Question

In the following mono-bromination reaction, the number of possible chiral product(s) is (are)...

Answer 1

6

Answer 2

We start with a chiral molecule at a tetrahedral carbon (C*) with four different groups: H, Br, CH₃, and CH₂CH₂CH₃. In free‐radical bromination at 300°C, Br• can abstract any hydrogen. Let the sites be:

1. At C (the chiral center):*

   • Replacing H gives C*(Br, Br, CH₃, CH₂CH₂CH₃) which is not chiral (two Br are identical).

2. On the CH₃ substituent:

   • All 3 H’s are equivalent giving one product.
   • The original chiral center remains so → product is chiral (1 product).

3. On the propyl group (CH₂CH₂CH₃):

   (a) At the CH₂ directly attached to C (C₁'):*

   • The two H’s are pro‐chiral; substitution creates a new stereocenter. Two configurations (diastereomers) are possible. → 2 chiral products.

   (b) At the middle CH₂ (C₂'):

   • Similarly, substitution gives a new chiral center yielding 2 diastereomers. → 2 chiral products.

   (c) At the terminal CH₃:

   • The three H’s are equivalent giving a CH₂Br group; no new stereocenter is formed, so only the original chiral center remains → 1 chiral product.

Adding these (excluding the achiral product from C*): 1 (from CH₃) + 2 (from C₁') + 2 (from C₂') + 1 (from terminal CH₃) = 6 chiral products.

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Minimal Explanation:

• Bromination occurs at 4 different sites (ignoring substitution at C* which gives an achiral gem-dibromide).
• Substitution at the CH₃ and terminal CH₃ gives one product each.
• Substitution at each CH₂ of the propyl leads to a new stereocenter giving 2 products each.

Total = 1 + 2 + 2 + 1 = 6 chiral products.