Show steps for the following conversions:
(a) $\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \longrightarrow \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3$
(c) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \longrightarrow \mathrm{CH}_2 \mathrm{ClCHClCH}$ Cl
(b) $\mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3 \longrightarrow \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$
To do syntheses, it is best to work backward while keeping in mind your starting material. As you do this, keep asking what is needed to make what you want. Always try to use the fewest steps.
(a) The precursors of the possible product are the corresponding alcohol, 1-butene, and 2-butene. The alcohol is a poor choice, because it would have to be made from either of the alkenes and an extra step would be needed. 2Butene cannot be made directly from the $1^{\circ}$ halide, but 1-butene can.
$$
\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\text { alc, } \mathrm{KOH}}{\longrightarrow} \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3
$$
(b) To ensure getting 1-butene, the needed precursor, use a bulky base for the dehydrohalogenation of the starting material.
$$
\mathrm{CH}_3 \mathrm{CHBrCH}_2 \mathrm{CH}_3 \stackrel{\mathrm{Me}_3 \mathrm{CO}^{-} \mathrm{K}^{+}}{\longrightarrow} \mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{CH}_3 \underset{\text { peroxides }}{\stackrel{\text { HBr }}{\longrightarrow}}-\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
$$
(c) The precursor for vic dichlorides is the corresponding alkene; in this case, $\mathrm{H}_2 \mathrm{C}-\mathrm{CHCH}_2 \mathrm{Cl}$, which is made by allylic chlorination of propene. Although free-radical chlorination of propane gives a mixture of isomeric propyl chlorides, the mixture can be dehydrohalogenated to the same alkene, making this particular initial chlorination a useful reaction.
$$
\begin{aligned}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \underset{\mathrm{Uv}}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{CH}_3 \mathrm{CHClCH}_3 \underset{\mathrm{KOH}}{\stackrel{\text { alc. }}{\longrightarrow}} \mathrm{CH}_2=\mathrm{CHCH}_3 \underset{\mathrm{uv}}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{CH}_2= & \mathrm{CHCH}_2 \mathrm{Cl} \\
& \downarrow \mathrm{Cl}_2 \\
& \mathrm{ClCH}_2 \mathrm{CHClCH}_2 \mathrm{Cl}
\end{aligned}
$$