Question

An alkane $C_7{H}_{16}$ is produced by the reaction of lithium di(3-pentyl)cuprate with ethyl bromide. The name of the product is:

• A. 3-methylhexane
• B. 2-ethylpentane
• C. 3-ethylpentane
• D. n-heptane.

Answer 1

Answer: (C)

3-ethylpentane

Lithium di(3-pentyl)cuprate is a Gilman reagent. It couples with alkyl halides to give alkanes.

$\underset{\text{Lithium di(3-pentyl)cuprate}}{[{\left(C{H}_{3}C{H}_2\right)}_{2}CH{]}_{2}CuLi}+\underset{\text{ethyl bromide}}{BrC{H}_{2}C{H}_3}⟶\underset{3-ethylpentane}{C_2{H}_5-\stackrel{\underset{|}{C_2{H}_5}}{CH}-C_2{H}_5}+{\left(C{H}_{3}C{H}_2\right)}_{2}CHCu+LiBr$

Hence the product is 3-ethylpentane.