There are two primary alkyl halides having the formula, C4H9Br. They are n − bulyl bromide
and isobutyl bromide.
CH3—CH2—CH2—CH2—Br CH3—CH—CH2—Br
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CH3
n-Butyl bromide lsobutyl bromide
Therefore, compound (a) is either n−butyl bromide or isobutyl bromide.
Now, compound (a) reacts with Na metal to give compound (b) of molecular formula,
C8H18, which is different from the compound formed when n−butyl bromide reacts with Na
metal. Hence, compound (a) must be isobutyl bromide.
2CH3CH2CH2CH2Br2Na/dry ether→(Wurtz reaction) CH3CH2CH2CH2CH2CH2CH2CH3+2NaBr
n-Butyl bromide n-Octane
CH3CHCH2Br2Na/dry ether→(Wurtz reaction ) CH3CHCH2CH2CHCH3 +2NaBr
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lsobutyl bromide CH3 CH3
(a) 2-5-Dimethylhexane
Thus, compound (d) is 2, 5−dimethylhexane.
It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence,
compound (b) is 2−methylpropene.
CH3—CH—CH2—Br →(Dehydrohalogenation)KOH(alc)∆
Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a).
Hence, compound (c) is 2−bromo−2−methylpropane.