JEE Advanced 2013 · Paper 2 · Chemistry Q28

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ChemistryMultiple correct+3 / -1Question 28 of 60

After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are)

Reaction I: acetone (1.0 mol) with Br₂ (1.0 mol) in aqueous NaOH. Reaction II: acetone (1.0 mol) with Br₂ (1.0 mol) in CH₃COOH.

Candidate compounds: P bromoacetone CH₃COCH₂Br; Q 1,1,1-tribromoacetone CH₃COCBr₃; R hexabromoacetone Br₃CCOCBr₃; S 1,3-dibromoacetone BrCH₂COCH₂Br; T sodium acetate CH₃COONa; U bromoform CHBr₃.

Reaction I: P and Reaction II: P

Reaction I: U, acetone and Reaction II: Q, acetone

Reaction I: T, U, acetone and Reaction II: P

Reaction I: R, acetone and Reaction II: S, acetone

Answer and worked solution

Official answer: C

Reaction I (base). Each bromination makes the next faster, so a molecule that starts reacting goes all the way to CH₃COCBr₃, which is cleaved by OH⁻ (haloform). With 1 mol Br₂ only ⅓ mol acetone reacts:
$\ce{CH3COCH3 + 3Br2 + 4NaOH -> CH3COONa + CHBr3 + 3NaBr + 3H2O}.$
The mixture holds T, U and unreacted acetone.

Reaction II (acid). Bromination via the enol slows after the first Br, so monobromination gives bromoacetone (P).

(C)

Topics:Named reactions

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JEE Advanced 2013 · Paper 2, Chemistry question 28