Consider the reaction: CH_3-CH_2-CH(F)-CH_3 x | vedclass

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Question

(C) In the $E2$ elimination reaction of alkyl fluorides with a strong base like $CH_3O^{-}$,the $Hofmann$ product (the less substituted alkene) is the

Vedclass · Accepted Answer

$CH_3-CH_2-CH=CH_2$

Vedclass · Answer

(C) In the $E2$ elimination reaction of alkyl fluorides with a strong base like $CH_3O^{-}$,the $Hofmann$ product (the less substituted alkene) is the major product.This occurs because the fluorine atom is a poor leaving group and highly electronegative,which imparts significant carbanion character to the transition state.The more stable carbanion-like transition state is formed by removing a proton from the less substituted carbon atom.Therefore,$CH_3-CH_2-CH=CH_2$ $(but-1-ene)$ is the major product.

Consider the reaction: $CH_3-CH_2-CH(F)-CH_3 \xrightarrow{CH_3O^{-}} X$. The alkene formed in major amount is:

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