The major product obtained on treatment of CH | vedclass

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Question

(B) The reaction of $CH_3CH_2CH(F)CH_3$ with a strong base like $CH_3O^-$ in $CH_3OH$ proceeds via an $E2$ elimination mechanism.Two possible alkenes

Vedclass · Accepted Answer

$CH_3CH=CHCH_3$

Vedclass · Answer

(B) The reaction of $CH_3CH_2CH(F)CH_3$ with a strong base like $CH_3O^-$ in $CH_3OH$ proceeds via an $E2$ elimination mechanism.Two possible alkenes can be formed: $CH_3CH=CHCH_3$ (but-$2$-ene) and $CH_3CH_2CH=CH_2$ (but-$1$-ene).According to Saytzeff's rule,the more substituted alkene is the major product because it is more stable.$CH_3CH=CHCH_3$ has two alkyl substituents on the double-bonded carbons,while $CH_3CH_2CH=CH_2$ has only one.Therefore,$CH_3CH=CHCH_3$ is the major product.

The major product obtained on treatment of $CH_3CH_2CH(F)CH_3$ with $CH_3O^-/CH_3OH$ is

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