[sparkle123]

A hydrocarbon X of molecular formula C6H14 was allowed to react with limited chlorine gas
in the presence of light. The resulting mixture contained two products with different boiling
points, as well as excess starting material. What is the systematic name of X?
(a) 2,2-dimethylbutane
(b) 2,3-dimethylbutane
(c) 2-methylpentane
(d) 3-methylpentane
(e) hexane

The answer is b), but I think it should be e), hexane, because 2-chlorohexane and 3-chlorohexane can be produced.
Could someone please show me why b) is correct? I can only get the one product 2-chloro-2,3-dimethylbutane.

[Arctic-Nation]

If we assume that radical formation occurs exclusively on tertiary carbon atoms, compounds a) and e) are out. Both c) and d) have only one tertiary carbon atom, so can give only one reaction product. However, compound b) has two tertiary carbon atoms, and so the products 2-chloro-2,3-dimethylbutane and 2,3-dichloro-2,3-dimethylbutane can be formed.

While compounds c), d), and e) are technically correct answers in that they will give two or even three (for c)) isomers upon radical chlorination, the boiling points of these isomers will most probably not be very different (e.g., 122°C for 2-chlorohexane and 123°C for 3-chlorohexane).

[Honclbrif]

This seems to be one of those problems which relies on the reader to determine the intent of the writer more than it really should. Working backward from the provided answer, when they say it yielded 2 products, they seem mean "it yielded 2 and only 2 products after monochlorination". To solve this one, you have to look at the equivalence of the various sites on the organic substrate and what happens when you modify each with a single Cl.

A) 2,2-dimethylbutane can be monochlorinated to yield 3 products
B) 2,3-dimethylbutane can be monochlorinated to yield 2 products
C) 2-methylpentane can be monochlorinated to yield 5 products
D) 3-methylpentane can be monochlorinated to yield 4 products
E) Hexane can be monochlorinated to yield 3 products