Explanation

(b)
Since, the alkyl halide RX gives 4, 5-diethyloctane, when reacts with Na, it must be $CH_{3}\ \left( CH_{2} \right)_{2}\ CH(Br)CH_{2}CH_{3}$ .
$2CH_{3}CH_{2}CH_{2}CH - Br\underset{Dry\ ether}{\overset{\ \ \ \ \ \ \ \ Na\ \ \ \ \ \ \ \ \ \ }{\rightarrow}}$
|
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ CH_{2}CH_{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ CH_{2}CH_{3}\ CH_{2}CH_{3}$
| |
$CH_{3}\left( CH_{2} \right)_{2}CH\ \ - \ \ CH - \left( CH_{2} \right)_{2} - CH_{3}$
The reaction is known as Wurtz reaction.