2-Methylbutane is monochlorinated by free radicals using relative rates per hydrogen of primary : secondary : tertiary $= 1 : 3.8 : 5$ . Ignoring stereoisomers (each constitutional isomer counted once), the major monochloro product and its approximate percentage are

Answer: A

2-Methylbutane has 6 H on the two equivalent

CH X_{3}

of the isopropyl end (-> 1-chloro-2-methylbutane), 3 H on the terminal

CH X_{3}

(-> 1-chloro-3-methylbutane), 2 secondary H (-> 2-chloro-3-methylbutane) and 1 tertiary H. Weights:

6

3

2 \times 3.8 = 7.6

1 \times 5 = 5

; total

21.6

. The secondary chloride is largest,

7.6/21.6 = 35.2 %

(the tertiary chloride is only

23.1 %