CH_3-CH_2-C(Br)(CH_2CH_3)-CH_2(CH_2CH_2CH_3) | vedclass

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(C) The reaction of $3-bromo-3-ethylhexane$ with $alc. KOH$ proceeds via an $E2$ elimination mechanism.The substrate has three types of $\beta$-hydrog

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(C) The reaction of $3-bromo-3-ethylhexane$ with $alc. KOH$ proceeds via an $E2$ elimination mechanism.The substrate has three types of $\beta$-hydrogens available for abstraction:$1$. From the $CH_3$ group of the ethyl chain.$2$. From the $CH_2$ group of the propyl chain.$3$. From the $CH_2$ group of the ethyl chain.According to $Zaitsev's$ rule,the major product is the most substituted alkene,which is also the most stable due to hyperconjugation (more $\alpha$-hydrogens).By eliminating a proton from the $CH_2$ group of the propyl chain,we form the most substituted alkene,which is $3-ethylhex-3-ene$. This product has the highest number of $\alpha$-hydrogens,making it the major product as shown in the solution image.

$CH_3-CH_2-C(Br)(CH_2CH_3)-CH_2(CH_2CH_2CH_3) \xrightarrow{alc. KOH} (A)$ (major)
$n-Pr = n-propyl$
Product $(A)$ is

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$CH_3-CH_2-C(Br)(CH_2CH_3)-CH_2(CH_2CH_2CH_3) \xrightarrow{alc. KOH} (A)$ (major)$n-Pr = n-propyl$Product $(A)$ is