The relative reactivity of  1^{\circ}, 2^{\circ}, 3^{\circ}° hydrogen’s towards chlorination is 1 : 3.8 : 5. Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.

Answers (1)

There are 9 primary, 2 secondary and 1 tertiary hydrogen atoms in 2-methylbutane and the relative reactivity of 1^{\circ}, 2^{\circ}, 3^{\circ}hydrogen atoms towards chlorination is 1:3:8:5.

The relative amount of product after chlorination = no. of hydrogen atom X relative reactivity.

Relative

Amount

1^{\circ} halide

9 × 1 = 9

2^{\circ}halide

2 × 3.8 = 7.6

3^{\circ} halide

1 × 5 = 5

Therefore, the total amount of monochloro product will be :

9 + 7.6 + 5 = 21.6             

Now, 1^{\circ} monochloro product% = 9 \times \frac{100}{21.6} =41.7 %

2^{\circ} monochloro product % = 7.6\times \frac{100}{21.6} = 35.2%

3^{\circ} monochloro product % = 5\times \frac{100}{21.6}=23.1%

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