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 Z is
(A)(see structure)✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)
Answer: (A)
Solution
Continuing from compound X (cinnamic acid: C6H5CH=CHCOOH):
X + H2/Pd−C reduces the double bond to give 3-phenylpropanoic acid: C6H5CH2CH2COOH.
Treatment with H3PO4 (dehydrating agent) causes intramolecular Friedel-Crafts acylation to form a cyclic ketone.
The product is 1-indanone (α-tetralone's 5-membered analog), which is compound Z.
Compound Z is 1-indanone.
Correct answer: (A)
JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Single correct option