An alkyl bromide reacts with sodium ethoxide | vedclass

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(B) The hydrogenation of the alkene gives $2-$methylbutane,which has the structure $CH_3-CH(CH_3)-CH_2-CH_3$. Since the alkyl bromide gives only one a

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$1-$Bromo$-2-$methylbutane

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(B) The hydrogenation of the alkene gives $2-$methylbutane,which has the structure $CH_3-CH(CH_3)-CH_2-CH_3$. Since the alkyl bromide gives only one alkene upon reaction with sodium ethoxide (dehydrohalogenation),we look for a structure that eliminates to form a single product. $1-$Bromo$-2-$methylbutane $(CH_3-CH_2-CH(CH_3)-CH_2Br)$ would undergo elimination to form $2-$methylbut$-1-$ene. $2-$Bromo$-2-$methylbutane would form a mixture of alkenes. $1-$Bromo$-2,2-$dimethylpropane cannot undergo $E2$ elimination easily as it has no $\beta-$hydrogen. Thus,the correct alkyl halide is $1-$bromo$-2-$methylbutane.

An alkyl bromide reacts with sodium ethoxide and ethanol to give only one alkene. Hydrogenation of this alkene yields $2-$methylbutane. What is the alkyl halide?

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