NCERT Solutions for Class 12 Chemistry Part 1 Chapter 12
Aldehydes, Ketones and Carboxylic Acids Class 12
Chapter 12 Aldehydes, Ketones and Carboxylic Acids Exercise Solutions
In text : Solutions of Questions on Page Number : 353
Q1 :
Write the structures of the following compounds.
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec-butyl ketone
(vi) 4-Fluoroacetophenone
Answer : (i)
(iii)
(v)
Q2 :
Write the structures of products of the following reactions; (i)
(iii)
Q3 :
Arrange the following compounds in increasing order of their boiling points. CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
Answer :
The molecular masses of the given compounds are in the range 44 to 46. CH3CH2OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH3CHO is more polar than CH3OCH3and so CH3CHO has stronger intermolecular dipole - dipole attraction than CH3OCH3┹”¦CH3CH2CH3has only weak van der Waals force. Thus, the arrangement of the given compounds in the increasing order of their boiling points is given by:
Q4 :
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions. (i)Ethanal, Propanal, Propanone, Butanone.
(ii)Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone.
Hint:Consider steric effect and electronic effect.
Answer :
The +I effect of the alkyl group increases in the order:
Ethanal < Propanal < Propanone < Butanone
The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is:
Butanone < Propanone < Propanal < Ethanal
The +I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron- donating -CH3group and the lowest in p-nitrobezaldehyde because of the presence of the electron-withdrawing - NO2group. Hence, the increasing order of the reactivities of the given compounds is:
Acetophenone < p-tolualdehyde < Benzaldehyde
Q5 :
Predict the products of the following reactions: (i)
(ii)
Answer : (i)
(iii)
Q6 :
Give the IUPAC names of the following compounds:
(i) PhCH2CH2COOH (ii) (CH3)2C=CHCOOH
(iii) (iv)
Answer :
(i) 3-Phenylpropanoic acid
(ii) 3-Methylbut-2-enoic acid
(iii) 2-Methylcyclopentanecarboxylic acid
(iv)
Q7 :
Show how each of the following compounds can be converted to benzoic acid.
(i) Ethylbenzene (ii) Acetophenone
(iii) Bromobenzene (iv) Phenylethene (Styrene)
Answer : (i)
(iii)
Q8 :
Which acid of each pair shown here would you expect to be stronger?
(i) CH3CO2H or CH2FCO2H (ii)CH2FCO2H or CH2ClCO2H
(iii) CH2FCH2CH2CO2H or CH3CHFCH2CO2H (iv)
Answer : (i)
The +I effect of -CH3group increases the electron density on the O-H bond. Therefore, release of proton becomes difficult. On the other hand, the -I effect of F decreases the electron density on the O-H bond. Therefore, proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H.
F has stronger -I effect than Cl. Therefore, CH2FCO2H can release proton more easily than CH2ClCO2H. Hence,CH2FCO2H is stronger acid than CH2ClCO2H.
Inductive effect decreases with increase in distance. Hence, the +I effect of F in CH3CHFCH2CO2H is more than it is in CH2FCH2CH2CO2H. Hence, CH3CHFCH2CO2H is stronger acid than CH2FCH2CH2CO2H.
Due to the -I effect of F, it is easier to release proton inthe case of compound (A). However, in the case of compound (B), release of proton is difficult due to the +I effect of -CH3group. Hence, (A) is a stronger acid than (B).
Exercise : Solutions of Questions on Page Number : 377
Q1 :
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:
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:
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:
(v) Hemiacetal:
General structure of a hemiacetal
(vi) Oxime:
(vii) Ketal:
Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals.
(viii) Imine:
(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.
To identify and characterize aldehydes and ketones, 2, 4 - DNP derivatives are used.
(x) Schiff's base:
Schiff's base (or azomethine) is a chemical compound containing a carbon-nitrogen double bond with the nitrogen atom connected to an aryl or alkyl group-but not hydrogen. They have the general formula R1R2C = NR3. Hence, it is an imine.
Aldehydes and ketones on treatment with primary aliphatic or aromatic amines in the presence of trace of an acid yields a Schiff's base.
Q2 :
Name the following compounds according to IUPAC system of nomenclature:
(i) CH3CH(CH3)CH2CH2CHO
(ii) CH3CH2COCH(C2H5)CH2CH2Cl
(iii) CH3CH=CHCHO
(iv) CH3COCH2COCH3
(v) CH3CH(CH3)CH2C(CH3)2COCH3
(vi) (CH3)3CCH2COOH
(vii) OHCC6H4CHO-p
Answer :
(i) 4-methylpentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-en-1-al
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethylhexan-2-one
(vi) 3,3-Dimethylbutanoic acid
(vii)
Q3 :
Draw the structures of the following compounds.
(i) 3-Methylbutanal (ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid
(vii) p,p'-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid
Answer : (i)
(iv)
(v)
(vii)
Q4 :
Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH3CO(CH2)4CH3 (ii) CH3CH2CHBrCH2CH(CH3)CHO
(iii) CH3(CH2)5CHO (iv) Ph-CH=CH-CHO
(v) (vi) PhCOPh
Answer :
(i) CH3CO(CH2)4CH3
IUPAC name: Heptan-2-one
Common name: Methyl n-propyl ketone
(ii) CH3CH2CHBrCH2CH(CH3)CHO IUPAC name: 4-Bromo-2-methylhaxanal
Common name: (γ-Bromo-α-methyl-caproaldehyde)
(iii) CH3(CH2)5CHO IUPAC name: Heptanal
(iv) Ph-CH=CH-CHO
IUPAC name: 3-phenylprop-2-enal Common name: β-Pheynolacrolein
(v)
IUPAC name: Cyclopentanecarbaldehyde
(vi)PhCOPh
IUPAC name: Diphenylmethanone Common name: Benzophenone
Q5 :
Draw structures of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Answer : (i)
(iii)
(iv)
(vi)
Q6 :
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)
(iv)
Q7 :
Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal (ii) 2-Methylpentanal
(iii) Benzaldehyde (iv) Benzophenone
(v) Cyclohexanone (vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde (viii) Butan-1-ol
(ix) 2, 2-Dimethylbutanal
Answer :
Aldehydes and ketones having at least one α-hydrogen undergo aldol condensation. The compounds (ii) 2- methylpentanal, (v) cyclohexanone, (vi) 1-phenylpropanone, and (vii) phenylacetaldehyde contain one or more α- hydrogen atoms. Therefore, these undergo aldol condensation.
Aldehydes having no α-hydrogen atoms undergo Cannizzaro reactions. The compounds (i) Methanal, (iii) Benzaldehyde, and (ix) 2, 2-dimethylbutanal do not have any α-hydrogen. Therefore, these undergo cannizzaro reactions.
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)
(vi)
Cannizzaro reaction (i)
(ix)
Q8 :
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)
(ii) On treatment with dilute alkali, ethanal gives 3-hydroxybutanal which on heating produces but-2-enal.
(iii)
Q9 :
Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
Answer :
(i)
(ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.
(iii)
(iv) Taking one molecule each of propanal and butanal in which propanal acts as an electrophile and butanal acts as a nucleophile.
Q10 :
An organic compound with the molecular formula C9H10O 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 C9H10O) forms 2, 4-DNP derivative and reduces Tollen's reagent. Therefore, the given compound must be an aldehyde.
The given reactions can be explained by the following equations.
Q11 :
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.
Q12 :
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer :
(i) When HCN reacts with a compound, the attacking species is a nucleophile, CN - . Therefore, as the negative charge on the compound increases, its reactivity with HCN decreases. In the given compounds, the +I effect increases as shown below. It can be observed that steric hindrance also increases in the same
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)
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, - CH3group 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, CH3CH(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 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
Q13 :
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
(b) Fehling's test
Aldehydes respond to Fehling's test, but ketones do not.
(c) Iodoform test:
(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:
But benzoic acid reacts with neutral FeCl3 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.
(v) Pentan-2-one and pentan-3-one can be distinguished by iodoform test.
Iodoform test:
(vi) Benzaldehyde and acetophenone can be distinguished by the following tests.
(a) Tollen's Test
(b) Iodoform test
(vii) Ethanal and propanal can be distinguished by iodoform test.
Iodoform test
Q14 :
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)
(iii)
(v)
Q15 :
How will you bring about the following conversions in not more than two steps?
(i) Propanone to Propene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone
(vi) Bromobenzene to 1-Phenylethanol
(vii) Benzaldehyde to 3-Phenylpropan-1-ol
(viii) Benazaldehyde to α-Hydroxyphenylacetic acid
(ix) Benzoic acid to m- Nitrobenzyl alcohol
Answer : (i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Q16 :
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.
(ii) Cannizzaro reaction:
The self oxidation-reduction (disproportionation) reaction of aldehydes having no α-hydrogens 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.
(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.
(iv) Decarboxylation:
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.
Q17 :
Complete each synthesis by giving missing starting material, reagent or products (i)
(ii)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
Answer :
(ii)
(iv)
(vii)
(ix)
(xi)
Q18 :
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.
For this reason, it does not form a cyanohydrin.
(ii)
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 - NH2 group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.
(iii)
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.
Q19 :
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 %
Therefore, the empirical formula of the compound is C5H10O. Now, the empirical formula mass of the compound can be given as:
5 × 12 + 10 ×1 + 1 × 16
= 86
Molecular mass of the compound = 86
Therefore, the molecular formula of the compound is given by C5H10O.
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-one.
Q20 :
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer :
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.
In the case of carboxylate ion, resonating structures I”² and II”² contain a charge carried by a more electronegative oxygen atom.
Further, in resonating structures I”² and II”², 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.