JEE Main 2017 · 8 Apr · Shift 1 · Chemistry Q56

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ChemistrySingle correct+4 / -1Question 56 of 90

The major product of the following reaction is :
$\ce{CH3CH(Br)CH2CH(Br)CH2CH3 ->[KOH, CH3OH][heat]}$

\ce{CH2=CHCH2CH=CHCH3}

\ce{CH2=CHCH=CHCH2CH3}

\ce{CH3CH=C=CHCH2CH3}

\ce{CH3CH=CH-CH=CHCH3}

Answer and worked solution

Official answer: D

Alcoholic KOH removes two molecules of HBr. The double bonds form at C2=C3 and C4=C5, giving the conjugated diene hexa-2,4-diene, $\ce{CH3CH=CH-CH=CHCH3}$ .

Topics:Elimination reactions

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JEE Main 2017 · 8 Apr · Shift 1, Chemistry question 56