10.21 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). 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. Give the structural formula of (a) and write the equations for all the reactions. 


There are two primary alkyl halides having the formula, C4H9Br. They are n − bulyl bromide
and isobutyl bromide. 
CH3CH2CH2CH2Br    CH3CHCH2Br
                                                             |
                                                            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.
2CH3CH2CH2CH2Br(Wurtz reaction)2Na/dry ether CH3CH2CH2CH2CH2CH2CH2CH3+2NaBr
n-Butyl bromide                                n-Octane
CH3CHCH2Br(Wurtz reaction )2Na/dry ether CH3CHCH2CH2CHCH3  +2NaBr
         |                                         |                  |        
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.
CH3CHCH2Br  (Dehydrohalogenation)KOH(alc)CH3C=CH2 +H Br
          |                                                               |
lsobutyl chloride                                                2-Methylpropene
      (a)                                                                   (b)
                                                                   
              
Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a).
Hence, compound (c) is 2−bromo−2−methylpropane.
                                                                      Br
                                                                       |
CH3CH=CH2   (Markovnikov addition)HBr  CH3 CHCH3
             |
2-Methlpropene                                       2-Bromo-2-methylpropane
          (b)                                                                           (c) 
                                                                       (an isomer of (a))