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| Chapter | 12. Aldehydes, ketones and Carboxylics Acids |
| Subject | Chemistry |
| Textbook | Chemistry II |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 12: Aldehydes, ketones and Carboxylics Acids.
NCERT Solutions for Class 12 Chemistry Chapter 12
Aldehydes, ketones and Carboxylics Acids Solutions
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) Cyanohydrins are organic compounds having the formula RR′C(OH)CN, where R and R′ can be alkyl or aryl groups.
In the presence of excess sodium cyanide (NaCN) as a precursor to using cyanohydrin, aldehydes and ketones react with hydrogen cyanide (HCN). Such reactions are classified as reactions to cyanohydrins.
The useful synthetic intermediates are Cyanohydrins.
Answer ii) The gem−dialkoxy alkanes at the terminal carbon atom consisting of two groups of alkoxy are called Acetal. One bond is connected to an alkyl group, and the other to a hydrogen atom.
The general structure of an acetal.
Hemiacetals are formed in the presence of dry HCl gas when aldehydes are treated with two monohydric alcohol equivalents. In addition, hemiacetals react with another alcohol molecule to yield acetal.
Answer iii) Semicarbazone comes from aldehydes and ketones, which is formed by the condensation reaction between a ketone or aldehyde and semicarbazide.
Semicarbarzones serve to identify and characterize aldehydes and ketones.
Answer iv) 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.
Answer v) Hemiacetal: Hemiacetals are α−alkoxyalcohols
The general structure of a hemiacetal. In the presence of dry HCl gas, aldehyde reacts with one single molecule of monohydrated alcohol.
Answer 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.
Answer 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.
Ketones react with ethylene glycol in the presence of dry HCl gas to give a cyclic product known as ethylene glycol ketals.
Answer 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.
Answer 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.
Answer 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. It is named after a scientist, Hugo Schiff.
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) Benzene-1,4-dicarbaldehyde
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)
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
Answer ii) CH3CH2CHBrCH2CH(CH3)CHO
IUPAC name: 4-Bromo-2-methylhaxanal
Common name: (γ-Bromo-α-methyl-caproaldehyde)
Answer iii) CH3(CH2)5CHO
IUPAC name: Heptanal
Answer iv) Ph-CH=CH-CHO
IUPAC name: 3-phenylprop-2-enal
Common name: β-Pheynolacrolein
Answer v) IUPAC name: Cyclopentanecarbaldehyde
Answer 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)
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)
Answer ii)
Answer iii)
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
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) Ethanal produces 3-hydroxybutanal gives butane-1, 3-diol on reduction when it is treated with dilute alkali.
Answer ii) Ethanal gives 3-hydroxybutanal which on heating produces but-2-enal when it is treated with dilute alkali.
Answer iii) When treated with Tollen’s reagent, But-2-enal produced in the above reaction produces but-2-enoic acid.
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) Taking two molecules of propanal, one which acts as a nucleophile and the other as an electrophile.
ii) Taking two molecules of butanal, one which acts as a nucleophile and the other as an electrophile.
iii) Taking one molecule each of propanal and butanal in which propanal acts as a nucleophile and butanal acts as an electrophile.
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.
Again, the compound undergoes cannizzaro reaction and on oxidation gives 1, 2-benzenedicarboxylic acid. Therefore, the −CHO group is directly attached to a benzene ring and this benzaldehyde is ortho-substituted. Hence, the compound is 2-ethylbenzaldehyde.
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.
All the given reactions can be explained by the following equations.
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) 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.
Answer 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, 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.
Answer 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 < 3,4-Dinitrobenzoic 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: 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.
(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.
Answer 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.
Answer iii) Phenol and benzoic acid can be distinguished by ferric chloride test.
Ferric chloride test: Phenol reacts with neutral FeCl3 to form an iron-phenol complex giving violet colouration.
But benzoic acid reacts with neutral FeCl3 to give a buff coloured ppt. of ferric benzoate.
Answer iv) Benzoic acid and Ethyl benzoate can be distinguished by sodium bicarbonate test.
Sodium bicarbonate test: Brisk effervescence is produced when acids react with NaHCO3 due to the evolution of CO2 gas. Benzoic acid being an acid responds to this test, but ethylbenzoate does not.
Answer 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.
Answer 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.
Answer 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.
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)
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)
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.
For example, acetylation of ethanol produces ethyl acetate.
Answer 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.
For example, when ethanol is treated with concentrated potassium hydroxide, ethanol and potassium ethanoate are produced.
Answer 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.
Answer 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.
Q17) Complete each synthesis by giving missing starting material, reagent or products
Answer)
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.
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.
Answer ii) Semicarbazide undergoes resonance involving only one of the two −NH2 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 −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.
Answer 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.
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 %
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 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.
The given reactions can be explained by the following equations:
Q20) 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:
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.
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