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JEE Advanced 2019 · Paper 2, Chemistry question 7

Choose the correct option(s) for the following reaction sequence

Consider Q, R and S as major products

4-(4-Methoxyphenyl)but-3-ynal (MeO-C₆H₄-C≡C-CH₂-CHO) with (i) Hg²⁺, dil. H₂SO₄, (ii) AgNO₃, NH₄OH, (iii) Zn-Hg, conc. HCl gives Q; Q with (i) SOCl₂, pyridine, (ii) AlCl₃ gives R; R with Zn-Hg, conc. HCl gives S.
Choose one or more options
Answer and worked solution

Official answer: C, D

Q. The starting material is Ar−C≡C−CHX2−CHO\ce{Ar-C#C-CH2-CHO} with Ar\ce{Ar} = 4-methoxyphenyl.

  • HgX2+\ce{Hg^{2+}}-catalysed hydration puts the C=O next to the aryl group: Ar−CO−CHX2CHX2−CHO\ce{Ar-CO-CH2CH2-CHO}.
  • Tollens' reagent oxidises the aldehyde to Ar−CO−CHX2CHX2−COOH\ce{Ar-CO-CH2CH2-COOH}.
  • Clemmensen reduction gives Ar−(CHX2)X3−COOH\ce{Ar-(CH2)3-COOH}: Q is 4-(4-methoxyphenyl)butanoic acid.

R. The acid chloride undergoes intramolecular Friedel-Crafts acylation ortho to the chain, giving 7-methoxy-1-tetralone.

S. Clemmensen reduction gives 6-methoxytetralin.

The options.

  • (A) has the wrong Q.
  • (B) has the wrong R.
  • (C) and (D) are both correct.

(C), (D)

Topics:Oxidation and reductionAlkynesAromatic hydrocarbons and electrophilic substitution
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