IIT-JEE 2008 · Paper 2 · Chemistry Q59

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ChemistrySingle correct+4 / -1Question 59 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 structure of compound I is

\ce{PhCH=C(CH3)Ph}

(the two Ph groups trans)

(E)-1,2-Diphenylpropene: Ph and H on one alkene carbon, CH₃ and Ph on the other, with the two phenyls on opposite sides.

\ce{CH3CH=CPh2}

\ce{PhCH=C(CH3)CH2Ph}

PhCH=C(CH₃)CH₂Ph: Ph and H on one carbon, CH₃ and CH₂Ph on the other.

\ce{CH3C(Ph)=CHCH3}

Answer and worked solution

Official answer: A

I is 1,2-diphenylpropene, $\ce{PhCH=C(CH3)Ph}$ (dehydration towards conjugation with both rings). The more stable isomer, with the two phenyl groups trans, is the major product.

(A)

Topics:Alcohols Geometrical isomerism

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IIT-JEE 2008 · Paper 2, Chemistry question 59