Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2/Pd−C, followed by H3PO4 treatment gives Z as the major product.
Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2/Pd−C, followed by H3PO4 treatment gives Z as the major product.
The compound Y is
(A)(see structure)
(B)(see structure)
(C)(see structure)✓ Correct
(D)(see structure)
Answer: (C)
Solution
The reaction sequence:
Benzene + CO/HCl with AlCl3/CuCl gives benzaldehyde (Gattermann-Koch reaction).
Benzaldehyde + Ac2O/NaOAc gives cinnamic acid (Perkin reaction), which is compound X: C6H5CH=CHCOOH.
X + Br2/Na2CO3 followed by heating with moist KOH at 473 K gives phenylacetylene (decarboxylation and elimination), which is compound Y.
Compound Y is phenylacetylene: C6H5C≡CH
Correct answer: (C)
JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Single correct option