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Q.

An alkyl bromide reacts with sodium to form 4, 5-diethyloctane. The alkyl bromide is

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a

${CH}_{3} {({CH}_{2})}_{2} {CHBrCH}_{2} {CH}_{3}$

b

${CH}_{3} {({CH}_{2})}_{4} {CH}_{2} Br$

c

${CH}_{3} {({CH}_{2})}_{3} {CHBrCH}_{3}$

d

${CH}_{3} {({CH}_{2})}_{2} {CH}_{2} Br$

answer is D.

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Detailed Solution

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$C H_{3} {(C H_{2})}_{2} C H B r C H_{2} C H_{3}$

$R - X \overset{Na / ether}{\to} R - R$

so given alkane divided by half and gets corresponding alkyl halide as following:

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