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XII, Question 12.1:, What is meant by the following terms? Give an example of the reaction in each case., (i) Cyanohydrin (ii) Acetal, (iii) Semicarbazone (iv) Aldol, (v) Hemiacetal (vi) Oxime, (vii) Ketal (vii) Imine, (ix) 2,4-DNP-derivative (x) Schiff’s base, Answer, (i), , Cyanohydrin:, , Cyanohydrins are organic compounds having the formula RR′C(OH)CN, where R and R′, can be alkyl or aryl groups., , Aldehydes and ketones react with hydrogen cyanide (HCN) in the presence of excess, sodium cyanide (NaCN) as a catalyst to field cyanohydrin. These reactions are known as, cyanohydrin reactions., , Cyanohydrins are useful synthetic intermediates., (ii) Acetal:, Acetals are gem−dialkoxy alkanes in which two alkoxy groups are present on the terminal, carbon atom. One bond is connected to an alkyl group while the other is connected to a, hydrogen atom., , 1
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When aldehydes are treated with two equivalents of a monohydric alcohol in the presence, of dry HCl gas, hemiacetals are produced that further react with one more molecule of, alcohol to yield acetal., , (iii) Semicarbarbazone:, Semicarbazones are derivatives of aldehydes and ketones produced by the condensation, reaction between a ketone or aldehyde and semicarbazide., , Semicarbazones are useful for identification and characterization of aldehydes and, ketones., (iv) Aldol:, A β-hydroxy aldehyde or ketone is known as an aldol. It is produced by the condensation, reaction of two molecules of the same or one molecule each of two different aldehydes or, ketones in the presence of a base., , (v), , Hemiacetal:, , Hemiacetals are α−alkoxyalcohols, , 2
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General structure of a hemiacetal, Aldehyde reacts with one molecule of a monohydric alcohol in the presence of dry HCl gas., , (vi), , Oxime:, , Oximes are a class of organic compounds having the general formula RR′CNOH, where R, is an organic side chain and R′ is either hydrogen or an organic side chain. If R′ is H, then, it is known as aldoxime and if R′ is an organic side chain, it is known as ketoxime., , On treatment with hydroxylamine in a weakly acidic medium, aldehydes or ketones form, oximes., , (vii) Ketal:, Ketals are gem−dialkoxyalkanes in which two alkoxy groups are present on the same, carbon atom within the chain. The other two bonds of the carbon atom are connected to, two alkyl groups., , 3
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Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product, known as ethylene glycol ketals., , (viii) Imine:, Imines are chemical compounds containing a carbon nitrogen double bond., , Imines are produced when aldehydes and ketones react with ammonia and its derivatives., , (ix), , 2, 4−DNP−derivative:, , 2, 4−dinitrophenylhydragones are 2, 4−DNP−derivatives, which are produced when, aldehydes or ketones react with 2, 4−dinitrophenylhydrazine in a weakly acidic medium., , 4
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(vi) 3,3-Dimethylbutanoic acid, (vii) Benzene-1,4-dicarbaldehyde, Question 12.3:, Draw the structures of the following compounds., (i) 3-Methylbutanal (ii) p-Nitropropiophenone, (iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one, , 6
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4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid, (vii) p,p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid, Answer, (i), , (ii), , (iii), , (iv), , (v), , (vi), , (vii), , (viii), , 7
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Question 12.6:, Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents., (i) PhMgBr and then H3O+, (ii)Tollens’ reagent, (iii) Semicarbazide and weak acid, (iv)Excess ethanol and acid, (v) Zinc amalgam and dilute hydrochloric acid, Answer, (i), , (ii), , (iii), , 10
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Compound (iv) Benzophenone is a ketone having no α-hydrogen atom and compound (viii), Butan-1-ol is an alcohol. Hence, these compounds do not undergo either aldol condensation, or cannizzaro reactions., Aldol condensation (ii), , (v), , (vi), , (vii), , Cannizzaro reaction (i), , 12
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(iii), , (ix), , Question 12.8:, How will you convert ethanal into the following compounds?, (i) Butane-1, 3-diol (ii) But-2-enal (iii) But-2-enoic acid, Answer, (i), , On treatment with dilute alkali, ethanal produces 3-hydroxybutanal gives butane-, , 1, 3-diol on reduction., , (ii), , On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating, , produces but-2-enal., , (iii), , When treated with Tollen’s reagent, But-2-enal produced in the above reaction, , produces but-2-enoic acid ., , 13
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Question 12.10:, An organic compound with the molecular formula C 9H10O forms 2, 4-DNP derivative,, reduces Tollens’ reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it, gives 1, 2-benzenedicarboxylic acid. Identify the compound., Answer, It is given that the compound (with molecular formula C 9H10O) forms 2, 4-DNP derivative, and reduces Tollen’s reagent. Therefore, the given compound must be an aldehyde. Again,, the, , compound, , undergoes, , cannizzaro, , reaction, , and, , on, , oxidation, , gives, , 1,, , 2benzenedicarboxylic acid. Therefore, the −CHO group is directly attached to a benzene, ring, , and, , this, , benzaldehyde, , is, , ortho-substituted., , Hence,, , 2ethylbenzaldehyde., , The given reactions can be explained by the following equations., , 15, , the, , compound, , is
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Question 12.11:, An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric, acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid, produced (B). (C) on dehydration gives but-1-ene.Write equations for the reactions, involved., Answer, An organic compound A with molecular formula C8H16O2 gives a carboxylic acid (B) and an, alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound A must be an ester., Further, alcohol C gives acid B on oxidation with chromic acid. Thus, B and C must contain, equal number of carbon atoms., Since compound A contains a total of 8 carbon atoms, each of B and C contain 4 carbon, atoms., Again, on dehydration, alcohol C gives but-1-ene. Therefore, C is of straight chain and, hence, it is butan-1-ol., On oxidation, Butan-1-ol gives butanoic acid. Hence, acid B is butanoic acid., Hence, the ester with molecular formula C8H16O2 is butylbutanoate., , All the given reactions can be explained by the following equations., , 16
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Hence, the given compounds can be arranged according to their increasing reactivities, toward HCN as:, Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde, (ii) After losing a proton, carboxylic acids gain a negative charge as shown:, , Now, any group that will help stabilise the negative charge will increase the stability of the, carboxyl ion and as a result, will increase the strength of the acid. Thus, groups having +I, effect will decrease the strength of the acids and groups having −I effect will increase the, strength of the acids. In the given compounds, −CH3 group has +I effect and Br− group, has −I effect. Thus, acids containing Br− are stronger., Now, the +I effect of isopropyl group is more than that of n-propyl group. Hence,, (CH3)2CHCOOH is a weaker acid than CH3CH2CH2COOH., Also, the −I effect grows weaker as distance increases. Hence, CH 3CH(Br)CH2COOH is a, weaker acid than CH3CH2CH(Br)COOH., Hence, the strengths of the given acids increase as:, (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH, (iii) As we have seen in the previous case, electron-donating groups decrease the, strengths of acids, while electron-withdrawing groups increase the strengths of acids. As, methoxy group is an electron-donating group, 4-methoxybenzoic acid is a weaker acid, 18
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than benzoic acid. Nitro group is an electron-withdrawing group and will increase the, strengths of acids. As 3,4-dinitrobenzoic acid contains two nitro groups, it is a slightly, stronger acid than 4-nitrobenzoic acid. Hence, the strengths of the given acids increase, as:, 4-Methoxybenzoic, , acid, , <, , Benzoic, , acid, , <, , 4-Nitrobenzoic, , acid, , < 3,4-Dinitrobenzoic acid, Question 12.13:, Give simple chemical tests to distinguish between the following pairs of compounds., (i), , Propanal and Propanone, , (ii), , Acetophenone and Benzophenone, , (iii) Phenol and Benzoic acid, (iv) Benzoic acid and Ethyl benzoate, (v), , Pentan-2-one and Pentan-3-one, , (vi) Benzaldehyde and Acetophenone, (vii) Ethanal and Propanal, Answer, (i) Propanal and propanone can be distinguished by the following tests., (a) Tollen’s test, Propanal is an aldehyde. Thus, it reduces Tollen’s reagent. But, propanone being a ketone, does not reduce Tollen’s reagent., , (b) Fehling’s test, Aldehydes respond to Fehling’s test, but ketones do not., Propanal being an aldehyde reduces Fehling’s solution to a red-brown precipitate of Cu2O,, but propanone being a ketone does not., , 19
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(c) Iodoform test:, Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon, atom respond to iodoform test. They are oxidized by sodium hypoiodite (NaOI) to give, iodoforms. Propanone being a methyl ketone responds to this test, but propanal does not., , (ii) Acetophenone and Benzophenone can be distinguished using the iodoform test., Iodoform test:, Methyl ketones are oxidized by sodium hypoiodite to give yellow ppt. of iodoform., Acetophenone being a methyl ketone responds to this test, but benzophenone does not., , (iii) Phenol and benzoic acid can be distinguished by ferric chloride test., Ferric chloride test:, Phenol reacts with neutral FeCl 3 to form an iron-phenol complex giving violet colouration., , But benzoic acid reacts with neutral FeCl 3 to give a buff coloured ppt. of ferric benzoate., , (iv) Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test., Sodium bicarbonate test:, Acids react with NaHCO3 to produce brisk effervescence due to the evolution of CO2 gas., Benzoic acid being an acid responds to this test, but ethylbenzoate does not., , 20
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(v) Pentan-2-one and pentan-3-one can be distinguished by iodoform test., Iodoform test:, Pentan-2-one is a methyl ketone. Thus, it responds to this test. But pentan-3-one not, being a methyl ketone does not respond to this test., , (vi) Benzaldehyde and acetophenone can be distinguished by the following tests., (a) Tollen’s Test, Aldehydes respond to Tollen’s test. Benzaldehyde being an aldehyde reduces Tollen’s, reagent to give a red-brown precipitate of Cu2O, but acetophenone being a ketone does, not., , (b) Iodoform test, Acetophenone being a methyl ketone undergoes oxidation by sodium hypoiodite (NaOI) to, give a yellow ppt. of iodoform. But benzaldehyde does not respond to this test., , (vii) Ethanal and propanal can be distinguished by iodoform test., Iodoform test, Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon, atom responds to the iodoform test. Ethanal having one methyl group linked to the, carbonyl carbon atom responds to this test. But propanal does not have a methyl group, linked to the carbonyl carbon atom and thus, it does not respond to this state., , 21
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Question 12.14:, How will you prepare the following compounds from benzene? You may use any inorganic, reagent and any organic reagent having not more than one carbon atom, (i) Methyl benzoate (ii) m-Nitrobenzoic acid, (iii) p-Nitrobenzoic acid (iv) Phenylacetic acid, (v) p-Nitrobenzaldehyde., Answer, (i), , (ii), , 22
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(iii), , (iv), , 23
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(ii), , (iii), , (iv), , (v), , (vi), , 25
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(vii), , (viii), , (ix), , Question 12.16:, Describe the following:, (i), , Acetylation, , (ii) Cannizzaro reaction, (iii) Cross aldol condensation, (iv) Decarboxylation, Answer, (i) Acetylation, The introduction of an acetyl functional group into an organic compound is known as, acetylation. It is usually carried out in the presence of a base such as pyridine,, dirnethylaniline, etc. This process involves the substitution of an acetyl group for an active, hydrogen atom. Acetyl chloride and acetic anhydride are commonly used as acetylating, agents., For example, acetylation of ethanol produces ethyl acetate., , 26
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(ii) Cannizzaro reaction:, The self oxidation-reduction (disproportionation) reaction of aldehydes having no, αhydrogens on treatment with concentrated alkalis is known as the Cannizzaro reaction., In this reaction, two molecules of aldehydes participate where one is reduced to alcohol, and the other is oxidized to carboxylic acid., For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and, potassium ethanoate are produced., , (iii) Cross-aldol condensation:, When aldol condensation is carried out between two different aldehydes, or two different, ketones, or an aldehyde and a ketone, then the reaction is called a cross-aldol, condensation. If both the reactants contain α-hydrogens, four compounds are obtained as, products., For example, ethanal and propanal react to give four products., , 27
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(iv) Decarboxylation:, Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form, hydrocarbons when their sodium salts are heated with soda-lime., , Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic, acids are electrolyzed. This electrolytic process is known as Kolbe’s electrolysis., Question 12.17:, Complete each synthesis by giving missing starting material, reagent or products (i), , 28
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(ii), , (iii), , (iv), , (v), , (vi), , (vii), , (viii), , (ix), , 29
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(x), , (xi), , Answer, (i), , (ii), , (iii), , (iv), , (v), , 30
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(vi), , (vii), , (viii), , (ix), , (x), , (xi), , Question 12.18:, 31
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Give plausible explanation for each of the following:, (i), , Cyclohexanone forms cyanohydrin in good yield but 2, 2, 6 trimethylcyclohexanone, , does not., (ii), , There are two −NH2 groups in semicarbazide. However, only one is involved in the, , formation of semicarbazones., (iii), , During the preparation of esters from a carboxylic acid and an alcohol in the, , presence of an acid catalyst, the water or the ester should be removed as soon as it is, formed., Answer, (i) Cyclohexanones form cyanohydrins according to the following equation., , In this case, the nucleophile CN− can easily attack without any steric hindrance. However,, in the case of 2, 2, 6 trimethylcydohexanone, methyl groups at α-positions offer steric, hindrances and as a result, CN− cannot attack effectively., , For this reason, it does not form a cyanohydrin., (ii), , Semicarbazide undergoes resonance involving only one of the two −NH 2 groups,, which is attached directly to the carbonyl-carbon atom., , Therefore, the electron density on −NH2 group involved in the resonance also decreases., As a result, it cannot act as a nucleophile. Since the other −NH 2 group is not involved in, , 32
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resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes, and ketones to produce semicarbazones., (iii), , Ester along with water is formed reversibly from a carboxylic acid and an alcohol in, , presence of an acid., , If either water or ester is not removed as soon as it is formed, then it reacts to give back, the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward, direction i.e., to produce more ester, either of the two should be removed., Question 12.19:, An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The, molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an, addition compound with sodium hydrogensulphite and give positive iodoform test. On, vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the, compound., Answer, % of carbon = 69.77 %, % of hydrogen = 11.63 %, % of oxygen = {100 − (69.77 + 11.63)}%, = 18.6 %, Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic, compound can be given as:, , Therefore, the empirical formula of the compound is C 5H10O. Now, the empirical formula, mass of the compound can be given as:, 5 × 12 + 10 ×1 + 1 × 16, , 33
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= 86, Molecular mass of the compound = 86, Therefore, the molecular formula of the compound is given by C 5H10O., Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again,, the compound forms sodium hydrogen sulphate addition products and gives a positive, iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone., The given compound also gives a mixture of ethanoic acid and propanoic acid., Hence, the given compound is pentan−2−ol., , The given reactions can be explained by the following equations:, , Question 12.20:, Although phenoxide ion has more number of resonating structures than carboxylate ion,, carboxylic acid is a stronger acid than phenol. Why?, Answer, Resonance structures of phenoxide ion are:, , 34
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It can be observed from the resonance structures of phenoxide ion that in II, III and IV,, less electronegative carbon atoms carry a negative charge. Therefore, these three, structures contribute negligibly towards the resonance stability of the phenoxide ion., Hence, these structures can be eliminated. Only structures I and V carry a negative charge, on the more electronegative oxygen atom., Resonance structures of carboxylate ion are:, , In the case of carboxylate ion, resonating structures I′ and II′ contain a charge carried by, a more electronegative oxygen atom., Further, in resonating structures I′ and II′, the negative charge is delocalized over two, oxygen atoms. But in resonating structures I and V of the phexoxide ion, the negative, charge is localized on the same oxygen atom. Therefore, the resonating structures of, carboxylate ion contribute more towards its stability than those of phenoxide ion. As a, result, carboxylate ion is more resonance-stabilized than phenoxide ion. Hence, carboxylic, acid is a stronger acid than phenol., , 35
