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Question

(B) The reaction between a tertiary alkyl halide,$(CH_{3})_{3}CCl$,and a strong base,$CH_{3}CH_{2}ONa$,in ethanol primarily undergoes an $E2$ eliminat

Vedclass · Accepted Answer

$CH_{2}=C(CH_{3})_{2}$

Vedclass · Answer

(B) The reaction between a tertiary alkyl halide,$(CH_{3})_{3}CCl$,and a strong base,$CH_{3}CH_{2}ONa$,in ethanol primarily undergoes an $E2$ elimination reaction.Since $(CH_{3})_{3}CCl$ is a tertiary halide,steric hindrance makes substitution difficult.The strong base $CH_{3}CH_{2}O^-$ abstracts a proton from the $\beta$-carbon,leading to the formation of an alkene.The reaction is: $(CH_{3})_{3}CCl + CH_{3}CH_{2}ONa \rightarrow CH_{2}=C(CH_{3})_{2} + CH_{3}CH_{2}OH + NaCl$.The major product is $2$-methylpropene,which corresponds to option $B$.

The major product of the reaction between $CH_{3}CH_{2}ONa$ and $(CH_{3})_{3}CCl$ in ethanol is

Similar Questions

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