Chemistry · Aldehydes, ketones and carboxylic acids · chapter 8 of 10

Class 12 Chemistry Chapter 8: Aldehydes, Ketones and Carboxylic Acids MCQs with answers

10 multiple-choice questions on Aldehydes, Ketones and Carboxylic Acids, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
In propanal, CH₃CH₂CHO, the hybridisation of the carbonyl carbon and the approximate bond angle around it are
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Answer: (a) sp², 120°

The carbonyl carbon forms three σ bonds from sp² orbitals and a π bond with oxygen from its p orbital; the three σ bonds lie in one plane about 120° apart.

Question 2
The IUPAC name of CH₃CH₂CH(CH₃)CH₂CHO is
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Answer: (c) 3-methylpentanal

The longest chain containing –CHO has five carbons, numbered from the aldehyde carbon (C-1); the methyl group is on C-3: 3-methylpentanal. '-one' is for ketones, and 3-ethylbutanal uses a shorter chain.

Question 3
Oxidation of 3-methylbutan-2-ol, CH₃CH(OH)CH(CH₃)₂, with acidified K₂Cr₂O₇ gives
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Answer: (b) 3-methylbutan-2-one

It is a secondary alcohol (–OH on C-2, which carries one H); secondary alcohols are oxidised to ketones: CH₃COCH(CH₃)₂, 3-methylbutan-2-one.

Question 4
Propanoyl chloride, CH₃CH₂COCl, is converted to propanal by
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Answer: (d) H₂ over Pd–BaSO₄

Rosenmund reduction: hydrogen over palladium on barium sulphate reduces an acyl chloride only as far as the aldehyde. LiAlH₄ and NaBH₄ go on to the alcohol; Zn–Hg/HCl reduces C=O to CH₂.

Question 5
Butanenitrile, CH₃CH₂CH₂CN, is treated with SnCl₂ and HCl, and the product is hydrolysed with H₃O⁺. The final product is
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Answer: (c) butanal

Stephen reaction: SnCl₂/HCl reduces the nitrile to the imine CH₃CH₂CH₂CH=NH, which hydrolyses to CH₃CH₂CH₂CHO, butanal.

Question 6
Methyl pent-3-enoate, CH₃–CH=CH–CH₂–COOCH₃, is treated with DIBAL-H and then with water. The organic product is
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Answer: (c) CH₃–CH=CH–CH₂–CHO

DIBAL-H reduces an ester to the aldehyde and leaves the C=C untouched, as it does with nitriles: pent-3-enal. It does not hydrogenate the double bond, and hydrolysis to the acid is not a reduction.

Question 7
Toluene can be converted to benzaldehyde, the oxidation stopping at the aldehyde stage, by treating it with
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Answer: (b) CrO₂Cl₂ in CS₂, then H₃O⁺

Etard reaction: chromyl chloride turns the methyl group into a chromium complex, C₆H₅CH(OCrOHCl₂)₂, which hydrolyses to C₆H₅CHO. Strong oxidants (KMnO₄, K₂Cr₂O₇) go on to benzoic acid; Cl₂/FeCl₃ chlorinates the ring.

Question 8
Cyclohexanecarbonyl chloride, C₆H₁₁COCl, reacts with dimethylcadmium. The product is
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Answer: (a) 1-cyclohexylethan-1-one, C₆H₁₁COCH₃

2R′COCl + R₂Cd → 2R′COR + CdCl₂: the acyl group joins the methyl group of the cadmium reagent, giving the ketone C₆H₁₁COCH₃. Dialkylcadmium stops at the ketone; it does not go on to the tertiary or secondary alcohol.

Question 9
Ethanenitrile, CH₃C≡N, is treated with ethylmagnesium bromide in dry ether and the product is hydrolysed. The compound obtained is
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Answer: (a) butan-2-one

CH₃C≡N + C₂H₅MgBr → CH₃C(=NMgBr)C₂H₅; H₃O⁺ hydrolyses the imine salt to CH₃COC₂H₅, butan-2-one. The nitrile carbon becomes the carbonyl carbon carrying both groups.

Question 10
2-Methylpent-2-ene, (CH₃)₂C=CH–CH₂CH₃, is treated with ozone and then with zinc dust and water. The products are
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Answer: (d) propanone and propanal

Ozonolysis cleaves the C=C and puts C=O on each carbon: (CH₃)₂C= gives (CH₃)₂C=O, propanone; =CH–CH₂CH₃ gives OHC–CH₂CH₃, propanal. Zinc dust prevents oxidation of the aldehyde to the acid.

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