Class 12 Chemistry Chapter 8: Aldehydes, Ketones and Carboxylic Acids worksheet with answer key

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.

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.

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.

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.

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.

Base removes a proton from HCN to give CN⁻, a much stronger nucleophile than HCN; CN⁻ attacks the electrophilic carbonyl carbon and the alkoxide formed takes a proton. Product: (CH₃)₂C(OH)CN, propanone cyanohydrin (2-hydroxy-2-methylpropanenitrile).

• base generates CN⁻, the stronger nucleophile, which attacks the carbonyl carbon (1)
• product (CH₃)₂C(OH)CN, the cyanohydrin (1)

(a) (CH₃)₂CH–COONa + NaOH → (CH₃)₂CH₂ + Na₂CO₃ (CaO, heat): propane, CH₃CH₂CH₃; decarboxylation removes one carbon. (b) Kolbe electrolysis joins two ethyl groups: CH₃CH₂–CH₂CH₃, butane.

• (a) propane, with the decarboxylation equation (1)
• (b) butane by Kolbe electrolysis (1)

Shake the mixture with saturated aqueous NaHSO₃: pentanal forms the addition compound CH₃(CH₂)₃CH(OH)SO₃⁻Na⁺, which is water soluble and passes into the aqueous layer; pentane does not react. The aqueous layer, treated with dilute mineral acid or alkali, gives back pentanal.

• NaHSO₃ adds to the aldehyde to give a water-soluble addition compound, separated from pentane (1)
• dilute acid or alkali regenerates pentanal (1)

The α-hydrogen, on C-2 next to the C=O, is acidic; its carbanion attacks the carbonyl carbon of a second molecule. Aldol: CH₃CH₂CH₂–CH(OH)–CH(CH₂CH₃)–CHO, 2-ethyl-3-hydroxyhexanal. On heating it loses water: CH₃CH₂CH₂–CH=C(CH₂CH₃)–CHO, 2-ethylhex-2-enal.

• the α-hydrogen on C-2 is the acidic one (1)
• aldol structure and name, 2-ethyl-3-hydroxyhexanal (1)
• dehydration product 2-ethylhex-2-enal (1)

1. CN⁻ approaches the electrophilic carbonyl carbon roughly perpendicular to the plane of its sp² orbitals and forms a C–C bond; the π electrons move onto oxygen. 2. A tetrahedral alkoxide intermediate, CH₃CH(O⁻)CN, forms. 3. The alkoxide takes a proton from the medium (HCN or water): CH₃CH(OH)CN, 2-hydroxypropanenitrile. The carbon changes from sp², trigonal planar, to sp³, tetrahedral.

• step 1: CN⁻ attacks the carbonyl carbon, π electrons to oxygen (1)
• steps 2 and 3: alkoxide intermediate protonated to CH₃CH(OH)CN (1)
• sp² trigonal planar becomes sp³ tetrahedral (1)

Route 1: CH₃CH₂CH₂Br + KCN → CH₃CH₂CH₂CN; CH₃CH₂CH₂CN → CH₃CH₂CH₂CONH₂ → CH₃CH₂CH₂COOH (H₃O⁺, heat). Route 2: CH₃CH₂CH₂Br + Mg → CH₃CH₂CH₂MgBr (dry ether); + CO₂ → CH₃CH₂CH₂COO⁻MgBr⁺; H₃O⁺ → CH₃CH₂CH₂COOH. Both give an acid with one carbon more than the alkyl halide.

• nitrile route with hydrolysis (1)
• Grignard route with CO₂ and acidification (1)
• both add one carbon (ascending the series) (1)