2-Methylbutane, $CH X_{3} - CH (CH X_{3}) - CH X_{2} - CH X_{3}$ , undergoes monochlorination with $Cl X_{2}$ under light. Assume the relative reactivity per hydrogen is primary : secondary : tertiary = 1 : 4 : 5. Counting all stereoisomers together as one constitutional product, calculate the percentage of 2-chloro-2-methylbutane in the monochloro mixture. Give the answer to the nearest integer.

Numerical answer type, tolerance 0.5.

Answer: 23

2-Methylbutane has 9 primary H (three

CH X_{3}

groups), 2 secondary H and 1 tertiary H. Relative amounts: primary

9 \times 1 = 9

, secondary

2 \times 4 = 8

, tertiary

1 \times 5 = 5

, total 22. The tertiary chloride is

5/22 \times 100 = 22.7%

, i.e. about 23.