JEE Chemistry Aldehydes Ketones and Carboxylic Acids MCQs Set 31

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Chapter-wise Objective Questions: Aldehydes Ketones and Carboxylic Acids

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Multiple Choice Questions

Question 1: The species \(\ce{^-CH2-CO-CH3}\) and \(\ce{CH2=C(O^-)-CH3}\) are:
(a) resonating structures
(b) tautomers
(c) geometrical isomers
(d) optical isomers
Show Answer & Explanation

Answer: (a) resonating structures

Explanation:
1. They differ only in the position of electrons (the negative charge on carbon or on oxygen), with the same atomic arrangement.
2. Tautomers differ in the position of an atom (a proton), so these are resonance structures of the enolate ion.

Teacher's Note:
1. Resonance structures differ in electron placement only; tautomers differ in proton placement.

Question 2: In the sequence \(\ce{CH3CN + 2[H] ->[HCl, ether] X ->[H2O/H+] Y}\), the compound Y is:
(a) acetaldehyde
(b) ethanamine
(c) acetone
(d) acetic acid
Show Answer & Explanation

Answer: (a) acetaldehyde

Explanation:
1. This is the Stephen reaction: reduction of the nitrile with \(\ce{SnCl2}\)/HCl gives the imine hydrochloride (X).
2. Hydrolysis of the imine gives acetaldehyde.

Teacher's Note:
1. Stephen reduction converts nitriles to aldehydes.

Question 3: An ester A (molecular formula \(\ce{C9H10O2}\)) is treated with excess \(\ce{CH3MgBr}\) and then with \(\ce{H2SO4}\) to give an olefin B. Ozonolysis of B gives a ketone of formula \(\ce{C8H8O}\) that shows the iodoform test. The ester A is:
(a) phenyl propanoate
(b) ethyl benzoate
(c) 4-methoxyacetophenone
(d) propyl benzoate
Show Answer & Explanation

Answer: (b) ethyl benzoate

Explanation:
1. The ketone \(\ce{C8H8O}\) giving the iodoform test is acetophenone, so B = \(\ce{PhC(CH3)=CH2}\).
2. B comes from dehydration of \(\ce{PhC(CH3)2OH}\), the product of two additions of \(\ce{CH3MgBr}\) to an ester of benzoic acid.
3. Of the options only ethyl benzoate (\(\ce{C9H10O2}\)) is such an ester.

Teacher's Note:
1. Esters react with two equivalents of Grignard reagent to give tertiary alcohols with two identical new groups.

Question 4: Phenylmethanol can be obtained by reducing benzaldehyde with:
(a) \(\ce{CH3Br}\) and Na
(b) \(\ce{CH3I}\) and Mg
(c) \(\ce{CH3Br}\)
(d) \(\ce{LiAlH4}\)
Show Answer & Explanation

Answer: (d) \(\ce{LiAlH4}\)

Explanation:
1. Reduction of benzaldehyde with \(\ce{LiAlH4}\) gives benzyl alcohol (phenylmethanol).
2. The other reagents are alkylating or Grignard-forming species and would not just reduce the aldehyde.

Teacher's Note:
1. NaBH4 and LiAlH4 are the usual hydride reducing agents for aldehydes and ketones.

Question 5: Which of the following will undergo self aldol condensation in the presence of dilute alkali? (i) phenylacetaldehyde (ii) propenal (iii) benzaldehyde (iv) propanal
(a) (i) and (iv) only
(b) (ii) and (iii) only
(c) (i) and (ii) only
(d) (iii) and (iv) only
Show Answer & Explanation

Answer: (a) (i) and (iv) only

Explanation:
1. Self aldol condensation needs an alpha hydrogen.
2. Phenylacetaldehyde and propanal have them.
3. Propenal (acrolein) and benzaldehyde have none.

Teacher's Note:
1. Check the alpha carbon; an sp2 alpha carbon in propenal carries no hydrogen on the carbon next to CHO.

Question 6: \(\ce{(CH3)2C=CHCOCH3}\) can be oxidised to \(\ce{(CH3)2C=CHCOOH}\) by:
(a) chromic acid
(b) NaOI
(c) Cu at 573 K
(d) \(\ce{KMnO4}\)
Show Answer & Explanation

Answer: (b) NaOI

Explanation:
1. The methyl ketone is converted by the haloform reaction (NaOI) into a carboxylate with loss of \(\ce{CHI3}\).
2. The C=C bond is not touched by hypoiodite, whereas chromic acid and \(\ce{KMnO4}\) would attack the double bond.

Teacher's Note:
1. Haloform reagents oxidise the COCH3 group selectively.

Question 7: Which of the following gives a carboxylic acid with fewer carbon atoms than the starting compound on oxidation with hot \(\ce{KMnO4/H+}\)?
(a) acetone
(b) propanal
(c) propan-1-ol
(d) butanal
Show Answer & Explanation

Answer: (a) acetone

Explanation:
1. Acetone is cleaved at the C-C bond next to the carbonyl and gives acetic acid (and \(\ce{CO2}\)), so carbon atoms are lost.
2. Propanal, propan-1-ol and butanal are oxidised to acids with the same number of carbons.

Teacher's Note:
1. Aldehydes and primary alcohols keep their carbon skeleton; ketones are cleaved.

Question 8: If formaldehyde is heated with concentrated KOH, the products are potassium formate and:
(a) methane
(b) methyl alcohol
(c) ethyl formate
(d) acetylene
Show Answer & Explanation

Answer: (b) methyl alcohol

Explanation:
1. Formaldehyde has no alpha hydrogen and undergoes the Cannizzaro reaction: \(\ce{2HCHO + KOH -> CH3OH + HCOOK}\).

Teacher's Note:
1. The alcohol product of the Cannizzaro reaction of HCHO is methanol.

Question 9: In the sequence toluene \(\ce{->[KMnO4]}\) A \(\ce{->[SOCl2]}\) B \(\ce{->[H2/Pd-BaSO4]}\) C, the product C is:
(a) benzoic acid
(b) toluene
(c) benzyl alcohol
(d) benzaldehyde
Show Answer & Explanation

Answer: (d) benzaldehyde

Explanation:
1. Oxidation of toluene gives benzoic acid (A).
2. Thionyl chloride converts it to benzoyl chloride (B).
3. The Rosenmund reduction (H2, Pd-BaSO4 poisoned) stops at the aldehyde stage, giving benzaldehyde (C).

Teacher's Note:
1. BaSO4 poisons the palladium so that the aldehyde is not over-reduced to alcohol.

Question 10: Acetaldehyde and acetophenone can be distinguished chemically by which of the following? (i) Benedict's test (ii) iodoform test (iii) Tollens' reagent test (iv) Fehling's solution test
(a) (i), (iii) and (iv) only
(b) (ii) only
(c) (i) and (ii) only
(d) (ii), (iii) and (iv)
Show Answer & Explanation

Answer: (a) (i), (iii) and (iv) only

Explanation:
1. Acetaldehyde is an aliphatic aldehyde and reduces Benedict's, Tollens' and Fehling's reagents; acetophenone does not.
2. Both compounds contain the \(\ce{CH3CO-}\) group and both give the iodoform test, so test (ii) does not distinguish them.

Teacher's Note:
1. A test that gives the same result for both compounds cannot be used to distinguish them.

Question 11: In the laboratory 3,3-dimethylbutan-2-one \(\ce{(CH3)3C-CO-CH3}\) is converted in steps: bromination (50% yield) to the alpha-bromoketone, reduction to the bromohydrin (60%), and ring closure to the epoxide (70%). The overall yield of the epoxide is:
(a) 21%
(b) 42%
(c) 18%
(d) 30%
Show Answer & Explanation

Answer: (a) 21%

Explanation:
1. Overall yield = product of the fractional yields = 0.50 x 0.60 x 0.70 = 0.21.
2. That is 21%.

Teacher's Note:
1. Multiply step yields, do not add them.

Question 12: Acetone reacts with excess formaldehyde in the presence of KOH to give \(\ce{(HOCH2)3C-CO-C(CH2OH)3}\). The number of moles x of HCHO consumed per mole of acetone is:
(a) 4
(b) 5
(c) 6
(d) 7
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. Each of the six alpha hydrogens of acetone is replaced by a \(\ce{-CH2OH}\) group through aldol addition to HCHO.
2. So six moles of HCHO are consumed.

Teacher's Note:
1. Formaldehyde has no alpha hydrogen, so it is only an acceptor in crossed aldol reactions.

Question 13: In the conversion \(\ce{CH3CHO + 4HCHO ->[conc. KOH] C(CH2OH)4}\) (pentaerythritol), the total number of aldol reactions and crossed Cannizzaro reactions that occur is:
(a) 3
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Acetaldehyde has three alpha hydrogens, so three aldol additions with HCHO give \(\ce{(HOCH2)3C-CHO}\).
2. The fourth molecule of HCHO then reduces this aldehyde to the alcohol in a crossed Cannizzaro reaction.
3. Total = 3 + 1 = 4.

Teacher's Note:
1. Pentaerythritol is a classic example of aldol addition followed by crossed Cannizzaro reaction.

Question 14: How many of the following give the silver mirror test with Tollens' reagent: benzaldehyde, acetaldehyde, acetone, formic acid, acetophenone, ethanol, butanal, glyoxal (ethanedial), cyclohexanecarboxylic acid?
(a) 4
(b) 5
(c) 6
(d) 7
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. Aldehydes (including formic acid, which has an aldehydic group) reduce Tollens' reagent.
2. Benzaldehyde, acetaldehyde, formic acid, butanal and glyoxal are positive.
3. Acetone, acetophenone, ethanol and cyclohexanecarboxylic acid are not.
4. Total = 5.

Teacher's Note:
1. Formic acid is the only carboxylic acid that gives Tollens' and Fehling's tests.

Question 15: How many of the following are reduced by \(\ce{NaBH4}\): benzaldehyde, benzamide, phthalide (a lactone), cyclohexanone, 3-methylbutan-2-one, propanal, ethyl butanoate, acetophenone, succinimide?
(a) 4
(b) 5
(c) 6
(d) 7
Show Answer & Explanation

Answer: (b) 5

Explanation:
1. Sodium borohydride reduces aldehydes and ketones.
2. Benzaldehyde, cyclohexanone, 3-methylbutan-2-one, propanal and acetophenone are reduced.
3. Amides, imides, esters and lactones are not reduced, so five compounds react.

Teacher's Note:
1. NaBH4 is selective for aldehydes and ketones; LiAlH4 also reduces esters, acids and amides.

Question 16: How many of the following compounds give a positive iodoform test: pentan-3-one, butan-2-one, acetone, acetaldehyde, propiophenone, acetophenone, cyclohexane-1,3-dione, 1,3-diphenylpropane-1,3-dione and cyclopentanone?
(a) 3
(b) 4
(c) 5
(d) 6
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. A positive test requires a \(\ce{CH3CO-}\) group (or \(\ce{CH3CH(OH)-}\)).
2. Butan-2-one, acetone, acetaldehyde and acetophenone have it.
3. The others have no methyl ketone group.

Teacher's Note:
1. Count the CH3-C=O units directly attached to carbon or H.

Question 17: How many of the compounds p-methoxybenzaldehyde, p-chlorobenzaldehyde, p-nitrobenzaldehyde, p-dimethylaminobenzaldehyde, p-hydroxybenzaldehyde and p-cyanobenzaldehyde show a higher rate of nucleophilic addition than p-methylbenzaldehyde?
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Electron-withdrawing groups make the carbonyl carbon more electrophilic.
2. p-Chloro, p-nitro and p-cyano compounds are more reactive than p-tolualdehyde.
3. The methoxy, dimethylamino and hydroxy groups donate electrons by resonance, so those compounds are less reactive.

Teacher's Note:
1. Reactivity towards nucleophiles rises with the electron-withdrawing power of the para substituent.

Question 18: How many reagents among Zn-Hg/HCl, \(\ce{LiAlH4}\), \(\ce{CHCl3 + NaOH}\), \(\ce{N2H4/OH^-}\), ethane-1,2-dithiol followed by \(\ce{H2/Ni}\), and \(\ce{SeO2}\) can convert \(\ce{R-CO-R}\) into \(\ce{R-CH2-R}\)?
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Zn-Hg/HCl (Clemmensen), hydrazine with base (Wolff-Kishner) and conversion into the dithioacetal followed by Raney Ni desulphurisation all reduce C=O to CH2.
2. \(\ce{LiAlH4}\) stops at the alcohol, \(\ce{CHCl3/NaOH}\) generates dichlorocarbene and \(\ce{SeO2}\) is an oxidant.

Teacher's Note:
1. Three standard C=O to CH2 routes: Clemmensen, Wolff-Kishner and thioacetal desulphurisation.

Question 19: A decalin-type bicyclic compound has two carbon atoms, each carrying two -COOH groups and each directly attached to a ketone carbonyl. How many moles of \(\ce{CO2}\) are released when one mole is heated?
(a) 2
(b) 3
(c) 4
(d) 6
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Each gem-dicarboxylic acid unit next to a ketone is a beta-keto acid and also a malonic acid derivative.
2. On heating, the first COOH is lost easily and the second one (still beta to the ketone) is lost as well.
3. Two such units release 2 x 2 = 4 moles of \(\ce{CO2}\).

Teacher's Note:
1. Beta-keto acids and malonic acids lose CO2 on mild heating through a cyclic six-membered transition state.

Question 20: How many of the following esters can undergo Claisen condensation: ethyl propanoate, ethyl formate, ethyl benzoate, ethyl cyclopentylacetate, diethyl oxalate and ethyl phenylacetate?
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. A Claisen self-condensation needs an ester with alpha hydrogen.
2. Ethyl propanoate, ethyl cyclopentylacetate and ethyl phenylacetate have them.
3. Ethyl formate, ethyl benzoate and diethyl oxalate have none.

Teacher's Note:
1. Diethyl oxalate and ethyl formate are useful partners in crossed Claisen condensations since they have no alpha hydrogen.

Question 21: A six-membered lactone (delta-valerolactone) is treated with X moles of \(\ce{CH3MgBr}\) followed by \(\ce{H+/H2O}\) to give the diol \(\ce{HO-(CH2)4-C(CH3)2-OH}\). The value of X is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. The first mole of Grignard reagent opens the lactone (attack at the ester carbonyl and loss of the alkoxide) to give a ketone.
2. The ketone reacts with a second mole of \(\ce{CH3MgBr}\) to give the tertiary alkoxide.
3. Hydrolysis gives the diol, so X = 2.

Teacher's Note:
1. Esters and lactones consume two equivalents of a Grignard reagent.

Question 22: How many different condensation products are possible when ethyl acetate and ethyl propanoate are heated together with sodium ethoxide in ethanol?
(a) 2
(b) 3
(c) 4
(d) 6
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Both esters have alpha hydrogens, so each can act as the enolate (nucleophile) or as the acceptor.
2. Two self-condensation products and two crossed-condensation products form: 2 x 2 = 4.

Teacher's Note:
1. In a crossed Claisen condensation between two enolisable esters, four products are possible.

Question 23: How many of the following esters show the \(\ce{A_{AL}1}\) mechanism of hydrolysis (acid catalysed, unimolecular, alkyl-oxygen fission): tert-butyl acetate, 1,1-diphenylethyl acetate, methyl benzoate, methyl 2,4,6-trimethylbenzoate, tert-butyl 4-nitrobenzoate and gamma-butyrolactone?
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. The AAL1 path needs an alkyl group that forms a stable carbocation, i.e. a tertiary or highly stabilised group on oxygen.
2. tert-Butyl acetate, 1,1-diphenylethyl acetate and tert-butyl 4-nitrobenzoate fit.
3. The methyl esters and the lactone do not form stable cations (methyl 2,4,6-trimethylbenzoate hydrolyses by the AAC1 acyl-oxygen path because of steric hindrance).

Teacher's Note:
1. Tertiary alkyl esters hydrolyse with alkyl-oxygen fission in acid.

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