Answer:
Two possible primary alkyl,
halides having four carbonatoms each are \[I\] and \[II\].
\[C{{H}_{3}}-C{{H}_{2}}-\underset{(I)}{\mathop{C{{H}_{2}}}}\,-C{{H}_{2}}-X\]
\[C{{H}_{3}}-\underset{\underset{\begin{smallmatrix}
\text{C}{{\text{H}}_{\text{3}}}
\\
(II)
\end{smallmatrix}}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{2}}-X\]
Since, the alkane \[{{C}_{8}}{{H}_{18}}\]
on monobromination yields a single isomer of tertiary alkyl halide, therefore
the alkane, \[{{C}_{8}}{{H}_{18}}\]must contain a tertiary hydrogen. This is
possible only if the starting primary alkyl halide has a tertiary
hydrogen(structure\[II\]).
\[C{{H}_{3}}-\underset{\underset{\underset{\begin{smallmatrix}
1-Halo-2-methylpropane \\
(Primary\,alkayl\,halide)
\end{smallmatrix}}{\mathop{C{{H}_{3}}}}\,}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{2}}-X\xrightarrow[(Wutrz\,reaction)]{Na/Dry\,ether}\]
\[\underset{2,5Dimethylhexane\,({{C}_{8}}{{H}_{18}})}{\mathop{C{{H}_{3}}-\underset{\underset{\text{C}{{\text{H}}_{\text{3}}}}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{2}}-\underset{\underset{\text{C}{{\text{H}}_{\text{3}}}}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{3}}\xrightarrow[hv]{B{{r}_{2}}}}}\,\]
\[\underset{2-Bromo-2,5-\dimethyl\,hexane\,(tertiary\,bromide)}{\mathop{C{{H}_{3}}-\underset{\underset{C{{H}_{3}}}{\mathop{|}}\,}{\overset{\overset{\text{Br}}{\mathop{|}}\,}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{2}}-\underset{\underset{\text{C}{{\text{H}}_{\text{3}}}}{\mathop{|}}\,}{\mathop{CH}}\,-C{{H}_{3}}}}\,\]
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