An alkyl bromide produces a single alkene whe | vedclass

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(C) The reaction of an alkyl bromide with sodium ethoxide $(C_2H_5ONa)$ in ethanol $(C_2H_5OH)$ is a dehydrohalogenation reaction (elimination reactio

Vedclass · Accepted Answer

$1-$bromo$-2-$methylbutane

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(C) The reaction of an alkyl bromide with sodium ethoxide $(C_2H_5ONa)$ in ethanol $(C_2H_5OH)$ is a dehydrohalogenation reaction (elimination reaction).For the alkyl bromide to produce a single alkene,it must have only one possible $\beta-$hydrogen elimination path.$1-$bromo$-2-$methylbutane $(CH_3-CH_2-CH(CH_3)-CH_2Br)$ undergoes elimination to form $2-$methylbut$-1-$ene $(CH_3-CH_2-C(CH_3)=CH_2)$.Hydrogenation of $2-$methylbut$-1-$ene $(CH_3-CH_2-C(CH_3)=CH_2 + H_2 \rightarrow CH_3-CH_2-CH(CH_3)-CH_3)$ yields $2-$methylbutane.Therefore,the correct alkyl bromide is $1-$bromo$-2-$methylbutane.

An alkyl bromide produces a single alkene when it reacts with sodium ethoxide and ethanol. This alkene undergoes hydrogenation and produces $2-$methylbutane. What is the identity of the alkyl bromide?

Similar Questions

Assertion $(A)$: Propene on addition with hydrogen bromide in the presence of peroxide gives $1-$bromopropane as the major product.
Reason $(R)$: $1-$bromopropane is the major product because it is formed through the stable carbocation.
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