.
- 174\. Which of the following statements is false about cellulose ?
- (a) It is a polymer of glucose molecules joined in β-1, 4 linkages
- (b) It is a major component of cotton
- (c) It is used in the manufacture of Dacron fibres
- (d) It is used in the manufacture of Rayon fibres.
- 175\. Which of the following carbohydrates is not a reducing sugar ?
- (a) Glucose (b) Sucrose
- (c) Fructose (d) Lactose.
- 176\. Sulphadrugs are all structurally related to :
- (a) Nitrobenzene (b) Sulphanilamide
- (c) Pyridine (d) Aniline.
- 177\. Sulphanilamide :
- (a) is a sulphur containing amino acid
- (b) is the compound from which other sulpha drugs are derived
- (c) is an orange coloured aniline dye
- (d) is commonly administered in aqueous solution.
- 178\. Which of the following has the lowest boiling point ?
- (a) CH3CH2OH (b) CH3CH2CH2SH
- (c) HOCH2CH2OH (d) CH3CH2CH2OH.
- 179\. Diethyl sulphide on hydrolysis with aqueous alkali gives :
- (a) methanol (b) methyl mercaptan
- (c) ethanol (d) ethyl mercaptan.
- 180\. Which of the property is not a property of thiols ?
- (a) They are all solids
- (b) They can be oxidised to disulphides
- (c) They have foul odours
- (d) The are weak acids.
- 181\. n-butyl bromide reacts with NaSH to give :
- (a) CH3CH2CH2SH (b) CH3SCH3
- (c) CH3CH2CH2CH2SH (d) CH3CH2SCH2CH3.
- 182\. Thiols are alcohol analogs in which the oxygen has been replaced by sulphur (e.g., CH3SH). Given the fact that the S – H bond is less polar than the O – H bond, which of the following statements comparing thiols and alcohols is correct ?
- (a) Hydrogen bonding forces are weaker in thiols
- (b) Hydrogen bondig forces are stronger in thiols
- (c) Hydrogen bonding forces would be the same
- (d) No comparison can be made without additional information.
- 183. Furan has :
- (a) weakly acidic property (b) strongly acidic property
- (c) weakly basic property (d) strongly basic property.
- 184\. Compound having two hetero atoms is :
- (a) Furan (b) Ooxazole
- (c) Pyrimidine (d) Pyridine.
- 185\. The isoelectric point of a protein is :
- (a) the pH at which the protein molecule has no charges on its surface
- (b) the pH at which a protein in solution has an equal number of positive and negative charges
- (c) the electric charges under isothermal conditions
- (d) none of these.
- 186\. Which of the following reactions is suitable for the preparation of α-amino acids ?
- (a) Schmidt reaction
- (b) Hofmann’s degradation of amides
- (c) Strecker’s synthesis
- (d) Reduction of nitro compounds.
- 187\. The primary structure of a protein refers to :
- (a) whether the protein is fibrous or globular
- (b) the amino acid sequence in the polypeptide chain
- (c) the orientation of the amino acid side chains in space
- (d) the presence or absence of an α-helix.
- 188\. The α-Helix is a common form of :
- (a) Primary structure (b) Tertiary structure
- (c) Secondary structure (d) None of these.
- 189\. The α-Helix is held in a coiled conformation partially because of :
- (a) Optical activity (b) Hydrogen bonding
- (c) Resonance (d) Delocalization.
- 190\. A compound gives a positive Tollen’s test but negative Ninhydrin test. It
is :
- (a) a protein (b) an amino acid
- (c) a monosaccharide (d) pyridine.
- 191\. The main method for the synthesis of quinolene is :
- (a) Scrop’s synthesis (b) Kolbe’s synthesis
- (c) Art-Estart Synthesis (d) Sendmayer Synthesis.
- 192\. By which of the following methods synthesis of quinolene is not possible ?
- (a) Dobner miller synthesis (b) Free lander synthesis
- (c) Scrop Synthesis (d) Kolbe’s synthesis.
- 193. Compound having two hetero atoms in six membered ring is :
- (a) Pyridine (b) Pypradine
- (c) Pyron (d) Pyrimidine.
- 194\. Succinic acid on dehydration by P2O5 gives :
- (a) Thiophene (b) Pyroll
- (c) Furan (d) Thiazole.
- 195\. Pyroll on complete reduction gives :
- (a) Pyrolidine (b) Pyroline
- (c) Butylamine (d) Propyl amine.
- 196\. Which of the following compounds will not be classified as lipids ?
- (a) fats (b) waxes
- (c) soaps (d) oils.
- 197\. Partial hydrogenation of vegetable oils in presence of Ni catalyst at 200°C gives :
- (a) Vanaspati ghee (b) Margrine
- (c) both of these (d) none of these.
- 198\. Synthetic detergents can be represented by the following general formula :
(a) RONa
(b) ROSO3Na
(c) RCOONa
(d) RCOOH.
199\. Quinolene on oxidation gives :
(a) phthalic acid
(b) oxalic acid
(c) cinnamic acid
(d) none of these.
200\. The nature of pyroll is :
(a) weak acidic
(b) strong basic
(c) weak basic
(d) strong acidic.
201\. Pyroll on treatment with alkyl halide gives :
(a) 2-methyl pyroll (b) 3-methyl pyroll
(c) 5-methyl pryoll (d) N-methyl pyroll.
202\. Pyroll on coupling with diazonium salt gives :
(a) 2-Phenyl azo pyroll
(c) 3-Phenyl azo pyroll
(b) Pyroll 2-aldehyde
(d) Pyroll 3-aldehyde.
203\. Vishler-Nepiyralski prepared :
(a) Isoquinolene
(b) Quinolene
(c) Indole
(d) Pyridine.
204\. Isoquinolene in comparison of quinolene is :
(a) strong acidic (b) weak acidic
(c) strong basic (d) weak basic.
B. Sc. III Chemistry II 31
- 205\. Compound having two hetero atoms in five membered ring is :
- (a) Imidazole
- (c) a and b both
- 206\. Thiophene on bromination gives :
- (a) 2-bromothiophene
- (c) 2,5-di bromothiophene
- 207\. Hetero atom is :
- (a) O
- (c) S
(b) Pyrazole
- (d) Pyroll.
- (b) 5- bromothiophene
- (d) 2, 3, 4, 5-tetra bromo thiophene.
(b) N
(d) All of these.
Answer Sheet
- 1\. (d), 2. (b), 3. (a), 4. (b), 5. (d), 6. (c), 7. (a), 8. (a), 9. (d), 10. (d), 11. (a),
- 12\. (a), 13. (b), 14. (a), 15. (a), 16. (a), 17. (a), 18. (c), 19. (d), 20. (a), 21. (c),
- 22\. (c), 23. (a), 24. (a), 25. (a), 26. (a), 27. (c), 28. (b), 29. (a), 30. (a), 31. (b),
- 32\. (c), 33. (b), 34. (b), 35. (d), 36. (a), 37. (a), 38. (a), 39. (a), 40. (a), 41. (a),
- 42\. (b), 43. (b), 44. (d), 45. (c), 46. (b), 47. (b), 48. (b), 49. (a), 50. (b), 51. (c),
- 52\. (b), 53. (a), 54. (c), 55. (b), 56. (c), 57. (a), 58. (d), 59. (d), 60. (a) 61. (c)
- 62\. (b), 63. (c), 64. (c), 65. (b), 66. (b), 67. (d), 68. (c), 69. (c), 70. (b), 71. (c)
- 72\. (c), 73. (d), 74. (a), 75. (a), 76. (d), 77. (c), 78. (d), 79. (a), 80. (a), 81. (c)
- 82\. (b), 83. (b), 84. (c), 85. (a), 86. (d), 87. (b), 88. (c), 89. (d), 90. (a), 91. (b) 92. (d), 93. (a), 94. (a), 95. (a), 96. (c), 97. (d), 98. (b), 99. (b), 100. (c), 101. (a) 102. (a), 103. (c), 104. (d), 105. (a), 106. (d), 107. (d), 108. (d), 109. (a)
- 110\. (c), 111. (d), 112. (a), 113. (a), 114.
- 119\. (b), 120. (a), 121. (a), 122. (a), 123.
- 128\. (c), 129. (d), 130. (b), 131. (c), 132.
- 137\. (a), 138. (d), 139. (d), 140. (a),
- 145\. (c), 146. (b), 147. (a), 148. (d),
- 153\. (a), 154. (b), 155. (a), 156. (d), 161. (d), 162. (a), 163. (b), 164. (b), 169. (b), 170. (c), 171. (c), 172. (d),
(c), 115. (b), 116. (b), 117. (a), 118. (a), (a), 124. (c), 125. (b), 126. (c), 127. (c), (a), 133. (c), 134. (b), 135. (a), 136. (d), 141. (a), 142. (c), 143. (b), 144. (a), 149. (c), 150. (c), 151. (a), 152. (d), 157. (c), 158. (a), 159. (b), 160. (b), 165. (b), 166. (b), 167. (a), 168. (c), 173. (d), 174. (c), 175. (b), 176. (b),
- [177. (b), 178. (a), 179. (b), 180. (a), 181. (a), 182. (a), 183. (c), 184.](#bookmark192)
- [185. (b), 186. (a), 187. (b), 188. (c), 189. (b), 190. (c), 191. (a), 192.](#bookmark193)
- [193. (d), 194. (c), 195. (a), 196. (b), 197. (a), 198. (b), 199. (a), 200.](#bookmark194)
- 201\. (d), 202. (a), 203. (a), 204. (c), 205. (c), 206. (c), 207. (d).
32 Vikas, 2009 (A. U. )
Very Short Answer Type Questions
Q. 1. What do you mean by spectroscopy ?
Ans. The branch of Chemistry which deals with the methods for analysis of structure determination for organic and inorganic compounds. The interaction of matter with electromagnetic radiation is studied in spectroscopy.
Q. 2. Arrange the following compounds in order of their increasing wavelength of UV absorption maxima (λmax).
- (a) Ethylene (b) Naphthalene (c) Anthracene (d) 1, 3 butadiene.
Ans. (a) < (d) < (b) < (c).
Q. 3. Using IR spectroscopy how will you distinguish between (CH3)3N and CH3CH2NCH3 ?
Ans. CH3CH2NCH3 will show a medium bond in the region 3310–3500 cm–1 due to N – H stretching. This bond will be absent in the IR spectrum of (CH3)3N.
Q. 4. Give suitable explanation for the observation that the VO – H bond appears near 3570 cm–1 whereas the VO–D bond near 2630 cm–1.
Ans. According to Hook’s law, the strecting frequency of a bond increases as the reduced mass of the bonded atoms decreases. Since hydrogen has lesser atomic mass than duterium, the VO–H frequency (near 3570 cm–1) is higher than the VO–D frequency (near 2630 cm–1).
Q. 5. How will you distinguish o-hydroxy benzaldehyde and m-hydroxy benzaldehyde with the help of IR spectroscopy ?
Ans. In salicylaldehyde due to intramolecular hydrogen bonding, νO–H and νC = O bonds are shifted to lower wave numbers. Since it is intramolecular, charge in concentration does not cause any shift in νO–Hand νC = O bonds. In
H CHO
_files/AU20B.20Sc.20III20Chemistry20II(OK)-14.png)o-hydroxy benzaldehye m-hydroxy benzaldehyde
the case of m-hydroxy benzaldehyde also νO–H and νC = O bonds occur at a lower wave number due to inter molecular hydrogen bonding. In this case νO– H and νC = O bonds shift to higher wave number on dilution with a non-polar solvent.
Q. 6. Which member of each of the following pairs of the structural isomers expacted to exhibit a single peak in its PMR spectrum ?
- (a) CH3 – CCl3 or Cl CH2 – CHCl2
- (b) CH3
C = CH2 or
CH3
- (c) ClCH2CH2Cl or CH3CHCl2
- (d) H2C = C = CH2 or CH3 – C ≡ CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-15.jpg)Ans. (a) CH3 – CCl3
(b)
- (c) ClCH2 – CH2Cl
- (d) H2C = C = CH2
- (e) Cl Cl
Q. 7. Give two properties of a compound to act as a dye.
Ans. (i) It must have a suitable colour.
- (ii) It must be capable of fixing itself to the material to be dyed.
- (iii) When fixed, it must have fastness properties.
Q. 8. Give the classification of dyes on the basis of structure.
Ans. (i) nitro-nitroso dyes (ii) azo dyes
- (iii) triphenyl methane dyes (iv) phthalein dyes
- (v) anthraquinone dyes (vi) Indigo dyes.
Q. 9. Give the classification of dyes on the basis of
application.
Ans. (i) Acidic Dyes
(ii) Basic Dyes
(iv) Vat Dyes
(vi) Ingrain Dyes
(viii) Reactive Dyes
- (iii) Direct Dyes
(v) Mordant Dyes
(vii) Disperse Dyes
Q. 10. Give structure of methyle orange.
Ans. CH3
NaO3S N = N N
CH3
Q. 11. Write the constitution of Indigo.
Ans. O
_files/AU20B.20Sc.20III20Chemistry20II(OK)-16.png)Indigo O
Q. 12. What are organometallic compounds ?
Ans. Organometallic compound are the compounds in which M – C bonds are present except carbonates e.g.
(C2H5)4Pb C2H5MgI CH3Li
Tetra ethyl lead Ethyl magnesium iodide Alkyl lithium
Q. 13. Organometallic compounds are highly reactive. Why?
Ans. Organometallic compound i.e. Grignard’s reagent is highly reactive because C – Mg bond is polar due to high electronegativity of carbon than Mg. Due to which C–atom acquires partial – ve and Mg atom acquire partial + ve charge.
Q. 14. What happens when acetone reacts with Grignard’s reagents ?
Ans. t-alcohols are formed
O|| δ– δ+
CH3 – C – CH3 + CH3MgI δ+
H O |
—\* CH3– C – CH3
|
CH3
OH
| OH
CH3– C – CH3 + Mg
|CH3 I
t-butyl alcohol
_files/AU20B.20Sc.20III20Chemistry20II(OK)-17.png)
Q. 15. Give the products of Grignard’s reagent and alkyl
lithium with CO2.
Ans. With CO2 Grignard’s reagent give carboxylic acid whereas alkyl lithium gives acetone.
O
|| δ– δ– δ+
δ C = O + CH3MgI
O
|| H
^ CH3 – C – O
OH
MgI
\---► CH3COOH + Mg Acetic acid I
OH
O
||
C = O + CH3Li
O
||
> CH3 – C – OLi
CH3Li
_files/AU20B.20Sc.20III20Chemistry20II(OK)-18.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-19.png)
OLi
OLi
CH
CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-20.png)C = O
3
– H2O ◄------
2H2O
CH, 1 / OH
3 C + 2LiOH
CH3 OH
Acetone
Q. 16. Give the reaction orthoformic ester.
between alkyl lithium and
Ans.
[OC HOC H](#bookmark197)
[– + 2525](#bookmark198)
δδ
CH3 – Li + H – C – OC2H5 Hexene\* H – C – OC2H5
[OC2H5OC](#bookmark199)
H2O
CH3CHO + 2C2H5OH
Acetaldehyde Ethanol
CH3
Q. 17. Give the IUPAC name of CH – δH.
CH3
CH3
Ans. IUPAC name of CH – δH is 2-propane thiol.
CH3
Q. 18. Boiling points of thiols are much lower than that of alcohols. Why ?
Ans. Boiling points of thiols are much lower than that of alcohols due to absence of H-bonding in thiols. Alcohols have higher boiling point due to presence of hydrogen bonding.
Q. 19. What happens when thioalcohols is oxidised by conc. HNO3 ?
Ans. Ethane sulphonic acid is formed .
HNO
C2H5SH + 3\[O\] –––––––3–→ C2H5SO3H
Ethane sulphonic acid
Q. 20. How picric acid is obtained from p-hydroxy benzene sulphonic acid ?
Ans.
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-21.png)SO3H
ON
OH
HNO3
_files/AU20B.20Sc.20III20Chemistry20II(OK)-22.png)SO3H
NO2 fuming
HNO3
OH
O2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-23.png)NO2
Picric Acid
NO2
Q. 21. Give the structures of quinolene and indole. Ans.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-24.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-25.png)Quinolene
H
Indole
Q. 22. Pyrrole is a weak acid. Explain.
Ans. Due to greater s-character of N–H bond the electronegativity of nitrogen is more hence the bonded pair of electron of the N – H bonds are strongly attracted by nitrogen and hydrogen can be eleminated as a protone.
Q. 23. Give the reaction of pyrrole with chloroform and KOH.
Ans. 2-formyl pyrrole is obtained. This reaction is called Reimer-Tiemann
reaction.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-26.png)\+ CHCl3 + KOH
H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-27.png)H
O
_files/AU20B.20Sc.20III20Chemistry20II(OK)-28.png)
Cl
2-formyl pyrrole 3-chloro pyridine
Q. 24. What happens when nicotinic acid is distilled with sodium hydroxide ?
Ans.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-29.png)
COOH
Distillation
\+ NaOH ---------" k I + Na2CO3 + H2O
Nicotinic acid Pyridine
Q. 25. Pyridine is mono acid tertiery base ? Express it with suitable equation.
Ans. Pyridine is mono acid tertiery base because on reaction with alkyl halide it form quaternary salt.
C5H5N + CH3I –––––––→ C5H5N+CH3I–
Formation of N-methyl pyridinium iodide clearly indicates that pyridine is a tertiery base.
Q. 26. Give Fisher indol synthesis.
Ans. CH3
_files/AU20B.20Sc.20III20Chemistry20II(OK)-30.png)
CH3– C – COOH
\+ ||
NH2 O
NH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-31.png)
C – COOH
o
_files/AU20B.20Sc.20III20Chemistry20II(OK)-32.png)
i z ö 'l
|| N
NH
Pyrubic acid phenyl hydrazone
–CO2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-33.png)H
H Indole
Q. 27. What are oils and fats ?
COOH
Ans. Oils and fats are triglucerides of higher fatty acids.
CH2OOCC15H31
|
CHOOCC15H31
|
CH2OOC15H31
Tripalmitin
(Simple gluceride)
CH2OOCC17H35
|
CHOOCC17H33
|
CH2OOCC15H31
Neo, palmito sterin (Mixed gluceride)
Q. 28. Write short note on saponification.
Ans. Alkaline hydrolysis of oils or fats in which glycerol and sodium salts of higher fatty acids are obtained, is known a soaponification reaction. Sodium salt of higher fatty acid is called soap.
CH2COOC17H35 CH2OH
||
CHCOOC17H35 + 3NaOH ^ CHOH + 3C17H35COONa
||
CH2COOC17H35 CH2OH Sodium Sterate
glycerol (soap)
Q. 29. Define acid value.
Ans. Acid value is defined as the number of milligrams of KOH required to completely soaponify one of oil or fat.
Q. 30. Give the chemical name of two detergents.
Ans. (a) Sodium Lauryl Sulphate.
- (b) Sodium-p-(n-dodecyl) benzene sulphonate.
Q. 31. What are carbohydrates ?
Ans. Carbohydrates are defined as polyhydroxyaldehydes or polyhydroxy ketones or substances which give those on hydrolysis and contain at least one chiral C-atom.
Q. 32. Give two examples each of mono, di and poly saccharides.
Ans. Mono saccharide – glucose, fructose Disaccharide– sucrose, maltose Polysaccharide – Starch, cellulose.
Q. 33. What are reducing and non-reducing sugars ?
Ans. Reducing sugars are those which reduces Tollen’s reagent and Fehling’s solution. Those give silver mirror with Tollen’s reagent and red ppt of Cu2O with Fehlings solution. Where are non-reducing sugar do not reduces Tollen’s reagent and Fehlings solution.
Q. 34. Give the reaction of phenyl hydrazine with glucose.
Ans. Glucose with excess of phenyl hydrazine gives glucose zone :
CHO + H2N.NHC6H5 CH = N - NHC6H5 CH = N.NHC6H5
CHOH
CHOH C6H5NH.NH2
C = O
1
(CHOH)3 |
' |
(CHOH)3 |
-NH3
–C6H5NH2
|
(CHOH) |
CH2OH
CH2OH Glucose phenyl
CH2OH Glucose
hydrazone
CH = N.NHC6H5 |
C6H5NH.NH2
C = N.NHC6H5
|
(CHOH)3 |
CH2OH
Glucose Zone
Q. 35. What are epimers ?
Ans. Epimers are a pair of distereomers that differ only in the configuration about a single carbon atom.. The phenomenon is called epimerisation.
structure of α-methyl glucoside and
Q. 36. Give the β-methyl glucoside.
Ans.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-34.png)α -methyl glucoside
_files/AU20B.20Sc.20III20Chemistry20II(OK)-35.png)β -methyl glucoside
Q. 37. How the presence of 5–OH groups in glucose is identified ?
Ans. Glucose on treatment with PCl5, acityl chloride gives penta-chloro glucose and glucose pentraacetate. Formation of these compounds indicates that glucose has 5 – OH groups.
Q. 38. What do you mean by muta rotation ?
Ans. A spontaneous change in specific rotation of solution of a optically active compound is called muta rotation.
Q. 39. Write a note on inversion of sugar.
Ans. It is dextro rotatory with a specific rotation \[α\]D = + 66.5°, but on hydrolysis it gives dextrorotatory glucose and leavo rotatory fructose. Since the leavo-rotation (– 92.4°) of fructose is more than dextro rotation of glucose
(52.5°), therefore the resulting solution becomes leavo rotatory. This net change in the sign of optical rotation from dextro to leavo is called inversion of sugar.
Q. 40. What happen when lactose is hydrolysed by the enzyme lactose ?
Ans. Lactose upon hydrolysis gives 1 molecule of glucose and 1 molecule of lactose.
H+
C12H22O11 + H2O---------»C6H12O6 + C6H12O6
Lactose Glucose Lactose
Q. 41. What are a-amino acids ?
Ans. Amino acids are compounds that contain at least one carboxylic group (- COOH) and one amino group (- NH2). In a-amino acids, the amino group is located on the a-carbon e.g.
O
| |
CH3 – CH – C – OH | NH2
Q. 42. What are essential amino acids ?
Ans. Essential amino acids are those which can not be produced in sufficient quantity by human body. They must be induced in diet. Examples are valine, phenyl alanine, tryptophan.
Q. 43. What is iso-electric point ?
Ans. The iso-electric point is the pH at which an amino acid exists completely as the zwitter ion.
Q. 44. What is electrophoresis ?
Ans. Electrophoresis is a technique of separation and purification of amino-acids on the differential movement of charged particles in an electric field.
Q. 45. What happens when p-amino acid is heated ?
Ans. Unsaturated acids are formed.
R - CH - CH2 - COOH-------> R - CH = CH - COOH
| a, p-unsaturated acid
NH2
Q. 46. What are peptides ?
Ans. Two amino acids can combine with the elemination of water. In this reaction carboxylic group of one amino acid combines with amino group of another.
O | | H2NCH2 – C –
OH + H
O
| | –H2O
NH – CH – C – OH → H2N CH2 –
O
| |
C – NH
O
| |
– CH – C – OH
| |
CH3 CH3
Glycine Alanine Peptide Linkage Glucyl alanine
Q. 47. How many purine bases are present in amino acids ?
Ans. Two purines that are constituents of nucleic acid are adonine and
guanine. O NH2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-36.png)Guanine
Adinine
Q. 48. What are polymers ?
Ans. Polymers are high molecular weight compounds whose structures are made up of a large number of simple repeating units.
Q. 49. How Nylon-6,6 is synthesized ?
Ans. It is prepared by condensing adipic acid and hexa methylene diamine.
O O
| | | |
HO – C – (CH2)4 – C – OH + H2N – (CH2)6 – NH2
252°C
O | |
– H2O
HO – C – (CH2)4 – C – NH – (CH2)6 – NH2
O
| |
HO – C – (CH2)4 – C – OH/H2N(CH2)4 – NH2
O O
| | | |
O | |
O | |
HO – C – (CH2)4 – C – \[– NH(–CH2)6 – NH – C – (CH2)4 – C –\]n
NH(CH2)6NH2
Nylon 6, 6
Q. 50. What is the composition of natural rubber ?
Ans. Natural rubber is polymer of isoprene. Its composition is (C5H5)n. It can be demonstrated as follows :
CH3 CH3
| Polymerisation |
n CH2 = C – CH = CH2----------→ \[CH2 – C = CH – CH2 –\]n
Isoprene Polyisoprene
Section ‘B’
Short Answer Type Questions
Q. 1. Draw the structure of a compound with each of the following molecular formulae that will show only one peak in its NMR spectrum :
(a) C3H6Cl2 (b) C5H12 (c) C2H6O (d) C4H6
Ans. Possible structures are :
Cl
(a) CH3–C–CH3
Cl
CH3
(b) CH3–C–CH3
CH3
(c) CH3–O–CH3
(d) CH3–––CCH3
Q. 2. Give a structure consistent with the following NMR data :
Molecular formula = C10H14
(a) Singlet at δ 1.30, 9H (b) Singlet at δ 7.28, 5H
Ans. The compound is :
b
_files/AU20B.20Sc.20III20Chemistry20II(OK)-37.png)C — CH3
tert-Butylbenzene
CH3
Q. 3. Write a short note on saponification value.
Ans. Saponification Value : It is a measure of fatty acids present as esters in oils and fats. It is defined as the number of milligrams of KOH required to saponify one gram of the oil or fat or number of milligrams of KOH required to neutralize the free acids resulting from the hydrolysis of one gram of an oil or fat. It is determined by refluxing a wighed amount of the oil or fat with known
excess of standard alcoholic caustic potash solution and back titrating the excess of alkali with standard acid solution, usually N/2 alkali solution is used. Saponification value tells about the approximate molecular weight of a fat or oil. One mole of a fat or oil reacts with three moles of KOH. If M is the molecular weight of the fat or oil, M grams of its will require 3 x 56 = 168 grams or 168,000 milligrams of KOH for saponification. Therefore,
168,000 M
Saponification number of fat or oil =
Knowing the saponification number experimentally, the value of molecular weight, M can be determined.
The saponification number is inversely proportional to the molecular weight. The smaller the saponification value, the higher the molecular wieght. The saponification value is characteristic of a particular fat or oil and serves for its identification.
Q. 4. How many signals (ignoring the splitting patterns) would you see in the NMR spectra of the following compounds ?
(a) Butanone (b) Cyclobutane
- (c) p-xylene (d) 2-Propanol
Ans. (a) Three signals; (b) One signal; (c) Two signals; and (d) Three signals.
Q. 5. What do you mean by a dye ?
Ans. Dye : A dye is coloured compound, normally used in solution which
is capable of being fixed to fabric.
It must be fast. A dye owes its colour to the presence of chromophore e.g., (NO, NO2, etc.) and its fixing properly to the acidic or basic auxochromes such as OH, SO3H, NH2 etc.
Dye = Chromogen containing an anxochrome
For example, nitrobenzene is a yellow compound, but not a dye, because
it contains only chromophore (nitro group) while para nitrophenol is a yellow
dye as it contains both chromophore and the auxochrome (the hydroxyle
group).
_files/AU20B.20Sc.20III20Chemistry20II(OK)-38.png)NO2
HO
_files/AU20B.20Sc.20III20Chemistry20II(OK)-39.png)NO2
Nitro Benzene
Para Nitrophenol
Q. 6. Give the chemistry of crystal voilet. Also give their methods of preparation and uses.
Ans. (a) Crystal violet : Its chemical name is hexamethyl parafuchsine or hexamethyl para rosaniline hydrochloride. It is obtained by heating Michler’s
ketone with dimethyl aniline in the presence of phosphoryl chloride or carbonyl chloride. If the latter compound is used, then crystal violet may be prepared directly by heating carbonyl chloride and dimethyl aniline.
2(CH3)2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-40.png)\+ POCl3 —
(CH3)2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-41.png)N(CH3)2 Michler's ketone
OH
(CH3)2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-42.png)N(CH3)2 POCl3
N(CH3)2
w
_files/AU20B.20Sc.20III20Chemistry20II(OK)-43.png)= N(CH3)2}C1
N(CH3)2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-44.png)\+ COCl2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-45.png)\[ 3)2
(CH3)2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-46.png)C= O =
N(CH3)2+}Cl
N(CH3)2
Crystal violet
Its weak acid solution is violet, its strongly acid solution is green and its very strongly solution is yellow. The colour change may be explained as follows :
In wealkly acid solution, the crystal violet has been found to exist as the highly charge ion (I). In this state, two-third of the charge will undergo oscillation in the horizontal direction. In strongly acid solution, it has been found to exist as the doubly charged ion (II). In this state, the whole unit of charge will undergo oscillation in the horizontal direction and, therefore, the colour deepens. It is important to remember that the vertical direction of oscillation gets inhibited due to the fixation of the long pair by proton addition. In very strongly acid solution it has been found to exist in form (III) having three charges. In this ion, relatively little resonance (with oscillation of charge) is possible and therefore the colour lightens.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-47.png)+
NMe2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-48.png)
\+ NH(Me2)
(III)
It is used to dye silk, wool and tannin-mordanted cotton to bluish violet colour but the colour is not fast to light. It is used in making indelible ink and pencils, in stamping pads etc. It is used as an indicator in the determination of hydrogen ion concentration of solution.
Q. 7. What are organometallic compounds ? Give
examples.
Ans. Organometallic Compounds : Organic compounds in which a metal is directly linked to carbon are termed Organometallic compounds.
C2H5 C2H5
Mg < 2 5 Mg < 2 5 Pb(C2H5)4
I C2H5
Ethyl magnesium Diethyl zinc Tetra ethyl
iodide lead
These may be largely ionic or largely covalent. The ionic character of the carbon-metal bond depends on the nature of the metal. It decreases in the order given below :
Na > Li > Mg > Al > Zn > Cd > Hg
Thus organometallic compounds of alkali and alkaline earth metals consist of ions or ion pairs.
– + – 2 + – – +
R Na R Mg X C6H5Li
Alkyl sodium Alkyl magnesium halid Phenyl lithium
(Girgnard reagent)
Q. 8. What happens when
- (i) Grignard reagent reacts with acetone.
- (ii) Grignard reagent reacts with ethyl orthoformate.
Ans. (i) Reaction with acetone (Preparation of t-butyl alcohol)–Grignard reagent on treatment with acetone gives t-butyl alcohol.
CH3 H
C–O
CH3 / I
3 CH3
CH3 CH3
C = O + Mg
CH3 Br
OH
Mg Br
H2O
H3C OH
- H3C^ C–OH + Mg
H3C Br
t-butyl alcohol
- (ii) When Ethyl-ortho formate reacts with ethyl ortho-formate acetaldehyde is obtained.
,OC2H5 CH3 /OC2I15
H.c4oC2H5+Mg/ --<H3CH(OC2H5)2+Mg/
XOC2H5 | XI
CH3CHO + 2C2HsOH Acetaldehyde
- Q. 9. A chemist prepared the two isomeric ketones (A) and (B), placed them in separate flasks, but forgot to label them. How could you differentiate the two by UV spectroscopy
O
II
CH3CH2CH=CHCCH3
(A)
O
II
CH3CH=CHCH2CCH3
(B)
Ans. In the UV spectra, (A) a conjugated system, would show λmax at higher wavelengths than (B), a nonconjugated compound.
Q. 10. What are polysaccharides ?
Ans. Polysaccharides : Polysaccharides give large number of monosaccharide units upon hydrolysis. A polysaccharide contains many monosaccharides joined with each other by glycosidic linkage in a long linear or branched structure. Accordingly, there can be a linear polysaccharide, or a branched-chain polysaccharide.
Characteristics of Polysaccharides : Polysaccharides are amorphous, tasteless and mostly insoluble in water. Some typical polysaccharides are, starch, dextrin, cellulose and glycogen. Polysaccharides are the most common carbohydrates in nature. Their main functions are food storage and structural.
Q. 11. Write a short note on glycosidic linkage.
Ans. Glycosidic Linkage : Disaccharides are formed by a condensation reaction between two monosaccharides, accompanied by the elimination of one molecule of water.
The hydroxyl groups of the hemiacetals of two monosaccharide molecules condense to form a disaccharide. A bond of the type – O – is formed between the two monosaccharide molecules. Such – O – bond between the two sugar units is called glycoside or glycosidic linkage. Formation of glycoside linkage during the formation of maltose is shown in Fig. Here, the glycoside linkage is between 1 and 4 C carbon atoms of the two glucose units.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-49.png)–H2O
GLUCOSE
GLUCOSE
(MONOSACCHARIDE
(MONOSACCHARIDE)
HOH C
H
HO
_files/AU20B.20Sc.20III20Chemistry20II(OK)-50.png)
H
Q. 12. What are amino acids ?
(GLYCOSIDIC OH BOND)
MALTOSE
(DISACCHARIDE)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-51.png)H
CH2OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-52.png)OH
Ans. Amino acids : The organic compounds having both amino and carboxylic groups in a molecule are known as aminoacids. It is present in
combined form as proteins. About two dozens of amino acids are prepared from proteins called as natural amino acids. The simplest amino acid is glycine (amino-acetic acid).
H2N.CH2.COOH
Glycine
The amino acids may be neutral, CH2.(NH2). COOH acidic, H2N.CH(COOH).CH2COOH (aspartic acid) or basic H2NCH(NH2)COOH (lysine) depending upon their nature.
Q. 13. What is the effect of heat on a-amino acids ?
Ans. Action of Heat : The behaviour of amino acids on heating is similar to that of hydroxy acids.
- (i) a-Amino acids lose two molecules of water between two molecules of the acid to form cyclic di-amides known as Diketopiperazines.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-53.png)
- (ii) p-Amino acids split out a molecule of ammonia and yield a, 0Unsaturated acids.
|nh2 h|
I I A
CH,—CH—COOH -> CH2 = CH—COOH + NH3
0-Aminoprypionic acid Acrylic acid
- (iii) y and 8 amino acids lose a molecule of water between NH2 and COOH of the same molecule to form inner anhydride called Lactams.
CH—CH, CH, — CH,
CH, C = O -A-> CH, C = O
y-aminobutyric acid
y-butyrolactam
_files/AU20B.20Sc.20III20Chemistry20II(OK)-54.png)Q. 14. Suggest the structure of a compound each with the following NMR structural features ?
- (a) An alcohol with two NMR peaks
- (b) A compound C5H10 with a single NMR peak
- (c) A compound C4H6 with a single NMR peak.
Ans. (a) Methyl alcohol; (b) Cyclopentane; (c) 2-Butyne.
Q. 15. Differentiate between primary and secondary structures of proteins.
Ans. Difference between primary and secondary structures of Proteins.
Primary structure
Secondary structure
1\. Amino acids are joined to-gether by peptide linkage.
It arises due to coiling of the polypeptide chain due to intramolecular hydrogen bonding between >C = O and — NH groups.
2\. The sequence in which the amino acids are arranged in any protein is called its Primary structure.
The sequence in which the amino acids are arranged in any protein is called its Primary structure.
Q. 16. Discuss the principle of IR spectroscopy. How will you distinguish between the following pairs of compounds on the basis of IR spectroscopy :
- (a) Ethyl alcohol and diethyl ether,
- (b) Acetic acid and ethyl acetate ?
Ans. (a) Ethyl alcohol shows a strong/broad absorption at 3200-3500 cm–1 (due to O – H bond) while diethyl ether does not; and (b) Acetic acid shows a strong/broad absorption at 2500 – 3500 cm–1 (due to O – H bond) while ethyl acetate does not.
Q. 17. How would you distinguish between the following pairs of compounds by IR spectroscopy ?
- (a) CH3CH2CH2N(CH3)2 and CH3CH2CH2NH2
- (b) CH3CH2CH2COOH and CH3CH2COOCH3
- (c) CH3CH2COCH3 and CH3CH2COOCH3
Ans. (a) CH3CH2CH2NH2 shows N – H absorption at 3300-3500 cm–1 while CH3CH2CH2N(CH3)2 does not.
- (b) CH3CH2CH2COOH shows O ‒ H absorption at 2500-3500 cm‒1 while CH3CH2COOCH3 does not.
- (c) CH3CH2COOCH3 shows C ‒ O absorption at 1050-1350 cm‒1 where as CH3CH2COCH3 does not.
Q. 18. Two compounds, (A) and (B), have the same molecular formula C2H6O. They have different IR spectra. Compound (A) shows a strong/broad absorption at 3400 cm–1, while compound (B) does not. Suggest formulae for (A) and (B) which account for the difference.
Ans. (A) is ethyle alcohol (CH3CH2OH); (B) is dimethyl ether (CH3OCH3). The strong absorption at 3400 cm‒1 in (A) is due to the O ‒ H bond.
Q. 19. The UV spectrum for a compound with formula C3H6O shows a weak absorption band at 280 nm. The NMR spectrum shows only one signal, a singlet. What is the structure of the compound ?
Ans. The weak UV absorption at 280 nm is characteristic of a carbonyl group (C = O), leaving C2H6 for the remainder of the original formula. This suggests two CH3 groups. The structure of acetone (CH3COCH3) fits all the data.
Q. 20. What do you mean by spectroscopy ? Give principle.
Ans. Spectroscopy : Spectroscopy is the branch of chemistry which deals with the methods for the analysis of structure determination for organic and inorganic compounds.
These methods have three major advantages over most chemical methods :
- (1) Spectroscopic methods are easier and faster to do than most chemical tests or reacitons.
- (2) Spectroscopic methods provide far more information about molecular structure. Practically all functional groups and structural features can be detected with very small amounts of sample.
- (3) Spectroscopic methods are non-destructive and, if necessary, the entire sample can be recovered.
There are four spectroscopic methods which are very widely used in Organic Chemistry. They are ultraviolet-visible, infrared, nuclear magnetic resonance and mass spectroscopy. The basis of these methods is electromagnetic radiation.
Principle of Spectroscopy
All organic compounds interact with electromagnetic radiation, that is, they absorb energy. When a molecule absorbs energy, a transformation occurs that may be either temporary or permanent. Lower-energy radiation may cause a molecular rotation or a bond vibration. Higher-energy radiation may cause the promotion of electrons to higher energy levels or bond cleavage.
Whether the transformation involves molecular rotation, bond vibration, or electronic transition, the molecule absorbs only the wavelength of radiation is selective for a particular transition which depends on the structure of the molecule. By measuring the absorption spectra of known compounds, we can correlate the wavelengths of energy absorbed with characteristic structural feature. This information is then use to determine the structure of unknown compounds.
Q. 21. What are soaps ? Give their types.
Ans. Soaps : Soaps are the metallic salts of higher fatty acids such as palmitic, stearic, oleic, etc. The sodium and potassium salts are the common soaps which are soluble in water and used for cleansing purposes. Soaps of other metals such as calcium, magnesium, zinc, chromium, lead, etc., are insoluble in water. These are not used for cleanising purposes but for other purposes (lubricants, driers, adhesives, etc.)
Ordinary soaps (soidum and potassium) are the products of hydrolysis of oils and fats with sodium hydroxide or potassium hydroxide. The oils and fats are mixed glycerides and thus soaps are mixtures of salts of saturated and unsaturated long chain carboxylic acids containing 12 to 18 carbon atoms. This process always yields glycerol as a bye-product.
CH OCOR CH OH
21 2
R1COONa +
CH2OCOR1 + 3NaOH → CHOH +
R2COONa
CH2OCOR1 CH2OH
+
R3COONa
Triglyceride Glycerol
Soap
Types of soaps
- (i) Hard soaps : These are obtained from cheap oils and fats using sodium hydroxide. These contain free alkali and are used for washing purposes.
- (ii) Soft soaps : These are obtained from good oils using potassium hydroxide. These do not contain free alkali and are used as toilet soaps, shaving creams, in shaving sticks and shampoo.
- (iii) Transparent soaps : These are formed by dissolving toilet soaps in alcohol and evaporating the filtrate. They contain glycerol.
- (iv) Medicated soaps : Toilet soaps containing some medicinal important substances are called medicated soaps.
- (v) Metallic soaps : These are soaps of metal other than sodium and potassium.
Q. 22. Write a short note on cleanising action of soap.
Ans. Cleanising action of soap : Skin is covered with a thin layer of oil or grease which catches dirt. The soap molecules have a long chain non-polar hydrocarbon tail that is oil soluble, and a negative head (COO‒) O
\----------------------H—I
CH3‒CH2‒CH2‒CH2‒CH2..............CH2‒C‒O‒
Polar head (water soluble) which is water soluble. As soap water is poured over the skin (or a dirty garment) the hydrocarbon tail of the soap molecule peg into it while the negative head is held in water. The grease layer is then dislodged from skin by rubbing or from garments by tumbling or stirring. Each grease globule thus separated is pincushioned by hydrocarbon tail with negative heads outward in water. The negative globules keep apart by mutual replusions and are said to have been emulsified. The emulsified grease globules bearing dirt can be readily washed with water.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-55.png)A SOAP MOLECULE EMULSIFID GREASE GLOBULE
Q. 23. Give the constitution of isoquinoline.
Ans. Constitution of Isoquinoline
- (i) Molecular formula : C9H7N.
- (ii) Tertiary Base. It reacts with one mole of CH3I and forms quaternary ammonium salt showing the presence of a tertiary nitrogen atom.
- (iii) It undergoes electrophilic substitution like benzene, hence shows the presence of aromatic nucleus (like benzene rings).
- (iv) On oxidation, forms a mixture of phthalic acid and 3, 4-pyridine-dicarboxylic acid (cinchomeronic acid).
This shows that it contains a benzene ring fused with pyridine rings at 3, 4-positions. Hence isoquinoline may have the following structure :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-56.png)(v) This structure is confirmed by the following synthesis.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-57.png)Q. 24. What do you mean by mutarotation ? Give its mechanism.
Ans. Mutarotation : Two crystalline forms of D-glucose have been isolated. These are α-D-glucose and β-D-glucose respectively. The specific rotation of freshly prepared aqueous solution of glucose crystallised from alcohol or acetic acid is 112° which on standing gradually falls to +52.5°. Similarly the specific rotation of freshly prepared aqueous solution of glucose crystallised from pyridine is +19° which on standing increase to +52.5°.
When either of these forms of D-glucose is dissolved in water and allowed to stand, a gradual change in specific rotation occurs. The specific rotation of α-form falls and that of β-form rises until a constant value of +52.5° is reached. This change in specific rotation of a solution of either form of glucose until a constant value is obtained, is called mutarotation.
Mutarotation has been explained by assuming that glucose exist in two forms, anomeric α- and β- varieties. Glucose obtained from alcoholic or acetic acid solution is the α-form with specific rotation \[α\]D = 110° while glucose obtained from pyridine solution is the β-variety with specific rotation \[α\]D = +19°. In an aqueous solution, the two varieties are present as an equimolecular mixture containing about 36% α- and 64% β- variety and formed by isomerisation of α- form into β- form and vice-versa.
Mechanism of Mutarotation—According to Lowry, mutarotation is possible only in presence of a solvent like water which can act both as an acid and a base. It appears that when mutarotation takes place, the ring opens and the recloses in the inverted position or in the original position. Lowry proposed that in water, the transformation occurred through the intermediate product aldehydrol.
H—C—OH H—C—(OH)2 HO—C—H
H—C—OH TT H—C—OH TT H—C—OH
| +H,0 | +H,0 |
HO—C—H O s „ „ HO—C—H HO—C—H O
- I - H.O I - H,0 I
H—C—OH
I
H—C---
I ch2oh a-glucose \[< = +110°
H—C—OH
I
H—C
I
CH,OH
Intermediate aldehydrol form
H—C—OH
I
H—C---
I
ch2oh
P-glucose \[< = +19°
\[α\]Dt of equilibrium mixture = + 52.5°
Q. 25. Write a short note on iso-electric point of amino acids.
Ans. Iso-electric Point : The aqueous solution of amino acids has a dipolar ion known as zwitter ion.
H2N.CH2.COOH^\[H2N.CH2.COO’+H+\]^H,N+.CH2COO Glycine (Ionised) zwitter ion
(un-ionised)
At lower pH (i.e., in more acidic solution the amino acids exist as positive ion.)
H3N+—CH2—COOH
while at higher pH (i.e., in more basic solution) the aminoacid exists as negative ion.
H3N+—CH2—COO‒
On the other hand, at some intermediate pH, the amino acid exists as a completely neutral zwitter ion having
H3N+—CH2—COO‒
equal positive and negative changes. When the aqueous solution of amino acid at this pH is placed in an electric field, it will not migrate towards any electrode. This pH is known as isoelectric point of the amino acid. It differs with different amino acids and is a characteristic property of it. The iso-electric point of glycine is 6∙0 aspartic acid \[HOOC.CH2.CH(NH2).COOH\] is 2∙8, lysine \[H2N.(CH2)4.CH(NH2)COOH\] is 9∙6 etc.
Q. 26. Give the various colour tests used for the identification of protein.
Ans. Colour Test of Protiens
- (i) Xantho-Protein Test : Protein solution on warming with conc. HNO2 forms yellow colour. This test is only due to tyrosin and tryptophan (amino acid with benzene ring) present in proteins. The yellow colour is produced by the nitration of benzene ring.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-58.png)
- (ii) Biuret Test : Protein solution on treatment with CuSO4 followed by alkalification by NaOH, produces violet colour. The colour is due to the coordination between Cu2+ and C = O ‒NH groups of peptide chains.
- (iii) Millon’s Test : On adding Millon’s reagent (Mercurous and mercuric nitrate in HNO3) to a protein solution, a white precipitate which turns brick red on heating is formed. This test is given by protein containing tyrosine.
- (iv) Hopkins-Cole Test : On adding conc. H2SO4 by the side of test-tube containing protein solution and glyoxalic acid. A violet layer appears at the juncture of the two liquids. The test is given by tyrosine containing protein.
- (v) Ninhydrin Test. When protein (containing α-amino acids) is boiled with a dilute solution of ninhydrin, a blue or violet colour is formed.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-59.png)NINHYDRIN
BLUE/VIOLET COMPLEX
Q. 27. What is the relation between amino acids and proteins ?
Ans. Proteins are built up of a large number of α-amino acid molecules interlocked by elimination of water between NH2 of one α-amino acid molecule and COOH of the another forming a peptide linkage. For illustration.
H R O H R' O H R" O
I I II ;.................., I I II ................... I I II
H—N—C—C— i OH—H i—N—C—C—ÖH + H i—N—C—C—OH
H H H
Peptide linkage
_files/AU20B.20Sc.20III20Chemistry20II(OK)-60.png)
Fig. A Protein fragment
On the other hand, when proteins are hydrolysed by acids, alkalins by proper enzyme, then a mixture of α-amino acids is obtained. This shows that proteins are made up of α-amino acids.
Protein gives Biuret test, which further proves that it contains peptide (—CONH) linkages. This linkage is formed by mutually reacting with amino (—NH2) group and a carboxylic group (—COOH).
A dipeptide formed by the ion of two molecules of α-amino acids still contains a —NH2 group at one end of the molecule and a —COOH group at the other end. These groups further reacts with other molecules of α-amino acids forming tri, tetra and ultimately polypeptides. Fischers on studying the properties of polypeptide, found it similar to proteins. This indicated that amino acids are unit of proteins. It is thought that simple proteins are amino acid molecules or their simple multiples.
Q. 28. Write a short note on denaturation of proteins.
Ans. Denaturation of Proteins :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-61.jpg)Native Form of The Protein
Energetically the most stable state of a protein is called its native state or native form. The native state of a protein is dictated by the amino acid sequence in the protein. Proteins are very sensitive to heat, acids, alkalies and even to the Denatured electrolytes. Properties of the globular Protein proteins change altogether on heating or
Fig. Denaturation of a protein
on treatment with acids/alkalies or electrolytes. On heating, water-soluble
globular proteins precipitate out due to the formation of water-insoluble fibrous proteins. The coagulated protein is called denatured protein.
The process which leads to the changes in the physical and biological properties of proteins without affecting its chemical composition, is called
denaturation of protein. Denaturation causes changes only in the secondary, and tertiary structures of proteins. Denaturation may be reversible in some cases. Denaturation is also caused by the following factors :
- (a) Change in the pH.
- (b) Increase in temperature.
- (c) Presence of acids, alkalies and/or salts.
- (d) Exposure to ultraviolet rays and X-rays.
The most common example of the denaturation of proteins are :
- (i) Boiling of egg : Boiling of an egg involves coagulation of the albumin present in the white of egg. When an egg is boiled in water, the soluble globular proteins present in the white of egg get converted into insoluble fibrous proteins.
- (ii) Preparation of cheese from milk : Milk protein, lactoalbumin gets coagulated (denatured) by the addition of an acid (lemon juice) to the hot milk.
The reversal of denaturation is known as renaturation or refolding. In case denaturation is effected by heat, renaturation (reversible denaturation) may carried out by very slow cooling (not rapid cooling). The process of this type of renaturation is known as annealing.
Q. 29. What are polymers ?
Ans. Polymers. Polymers are high molecular weight compounds whose structure is made up of a large number of simple repeating units. The repeating units are generally obtained from low molecular weight simple compounds known as monomers. The phenomenon is known as polymerisation. The formation of polythene from ethene is an example of polymerisation.
Repeating Units
CH2=CH2 JWEEl^il^ —CH—CH2—CH—CH2—CH—CH2—
Ethylene Polyethylene
(Monomer) (Polymer)
Polymers which are formed from only one kind of monomer are called Homopolymers, and which are prepared from more than one kind of monomer are called copolymers.
The properties of polymers are governed almost entirely by the bult of the polymer molecule rather than end groups.
Polymers are said to be linear if the repeating units are joined together like links in a chain. The chains may be branched, or joined together by crosslinks or both. More extensive cross-linking polymer is shown in figure given below :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-62.jpg)BRANCHED POLYMER CROSS-LINKED POLYMER
Polymer structure. Each circle represents a single repeating unit.
Q. 30. How the following polymers are synthesised ? Give their important uses.
- (i) Nylon (ii) Cellulloid.
Ans. (i) Nylon-6, 6 : It is obtained by heating adipic acid with hexamethylene diamine under nitrogen at 200°C.
HOOC(CH2)4CO|OH + h| NH(CH2)6NH2
Heat
▼
HOOC(CH2)4 CONH(CH2)5 +h2o
further reaction at each end CO(CH2)4 CONH(CH2)5NH^
NYLON-6,6
It is used in making fibres, for clothing and carpenting, filament for fishing lines and ropes, bristles for brushes, molded objects such as gearls and bearings.
- (ii) Celluloid : It was the first synthetic plastic. It is prepared by heating cellulose dinitrate with camphor to about 75°C in presence of some ethyl alcohol. It is colourless, tough and soluble in acetone. It can be coloured by mixing dyes and can be easily moulded to any shape when hot. It is used for making toys, bags, bangles, combs, balls and other decorative pieces. It is not explosive but is highly inflamable.
H2SO4
(C6H10O5)n+HNO324 \[C6 H8O5NO2)2\]n
Cellulose Cellulose Dinitrate
Q. 31. Predict the signal pattern of the CH3 protons in the NMR spectra of the following compounds :
O
- (a) CH3CHBr2 (b) CH3–C–OH
OH
- (c) CH3CH2Br (d) CH3–CH–CH3
Ans. Apply the (n + 1) rule. (a) The CH3 group has one neighbouring proton and would therefore be a doublet; (b) The CH3 group has no neighbouring proton, and would therefore be a singlet; (c) The CH3 group has two neighbouring equivalent protons and would therefore be a triplet; and (d) The two CH3 groups are equivalent and have one neighbouring proton. The CH3 signal would therefore be a doublet.
Q. 32. How will you distinguish between 1, 3-pentadiene and 1, 4-pentadiene by UV spectroscopy ?
Ans. 1, 3-Pentadiene will have higher λmax than 1, 4-pentadiene. This is because 1, 3-Pentadiene is conjugated; whereas 1, 4-pentadiene is not.
Q. 33. How will you distinguish between benzene and anthracene by UV spectroscopy ?
Ans. Anthracene will have higher λmax than benzene. This is because anthracene is more conjugate than benzene.
Q. 34. Two isomeric dienes (A) and (B) having the molecular formula C5H8 absorb at λmax 223 nm and λmax 178 nm respectively. Write the structures of the two isomers.
Ans. (A) is CH2 = CH ‒ CH = CH ‒ CH3 ; 1, 3-Pentadiene
(B) is CH2 = CH ‒ CH2 ‒ CH = CH2; 1, 4 Pentadiene
Q. 35. Arrange the following compounds in the increasing order of their UV absorption maxima :
- (a) Ethylene (b) Naphthalene
- (c) Anthracene (d) 1, 3-Butadiene
Ans. (a) < (d) < (c) < (b).
Q. 36. Give a structure consistent with the following NMR data :
Molecular formula = C3H5Cl3
- (a) Singlet at δ 2.20, 3H (b) Singlet at δ 4.02, 2H
b a
Ans. The compound is CH2Cl ‒ CCl2 ‒ CH3.
Q. 37. The NMR spectrum of compound (A), C5H12, gives only one signal, a singlet. What is the structure of (A) ?
Ans. The molecular formula, C5H12, indicates that (A) is an alkane. The single signal in the NMR spectrum indicates that all hydrogens are equivalent. The only possible C5H12 isomers that fits the data is (CH3)4C (2, 2-Dimethylpropane).
Q. 38. Suggest a structure consistent with the following NMR data : Molecular formula : C9H12
- (a) Singlet at δ 6.78, 3H (b) Singlet at δ 2.25, 9H
Ans. The compound is : a
_files/AU20B.20Sc.20III20Chemistry20II(OK)-63.png)
Mesitylene
Q. 39. How would you distinguish between the following pair of compounds by NMR spectroscopy ?
O
() Il
- (a) CH3‒C‒CH3
O
() II
- (c) CH3‒C‒CH3
O
.
and CH3CH2‒C‒H
O
d
and CH3‒C‒OCH3
Ans. (a) CH3COCH3 will give only one signal, while CH3CH2CHO will give three signals; and (b) CH3COCH3 will give two signals.
Q. 40. The NMR spectrum of compound (A) C2H6O, shows one signal only, a singlet. Deduce the structure of (A).
Ans. (A) is CH3 ‒ O ‒ CH3.
Q. 41. (a) Mixture of furan and ammonia is treated with steam in presence of Al2O3.
- (b) Pyroll is heated with chloroform and caustic potash.
- (c) Pyroll reacts with diazonium salts.
Ans. (a) Pyroll is obtained
CH—CH
Il II
HC CH
\+ NH,
A1,O: stan
_files/AU20B.20Sc.20III20Chemistry20II(OK)-64.png)N
H Pyroll
- (b) Pyroll 2-aldehyde is obtained
_files/AU20B.20Sc.20III20Chemistry20II(OK)-65.png)
CHCl3 + 3KOH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-66.png)CHO + 3KCl + 2H2O
H
- (c) Coupling takes place at 2-position in weak acid solution but in alkaline medium coupling takes place at 2 and 4 position and bi-diazonium compound is obtained.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-67.png)H + Cl N2 C2H5
Weak acidic -----------►
Solution
_files/AU20B.20Sc.20III20Chemistry20II(OK)-68.png)N=N—C6H5 + HCl
H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-69.png)\+ 2 C6H5N=N-Cl
N
H
Alkaline solution ‒2HCl
C6H5 -N=N- —N=N-C6H6
N
H
Q. 42. What do you mean by oils and fats ? Give the difference between them.
Ans. Oils and Fats : Oils and fats are the triesters of glycerols (a trihydric alcohol) with long chain monocarboxylic acids (usually 12 to 20 carbons). These esters are known as Triglycerides or Triacyl glycerols or Glyceryl esters. The following general structure represents an oil or a fat.
CH2‒O‒COR1
CH‒O‒COR2
CH2‒O‒COR3
Glycerol Acyl group part (Acid part)
The three acyl groups present in the molecule may be same or different. When all the three are same, the triester is known as simple glyceride (i.e., the three OH groups of glycerol are esterified with the same acid) and if two or three are different, the triester is known as mixed glyceride (i.e., the three OH groups of glycerol are esterified with two or three different acids).
CH ‒O‒COR
2
CH2OCOR1 CH2OCOR1
CH‒O‒COR
CHOCOR CHOCOR
22
CH2‒O‒COR
CH2OCOR3 CH2OCOR3
Simple glyceride
V
Mixed glycerides
The monocarboxylic acids that form ester chains in natural oils and fats may be saturated or unsaturated.
Difference between oils and fats–Oils and fats belong to the same chemical group, yet they are different in their physical state.
- 1\. Oils are liquids at ordinary temperature (below 20°C) while fats are semi-solids or solids (their melting points are more thant 20°C). A substance may be classed as fat in one season and oil in another season or the same glyceride may be solid at a hill station and liquid in plains. Thus, this distincition is not well founded as the melting points depend on climate and weather.
- 2\. The difference in oils and fats is actually dependent on the nature of monocarboxylic acids present in the glyceride. Oils contain large proportion of the glycerides of lower carboxylic acids (e.g., butyric acid, caprylic acid and caproic acid) and unsaturated fatty acids e.g., oleic, linoleic and linolenic acids) while fats contain a large proportion of glycerides of higher saturated carboxylic acids (e.g., palmitic, stearic acids).
Lard (fat of hogs) is a solid fat and its composition in terms of fatty acids produced on hydrolysis is approximately 32% palmitic acid, 18% stearic acid, 45% oleic acid and 5% linolenic acid. Olive oil on the other hand, contains 84% oleic acid, 4% linoleic acid, 9% palmitic acid and 3% stearic acid.
Q. 43. Write short notes on the following :
- (a) Hydrogenation and Hydrogenolysis
- (b) Hydrolysis of oils and fats
- (c) Rancidification.
Ans. (a) Hydrogenation : The vegetable oils containing glycerides of unsaturated fatty acids undergo catalytic hydrogenation with hydrogen at low pressure in presence of finely divided nickel. This results in the formation of saturated glycerides which are solid fats at room temperature. The hydrogenation process is called Hardening of oils.
ch2ococ
17H33 CH2OCOC17H35
Li । Z 1/
ch2ococ
17^33 ^ ' CH2OCOC17H33
ch2ococ
17H33 CH2OCOC17H35
Olein (liquid-oil) Stearin (A solid fat)
Partial hydrogenation of oils is used for the manufacture of vegetable Ghee. Complete hydrogenation would produce a hard brittle fat.
Hydrogenolysis : If excess of hydrogen, under pressure, is passed through oil or fat in the presence of copper-chromium catalyst, it is converted into glycerol and higher aliphatic alochols. This process is termed hydrogenolysis.
CH2OCOC17H35
CH2OCOC17H35
CH2OCOC17H35
6H2
200 atm.
CH OH 2
CHOH
CH2OH
\+ 3C17H35CH2OH
Oetadecyl alcohol
Glycerol
- (b) Hydrolysis : They are hydrolysed by heating with superheated steam, acids or alkalies.
- (i) By superheated steam : Under a pressure of about 8 atmospheres at 170°C in the presence of lime, zinc oxide or magnesia, oils and fats are hydrolysed. On cooling, free fatty acids separate with some calcium soap (or zinc or magnesium soap).
CH2‒O‒CO‒C17H35
CH‒O‒CO‒C17H35+ 3H2O
CH2‒O‒CO‒C17H35
Stearin
CH2OH
CHOH + 3C17H35COOH CH OH Stearic acid
Glycerol
- (ii) Base-hydrolysis : When fats and oils are heated with NaOH (or KOH) solution, the products are glycerol and sodium or potassium salts of fatty acids. The sodium or potassium salts of fatty acids are called soaps and the base-hydrolysis of fats or oils is termed as saponification.
CH O‒COR
2
CH OH
2
CHO‒CO‒R + 3NaOH
CHOH
CH2O‒COR
CH2OH
Fat or oil
Glyceol
\+ 3RCOONa
Salt of fatty acid (Soap)
- (iii) Enzyme hydrolysis : Enzyme like lipase, when added to an emulsion of fat in water, hydrolyses it into acid and glycerol in about two or three days.
- (c) Rancidification : On long storage in contact with air and moisture, oils and fats develop unpleasant smell. The process is known as rancidification. It is believed that rancidification occurs due to following chemical changes.
- (i) Enzyme hydrolysis : Producing bad smelling lower fatty acids.
- (ii) Oxidation of unsaturated acids : Producing aldehydes and ketones.
Q. 44. What do you mean by invert sugar ?
Ans. Invert Sugar : Canesugar (sucrose) is dextrorotatory. Specific rotation of a freshly prepared solution of canesugar is +66.5°. On hydrolysis, it gives dextrorotatory glucose (α = +52.7°) and laevorotatory fructose (α = ‒ 92.4°).
C12H22O11 + H2O -----→ C6H12O6 + C6H12O6
sucrose glucose fructose
(canesugar) dextrorotatory laevorotatory
(α = + 66.5°) (α = + 52.7°) (α = ‒ 92.4°)
The resulting solution which contains equimolar amounts of glucose and fructose is laevorotatory with specific rotation of ‒39.7°. Thus, the sign of rotation changes after hydrolysis. That is why, hydrolysis of canesugar is commonly called inversion of sugar. The equimolar mixture of glucose and fructose is called invert sugar.
Q. 45. Write a short note on epimerisation.
Ans. Epimerisation—G-glucose when warmed with dilute alkali solution rearranges to give a mixture of D-glucose, D-mannose and D-fructose.
CHO
CHO
CH2OH
|
H — C — OH
|
X-- HO — C — H
|
X-- C = O
|
(CHOH)3
| (CHOH)3
|
(CHOH)3
|
CH2OH
D-glucose
|
CH2OH
D-mannose
|
CH2OH
D-fructose
The same mixture is obtained if the starting material is fructose or mannose. The above equilibrium is established, viz. the endiol starting from any of these three hexoses. It can be shown as follows :
CHO |
HO — C — H CH2OH
|| ||
H — C — OH
C — OH X--C = O
glucose
trans-diol fructose
HO — C — H CHO
|| ||
HO — C X-- HO — C — H
cis-diol mannose
D(+) glucose and D(+) mannose are epimers. Epimers are a pair of distereomers that differ only in the configuration about a single carbon atom. This phenomenon is called epimerisation.
This rearrangement reaction of a monosaccharide in weakly alkaline solutions to give a mixture of isomeric sugar is known as Lobry de Bruyn Van Ekestein rearrangement.
O\\ /H
C
O\\ /H
C
H—C—OH
HO—C—H
HO—C—H
HO—C—H
H—C—OH
H—C—OH
H—C—OH
H—C—OH
CH2OH
CH2OH
Epimers
Q. 46. Give the mechanism of osazone formation.
Ans. Mechanism of Osazone Formation : We are familiar by the fact that when glucose reacts with excess of phenyl hydrazine, glucosazone is formed. The reaction takes place as follows :
CH2OH CH2OH +C6H5NHNH2
(II molecule)
(CHOH)3 (CHOH)3
–C6H5NH2,–NH3
CHOH CHOH
CH OH 2
(CHOH)3
C = O
CHO + H2N.NHC6H5 CH = N.NHC6H5
CH = N.NHC6H5
(I molecule)
C6H5NHNH2
(III molecule)
CH OH 2
(CHOH)3
C = N.NHC H 65
CH = N.NHC6H5 glucosazone
Mechanism—Waygand suggested that osazone is produced as Amadari rearrangement as given ahead :
CH = O
C6H5NHNH2
CH = N.NHC6H5
CH
= N.NHC6H5
CHOH
on -
—^ row
1st mol.
C–OH ^—
C–OH
glucose
phenyl hydrazone of glucose
/
CH–NH.NHC H CH .NH.NHC H 65 2 65
CH–NH.NHC6H5 CH = N.NHC6H5
–C6H5NH2
CH = N.NHC H CH = NH.NH
6 5 or
CH = NH C = N.NHC6H5
CH = NH.NHC H 2 65
C H NHNH
6 5 2C = O
2nd mol.
C6H5NHNH2 3rd mol.
CH = N.NHC6H5
C = N.NHC6H5
Glucosazone
Q. 47. Write short note on reformat sky reaction.
Ans. Reformatsky Reaction : This reaction involves the interaction of an a-bromoester with carbonyl compound (aldehyde or ketone) in presence of metallic zinc. An intermediate organozinc compound is first formed. This then adds on to the carbonyl group of aldehyde or ketone producing P-hydroxyester. These can be redily dehydrated to a, P-unsaturated acids by heating with concentrated sulphuric acid. It is sometimes necessary to activate zinc by adding a few crystals of iodine or mercuric bromide or copper.
(i) Formation of organozinc compound,
benzene $+ $
Br—CH2—COOC2H5 + Zn I BrZn—CH2COOC2H5
ethyl bromoacetate Organozinc compound
(ii) Addition to carbonyl group,
5+
ZnBr
CH3\\ 5+
> C = O + CH2-COOC2H5 ^
CH3
–+
O ZnBr
CH3
3 C–CH2–COOC2H5
CH
3 Intermediate adduct
(iii) Acid hydrolysis of the adduct,
[OZnBrOH](#bookmark208)
[CH3 +](#bookmark209)
[3 C–CH2 –COOC2H5 —H—2O—/H3](#bookmark210)
CH3
CH3
OH
5 + Zn Br
ethyl -hydroxy isovalerate
(iv) Dehydration to form a, ß-unsaturated ester,
OH
CH,\\ I H OS A
>C-CH, -COOCÄ - 4' >
CHf
_files/AU20B.20Sc.20III20Chemistry20II(OK)-70.png)
C = CH-COOC2H5+ H2O
ethyl ß, ß-dimethylacylate
It may be noted that the formation of organozinc compound is to be preferred for the above reaction where ester group need to be protected. Grignard reagent if used would react with ester group bonded to magnesium, that the end-product then could not be an unsaturated ester.
Q. 48. What do you mean by thioethers ? How they are named?
Ans. Thioethers : Sulphur analogs of ethers are termed as thioethers. Just as ethers are regarded as derivatives of water, thioethers could be considered as derived from hydrogen sulphide.
H—O—H--— > R—O—R
\+ 2R
H—S—H--^ R—S—R
\+ 2R Thioether
In thioether sulphide group (–S–) is treated as functional group.
The common names of thioethers are obtained by writing the names of alkyl groups followed by the word sulphide.
CH3 – S – CH3 CH3 – S – C2H5
Dimethyl sulphide Ethyl methyl sulphide
In IUPAC system, thioethers are named as alkyl thioaklane. Smaller alkyl group bonded to sulphur atom is regarded as the substituent.
SCH3
|
CH3—S—CH2CH2CH3 CH3—CH—CH2CH3
1-methyl thio-propane 2-methyl thiobutane
Q. 49. Give the oxidation reactions of thioalcohols.
Ans. Oxidation–The most significant difference in the chemical behaviour of thiols and alcohols is the ease with which thiols are oxidised.
(i) Mild oxidising agents such as halogens or H2O2 convert it to disulphides.
2C2H5SH + I2--------> C2H5 - S - S - C2H5 + 2HI
2C2H5SH + H2O2------> C2H5 - S - S - C2H5 + 2H2O
This conversion (Thiol----> Disulphide) is extremly important in biochemistry, where disulphide bridges form the cross links between protein chains that helps stabilizing the three dimensional conformations of proteins.
Protein — SH + HS — Protein-----> Protein —S—S— Protein
A cross-linked protein
- (iii) Strong oxidising agents such as KMnO4 or conc. HNO3 convert these into sulphonic acids.
KMnO
CHSH + 3\[O\]----------» CHSO3
- 2 5 HNO 2 5 3
Q. 50. What are thioalcohols ? Give two methods for their preparation.
Ans. Thiols or Thioalcohols : The sulphur analogs of alcohols are called thiols. Just as alcohols are regarded as alkyl derivatives of water, thiols could be considered as derived from hydrogen sulphide.
H—O—H
\--H-» R—O—H +R Alcohol
H—S—H
\--H-» R—S—H +R Thiol
In thiols—SH group is functional group. It is called sulphydryl, Mercapto or thiol group. Like hydrogen sulphide thiols are weakly acidic. They react with mercuric ions to form insoluble salts. Therefore, thiols were given the name mercaptans.
Preparation
- (i) From Grignard’s reagent : Grignard reagent with sulphur followed by acidic hydrolysis gives thiols.
S H O/H+
CH3CH2MgBr-------» CH3CH2SMgBr---------» CH3CH2SH
Propene Ethane thiol
- (ii) Hydrogen sulphide on addition to alkenes in presence of sulphuric acid as catalyst gives secondary thiols. This reaction follows Markonikov rule.
CH3
H SO |
CH3—CH = CH2 + H — SH------» CH3—CH2—SH
Propene 2-propane thiol
Q. 51. Give the reaction of aldehyde and ketones over thiols.
Ans. Reaction : with aldehydes and ketones–Thioacetals and thioketals are obtained. This reaction occurs in presence of HCl.
O SC2H5
II HC1 ।
CH3-C-H + 2C2H5SH —Cl» CH3-C-H
SC2H5
Diethyl methyl mercaptal
O SCH
H HCl „ J 25
CH3–C–CH3 + 2C2H5SH —HC→l CH3–C–CH3
SC2H5
Diethyl dimethyl mercaptal
Diethyl dimethyl mercaptal on oxidation with KMnO4 gives sulphonal which is an important hypnotic.
SC2H5 SO2C2H5
CH3–C–CH3 + 4\[O\] →CH3–C–CH3
SC2H5 SO2C2H5
Sulphonal
Q. 52. Write a short note on Sulphaguanidine.
Ans. Sulphaguanidine—Guanidine contains an imidine group and is the imido analogoue of urea. Its structure is as follows :
NH2
||
H2N — C — NH2
Guanidine
Sulphaguanidine is synthesised from acetanilide as given ahead :
NHCOCH3
_files/AU20B.20Sc.20III20Chemistry20II(OK)-71.png)Acetanilide
NHCOCH3
_files/AU20B.20Sc.20III20Chemistry20II(OK)-72.png)
NH
SO2 Cl – HI NH–C–NH2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-73.png)SO2NH–C–NH2
NH2
Hydrolysis
dil HCl V NH
SO2NH–C–NH2
Sulpha guanidine
Because it is absorbed by intestinal tracts in very least amount, therefore its maximum amount is used in the cure of bacillary dysentery.
Q. 53. What do you mean by sulphonation ? Give its mechanism.
Ans. Sulphonation : Reactions of aromatic hydrocarbons with fuming H2SO4 to yield sulphonic acid, is called sulphonation. In this reaction sulphonated arenes are obtained.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-74.png)
Mechanism : The mechanism of the reaction involves following steps :
Step 1. In this step, electrophile is formed. In this reaction, the electrophile is sulphur trioxide (SO3). In conc. H2SO4, SO3 is produced as follows :
2H2SO4 ====== SO3 + H3O+ + HSO4–
In fuming H2SO4, this step is unimolecular because the dissolved SO3 reacts directly.
Step 2. Electrophile attacks the benzene ring to give a carbonium ion.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-75.jpg)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-76.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-77.png)+O
O
S – O– –––
_files/AU20B.20Sc.20III20Chemistry20II(OK)-78.png)
Intermediate
The intermediate carbonium ion is resonance hybrid of following structures. It is stabilized.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-79.png)H
SO3–
Step 3. In this step, a proton is removed :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-80.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-81.png)SO3 + HSO4 –→
_files/AU20B.20Sc.20III20Chemistry20II(OK)-82.png)SO3
\+ H2SO4
(from step 1)
Step 4.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-83.png)SO3–
\+ H3O+
(from step 2)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-84.png)SO3H
\+ H2O
Benzenesulphonic acid
Q. 54. What are heterocyclic compounds? Explain with suitable example.
Ans. (a) Heterocyclic compounds : Those cyclic compounds in which other atoms except carbon are present are termed as heterocyclic compounds.
The atoms of other elements are known as hetero atoms. Nitrogen (N), Sulphur (S) and oxygen are treated as hetero atoms. Some examples are :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-85.png)Furan Thiphene Pyrrole Pyridine Quinoline
Q. 55. (a) Pyroll reacts with hydroxyl amine.
- (b) Pyridine reacts with mercuric and platinum salts.
- (c) Pyridine is treated with halogens.
Ans. (a) Pyroll ring breaks down and succinyl aldehyde oxime is obtained.
CH = CHx CH2-CH=N-OH
| >NH +2NH2OH---► | +NH3
CH = CH / CH2-CH=N-OH
Succinyl aldehyde oxime
- (b) Pyridine gives double salts with chlorides of mercuric and platinum.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-86.png)\+ HgCl2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-87.png)
N HgCl2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-88.png)\+ P + Cl4 +2HCl
♦ (C5 H5 N. P+Cl4.HCl)2
- (c) (i) At ordinary temperature addition reaction takes place and corresponding dihalide is obtained. With Br2 C5H5N+Br, Br– is obtained.
- (ii) At 300°C and in presence of a catalyst it gives a mixture of 3-bromo pyridine and 3, 5-dibromo pyridine.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-89.png)300°C/Br2
Charcoal
_files/AU20B.20Sc.20III20Chemistry20II(OK)-90.png)
+
_files/AU20B.20Sc.20III20Chemistry20II(OK)-91.png)
3-bromo puridine
3, 5-dibromo pyridine
- (iii) At 500°C it give a mixture of 2-bromo and 2, 6-dibromo pyridine.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-92.png)
2-bromo pyridine 2, 6-dibromo pyridine
Q. 56. What happens when ?
- (a) Thiophene reacts with acetyl chloride in presence of SnCl4.
- (b) Thiophene is shaked with conc. H2SO4 and a crystal of its atom.
Ans. (a) 2-Acetyl thiophene is obtained
_files/AU20B.20Sc.20III20Chemistry20II(OK)-93.png)
\+ Cl. CO CH3 SnCl4
_files/AU20B.20Sc.20III20Chemistry20II(OK)-94.png)CO. CH3 + HCl
2-Acetyl thiophene
- (b) When thiophene is shaked with conc. H2SO4 and ancrystal of isotin, indophenin dye is produced due to which a blue colour is obtained.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-95.png)II II
\+ HC CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-96.png)
CH—CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-97.png)–2H2O
CH CH
HC CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-98.png)Indophenin
CC
S
_files/AU20B.20Sc.20III20Chemistry20II(OK)-99.png)
Q. 57. What is Grignard’s reagent. How it is prepared ?
Ans. Girgnard reagent–The alkyl magnesium halides. R — Mg — X were introduced by the French chemist, Victor Grignard, in 1900 and are termed as Grignard reagent. In these compounds a metal atom is directly linked with carbon atom.
/CH, CH3
Mg<( Mg<C
I Br
Methyl magnesium iodide Ethyl magnesium bromide
These are extremely valuable synthetic reagents and are being used on a large scale at present.
Method of Preparation of Grignard’s Reagent
A Grignard reagent is prepared in laboratory by the interaction of dry Mg-turnings with alkly halide in presence of dry alcohol-free ether.
RX + Mg -----→ RMgX
It has been found that for a given alkyl radical the ease of formation of Grignard reagent is Iodide > Bromide > Chloride
Similarly for a given halogen atom, the ease of formation is greater with smaller alkyl group than with bigger ones, e. g. :
CH3I > C2H5I > C3H7I
Since tertiary alklyl iodides are readily decomposed to yield olifins, tertiary alkyl chlorides are used in their place.
Q. 58. How organozinc compounds are prepared ?
Ans. Organozinc Compounds : Organozinc compounds are less reactive thang Grignard reagents (RMgX) and are sometimes used for organic synthesis in place of Grignard reagent. Frankland discovered two derivatives of zinc.
CH3 –Zn – CH3 C2H5 – Zn – C2H5
dimethylzinc diethylzinc
Dialkylzincs are prepared by heating alkyl iodides with zince in an atmosphere of CO2, and then distilling the product (alkylzinc iodide) in an inert atmosphere of CO2.
CO2
C2H5 – I + Zn ---→ C2H5 – ZnI
ethyl iodide ethylzinc iodide
CO2
2C2H5 – ZnI -----→ C2H5 – Zn – C2H5 + ZnI2
distil diethylzinc
Q. 59. How many NMR signals do you expect from each of the following compounds ? Indicate also the splitting pattern of the various signals.
- (a) CH3OCH3 (b) CH3OCH2CH3 (c) CH3CH2OH
Ans. (a) One signal (singlet); (b) Three signals (singlet, quartet, triplet); and (c) Two signal (singlet, quartet).
Q. 60. How will you distinguish between the three dibromobenzenes by their NMR spectra ?
Ans. o-Dibromobenzene will show two peaks; m-dibromobenzene will
_files/AU20B.20Sc.20III20Chemistry20II(OK)-100.jpg)Br
ba o-Dibromobenzene (two peaks)
1 : 1
aa
p-Dibromobenzene (one peaks)
cb m-Dibromobenzene (three peaks)
1 : 2 :1
Q. 61. Write a note on auxochrome.
Ans.
Auxochrome
These are also called colour intensifier. These are the groups which deepen the colour of the group present in chromogen. Some auxochromes are :
—NH2, —NHR, —NR2, —OH, —OR etc.
Auxochromes not only deepen the colour but also provide dyeing property to chromophore. For example, Azobenzene is a chromogen, but not a dye. If by any mean —OH group or —NH2 group introduced in para position, then it will behave as a dye.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-101.png)
p-hydroxy azobenzene (dye)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-102.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-103.png)/>-amino azobenzene (dye)
Q. 62. What do you mean by chromophore and chromogen ?
Ans. (i) Chromophore : In every coloured substance there is at least one unsaturated group or a group having multiple bonding due to which the compound is coloured. Such types of unsaturated groups are termed as chromophores. Some chromophores are :
>C=C,
_files/AU20B.20Sc.20III20Chemistry20II(OK)-104.png)—SO3H, etc.
—N = N—
—N = O,
- (ii) Chromogen : These are also known as colour generator compounds, containing a chromophore are known as chromogen. In Nitrobenzene (C6H5NO2), —NO2 is chromophore. It is a liquid having yellow colour. Similarly, m-di-nitrobenzene is a substance of yellow coloured.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-105.png)
m-di-nitrobenzene (yellow colour)
Q. 63. Select chromophore, chromogen and auxochrome from the following :
(i)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-106.png)
(ii)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-107.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-108.png)
CH3
ch3
Ans. (i) Chromophore :
Chromogen
_files/AU20B.20Sc.20III20Chemistry20II(OK)-109.png)
yCH
Auxochrome : p—N\\„u
- (ii) Chromophore :
Chromogen
_files/AU20B.20Sc.20III20Chemistry20II(OK)-110.png)
/CH3
Auxochrome :
Section ‘C’
Essay Type Questions
Q. 1. Give the α-helix structure and β-flat sheet and β-pleated sheet structure for proteins.
Ans. (1) α-Helix structure : This type of structure becomes possible when the R-groups of the amino acids are quite large. The polypeptide chain coils up to form a spiral structure, and the turns of the coil are held together by N–H .... O = C hydrogen bonds. Each –NH group is hydrogen bonded to the oxygen of the carbonyl group of the third amono acid. There are 3.6 amono acid units per turn of the helix, and the pitch of the helix (totat rise of the helix per turn) is 0.54 nm.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-111.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-112.png)
Fig. The α-helix structure Fig. The flate sheet structure
of proteins (secondary structure) of proteins.
The α-helix can be right handed of left handed. It has been found that the right handed α-helix structure resulting from the naturally-occuring L-amino acids is more stable.
Many firbous protiens such as α-keratin in hair, nails, wool, skin and myosin in muscles have α-helix structure. Stretching property of human hair is due to the helical structure of α-keratin in hair.
- (2) β-Flat sheet and β-pleated sheet structure : The flat sheet structure becomes possible in proteins containing amino acids with small R-group or, with H. In this structure, the long polypeptide chains lie side by side in a zigzag manner with alternate R-group on the same side. Each polypeptide chain is held to the two adjacent chains (one on each side) by intermolecular hydrogen bonds. This arrangement leads to a flat sheet structure. The flat sheet structure of protein is shown in Fig.
When the R-group is of moderate size, then the polypeptide chains contract slightly to accommodate the R-group. The adjacent polypeptide chains are held together by N—H.....O = C hydrogen bonds. This gives rise to a pelated sheet structure as shown below in Fig. These pleated sheets may be parallel or aniparallel.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-113.png)
(INDIVIDUAL POLYPEPTIDE CHINS)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-114.png)
Fig. The β-pleated sheet structure in proteins.
The silk protein firboin has β-pleated sheet structure. It is due to the pleated structure that silk fibre cannot be stretched. However, it can be bent/ folded easily.
Q. 2. Give the constituent of nucleic acids.
Or
What do you mean by nucleosides and nucleotides.
Or
Write short note on nucleosides and nucleotides.
Ans. Constituents of Nucleic acids : Nucleic acids contain the following three constituents :
- (a) a pentose sugar (ribose or deoxyribose) :
- (b) a nitrogen containing heterocyclic base : a purine or a pyrimidine base;
- (c) a phosphate group.
Sugar units in nucleic acids : The structure of the two pentose sugars present in nucleic acids are given below :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-115.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-116.png)
Ribose 2-Deoxyribose
Both these sugars can exist in α- and β-forms.
Nitrogen bases in nucleic acids : There are two types of bases in nucleic acids. These bases are purines and pyrimidines. The names and structures of the bases in nucleic acids are given below :
Pyrimidine bases : Uracil (U), Thymine (T), Cytosine (C)
Purine bases : Adenine (A), Guanine (G)
The structural formulae of these bases are as follows :
Purines :
N
NH2
C
C'N%
C CH
O
HC ^CC^N H
C
^ \\C/N
HN C N CH
NC C \\
H
Adenine (A)
Pyrimidines :
Guanine (G)
N
O = C
NH2
O
O
C
C
C
N
H
CH
CH
HN
O = C
N
H
C—CH
3
HN
CH
CH
O = C
CH
N
H
Cytosine (C)
Thymine (T)
Uracil (U)
Phosphate group : Phosphate group is present at 5C of the sugar unit. The repeat units in the nucleic acid chain are linked through phosphate groups.
The nucleosides and nucleotides : Nucleic acids contain a pentose sugar, a nitrogen contianing heterocyclic base, and a phosphate group. These are bonded to each other in an ordered way. The smaller segments of a nucleic acid chain are called nucleosides and nucleotides. These are described below.
- (1) Nucleosides : The Base-Sugar unit in any nucleic acid chain is called a nucleoside. Thus, a nucleoside contains a pentose sugar and a nitrogen base. In a nucleoside, 1-position (N atom) of the pyrimidine base, or 9-position (N atom) of a purine base is linked to 1C of the sugar by a β-linkage.
The nucleosides are named after the name of the base, attached at carbon atom number-1 of the sugar unit. Thus, the nucleoside consisting of the sugar ribose, and the base adenine is called adenosine. Depending upon the sugar present, there are two types of nucleosides, viz.,
Sugar : Ribose Nucleoside : Ribonucleoside
Sugar : Deoxyribose Nucleoside : Deoxyribonucleosides
- (2) Nucelotides : The Base-Sugar-Phosphate unit is called a nucleotide. Thus, a nucleotide contains all the three components of nucleic acids. The phosphate group is attached to one of the – OH groups (generally, 5C – OH) of the sugar unit of the nucleoside by an ester bond. Thus, nucleotides are the
phosphate esters of nucleosides. A nucleotide can be described by the structure
shown in fig.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-117.png)
Nucleotides are named on the basis of sugar present in it. Since, there are only two sugars present in nucleic acids, hence the corresponding nucleotides are,
Sugar : Ribose Nucleotide : Ribonucleotide
Sugar : Deoxyribose Nucleotide : Deoxyribonucleotide
A nucleic acid can be considered as a polymer of nucleotides, which are linked as shown below.
Base Base Base
| | |
.... Sugar-Phosphate ...Sugar-Phosphate ...Sugar-Phosphate The nucleotide having only one phosphate group is termed monophosphate. It may, however, link itself with one or two additional phosphate groups. Accrodingly, the nucleotide is termed as diphosphate or triphosphate. The formation of adenosine phosphate is shown below.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-118.jpg)Ribose
Adenosine (nucleoside)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-119.jpg)OH OH
Adenosine-5' phosphate (AMP) (nucleoside)
-h2o -h2po4
O O O 0 0
...... n H,PO4 h II n
HO P O P O P O CH, £! Basc.\_2—?HO P O P O CH, £1 Base
¿H ¿H ¿H Q ¿H ¿H
OH OHH OH OH H
Adenosine-5' triphosphate (ATP) Adenosine-5' diphosphate (ADP)
Q. 3. What are proteins ? How they are classified ? Discuss in brief.
Ans. (a) Proteins. It is a class of most important compounds of living organisms. These are complex mitrogenous substances found in protoplasms of animals and plant cells. They have high molecular weights (10,000 to fifty lacs), no definite m. p. are optically acitve and colloidal in nature. They are built of a large number α-amino acid molecules, joined together by peptide linkage (—CONH—), formed by the elimination of water molecule between NH2, of one amino acid molecule and COOH of the other.
Classification. Proteins are classified either by (i) chemical composition or (ii) molecular shape.
- (i) Classification Based on Chemical Composition :
According to this, they are further sub-divided into (a) simple proteins, and (b) conjugated proteins.
- (a) Simple Proteins. They are made of chains of amino acid molecules only, linked by amide linkage.
Examples are :
Albumins : Found in Egg serum and milk
Globulins : Found in Tissues and serum
Glutelines : Found in wheat and rice
- (b) Conjugated Proteins. They are simple proteins united with nonprotein factor (called as prosthetic group)
Examples are :
Glyco-proteins : Mucin in saliva and prosthetic group is carbohydrate.
Chromo-proteins : Haemoglobin in red blood cells and prosthetic group in coloured ion pigment.
Phospho-protein : Casein in milk and prosthetic group in phosphoric acid.
- (ii) Classification Based on Molecular Shape :
It is sub-divided in two (a) Fibrous proteins, and (b) Globular proteins.
- (a) Fibrous Protein. In it, the polypeptide chains coil about on another and are joined by strong inter chain hydrogen bonds and thus assume the shape of large linear fibres, which are responsible for providing connections, support and structure in living organisms. Insoluble in water.
Examples are :
Keratin : Found in skin, hair and nails.
Myosin : Found in muscles.
Collagen : Found in tendons, cartilages and bones.
- (b) Globular Proteins. They are spherical made of polypeptide chains coiled back and forth forming a compact globular molecule. The polypeptide chains are joined together by weak interchain hydrogen bonds and strong intramolecular hydrogen bonds. Soluble in water acids bases and salt solution. They maintain and regulate life processes in living organisms.
Examples are :
Enzymes : Pepsin (stomach) regulate digestion.
Insulin : Secreted by Pancreas helps in glucose metabolism.
Haemoglobin : Found in Blood, carries oxygen from lungs to different parts of the body.
Antibodies : Gamaglobulin (Blood)—checks infection from outside.
Q. 4. What are condensation polymers? Give their mechanism.
Ans. Condensation Polymerisation. It is also known as step-reaction polymerisation. It involves a series of reactions each of which is independent, of the previous one. In it, the monomer happens to react at more than one functional group.
When an acid that contains more than one —COOH group reacts with an amine that contains more than one —OH group then the products are polyamides and polyesters. For example :
HOOC.(CH2)4.CO |oh + hJnH(CH2)6NH2
Adipic acid A | - H2O Hexamethylene
diammine
HOOC.(CH2)4.CONH.(CH2)6NH2
Polyamide (Nylone 66)
In condensation polymerisation, the monomer molecular combined with the loss of simple molecules, like water, alcohol etc.
Mechanism : In it, when n-molecules of dihydric alcohol react with n-molecules of dihasic acid, the following condensation polymer is obtained.
nHO—CH,—CH2—OH + nHO.OC.(CH,)„.COOH
HO—(—CH,—CH,—O.CO(CHJ,—COO—\],—H + (2n-l) H,0
Condensation polymerisation can also occur with monomers having two functional groups in it. For example, anhydroxy carboxyllic acid, di-basic acid.
In condensation polymerisation simple molecules like water, alcohol, NH3 etc. are separated during reactions.
The condensation polymer formed from a monomer or two different monomers is always linear type. If both reactants contain two or more functional
groups, then the polymer so formed will be three dimensional one. For example, the polymer formed by the condensation of glycerol with a dicarboxylic acid like phthalic acid. It is called is Glyptol.
O
_files/AU20B.20Sc.20III20Chemistry20II(OK)-120.png)O
PHTHALIC ACID
—OH
O
H + H O
O
_files/AU20B.20Sc.20III20Chemistry20II(OK)-121.png)
HO—C
PHTHALIC ACID
_files/AU20B.20Sc.20III20Chemistry20II(OK)-122.png)C = O
C = O
I
PHTHALIC ACID
II
O
_files/AU20B.20Sc.20III20Chemistry20II(OK)-123.png)
O
II
C—O—
O
I —nH2O
C—O—CH2—CH —CH2—O
OO
_files/AU20B.20Sc.20III20Chemistry20II(OK)-124.png)O
GLYPTOL
_files/AU20B.20Sc.20III20Chemistry20II(OK)-125.png)
Q. 5. What do you mean by formaldehyde resins.
Ans. Formaldehyde resins : Formaldehyde resins are typical thermosetting plastics. This class of plastic includes phenol-formaldehyde, ureaformaldehyde and melamine-formaldehyde resins. On heating, these resins become highly cross-linked thereby forming infusibile, and insoluble product.
- (a) Phenol-Formaldehyde resin : Quite often the terms resin and plastic are mis-stated for each other. In fact, the two are different, at least in terms of the working temperature. For example, polystyrene behaves like a plastic above 65°C, but as a resin below 65°. Phenol and its derivatives copolymerize with aldehydes to form useful resins. The best known is bakelite produced by the
condensation of phenol with formaldehyde in the presence of an acid or basic catalyst. When phenol is treated with formaldehyde in acidic medium, polymer obtained is known as novalak. In alkaline medium, the polymer obtained is known as resol. Both of these are linear polymers. True phenol-formaldehyde resin is bakelite.
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-126.png)H
\+ C = O +
H
phenol
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-127.png)o-hydroxybenzyl alcohol
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-128.png)
formaldehyde
Formation of linear polymer :
OH
CH2OH
n
_files/AU20B.20Sc.20III20Chemistry20II(OK)-129.png)
–H2O
OH
OH
CH2OH p-hydroxybenzyl alcohol
polymerization
_files/AU20B.20Sc.20III20Chemistry20II(OK)-130.png)CH2
linear polymer
Formation of cross-linked polymer :
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-131.png)
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-132.png)
Co-polymerization
– H2O
CH2OH
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-133.png)
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-134.png)
OH
CH2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-135.png)
CH2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-136.png)OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-137.png)OH
CH2
CH2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-138.png)OH
part of a bakelite molecule (cross-linked)
Properties and Uses : Cross-linked bakelite is a thermosetting polymer. Soft bakelites with low degree of polymerization are used as bonding glue for laminated wooden planks and in varnishes and lacquers.
High degree of polymerization leads to the formation of hard bakelite which are used for making combs, fountain-pen barrels, phonograph records, electrical goods, formica table-tops and many other products. Sulphonated bakelites are used as ion-exchange resins.
Phenol-formaldehyde resins are used for the manufacture of large scale gear wheels for steel rolling and paper mills, protective castings, and electrical equipments. They can be laminated with paper, leather in order to obtain products having great resistance to impact.
The only disadvantage in these resins is that these are usually dark coloured and hence can’t be used for fabricating white or bright coloured articles.
- (b) Urea-Formaldehyde resin : Urea condenses with aqueous formaldehyde in presence of a basic catalyst e.g., ammonia to form dimethylolurea. Dimethylol-urea is water-soluble colourless product. On heating it under pressure, it undergoes polymerization to give a thermosetting plastic.
O
NH2CONH2 + 2HC HOH2C–NH.CO.NH–CH2OH
ure formaldehyde dimethylol-urea
–CH2–N–CO–NH–CH2NHCO–N–CH2NHCONH-polymerization
– nH2O
–CH2NHCON–CH2–N–CONHCH2NHCO–N–CH2NHCOH–
CH2 CH2
–N–CH2NHCO–N–CH2NHCONHCH2–N–CONH– ure-formaldehyde resin
(c) Melamine-Formaldehyde resin : Melamine and formaldehyde copolymerize to give another polymer called melamine resin.
H2N N
NH2 melamine
H2 H2N NH2
\+ HCHO + HCHO
formaldehyde formaldehyde
NH2
resin intermediate
_files/AU20B.20Sc.20III20Chemistry20II(OK)-139.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-140.png)
polymerization –H2O
H2N
_files/AU20B.20Sc.20III20Chemistry20II(OK)-141.png)NH2
\+ HCHO formaldehyde
NH2
malamine polymer
Q. 6. Explain the mechanism of addition polymerization.
Ans. Addition Polymers (Chain Growth Polymers) : Addition polymers are formed by combination of alkene monomers to produce a single huge molecule only. These reactions are catalysed by peroxides or acids. The reactions require pressures of 1,000 atmospheres at 200°C. Much lower temperatures and pressures can be used with so called Ziegler Catalysis which consists of a trialkyl aluminium and titanium tetrachloride in an inert solvent.
Its examples are polyethylene, polyvinyl chloride, orlon, teflon, polystyrene etc.
Free Radical Mechanism of Addition polymerisation : Free-radical polymerisation is catalysed by organic peroxides or other reagents which decompose to give free radicals. Following steps are involved :
- (1) Chain Initiation : Organic peroxides undergo homolytic fission to form free radicals.
This bond
O
breaks O
R C () } () C R
r o q
h
\* R—C—O
—>R + CO2\]
PEROXIDE
UNSTABLE RADICAL
- (2) Chain Propagation : Free radical produced in the above step adds to an alkene molecule to form a new free radical.
R. + CH2^CH2 ^ R—CH2—CH2
This free radical can attack another alkene molecule and so on.
rch2ch>^ch2—ch2—ch2^ rch2ch2ch2ch2
RCH2CH2+ nCH2= CH2--► R(CH2CH2)2CH2CH2
2R(CH2CH2)nCH2CH2 ---► R(CH2CH2)nCH2CH2
CH2CH2(CH2CH2)nR
- (3) Chain Termination : The above chain reaction comes to halt when two free radical chains combine.
Benzoyl or t-butyl peroxides are common reagents for free radical polymerisation.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-142.png)(CH3)3C—O—O—C(CH3)3 -A^ 2(CH3)3CO.---► (CH3)2C=O + ch3
Phenyle or methyle free radicals will then initiate free-radical polymerisation.
Free-radical Polymerisation can also be promoted mixture of ferrous sulphate and hydrogen peroxide (FeSO4 + H2O2). These two compounds react to produce hydroxyle radicals (HO) which act as chain initiators. Ferrous sulphate and hydrogen peroxide mixture is used as catalyst in the manufacture of Orlon and Teflon.
Q. 7. How following polymers are synthesised ? Give their important application.
- (i) Polythene (ii) Polystyrene (iii) Terylene
Ans. (i) Polyethylene (Polythene) : It is obtained by polymerising ethylene. It is used in the manufacture of house-ware such as buckets and dustbins, carpet backing, packaging materials and cable insulation.
Polymerisation -
nCH2 = CH2 O high \* \[—CH2—CH2—\]n- Extended Lines
2 g or sigma bond
Ethylene Temp. and Polythylene
Pressure
Notice that the isomer (ethylene) contains a double bond and the polymer does not. The electrons of the monomer π-bond have moved and are used to link one monomer unit to another by σ-bonds as indicated by extende lines in abbreviated polymer formula. The backbone of the polymer consists of the carbon atoms that originally formed the double bonds. Nothing is lost. The monomers simply add to each other.
- (ii) Polystyrene : It is obtained by polymerising styrene (C6H5CH=CH2). It is used in the manufacture of food containers, cosmetic bottles, telivision cabinet, plastic cups, packaging material and toys etc.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-143.png)
n
CH = CH2 STYRENE
—CH—CH2— n-POLYSTYRENE
- (iii) Terylene : (Dacron) is the most important polyester. It is obtained by heating ethylene glycol with dimethyl terephthalate at 300°C in the presence of metal oxides. The polymer melt can be spum into fibres and combined with such naturally occurring fibres as cotton. The combination of cotton and polyester yields a fabric (e.g. 60/40 cloth) that dries quickly without wrinkling, yet still retains the coolness and comfort of 100% cotton.
HOCH2 CH2 O H + CH2 O CO
_files/AU20B.20Sc.20III20Chemistry20II(OK)-144.png)COOCH3
Heat
HOCH2 CH2 OOC
_files/AU20B.20Sc.20III20Chemistry20II(OK)-145.png)COOCH3+ CH3OH
further reaction at each end
—OCH2 CH2 OOC
_files/AU20B.20Sc.20III20Chemistry20II(OK)-146.png)COO— n
TERYLENE(DACRON)
Ethylene glycol is manufactured by catalytic oxidation of ethylene to ethylene oxide which is then treated with dilute acid.
O2/Ag \\ h+/H?O
ch2 = CH,^^ ch2 —ch2--^ HOCH2CH2OH
ETHYLENE ETHYLENE GLYCOL
Dimethyl terephthalate is obtained by catalytic oxidation of p-xylene followed by esterification.
Q. 8. What do you mean by ultra-violet spectroscopy ? Give its applications in organic chemistry.
Ans. Ultraviolet-Visible Spectroscopy
In ultraviolet-visible (UV-Vis) spectroscopy, the 200-750 nm region of the ultraviolet spectrum is used. This includes both the visible region (400-750nm) and near ultraviolet region (200-400 nm). Radiation of these wavelengths is sufficiently energetic to cause the promotion of loosely held electrons, such as non-bonding electrons or electrons involved in a π-bond to higher energy levels. For absorption in this particular region of ultraviolet spectrum, the molecule must contain conjugated double bonds. If the conjugation is extensive, the molecule will absorb in the visible region.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-147.png)Fig. A sample UV-Vis spectrum
The ultraviolet-visible spectrum is composed of only a few broad bands of absorption (Fig.). The wavelength of maximum absorbance is referred to as λmax. The following points should be kept in mind while interpreting a UV-Vis spectrum.
- (1) Nonconjugated alkenes show an intense absorption below 200 nm and is therefore inaccessible to most commonly used UV spectrometers. For example, ethylene has λmax = 171 nm. This absorption comes from the light-induced promotion of a π-electron to the next higher energy level.
- (2) Nonconjugated carbonyl compounds have a very weak absorption band in the 200-300 nm region. This band arises from excitation of one of the nonbonding electrons (from an unshared pair) to the next higher energy level. For example, O
CH3– C – CH3
Acetone Cyclohexanone
279 nm 291 nm
- (3) When a molecule contain two or more nonconjugated carbon-carbon double bonds, the UV spectrum is that expected of a simple alkene. However, when the double bonds are conjugated, λmax is shifted to longer wavelengths. For example,
CH2 = CH – CH2 – CH2 – CH = CH2
- 1, 5 - Hexadiene 178 nm
CH3 – CH = CH – CH = CH – CH3
- 2, 4-Hexadiene 227 nm
- (4) Conjugation of a carbon-carbon double bond and a carbonyl group shifts the λmax of the both groups to longer wavelengths. For example,
(CH3)2C=CH–CH2CH3 180 nm
O
II
(CH3)2CH–CH2–C–CH3 283 nm
O
(CH3)2C=CH–C–CH3 230 nm (for C=C)
327 nm (for C=O)
(5) As the number of double bonds in conjugation increases, λmax also
increases. For example,
CH2 = CH – CH = CH2
1, 3-Butadiene
215 nm
_files/AU20B.20Sc.20III20Chemistry20II(OK)-148.png)
Benzene
257 nm
Summary of UV-Vis Spectroscopy : (1) Absorption of ultraviolet-visible radiation (200-700 nm) causes electrons within molecules to be promoted from one energy level to a higher electronic energy level.
- (2) If an organic compound does not absorb UV-Vis radiation, it means that the compound does not contain conjugated double bonds.
- (3) If an organic compound absorbs UV-Vis radiation, it means that the compound contains a carbonyl group or conjugated double bonds. For example, conjugated dienes, carbonyl compounds, and aromatic compounds all absorb in the UV-Vis region.
Q. 9. (a) Give the Skrup’s synthesis of quinoline. Also discuss its following properties :
- (i) Oxidation
- (ii) Reduction
- (iii) Nitration
- (b) Describe the structure of quinoline.
Ans. (a) Skrup Synthesis for Quinoline
Quinoline is obtained by Skrup synthesis. In this method, a mixture of aniline and glycerol is heated in the presence of sulphuric acid and nitrobenzene (mild oxidising agent). The reaction is exothermic and tends to become violent. Ferrous sulphate is generally added to make the reaction less violent.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-149.png)
ch2oh
CHOH
CH2OH
h2so4
C6H5NO2Ak
_files/AU20B.20Sc.20III20Chemistry20II(OK)-150.png)QUINOLINE
ANILINE GLYCEROL
Reactions of Quinoline
- (i) Oxidation : With alkaline KMnO4, the benzene ring of quinoline is oxidised to form quinoline acid and oxalic acid.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-151.png)alk KMnO\*
QUINOLINE
Vikas, 2009 (A. U. )
COOH COOH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-152.png)COOH COOH
QUINOLINE ACID OXALIC ACID
(ii) Reduction : With Zn and HCl forms tetrahydroquinoline (II) while with red P and HI forms decahydroquinoline (III).
CH CH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-153.png)CH,
N
H
(iii)
(iii) Nitration : With conc. HNO3 and conc. H2SO4, a mixture of 5-and
CH,
8-nitroquinolines are formed.
NO,
_files/AU20B.20Sc.20III20Chemistry20II(OK)-154.png)NO
QUINOLINE 5-Nitro- 8-Nitro-
isoquinoline isoquinoline
- (b) Structure of Quinoline
- (1) It has the molecular formula C9H7N.
- (2) It gives the electrophilic substitution reactions like benzene and forms salts with acids just like pyridine.
- (3) On oxidation with alkaline potassium permanganate forms 2, 3-pyridine dicarboxylic acid. Form analogy of oxidation of naphthalene to form phthalic acid, it is argued that quinoline is made of a benzene ring fused with pyridine in 2,3-positions.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-155.png)
X\\/COOH
V^COOH
NAPHTHALENE PHTHALIC ACID
_files/AU20B.20Sc.20III20Chemistry20II(OK)-156.png)QUINOLINE 2,3-PYRIDINE DICARBOXYLIC
ACID
- (4) The structure of quinoline is confirmed by Skraup synthesis given above.
- (5) Quinoline is considered as a resonance hybrid of three canonical
forms.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-157.png)
◄---------CANONICAL FORMS---------► Resonance Hybrid
- (6) Quinoline is a 10π electron system like nephthalene. It obeys Huckel rule (n – 2 in 4n + 2) which proves its aromatic character.
Q. 10. What do you mean by IR spectroscopy ? Give the applications to organic compounds.
Ans. Infrared Spectroscopy
An infrared (IR) spectometer subjects a compound to infrared radiation in the 5000-667 cm–1 (2µm) range. Although this radiation is weak, it does supply sufficient energy for bonds in the molecule to vibrate by Stretching or Bending. The atoms of a molecule can be considered as linked by springs that are set in motion by the application of energy. As the molecule is subjected to the individual wavelengths in the 5000-667 cm–1 range, it absorbes only those possessing exactly the energy required to cause a particular vibration. Energy absorptions are recorded as bands (peaks) on chart paper.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-158.png)Stretching vibrations Bending vibrations
Fig. Molecular vibrations caused by infrared radiation.
Since different bonds and functional groups absorbed at different wavelengths, an infrared spectrum is used to determine the structure of organic molecules. For example, carbon-carbon triple-bond is stronger than a carboncarbon double bond and requires a shorter wavelength (greater energy) to stretch. The same considerations apply to carbon-oxygen and carbon-nitrogen bonds.
C ≡ C 2100-2200 cm–1
C = C 1620-1680 cm–1
C = O C ≡ N
1690-1750 cm–1 2210-2260 cm–1
C – O C – N
1050-1400 cm–1 1050-1400 cm–1
Thus, from the position of an absorption peak, one can identify the group that cause it. Fig. shows the general areas in which various bonds absorb in the infrared.
Vikas, 2009 (A. U. )
WAVE NUMBERS Cm–1
5000 3000 2000 1400 1200 11001000 900 800 700
100
100 I----------1------------1----------1-----1—i------1------1------1----------r
60
O–H and
N–H stretching
C–H stretching
C=O stretching C–O stretching
80
40
20
C=N and stretching
C=N stretching
C=C stretching
C–N stretching
C–C stretching
60
N–H bending
N–H bending
40
N–H bending
C–H bending
20
0
o–H bending I
0
2 3 4 5 6 7 8 9 10 11 12 13 14 15
Fig. Area of absorption for various bonds in the infrared.
An infrared spectrum is usually studied in two sections :
- (1) Functional Group Region : The area from 5000 cm–1 to 1300 cm–1 is called the functional group region. The bands in this region are particularly useful in determining the type of functional groups present at the molecule.
- (2) Fingerprint Region : The area from 1300 cm–1 to 667 cm–1 is called the fingerprint region. A peak-by-peak match of an unknown spectrum with the spectrum of the suspected compound in this region can be used, much like a fingerprint, to confirm its identity. Table given below shows some characteristic infrared absorption bands. Fig. shows some examples of infrared spectra.
Summary of IR Spectroscopy : (1) Absorption of infrared radiation causes covalent bonds within the molecule to be promoted from one vibrational energy level to a higher vibrational energy level.
- (2) Stronger bonds require greater energy to vibrate (stretch or bend). Therefore, such bonds absorb infrared radiation of shorter wavelengths.
- (3) Different functional groups absorb infrared radiation at different wavelengths, and their presence or absence in a molecule can be determined by examination of an IR spectrum.
- (4) No two compounds have exactly identical infared spectra.
Table : Characteristic of IR Absorption Bands
Range in cm–1
Bend Remark
1050-1400
C – O (in ethers, alcohols, esters)
1050-1400
C – N (in amines)
1315-1475
C – H (in alkanes)
1340-1500
NO2 (two peaks)
1450-1600
C = C (in aromatic rings; several peaks)
1620-1680
C = C (in alkenes)
1630-1690
C = O (in amides)
1690-1750
C = O (aldehydes, ketones, esters)
1700-1725
C = O (in carboxylic acids)
1770-1820
C = O (in acid chlorides)
2100-2200
2210-2260
2500
2700-2800
2500-3000
3000-3100
3330
3020-3080
2800-3000
3300-3500
3200-3600
3600-3650
Wave numbers in cm
C ≡ N
C ≡ N
S – H
C – H (of aldehyde group)
O – H (of COOH group)
C – H (C is part of aromatic ring)
C – H (C is part of C ≡ C)
C – H (C is part of C = C)
C – H (in alkanes)
N – H (in amines, amides) O – H (in H-bonded ROH) O – H
(A) Wave Numbers in cm–
5000 4000 3000 2000 1500 1400 1200 1100 1000 900 800 700
100
_files/AU20B.20Sc.20III20Chemistry20II(OK)-159.png)
s 80
1 60
I 40
à? 20
2 3 4 5 6 7 8 9 10 11 12 13 14 15
Wavelength in Microns
Wave Numbers in cm–
_files/AU20B.20Sc.20III20Chemistry20II(OK)-160.png)100
5000 4000 3000 2000 1500 1400 1200 1100 1000 900 800 700
_files/AU20B.20Sc.20III20Chemistry20II(OK)-161.jpg)2 3 4 5 6 7 8 9 10 11 12 13 14 15
Wavelength in Microns
80
60
40
20
Fig. Some examples of IR spectra. (A) is Methanol. (B) is Butanone. (C) is Isobutylamine
Q. 11. Give the modern theories of colour and constitution.
Ans. Theories of Colour and Constitution : There are two theories :
- (i) Chromophore-Auxochrome Theory— (Witt theory)
- (a) The colour of organic compounds is due to the presence of certain multiple bonded group called chromophore. Some of the chromophores are
—N = O —NO2 —N = N—
Nitroso Nitro Azo
To make a substance coloured, the chromophore has to be conjugated with an extensive system of alternate single and double bonds as exists in aromatic rings. Thus, nitromethane is colourless while 1-nitronaphthalene is yellow. NO,
ch3no2
Nitromethane (Colourless)
1 -Nitronaphthalene (yellow)
- (b) Certain groups, while not producing colour themselves, when present alongwith chromophores in an organic substance, intensify colour. Such colour assisting groups are called auxochromes. They are acidic and basic both, as given below :
Acidic —OH —SO3H —COOH
Basic —NH2 —NHR —NR2
For example, azobenzene is red in colour, while p-hydroxy azobenzene is brilliant red.
AUXOCHROME
CHROMOPHORE
_files/AU20B.20Sc.20III20Chemistry20II(OK)-162.png)AZOBENZENE (Red)
p-HYDROXYAZOBENZENE (Bright Red)
- (ii) Modern Theory of colour : When a molecule absorbs light, the bonding electrons absorb energy and are promoted to orbitals of higher energy, the energy absorbed by an electron in such a transition is equal to difference of the energy of MO in the ground state and the energy of MO to which the electron is promoted, the relative energies of the MOs in the ground state are :
π delocalized > π > σ
Therefore, to promote an electron to the same higher energy orbital, maximum energy will be required for σ electrons, less for x electron and still less for delocalized electron. Further, the smaller the energy required for promotion,
B. Sc. III Chemistry II 95 the longer will be the wavelength of light absorbed. For example, methane having g bonds only, on excitation absorbs large energy and smaller wavelengths in the far ultraviolet region. Hence it is colourless. Ethylene has a n bond and absorbs less energy and longer wavelengths still fall in the near ultraviolet region and ethylene is colourless.
INCREASING ENERGY
HIGH ◄-----------------LOW
ULTRAVIOLET
H m
o
>
।
5
o
o
>
1
i o
1
RED । ।
INFRARED j
400 450 500 550 600 650 700 750
WAVELENGTH (nm)
In a compound having a highly conjugated system of single double bonds, the carbon p-orbitals overlap and cause extensive delocalization of the n-electrons.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-163.png)
(a) Conjugated double bonds; (b) Overlapingp-orbitals cause n electron delocalization
The delocalized electrons in such a system being in a higher energy orbital than c and n orbital, require less energy for excitation. They absorb longer wavelengths which fall in the range of the visible spectrum and hence the compound shows colour. Thus p-carotene, the orange pigment of carrots, with eleven conjugated double bonds, has a highly n-delocalized electron system covering 22 carbon atoms.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-164.png)p-CAROTENE (ORANGE)
Benzene has a ring structure made of three conjugated double bonds.
The carbon p-orbitals are involved in forming 6n delocalized electron system. It
absorbs wavelengths in the ultraviolet region and is colourless. When it is conjugated with chromophoric groups which contain double bonds, electron delocalization, is increased and the compound is coloured. Electron delocalization is further enhanced by groups such as —OH and —NH2 which can donate unshared electron pair for still greater extension of the conjugated system. Thereby the absorption is shifted to a longer wavelength (toward red), making the colour more intense.
The extended chromophoric systems delocalized by resonance are responsible for the colour of many dyes and indicators.
Q. 12. Classify the dyes on the basis of their application.
Ans. Classification of Dyes on the basis of application
- (1) Direct Dyes : These can be applied to a fabric by direct immersions in a water solution of the dye. It contains acidic or basic auxochrome which combines with the opposite polar group present in the chemical structure of the fibre. Wool and silk are readily dyed by this method. Martins yellow, a typical direct dye, has the acidic auxochrome —OH which interacts with the basic —NH2 group of wool or silk.
Fibre —NH2 + HO —Dye → fibre—NH3+...........O– – Dye
(Wool or silk)
- (2) Vat-Dyes : These dyes are insoluble in water but on reduction with sodium hydrosulphide in a Vat form a soluble compound which has a great affinity for cotton and other cellulose fibres. The cloth is soaked in the solution of a reduced dye and then hung in air, or treated with oxidants (perboric acid). As a result, the colourless compound is oxidised back to the insoluble dye which is now bound to the fabric. Indigo is a typical vat dye.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-165.png)Red
Oxd.
INDIGO (Blue)
OH H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-166.png)H OH
INDIGO WHITE (Colourless)
- (3) Mordant Dyes : These dyes have no natural affinity for the fabric and are applied to it with the help of oxides of Al, Cr, Cu and Ca. These salts are called mordants. A fibre such as cotton is first treated with a mordant and then with a dye solution. The mordant forms an insoluble co-ordination complex between the fibre and the dye and binds the two.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-167.png)
A mordant action of Alizarin
Alizarin is a typical mordant dye. It forms different colours depending upon the metal ion used. For example, with Al3+, alizarin gives a rose-red colour, with Ba3+, a blue colour. The mordant dyeing is most suitable for wool and nylon.
- (4) Azoic Dyes (Ingrain Dyes) : Ingrain dyes are dyes synthesised directly in the fabric. Azo dyes are typical ingrain dyes. The cloth is first soaked in the solution of a coupling reagent (usually a phenol or naphthol). Then it is immersed in the solution of an appropriate diazonium salt. The dye that is developed in the fabric binds pass to it via auxochromes. The ingrain dyeing is particularly suitable for cotton.
- (5) Disperse Dyes : These dyes are insoluble in water but can be dispersed in a colloidal form in water. The fabric is immersed in the colloidal dispersion of the dye. The fine dye particles are absorbed into the crystal structure of the fabric. Disperse dyes are used from modern synthetic fabrics such as nylon, orlon, polyesters and cellulose acetate.
Q. 13. How the following dyes are synthesized? Explain in brief.
- (a) Malachite Green (b) Methyl Orange.
- (c) Congo Red (d) Phenolphthalein
Ans. (a) Malachite Green : It is a triaryl methane dye. In it, the central carbon is bonded to three aromatic rings one of which is in the quinonoid form (the chromophore). The auxochromes are —N(CH3)2.
It is obtained by treating one mole of benzaldehyde with two moles of N, N-Di-methylaniline in the presence of conc. H2SO4 to give Leaco base (i.e., a colourless form). Oxidation of the leuco base with led oxide followed by treatment with HCl forms the dye.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-168.png)BENZALDEHYDE
_files/AU20B.20Sc.20III20Chemistry20II(OK)-169.png)(COLOURLESS)
MALACHITE GREEN
It is used as a direct dye for wool and silk. It fades in light.
- (b) Methyl Orange : Indicator Methyl orange is prepared by coupling reaction. When diazonium say of sulphanilic acid is coupled with N, N-dimethyl aniline, methyl orange is obtained.
NaO3S—
_files/AU20B.20Sc.20III20Chemistry20II(OK)-170.png)—NH2
NaNO2 2HCO
\* —NaO3S—
_files/AU20B.20Sc.20III20Chemistry20II(OK)-171.png)
Sod. Salt of Sulphanilic acid
—N NCl + NaCl + 2H2O
NaO3S—
_files/AU20B.20Sc.20III20Chemistry20II(OK)-172.png)
N = NCl + H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-173.png)CH3 NaOH
CH3
N, N dimethyl anilene
NaO3S—
_files/AU20B.20Sc.20III20Chemistry20II(OK)-174.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-175.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-176.png)
CH3
CH3
Methyl Orange
It is used as an indicator in acid – alkali titration.
- (c) Congo Red : It is an azo-dye and contains two azo groups.
It is obtained by coupling tetrazotized benzidine (1) with two molecules of naphthionic acid (2).
It is a direct dye. Its sodium salt is used for dyeing cotton red from aqueous solutions. It is also used as an indicator, being red in alkali and blue in acid solution.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-177.png)
(e) Phenolphthalein : It is also a triaryl methane dye. It is prepared by heating phthalic anhydride (1 mole) and phenol (2 mole) in the presence of any hydrous zinc chloride at 120°C. OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-178.png)
PHTHALIC PHENOL PHENOLPHTHALEIN
ANHYDRIDE
It is used as acid-base indicator. It gives red coloration in basic medium while remain colourless in acid medium.
Q. 14. What do you mean by synthetic detergents ? Explain with suitable examples.
Ans. Synthetic Detergents : The synthetic detergents or syndets are substitutes of soaps. They have cleansing power as good or better than ordinary
C12H25
OSO3Na
C15H31
COONa
Hydrophobic Hydrophilic
part part
Hydrophobic Hydrophlic
part part
Sodium lauryl sulphate (detergent)
Sodium palmitate (soap)
soaps. Like soaps, they contain both hydrophilic (water soluble) and hydrophobic (oil-soluble) parts in the molecule.
CH3(CH2)10CH2OH + HOSO3H ► CH3(CH2)10CH2OSO2OH
Lauryl alcohol Lauryl hydrogen sulphate
NaOH
CH3(CH2)10CH2OSO2Na
Sodium lauryl sulphate
(detergent)
Some of the detergents used these days are given below :
- (i) Sodium alkyl sulphates : These are sodium salts of sulphuric acid esters of long chain aliphatic alcohols containing usually 10 to 15 carbon atoms. The alcohols are obtained from oils or fats by hydrogenolysis.
CH3(CH2)10CH2OH + HOSO3H ► CH3(CH2)10CH2OSO2OH
Lauryl alcohol Lauryl hydrogen sulphate
NaOH
CH3(CH2)10CH2OSO2Na
Sodium lauryl sulphate
(detergent)
The other example is sodium cetyl sulphate, C16H33OSO2ONa. Unlike ordinary soaps, they do not produce OH– ions on hydrolysis and thus can be safely used for woollen garments.
- (ii) Sodium alkyl benzene sulphonates : Sodium p-dodecyl benzene sulphonate acts as a good detergent. It is most widely used since 1975.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-179.png)
Sodium dodecyl benzene sulphonate
- (iii) Quaternary ammonium slats : Quaternary ammonium salts with long chain alkyl group have been used as detergent, e.g., trimethyl stearyl
ammonium bromide.
Br
CH3N
C18H37
B. Sc. III Chemistry II 101
- (iv) Sulphonates with triethanol ammonium ion in place of sodium serve as highly soluble materials for liquid detergents.
R— —O–SO2–(CH2CH2OH)3NH
- (v) Partially esterified polyhydroxy compounds also act as detergents.
CH2OH
C17H35COOCH2–C–CH2OH
CH2OH
Pentaerythritol mono stearate
Detergents are superior cleansing agents due to following properties :
- (i) These can be used both in soft and hard waters as the calcium and magnesium ions present in hard water form soluble salts with detergents. Ordinary soap cannot be used in hard water.
- (ii) The aqueous solutions of detergents are neutral. Hence these can be used for washing all types of fabrics without any damage. The solution of ordinary soap is alkaline and thus cannot be used to wash delicate fabrics.
Q. 15. Give the chemistry of following dyes. Also give their methods of preparation and uses.
- (a) Fluorescene
- (b) Alizarin
- (c) Indigo.
Ans. (a) Fluorescene : It is obbtained by heating a mixture of phthalic anhydride and resorcinol in the malor ratio 1 : 2 with conc. H2SO4 or anhydrous
_files/AU20B.20Sc.20III20Chemistry20II(OK)-180.png)O =
_files/AU20B.20Sc.20III20Chemistry20II(OK)-181.png)Fluroescene
OH
OH
It is dark red coloured powder, insoluble in water but give red-brown solution on dissolving in alkalies which on dilution give yellow-green fluroescene. It is used in eye surgery. Its sodium salt is called uranin which dyes wool and silk in acidic medium.
(c) Alizarin : It is synthesized from phthalic anhydride and benzene follows :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-182.png)SO3H
O
Fuming H2SO4 180°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-183.png)O
2-anthraquinone sulphonoc acid
(i) NaOH/KClO3/Fuse.
(ii) H+
OO
_files/AU20B.20Sc.20III20Chemistry20II(OK)-184.png)
O
Alizarin
Alizarin is red crystalline solid insoluble in water but soluble in alcohol and alkali. It is mordant dye and combines with mordants, i.e. metallic hydroxides to form coloured insoluble compounds called lakes. The colour of the lake depends upon the mordant, i.e., cation used. The colours of the lakes along with the respective mordants are given as follows :
Violet
Brown black
Red
Mordant used
Ca2+, Ba2+, Pb2+
Sr2+
Mg2+, Sn4+
Al3+
Cr3+
Colour of the lake
Bluish-red
Red-violet
Violet
Red
Brownish-red
Fe2+
Fe3+
Sn2+
When it is applied to wool with aluminium mordant, it gives the well known turkety red. Alizarin, when converted to its calcium slats, forms a bluish-red powder used as pigment.
- (d) Indigo : It is the oldest dye. It occurs in the plants of indigofera group in the form of glucoside the indican. It is also known as indigotin. It is prepared by the following methods :
- (i) Reduction of isatin chloride (obtained by the aciton of phosphorus pentachloride on isatin) with zinc dust in glacial acetic acid yields indoxyl which upon oxidation in air gives indigotin.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-185.png)2 moles \[O\]
_files/AU20B.20Sc.20III20Chemistry20II(OK)-186.png)Indoxyl
_files/AU20B.20Sc.20III20Chemistry20II(OK)-187.png)Indigo
- (ii) It can be synthesized from anthranilic acid and chloroacetic acid by the following steps :
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-188.png)
Anthranilic
I
C=O
NH2
ClCH2COOH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-189.png)
OH C =
NH2 N-phenyl glycine -o-carboxylic acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-190.png)NaOH
COOH NaNH2'D
(–H2O)
acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-191.png)H
Indoxylic acid unstable
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-192.png)H
Indoxyl (2 moles)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-193.png)Indigo
Indigo is dark blue coloured powder. It is insoluble in water. It may be reduced with alkaline sodium hyposulphite to a colourless form the indigotin-white which is insoluble in alkali. It is alkaline solution which is applied to the fibre and then the fibre is exposed to air when the original blue colour of indigotin is regenerated in the cloth.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-194.png)= C
H
O
H
ONa
Indigotin white (colourless)
Q. 16. Give preparation and properties of benzene sulphonic acid.
Ans. Aromatic sulphonic acids are prepared by following methods :
- (a) By direct sulphonation : Arenes reacts with fuming H2SO4 at room temperature to give sulphonic acid. It can also be carried out with conc. H2SO4 alone but such reactions are slow and heating is generally required.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-195.png)\+ H2SO4 Fuming
_files/AU20B.20Sc.20III20Chemistry20II(OK)-196.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-197.png)\+ cone. H2SO4
Benzene sulphonic acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-198.png)p-toluene sulphonic acid
- (b) From sulphonyl chlorides : Arenes react with excess of chlorosulphonic acid to form sulphonyl chlorides. These on acid hydrolysis yield sulphonic acids.
ClSO H H O/H+
ArH 3 ArSO2Cl 2 AnSO3H
Arene Sulphonyl Arene sulphonic
chloride acid
- (c) Thiophenol on oxidation with alkaline KMnO4 gives sulphonic acid. SH SO4H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-199.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-200.png)
KMaO4
Physical properties : These are white crystalline solids which are highly hygroscopic and form syrups in moist air. That is why they are invariably used in the form of their salts. These decompose on heating therefore they have no sharp melting points. These are extremely soluble in water but sparingly soluble in organic solvents.
Chemical Properties :
- 1\. Acidic Nature : These are strong acids and are completely ionised in solution.
C6H5SO3H -----→ C6H5SO3– + H+
Sulphonation
(stable)
The high acidity of sulphonic acids is due to two reasons :
- (a) The two S = O bonds are polarised since sulphur atom has higher electronegativity than oxygen atoms. The electron withdrawing inductive effect so caused, facilitates the relase of proton.
8^ 5+
S = 0
_files/AU20B.20Sc.20III20Chemistry20II(OK)-201.png)
- (b) The sulphonate anion once formed is resonace stabilized.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-202.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-203.png)
O
O
Both above factors are more pronounced in case of sulphonic acids than carboxylic acids. Thus Ka for benzoic acid is 6.3 x 105, while for benzene sulphonic acid it is very large. Electron withdrawing substituents at the o- and p-position increase the acidity. For example, 2, 4-dinitrobenzene sulphonic acid is a stronger acid than sulphuric acid.
The acidic character of sulphonic acid is evident by the fact that sulphonic acids react with alkali metals and alkalies like sodium carbonate, NaHCO3, NaOH etc.
2C6H5SO2OH + 2Na -----→ 2C6H5SO2ONa + H2
Sod. benzene sulphonate
C6H5SO2.OH + NaOH ------→ C6H5SO2ONa + H2O
C6H5SO2OH + NaHCO3 ----→ C6H5SO2ONa + H2O + CO2
- 2\. Reaction with PCl5 or SOCl2 : Corresponding sulphonyl chloride is obtained.
heat
C6H5SO2.OH + PCl5 ---e→at C6H5SO2Cl + HCl + POCl3
C6H5SO2.OH + SOCl2---→ C6H5SO2Cl + HCl + SO2
It is interesting to note that unlike carboxylic acid, sulphonic acids do not react with alcohols and ammonia to form ester and amides respectively. Hence esters and amides of sulphonic acids are prepared indirectly by the action of alcohols and amines respectively on sulphonyl chloride.
C6H5SO2.Cl + H.NH2---→ C6H5SO2NH2 + HCl
Benzene sulphonamide
- 3\. Dehydration : Sulphonic acid, like carboxylic acids when heated with excess of P2O5 give sulphonic anhydride.
C6H5SO2OH
+
C6H5SO2OH
C6H5SO
PO
2 5 >
C6H5SO2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-204.jpg)O
- 4\. Replacement of sulphonic group by hydrogen (Desulphonation or hydrolysis) : When sulphonic acids are heated with a mineral acid at 150-170°C in presence of super heated steam, the – SO3H group is replaced by hydrogen atom to form arene.
C6H5SO3H + H2O----> C6H6 + H2SO4
Benzene sulphonic acid Benzene
(non-volatile) (volatile)
Since the reaction is the reverse of sulphonation, it is known as desulphonation. Further since desulfonation is a reversible process, its mechanism follows the reverse path of the mechanism of the sulphonation.
- 5\. Replacement of SO3H group by —OH : When the sodium salt of a sulphonic acid is fused with sodium hydroxide and the product is neutralized with an acid —SO3H group is replaced by —OH group forming phenol.
C6H5SO3Na + NaOH----> C6H5OH + Na2SO3
- 6\. Replacement of —SO3H group by —CN : When the sodium salt of a sulphonic acid is fused with sodium cyanide, —SO3H group is replaced by —CN group forming a nitrile.
C6H5SO3Na + NaCN----> C6H5CN + Na2SO3
Benzonitrile
The nitrile formed can be easily hydrolysed or reduced to form carboxylic acid or amine respectively.
reduction hydrolysis
C6H5CH2NH2«--------C6H5CN --“---“^ C6H5COOH
Benzyl amine Nitrile Benzoic acid
- 7\. Replacement of —SO3H group by NH2 group : When the soidum salt of a sulphonic acid is fused with sodamide, amines are formed.
C6H5SO3Na + NaNH2-----> C6H5NH2 + Na2SO3
- 8\. Replacement of —SO3H group by halogen : Sulphonic acid present in the o-and p-position to an —OH or —NH2 group can be easily replaced by a halogen atom.
NH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-205.png)SO3H
Bromine Water
_files/AU20B.20Sc.20III20Chemistry20II(OK)-206.png)Br
Br
Sulphanilic acid 2, 4, 6-tribromoaniline
- 9\. Reactions of benzene ring : The —SO3H group is deactivating and metal director. Therefore, the sulphonic acid undergoes the usual substitutions such as halogenation, nitration and sulphonation under more difficult conditions, than the benzene itself. The incoming occupies the meta positon.
SO3H
NO2 m-nitrobenzene sulphonic acid
SO3H
HNO3
H2SO4,
_files/AU20B.20Sc.20III20Chemistry20II(OK)-207.png)
Br2
FeBr3
fuming H2SO4
200°C
SO3H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-208.png)
Br m-nitrobenzene sulphonic acid
SO3H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-209.png)SO3H
m-benzene disulphonic acid
from Grignard’s
Q. 17. How followings are prepared
reagents :
- (a) Primary, Secondary and Tertiary alcohols
- (b) Alkanes
- (c) Esters
- (d) Ethyl, methyl ketone.
Ans. (a) With HCHO Methyl magnesium iodide gives
H–CH = O +
CH3 H OH
Mg CH3CH2 O MgBr
I dil. IICl
OH
CH3CH2OH + Mg Z
Ethanol I
If in place of HCHO, CH3CHO is used, isopropyl alcohol is obtained as the product.
CH3CH = O +
CH3
Mg < ICH3
CH3
CH3
H OH
CH–O Mgl
I H2O
CH3 OH
CHOH + Mg
Isopropyl alcohol (Secondary)
With 2-butanone tert. alcohol is obtained
CH3CH2
CH3
>C
CH3
C2H5
\---► CH3
CH3
_files/AU20B.20Sc.20III20Chemistry20II(OK)-210.png)H C–O
OH
Mgl
C2H5
OH
I
CH3 -A C–OH + Mg <
CH3
tert. alcohol
- (b) Preparation of alkanes : When methyl magnesium bromide is treated with an alkyl halide (compound containing reactive halogen atom) alkanes are obtained. Under such condition propane is produced when ethyl halide is treated with methyl magnesium bromide.
CH3MgBr + BrC2H5-----→ CH3CH2CH3 + MgBr2
Propane
With n-propyl bromide n-butane is obtained.
CH3MgBr + CH3CH2CH2Br --→ CH3CH2CH2CH3 + MgBr2
n-butane
- (c) Preparation of esters : When 1 mole of Grignard reagent reacts with 1 mole of ethyl chloroformate, formation of ester takes place.
C2H5|Mgl + C1|COO C2H5 -^ C2H5COOC,H5 + Mg<^
Ethyl magnesium Chloroformic Ethyl propionate iodide ester (ester)
- (d) Ethyl methyl ketone : Ethyl methyl ketone is produced when ethyl cyanide is used in palce of methyl cyanide.
CH3 O
> C = C2H5
H2
N –MgBr
H OH
CH3
C2H5CN + Mg ethyl cyanide Br
2H2O
CH3 OH
CH 3 C = O + NH3 + Mg
C2H5 I
ethyl methyl ketone
Q. 18. How pyridine is synthesized from acroline ? Give its
properties and structure.
Ans. Synthesis of pyridine : It is synthesised from acroline as follows :
Reactions of pyridine
_files/AU20B.20Sc.20III20Chemistry20II(OK)-211.png)K2Cr2O7H+ \[O\]
N
3-methyl pyridine
_files/AU20B.20Sc.20III20Chemistry20II(OK)-212.png)COOH
3-pyridine carboxylic acid
NH
2 CH=CH. CHO D 3
CaO/NaOH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-213.png)N
Pyridine
- (i) Salt formation : It forms salts with acids. Hence it is basic in nature.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-214.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-215.png)
Pyridineum hydrochloride
- (ii) Electrophilic Substitution : Substitution takes place at 3-position corresponding to nitrogen atom.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-216.png)
HNO3
H2SO4, 300°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-217.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-218.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-219.png)
H2SO4 SO3,200oC
3-Nitropyridine
_files/AU20B.20Sc.20III20Chemistry20II(OK)-220.png)
3-pyridine sulphonic acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-221.png)
Br2 300°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-222.png)Br
3-bromo pyridine
- (iii) Nucleophilic displacement : Nucleophilic substitution takes place
at position-2 in pyridines.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-223.png)
NaNH2 100°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-224.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-225.png)NH2
2-Amino pyridine
_files/AU20B.20Sc.20III20Chemistry20II(OK)-226.png)
C6H5Li r
100°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-227.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-228.png)C6H5
2-Phenyl pyridine
- (iv) Reduction : Pyridine on reduction gives piperadine.
[Na,+ C,H5OH +3H, I ' . ~ .---- ------—► Reduction ►](#bookmark239)
[H-Nz\\ /](#bookmark240)
[N “N](#bookmark241)
H
Constitution :
- (1) Molecular formula : C5H5N
- (2) Presence of tertiary nitrogen atom : Pyridine gives quaternary ammonium salt on reaction with methyl iodide. It confirms the presence of tertiary nitrogen atom in pyridine. –
C5H5N + CH3I --------\* \[C5H5N CH3\]+I
Pyridine methyle iodide N-methyl Pyridinium iodide
- (3) Resemblence with benzene :
Similar to benzene.
- (i) It gives helogenation, nitration and sulphonation reactions.
- (ii) Shows coupling reactions with diazonium salts obtained from amino derivatives.
- (iii) Its hydroxy derivatives show phenolic properties.
- (iv) Presence of cyclic structure : Pyridine on reduction gives piperadine by using 6-H atoms as benzene by using 6-H atoms gives cyclohexane.
C5H5N + 3H2
Pyridine
C6H6 + 3H2
Ni 200°C
\-------►
C5H11N Piperadine
C6H12 Cyclohexane
Based on above facts following cyclic structure is given to pyridine.
Piperadine can also be synthesised as follows :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-229.png)5 Chloro Pentylamine H
Piperadine
Because piperadine is a cyclic compound, hence pyridine should also be a cyclic compound because piperadine can be converted into pyridine.
- (v) Corner’s Formula : Based on all above facts Corner assigned following cyclic structure for pyridine similar to benzene in 1869.
CH CH
HCz^^XCH HCzi^XCH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-230.png)Pyridine
_files/AU20B.20Sc.20III20Chemistry20II(OK)-231.png)CH
Benzene
Confirmation by Synthesis : It can be synthesised by passing the mixture of acetylene and HCN from red hot tube.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-232.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-233.png)
H-C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-234.png)
C-H
H-C^^C-H N
Pyridine
Molecular Orbital Structure : All the atoms of pyridine (5c and 1M) are sp2 hybridised. Two sp2 hybridised orbitals of every atom ovelaps by neighbouring sp2 hybridised orbital of other atom and form C-C and C-N bond.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-235.png)Q. 19. How isoquinoline is synthesised from Bischler-Naiperalski method ? Discuss the following properties of isoquinoline :
- (i) Basic nature of nitrogen
- (ii) Reduction
- (iii) Oxidation
- (iv) Electrophilic substitution
- (v) Nucleophilic substitution
Ans. Isoquinoline
Preparation : (i) By heating oxime of cinnamaldehyde with P2O5 (Blockmann’s rearrangement).
_files/AU20B.20Sc.20III20Chemistry20II(OK)-236.png)CTNNAMAL DOXIME QUINOLINE
- (ii) By Bischler-Naiperalski Method. It is prepared from 2-
phenylethylamine by the following steps :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-237.png)2 - PHENYL ETHYL FORMYL H
AMINE CHLORIDE N-FORMYL-2-PHENYL ETHYL AMINE
_files/AU20B.20Sc.20III20Chemistry20II(OK)-238.png)Properties : (i) Basic Nature and Tertiary Nitrogen. From salts with acids and quaternary salts with CH3I.
C9H7N + HCl —→ \[C9H7N+H\]Cl–
Quinoline Quinoline
C9H7N + CH3I —→ \[C9H7N+CH3\]I–
N-Methylquinolinium Iodide
- (ii) Reduction. (a) With Na and liquid NH3 forms, 1, 2-dihydro-isoquinoline (II).
- (b) With Sn and HCl forms 1, 2, 3, 4-tetrahydro-isoquinoline (III).
- (c) On catalytic hydrogenation froms deca-hydroisoquinoline (IV).
_files/AU20B.20Sc.20III20Chemistry20II(OK)-239.png)3
(iii)
2 NH
Sn + HCl
_files/AU20B.20Sc.20III20Chemistry20II(OK)-240.png)Na +NH3
(i)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-241.png)(ii)
2 ZNH
1, 2, 3, 4, - Tetra hydro Isoquinoline
Isoqunoline
1, 2 - Dihydro-Isoquinoline
Catalynal
H2
_files/AU20B.20Sc.20III20Chemistry20II(OK)-242.png)4
3
6
7
9
5
10
8
12,NH
(iv)
Decahydro Isoquinoline
- (iii) Oxidation. With permanganate it forms a mixture of phthalic acid and cinchomeronic acid.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-243.png)Isoquinoline
_files/AU20B.20Sc.20III20Chemistry20II(OK)-244.png)Phthalic acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-245.png)Cinophytholic acid
- (iv) Electrophilic Substitution :
- (i) Nitration : With conc. HNO3 and conc. H2SO4, forms a mixture of 5-and 8-nitro isoquinoline.
- (ii) Sulphonation : With conc. H2SO4 forms 5- and 8-derivatives of sulphonic acid of isoquinoline.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-246.png)(iii) Nucleophilic Substitution : With soda-amide forms 1-
_files/AU20B.20Sc.20III20Chemistry20II(OK)-247.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-248.png)
Q. 20. Give the preparation and important properties of thiophene.
Ans. (i) Thiophene : It is obtained as follows :
- 1\. By heating sodium succinate with P2S3.
CH, COO Na+
CH = CH
I
CH = CH
Thiophene
I \_ + p^s5
CH, COO Na’
Sodium Succinate
- 2\. By heating n-butane with S at 600°C :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-249.png)Thiophene
CH3 CH, CH, CH3 + 4S
»-butane
Reactions : In thiophene electrophilic substitution takes place at
position 2.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-250.png)
H2SO4
_files/AU20B.20Sc.20III20Chemistry20II(OK)-251.png)SO3H
Thiophene 2-sulphonic acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-252.png)
HNO3
(CH3CO)2 O.10°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-253.png)
2-Nitrothiophene
I2 NaO
2-Iodothiophene
(ii) Furan : It is obtained as following form Pentosens (poly saccharide having 5e-atoms) by acid hydrolysis in following steps.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-254.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-255.png)
(Cs^OJ,
\+ CHO
H2O/H I H2SO4
(CHOH)3 — 3H2O \* 1 (II)
ch2oh
Aldopentose
_files/AU20B.20Sc.20III20Chemistry20II(OK)-256.png)
mn „ Ag2°lsteam
(Hl) ▼ \_co2
CH=CH v
J>0 Furan
CH=CHX
Important reactions : Like pyroll, furan gives electrophilic substitution reaction at position-2.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-257.png)HNO3 (CH3CO)2 O.10°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-258.png)NO2
2- Nitrofuran
_files/AU20B.20Sc.20III20Chemistry20II(OK)-259.png)
SO3 Pyridine,100°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-260.png)SO3H
- 2- Furan sulphonic acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-261.png)
(CH3CO)2O
BF3 , 0°C
_files/AU20B.20Sc.20III20Chemistry20II(OK)-262.png)C-CH3
- 2- Acetyl furan
_files/AU20B.20Sc.20III20Chemistry20II(OK)-263.png)
DIOXANE , 0°C
Br
_files/AU20B.20Sc.20III20Chemistry20II(OK)-264.png)2- Bromofuran
_files/AU20B.20Sc.20III20Chemistry20II(OK)-265.png)
_files/AU20B.20Sc.20III20Chemistry20II(OK)-266.png)
Tetrahydrofuran
Q. 21. How would you convert the following :
- (a) Aldose into ketose or glucose into fructose
- (b) Ketone into aldose or fructose into glucose
- (c) Aldosein to next higher aldose
- (d) Aldose into the next lower aldose.
Ans. (a) Conversion of glucose into fructose (Aldose into ketose) : For this conversion glucose is first warmed with excess of phenyl hydrazine, when glucosazone is formed. It on treatment with dil. HCl gives glucosone on hydrolysis. Glucose on reduction with Zn-dust and acetic acid gives fructose.
CHO
CH = N.NHC H
65
2H2O
CHOH
C = N.NHC6H5
–2C6H5NHNH2
(CHOH)3
(CHOH)3
CH2OH
CH2OH
Glucose
Glucosazone
CH = O
CH2OH
C = O
\[H\]
CO
Zin-dust AcOH
(CHOH)3
(CHOH)
CH2OH
CH2OH
Glucoson
Fructose
' 3
- (b) Fructose into glucose (ketose into aldose) : Fructose on catalytic reduction gives hexitol (a mixture of sorbitol and mannitol) which on oxidation
ch2oh
ch2oh
COOH
co------1
H2 /Ni
►CHOH -
\[0\] .
-h2o
(CHOH)2 o
CO \*
hno3 '
Ullvll .
Heat
(CHOH)3
(CHOH)3
(CHOH)3
Ln
ch2oh
ch2oh
ch2oh
ch2oh
Fmctose
Hexitol
Hexonic acid
y-lactone ¡2H
CHO-(CHOH)4-CH2OH
glucose
gives hexonic acid (a mixture of gluconic and mannonic acid). This loses water on heating to give γ-lactone which is subsequently reduced with Na— Hg in faintly acid solution to glucose.
- (c) Aldose into next higher aldose : This can be performed by means of Killiani reaction. In this conversion, the aldopentose is dissolved in dilute HCN and the cyanohydrine obtained is hydrolysed with aqueous barium hydroxide. The mixture is acidified with a calculated quantity of H2SO4 and BaSO4 precipitate is filtered off. A polyhydroxy acid with carbon atoms is obtained in aqueous solution. This solution is evaporated to dryness when γ-lactone is obtained which on reduction with Na—Hg is faintly acid solution yields an aldohexose.
CHO
CN COOH
HCN
(CHOH)3-----'
1 (i) Ba (OH), 1
\- CHOH------- CHOH
I (ii) H2SO4 I
ch2oh
ch2oh ch2oh
Aldopentose
Cyanohydrine Polyhydioxy acid
(6C-atoms)
CO
CHO
CHOH
CHOH
CHOH
CHOH
evaporate to 1 Na/Hg
dryness,-H2O ^
CHOH
CHOH
CHOH
CH2OH
CH2OH
y-lactone
Aldohexose (6C-atom)
- (d) Aldose into next lower aldose : It is performed by Wohl’s method. In this method, the aldohexose is treated with hydroxylamine and the oxime
CHO
1 NH OH
NOH
CN
(CH2CO)2O 1 AgOH
CN
CHOH 2
CHOH
2 2 CHO.OCCH
3
CHOH
(CHOH)3
(CHOH)3
(CHOCOCH3)3
(CHOH)3
CH2OH
CH2OH
CH2OCOCH3
CH2OH
Aldohexose
Oxime
Cyano compound
CHO
(CHOH)2
CH2OH
Aldopentose
produced is heated with acetic anhydride. In this way, oxime is dehydrated to the cyano compound whereas the hydroxyl groups get acetylated. The acetyl derivative is warmed with ammoniacal AgNO3 which remove the acetyl groups by hydrolysis and eliminates a molecule of HCN. And aldopentose is obtianed as a result of these changes.
Q. 22. Give the Cyclic structure of D-glucose.
Ans. Structure of glucose : Based on the analytical data, Baeyer in 1876, suggested the following open chain formula for glucose. This opem chain formula of glucose explains most of the reactions given by glucose but fails to explain the following facts :
- (i) Although it contains an aldehyde group, but it does not react with NH3 and NaHSO3.
- (ii) There are two forms of glucose : α-glucose and β-glucose. α- and β-forms showing specific rotation of +110° and + 19.7° respectively.
- (iii) In aqueous solution, both the forms of glucose show mutarotation. α-form undergoes mutarotation from 110° to 52.5, while β-form shows a mutaroation from 19.7° to 52.5°
Cyclic (ring) Structure of Glucose
To account for these facts, Tollen (1883) suggested a ring formula for glucose. Later Haworth, and others (1926) have shown that the ring formula is a six-membered amylene-oxide ring. The two stereo isomers (each mirror image of the other) are represented by the two structures.
The two structures shown above are called as Fischer Projection Formulae.
R ZOH
G------
^HO
H—C—OH
H—2C—OH
|
HO—C—H (
9 # HO—3C—OH
H—C—OH
H—4C—OH
H—C-----
H—5C—OH
ch2oh
6CH2OH
a- D-glucose
D-glucose
\[a\]D = + 110°
HOx/H
C------
H—C—OH
HO—C—H O
H—C—OH
H—C-----
ch2oh a- D-glucose \[a\]D = + 19.7°
The α-glucose and β-glucose, differ only in the orientation of the hydroxyl group at carbon atom-1. Such pairs of optical isomers which differ in the orientation of H and OH groups only at carbon atom-1, are called anomers. The 1C atom is called anomeric carbon atom (or glycosidic carbon).
The two forms of glucose, α- and β-glucose exist in separate crystalline forms and have different melting points and optical rotations. However, when either form of glucose is dissolved in water, it gets converted into the other form and an equilibrium mixture is formed together with a very small amount of open chain form i.e.,
α-Glucose ==== Open chain form === β-Glucose
(36%) (0.02%) (64%)
Pyranose structure of glucose : The cyclic structures of monosaccharides can be better described by Haworth Projection Formulae.
Haworth (1926) proposed a six-membered cyclic structure for glucose base on the structure of a heterocyclic compound pyran. Such a structure is known as the pyranose structure. This structure is supported by X-ray analysis. Pyranose structures for the two isomers of glucose are given below in Fig.
H
I
4C
i \\
OH
6CH2OH 5C H2OH
/1
H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-267.png)
3G
I
OH
I
2
H
I
OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-268.png)OH
H
CH2OH
_files/AU20B.20Sc.20III20Chemistry20II(OK)-269.png)OH
OH
α-GLUCOPYRANOSE
_files/AU20B.20Sc.20III20Chemistry20II(OK)-270.jpg)
6CH2OH 5c H2OH
H I
I H
4C
I \\ OH
OH\\ I
3C----
I
H
_files/AU20B.20Sc.20III20Chemistry20II(OK)-271.png)CH2OH
HH2 OH
OH H
OH H
H OH
OH
β-GLUCOPYRANOSE
Fig. Pyranose structures for α- and β-glucose.
Q. 23. Discuss the method of preparation and important properties of α-amino acid.
Ans. Preparation of α-Amino Acids
- (i) Amination of α-Halo acids : α-Chloro or α-bromo acids on treatment with aqueous ammonia give the respective amino acids.
Cl—CH2COOH + 2NH3 → H2N—CH2COOH + NH4Cl
Chloroacetic acid Aminoacetic acid
- (ii) Gabriel Phthalimide Synthesis : The treatment of an α-halo easter with potassium phthalimide and subsequent hydrolysis of the product gives α-amino acid.
_files/AU20B.20Sc.20III20Chemistry20II(OK)-272.png)Potassium phthalimide
NiK + Cl-i-CH,COOC,H. chloroacetic ester
-KC1
_files/AU20B.20Sc.20III20Chemistry20II(OK)-273.png)
C2H5OH +
_files/AU20B.20Sc.20III20Chemistry20II(OK)-274.png)COOH
\+ H2N-CH2COOH COOH Aminoacetic acid
_files/AU20B.20Sc.20III20Chemistry20II(OK)-275.png)N —CH2COOC2H5 (int-compd.)
2H2o|
(HC1)
Phthalic acid
- (iii) Streeker synthesis : The treatment of an aldehyde with hydrogen cyanide and ammonia followed by hydrolysis of the aminonitrile thus formed, yields α-amino acid.
HCN ?H NR T2
CH3—C=0--->CH3—C—CN--^> CH.—C—CN
[1 1 (-H.O)](#bookmark254)
[H H 1 J 7H](#bookmark255)
Acetaldehyde Cyanohydrin Aminonitrile
NH2
[CH.—C—CN + 2H2O^CH.—C—COOH + H2 HH](#bookmark256)
oc-aminopropionic acid
- (iv) Koop synthesis : The treatment of α-keto acids with ammonia forms an imine which on catalytic reduction gives amino acid.
O NH NH2
„ J! +NH3 II H2
CH3—C—COOH -H O CH3—C—COOH -Pd2 CH2—C—COOH
2
a-Ketopropionic acid Imine H
a-Aminopropionic acid
- (v) From Malonic Easter :
Synthesis of Alamine : (α-aminopropionic acid) :
NaOC.Hs CH.Br
CH2(COOC2H5)2-----2-^> Na.CH(COOC2H5) —2—►
Malonic Easter Sodio-deriv.
- (i) KOH Br.
CH3-CH(COOC2H5)2 ^)HC1 ► CH3-CH(COOH)2 - A
Methyl Deriv.
Methyl Malonic Acid
A NH3
CH3.C(Br).(COOH)2 CH3.CH(Br).(COOH)--- oc-Bromomethyl malonio oc-Bromopropionic acid
^ CH3—(NH2)2COOH
a-Aminopropionic acid (Alanine)
- (vi) By the reduction of oxime and hydrazone of α-Ketoacids and Amino
acids : NOH
H.NOH 11
- (a) CH.CO—COOH 2 ► CH3—C—COOH
Pyruvic acid Oxime
Na/Hg ।
\---CH3—CH—COOH
Alcohol Alamine
nh2
H,NNH, i
- (b) CH3CO.COOH —---^-CH^C-COOH^1\*
Hydrazone
NH2
(Glycine) CH .CH—COOH
- (II) Important Reactions : (A) Reactions due to COOH group : It gives the usual reactions of carboxylic acids.
- (i) Salt formation with bases :
H2N—CH2COOH + NaOH → H2N—CH2COO–Na+ + H2O
Aminoacetic acid Sod. aminoacetate
- (ii) Easter formation with alcohol in presence of dry HCl :
N,H—CH,COO + C?HSOH H,N —CH,COOC,HS
\- " (HCI) ‘ ‘
\+ H,O
AgOH
H3N—CH2COOC2H5 + OH —--► H2N—CH2COOC2H5
Ethyl aminoacetate
- (iii) Decarboxylation by boiling with barium hydroxide solution :
boil
H2N—CH2COOH + Ba(OH)2 -----► H2N—CH3 + BaCO3 + H2O
Methylamine
- (iv) Reduction with lithium aluminium hydride :
O
" LiAlH,
H2N—CH2C—0H+4H----^H2N—CH2CH2—OH+H2O
Aminoacetic acid 2-Aminoethanol
- (B) Reactions due to NH2 groups : At the NH2 group, amino acids respond to all the usual reactions of primary amines.
- (i) Reaction with strong acids to form salts.
CH—NH2 CH—N’H3C1
I + HC1 —► । COOH COOH
Glycine Glycine Hydrochloride
- (ii) Alkylation with alkyl halides in basic solution :
ch2—nh2
COOH
CH3Br ^^
CH2—NH—CH3
COOH
\+ HBr
Glycine Methyl bromide N-Methylglycine
- (iii) Acylation with acid chlorides in basic solution :
CH2—NH2
CH2—NH—COCH,
COOH
I + CH3COC1^^>
COOH
Glycine Acetyl chloride Acetyl glycine
- (iv) Reaction with nitrous acid to form hydroxy acid :
_files/AU20B.20Sc.20III20Chemistry20II(OK)-276.png)
- (v) Condensation with formaldehyde :
CH2—N\[H2 ch2—n=ch2
I +0\]=ch2—>1
COOH COOH
Glycine N-Methylene Glycine