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Free questions · JEE Advanced · Chemistry

Aldehydes and Ketones: questions with solutions

3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.

Question 1Level 2 · ExamSingle correct
Which one of the following reagents or sequences converts benzene into benzaldehyde as the main organic product? Exactly one option is correct.
ACO and HCl in the presence of anhydrous AlCl (with a little CuCl)
BCrOCl in CS, then HO
CCHCl with anhydrous AlCl, then hot alkaline KMnO followed by acidification
DCl with anhydrous FeCl, then Mg in dry ether, then CO, then HO
Show answer and solution

Answer: A

(A) Gattermann–Koch formylation: CO + HCl behave as formyl chloride (HCOCl) and, with anhydrous AlCl/CuCl, substitute a ring H by –CHO: CH + CO + HCl → CHCHO + HCl. Correct. (B) Chromyl chloride (Etard reaction) oxidises a side-chain CH (toluene) to CHO; benzene has no side chain, so no benzaldehyde. (C) Friedel–Crafts methylation gives toluene, but hot KMnO oxidises the CH all the way to –COOH: the product is benzoic acid. (D) Chlorobenzene → CHMgCl → carboxylation gives benzoic acid, again not the aldehyde. Answer: (A). Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
Toluene is subjected to each of the following treatments. Which treatment does NOT give benzaldehyde as the main organic product? Exactly one option is correct.
ACrOCl in CS, then HO
BCrO in acetic anhydride, then warming the isolated product with dilute aqueous acid
CCl (2 mol per mol of toluene) in sunlight, then HO at 373 K
DCl (2 mol per mol of toluene) with anhydrous FeCl in the dark, then HO at 373 K
Show answer and solution

Answer: D

(A) Etard reaction: the chromium complex formed on the CH group is hydrolysed to CHCHO. Gives benzaldehyde. (B) CrO/(CHCO)O converts the CH group into benzylidene diacetate CHCH(OCOCH); acid hydrolysis of this gem-diacetate releases CHCHO. Gives benzaldehyde. (C) Free-radical side-chain chlorination gives benzal chloride CHCHCl; hydrolysis of the gem-dihalide at 373 K gives CHCH(OH) → CHCHO + HO. Gives benzaldehyde. (D) Lewis-acid conditions in the dark give ring substitution (2- and 4-chlorotoluenes, then dichlorotoluenes). Aryl C–Cl bonds are not hydrolysed by water at 373 K and the CH group is untouched, so no benzaldehyde forms. Trap: the same reagent (Cl) gives side-chain or ring substitution depending on the conditions (light vs Lewis acid). Answer: (D). Difficulty: Very hard
Question 3Level 2 · ExamMatching list
Match each conversion in List-I with the reagent(s) in List-II that bring it about in good yield. Exactly one option is correct.
List-I
  1. (P) CHCOCl → CHCHO
  2. (Q) CHCN → CHCHO
  3. (R) CH → CHCHO
  4. (S) CHCOOCH → CHCHO
List-II
  1. (1) SnCl and HCl, then HO
  2. (2) DIBAL-H at low temperature (about −78 °C), then HO
  3. (3) H over Pd on BaSO (partially poisoned)
  4. (4) CO and HCl with anhydrous AlCl/CuCl
  5. (5) CrOCl in CS, then HO
AP→3, Q→1, R→5, S→2
BP→3, Q→2, R→4, S→1
CP→3, Q→1, R→4, S→2
DP→1, Q→3, R→4, S→2
Show answer and solution

Answer: C

(P) Acid chloride → aldehyde: Rosenmund reduction, H over Pd/BaSO; the poisoned catalyst stops the reduction at the aldehyde → (3). (Q) Nitrile → aldehyde: Stephen reduction, SnCl/HCl gives the imine (as its salt), which is hydrolysed to the aldehyde → (1). (DIBAL-H also reduces nitriles to aldehydes, but (2) is needed for S and every option gives each List-II entry to one item only.) (R) Benzene → benzaldehyde: Gattermann–Koch, CO/HCl/AlCl/CuCl → (4). Chromyl chloride (5) needs a CH side chain (toluene), so it fits none of P–S. (S) Ester → aldehyde: DIBAL-H at low temperature → (2). SnCl/HCl does not reduce an ester. P→3, Q→1, R→4, S→2. Answer: (C). Difficulty: Hard

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