JEE Advanced 2018 · Paper 1 · Chemistry Q15

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ChemistrySingle correct+3 / -1Question 33 of 54

Passage

Treatment of benzene with CO/HCl in the presence of anhydrous AlCl $_3$ /CuCl followed by reaction with Ac $_2$ O/NaOAc gives compound X as the major product. Compound X upon reaction with Br $_2$ /Na $_2$ CO $_3$ , followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H $_2$ /Pd-C, followed by H $_3$ PO $_4$ treatment gives Z as the major product.

The compound Y is

3-Hydroxy-2-bromo-3-phenylpropanoic acid, PhCH(OH)-CH(Br)-CO$_2$H

PhCHBr-CHBr-COBr (2,3-dibromo-3-phenylpropanoyl bromide)

Answer and worked solution

Official answer: C

Making X.

Gattermann-Koch ( $\ce{CO/HCl}$ , $\ce{AlCl3/CuCl}$ ) formylates benzene to $\ce{PhCHO}$ .
The Perkin reaction ( $\ce{Ac2O/NaOAc}$ ) gives cinnamic acid, $\ce{PhCH=CHCOOH}$ . This is X.

Making Y.

$\ce{Br2}$ adds across the C=C. In $\ce{Na2CO3}$ the dibromo acid loses $\ce{CO2}$ and $\ce{Br-}$ , giving β-bromostyrene, $\ce{PhCH=CHBr}$ .
Moist KOH at 473 K eliminates HBr, giving phenylacetylene, $\ce{PhC#CH}$ . This is Y.

(C)

Topics:Aldol and related condensations Named reactions Elimination reactions

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JEE Advanced 2018 · Paper 1, Chemistry question 15