Skip to content
Question of 87

Q.(a) Complete the reaction: CH3–CO–CH3 --Ba(OH)2--> A --Δ--> B.

(b) Distinguish, with the help of a chemical reaction: Formic acid and Acetic acid.
(c) Complete the conversion: Benzaldehyde to Benzophenone. OR
(a) An organic compound A (C8H16O2), on hydrolysis with dilute H2SO4, produces a carboxylic acid (B) and an alcohol (C). Oxidation of 'C' by chromic acid gives compound B. Dehydration of C produces But-1-ene. Identify A, B, C and write the equations of the relevant reactions.
(b) What is Tollens' reagent?
(c) CH3–CH2–CH2–CO–CH3 --concentrated HNO3, Δ--> A + B. Identify compounds A and B.
Tripura TbseHigher Secondary (+2 Stage) Examination 2024Subjective· 5mImportance★★★★★
0% · 0/87 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Answering the primary (a)(b)(c) parts: acetone undergoes a base-catalysed aldol condensation (giving diacetone alcohol, which dehydrates to mesityl oxide); formic acid is distinguished from acetic acid by the Tollens' silver-mirror test (only formic acid, having an aldehyde-like reducing carbon, gives it); and benzaldehyde is converted to benzophenone via benzoic acid and benzoyl chloride, then a Friedel-Crafts acylation with benzene.

  1. 2 CH3COCH3 --Ba(OH)2 (base catalyst)--> (CH3)2C(OH)–CH2–CO–CH3 (A, diacetone alcohol, 4-hydroxy-4-methylpentan-2-one), formed by base-catalysed aldol addition (the enolate of one acetone molecule attacks the carbonyl carbon of another). A --Δ (heat, dehydration)--> (CH3)2C=CH–CO–CH3 (B, mesityl oxide, 4-methylpent-3-en-2-one), by loss of water (E1cb-type dehydration of the β-hydroxy ketone).
  2. Distinguishing formic acid (HCOOH) from acetic acid (CH3COOH): formic acid has an aldehyde-like H directly bonded to the carbonyl carbon (H–COOH), giving it a reducing character; it reduces Tollens' reagent, depositing a bright silver mirror on the test-tube wall, and also decolourises acidified KMnO4 rapidly. Acetic acid (CH3–COOH) has no such H on the carbonyl carbon, does not reduce Tollens' reagent (no mirror forms), and does not decolourise KMnO4 under the same mild conditions.
  3. Benzaldehyde to benzophenone: C6H5CHO --oxidation, e.g., K2Cr2O7/H2SO4--> C6H5COOH (benzoic acid) C6H5COOH --PCl5 (or SOCl2)--> C6H5COCl (benzoyl chloride) C6H5COCl + C6H6 --anhydrous AlCl3 (Friedel-Crafts acylation)--> C6H5–CO–C6H5 (benzophenone) + HCl OR (a) Compound A (C8H16O2) on acidic hydrolysis gives acid B and alcohol C; oxidation of C by chromic acid gives B (so B and C share the same carbon skeleton); dehydration of C gives but-1-ene, so C is a 4-carbon primary alcohol. C = butan-1-ol, CH3CH2CH2CH2OH Oxidation of C (primary alcohol) by chromic acid gives the corresponding acid: B = butanoic acid, CH3CH2CH2COOH A, the ester of B and C: A = butyl butanoate, CH3CH2CH2COO–CH2CH2CH2CH3 (C8H16O2, matching the given formula) Equations: Hydrolysis: CH3CH2CH2COOCH2CH2CH2CH3 (A) + H2O --dil. H2SO4--> CH3CH2CH2COOH (B) + CH3CH2CH2CH2OH (C) Oxidation: CH3CH2CH2CH2OH (C) --CrO3/H+--> CH3CH2CH2COOH (B) Dehydration: CH3CH2CH2CH2OH (C) --conc. H2SO4, Δ--> CH3CH2CH=CH2 (but-1-ene) + H2O …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.