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(B) The reaction is an $E2$ elimination reaction of a secondary alkyl chloride with a strong base $(NaOEt)$.According to Saytzeff's rule,the major pro

Vedclass · Accepted Answer

$CH_3-C(CO_2CH_2CH_3)=CH-CH_3$

Vedclass · Answer

(B) The reaction is an $E2$ elimination reaction of a secondary alkyl chloride with a strong base $(NaOEt)$.According to Saytzeff's rule,the major product is the more substituted alkene,as it is more stable.In the given substrate,$CH_3-CH(Cl)-CH_2-CH_3$ (with a $CO_2Et$ group),the elimination of $HCl$ can occur from either the $CH_3$ group or the $CH_2$ group.Elimination from the $CH_2$ group leads to a trisubstituted alkene $(CH_3-C(CO_2Et)=CH-CH_3)$,which is more stable than the disubstituted alkene $(CH_3-CH_2-C(CO_2Et)=CH_2)$ formed by elimination from the $CH_3$ group.Therefore,the major product is $CH_3-C(CO_2CH_2CH_3)=CH-CH_3$.

The major product of the following reaction is:

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