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Isobutane on bromination forms tertiary butyl bromide (A) and isobutyl bromide (B). The percentage of A and B respectively when free radical bromination order is 1 : 82 : 1600 on $1^{0}, 2^{0} a n d 3^{0}$ hydrogen atoms is

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0.6% ,99.4%

99.4% ,0.6%

80% ,20%

50% ,50%

answer is A.

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Detailed Solution

$3^{0} = \frac{1600}{1609} \times 100 = 99.4 %$

$\begin{array}{l} 1^{0} H = 9 \times 1 \\ 3^{0} H = 1 \times 1600 \end{array}$

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