IIT-JEE 2008 · Paper 2 · Chemistry Q60

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ChemistrySingle correct+4 / -1Question 60 of 66

Passage

Paragraph for Question Nos. 58 to 60

A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K. Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M.

M = $\ce{CH3C(Ph)=CH-CO-Ph}$ (dypnone; the CH₃ and the COPh group drawn on the same side of the C=C, Ph and H opposite).

M: 1,3-diphenylbut-2-en-1-one (dypnone), H₃C(Ph)C=CH-C(=O)Ph.

The structures of compounds J, K and L, respectively, are

\ce{PhCOCH3}

\ce{PhCH2COCH3}

and

\ce{PhCH2COO^-K+}

PhCHO,

\ce{PhCH2CHO}

and

\ce{PhCOO^-K+}

\ce{PhCOCH3}

\ce{PhCH2CHO}

and

\ce{CH3COO^-K+}

PhCHO,

\ce{PhCOCH3}

and

\ce{PhCOO^-K+}

Answer and worked solution

Official answer: D

Ozonolysis of I gives J = PhCHO and K = $\ce{PhCOCH3}$ .

With KOH. Benzaldehyde (no α-H) undergoes the Cannizzaro reaction to benzyl alcohol and L = $\ce{PhCOO-K+}$ ; acetophenone undergoes aldol self-condensation to dypnone (M).

(D)

Topics:Named reactions Aldol and related condensations

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IIT-JEE 2008 · Paper 2, Chemistry question 60