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Consider the following two reactions: Reaction 1: C6H5-CH2-CH2-CH3 is treated with Br2 under UV light (hv). Reaction 2: C6H5-CH=CH-CH3 is treated with HBr in the presence of benzoyl peroxide. What are the major products of Reaction 1 and Reaction 2 respectively?
- C6H5-CH(Br)-CH2-CH3 and C6H5-CH2-CH(Br)-CH3
- C6H5-CH2-CH(Br)-CH3 and C6H5-CH(Br)-CH2-CH3
- C6H5-CH(Br)-CH2-CH3 and C6H5-CH(Br)-CH2-CH3
- C6H5-CH2-CH2-CH2Br and C6H5-CH2-CH2-CH2Br
Correct answer: C6H5-CH(Br)-CH2-CH3 and C6H5-CH2-CH(Br)-CH3
Solution
Free radical bromination at the benzylic position is highly selective, giving C6H5-CH(Br)-CH2-CH3. Anti-Markovnikov HBr addition (peroxide) places Br at the less hindered end: Br goes to C3 (the CH3-bearing carbon of the double bond? — re-examine). In Ph-CH=CH-CH3, Br radical adds to the less substituted carbon (the CH end, i.e., the one bearing Ph is more stable radical site on the other carbon), giving Ph-CH2-CH(Br)-CH3.
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