2-bromopentane is heated with potassium ethox | vedclass

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Question

(A) When $2-$bromopentane is heated with potassium ethoxide in ethanol,it undergoes a dehydrohalogenation reaction (an $E2$ elimination reaction).Pota

Vedclass · Accepted Answer

trans-pent$-2-$ene

Vedclass · Answer

(A) When $2-$bromopentane is heated with potassium ethoxide in ethanol,it undergoes a dehydrohalogenation reaction (an $E2$ elimination reaction).Potassium ethoxide $(C_2H_5OK)$ acts as a strong base,which abstracts a proton from the $\beta$-carbon.According to Zaitsev's rule,the more substituted alkene is the major product.$2-$bromopentane has two types of $\beta$-hydrogens,leading to the formation of pent$-1-$ene and pent$-2-$ene.$Pent-2-ene$ is more substituted and thus more stable than $pent-1-ene$.Between the geometric isomers of $pent-2-ene$,the $trans$ isomer is more stable than the $cis$ isomer due to reduced steric hindrance.Therefore,$trans-pent-2-ene$ is the major product.Thus,option $(A)$ is the correct answer.

$2-$bromopentane is heated with potassium ethoxide in ethanol. The major product is

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