In the reaction C_2H_5CH = CH_2 + H - X right | vedclass

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(C) The reaction of $1-$butene $(C_2H_5CH = CH_2)$ with hydrogen halide $(HX)$ follows Markovnikov's rule.According to Markovnikov's rule,the negative

Vedclass · Accepted Answer

$C_2H_5 - CHX - CH_3$

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(C) The reaction of $1-$butene $(C_2H_5CH = CH_2)$ with hydrogen halide $(HX)$ follows Markovnikov's rule.According to Markovnikov's rule,the negative part of the addendum $(X^-)$ attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms.In $C_2H_5 - CH = CH_2$,the $CH$ group has one hydrogen and the $CH_2$ group has two hydrogens.Therefore,the $X$ atom attaches to the $CH$ group,forming $C_2H_5 - CHX - CH_3$ as the major product.This is the more stable product because it involves the formation of a more stable secondary carbocation intermediate.

In the reaction $C_2H_5CH = CH_2 + H - X \rightarrow$ Product,what is the major product?

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