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Question

(A) The reaction is a dehydrohalogenation of an alkyl halide using alcoholic $KOH$ (an elimination reaction,specifically $E2$ mechanism).The substrate

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$Ph-CH_2-CH=C(CH_3)_2$

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(A) The reaction is a dehydrohalogenation of an alkyl halide using alcoholic $KOH$ (an elimination reaction,specifically $E2$ mechanism).The substrate is $Ph-CH_2-CH(Cl)-CH(CH_3)_2$.Elimination occurs by removing the $Cl$ atom from the $\alpha$-carbon and a hydrogen atom from an adjacent $\beta$-carbon.There are two possible $\beta$-carbons:$1$. The $\beta$-carbon towards the phenyl group $(Ph-CH_2-)$: Removing $H$ from here gives $Ph-CH=CH-CH(CH_3)_2$.$2$. The $\beta$-carbon towards the isopropyl group (the $CH$ of the isopropyl group): Removing $H$ from here gives $Ph-CH_2-CH=C(CH_3)_2$.According to Zaitsev's rule,the more substituted alkene is the major product.$Ph-CH_2-CH=C(CH_3)_2$ is a trisubstituted alkene,while $Ph-CH=CH-CH(CH_3)_2$ is a disubstituted alkene.Therefore,$Ph-CH_2-CH=C(CH_3)_2$ is the major product.

The main product of the following reaction will be:

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