Numerical problems

Problem 1 :

Explain why isobutane on bromination mainly gives tertiary butyl bromide ? or Discuss the mechanism of the following reaction.(DDUGU)

Solution : When a bromine molecule attacks on isobutane two types of alkyl free radicals are formed:

Tertiary butyl free radical;

Since tertiary butyl free radical is more stable than isobutyl free radical (due to more +I effect caused by greater number of methyl groups). It causes tertiary butyl free radical forms bromide more easily as compared sobutyl free radical .The yield of the product in this case depends on the nature of hydrogen abstracted. Practically, it was found that tertiary hydrogen atom is easily removed than secondary hydrogens and sec. hydrogen can replaced more easily than primary hydrogens. The expected ratio of isobutyl bromide and Tertiary butyl bromide in the above reaction can be calculated as follows:

Similarly, In the reaction;

or Sec butyl bromide = 54.6 x n.Butyl bromide

Problem 2:

The relative rate of abstraction of per H atoms of tertiary, secondary and primary alkane during horabon is 5.0:3.8: 1.0. Determine the percentage composition of the products formed in the chlorination -Butane. Saison : n-Butane has six primary and four secondary hydrogen atoms. On chlorination, it forms n-butyl onde formed by the abstraction of primary hydrogen atom) and secondary butyl chloride (formed by the traction of secondary hydrogen atom).

Ratio of the two products formed can be determined by the following general formula :

Hence, weight reactivity = 6.0 + 15.2 = 21.2

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