Two primary alkyl halides with molecular formula C4H9Br are possible. They are n-butyl bromide and isobutyl bromide.
When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Hence, compound a is isobutyl bromide and compound d is 2,5-dimethylhexane.
2CH3CH2CH2CH2Br+2NaWurtz reaction−−−−−−−−−→CH3CH2CH2CH2CH2CH2CH2CH3(n−octane)
2CH3CH(CH3)CH2Br+2Nawurtz reaction−−−−−−−−−→CH3CH(CH3)CH2CH2CH(CH3)CH3(2,5−dimethylhexane)
(a) reacted with alcoholic KOH to give compound (b).
CH3CH(CH3)CH2BralcKOH−−−−−→CH3C(CH3)=CH2(2−methyl−1−propene)
Compound (b) is reacted with HBr to give (c) which is an isomer of (a).
CH3−C(CH3)=CH2HBr−−−−−−−−−−−−−→Markownikoff′s ruleCH3−CBr(CH3)CH3(tert−butylbromide)