2-Methylbutane is monochlorinated with $Cl X_{2}$ and light at room temperature. Take the relative reactivity per hydrogen as primary : secondary : tertiary $= 1 : 3.8 : 5$ . What is the percentage of 2-chloro-3-methylbutane, $CH X_{3} CHClCH (CH X_{3}) X_{2}$ , in the monochloro product mixture (count constitutional isomers only; give the answer to one decimal place)?

Numerical answer type, answer in %, tolerance 0.2.

Answer: 35.2 %

2-Methylbutane has 9 primary H, 2 secondary H (the

CH X_{2}

) and 1 tertiary H. Weights:

9 \times 1 = 9

2 \times 3.8 = 7.6

1 \times 5 = 5

; total

21.6

. Substitution at the secondary

CH X_{2}

gives 2-chloro-3-methylbutane, so its share is

\frac{7.6}{21.6} \times 100 = 35.2%