An alkane `C_(8)H_(18)` is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.

Explanation

(i)We know that Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Since here Wurtz reaction of a primary of a primary alkyl halide gives an alkane with M.F. `C_8H_18`, therefore , the `1^@` alkyl halide must contain four carbon atoms. Now the two possible primary alkyl halides having four carbon atoms each are I and II.
`underset"I"(CH_3CH_2CH_2CH_2-X) " " underset"II"(CH_3-oversetoverset(CH_3)|CH-CH_2X)`
(ii)Since the alkane `C_8H_18` on monobromination yields a single isomer of a tertiary alkyl halide, therefore, the alkene `C_8H_18` must contain a tertiary hydrogen. This is possible only if the starting primary alkyl halide has a tertiary hydrogen. Out of I and II , only primary alkyl halide (II) has a tertiary hydrogen. In other words, the starting primary alkyl halide is 1-halo-2-methylpropane (II). If this is so, then the alkene with M.F. `C_8H_18` must be 2,5-dimetylhexane (III) and the monobromoderivative must be 2-bromo-2,5-dimethylhexane (IV).