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The compound 2-methylpent-3-ene (CH3-CH=CH-CH(CH3)-CH3) is treated with NBS in CCl4 under UV light. Which of the following products CANNOT be formed in this reaction?

1. CH3-CH=CH-CH(CH2Br)-CH3
2. BrCH2-CH=CH-CH(CH3)-CH3
3. CH3-CHBr-CH=CH(CH3)-CH3
4. CH3-CH=CH-CBr(CH3)-CH3

Correct answer: CH3-CH=CH-CH(CH2Br)-CH3

Solution

In CH3-CH=CH-CH(CH3)-CH3 the double bond is between C2 and C3. Allylic positions are C1 (the CH3 attached to C2) and C4 (the CH bearing the methyl at C4). Abstracting H from C1 gives radical at C1 delocalised to C3, producing BrCH2-CH=CH-CH(CH3)-CH3 (option B) or CH3-CH=CH-CBr(CH3)... wait, the allylic radical from C1 spans C1-C2=C3 giving product at C1 or C3 but not C4. Abstracting H from C4 gives radical spanning C4-C3=C2, yielding bromination at C4 (option D) or C2 (option C). The CH2Br group in option A would require abstraction of a hydrogen from the methyl substituent at C4, which is NOT allylic (it is two bonds away from the double bond), so this product cannot be formed by NBS allylic bromination.

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