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A compound P with molecular formula C9H10 gives benzaldehyde and an aliphatic aldehyde on reductive ozonolysis. The aliphatic aldehyde gives a positive iodoform test. When P is reacted with HBr in CCl4 (Markovnikov addition via the most stable carbocation), how many major organic products are formed?

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

P is 1-phenylprop-1-ene (PhCH=CHCH3). Reductive ozonolysis gives PhCHO (benzaldehyde) and CH3CHO (acetaldehyde; iodoform-positive). Reaction with HBr: most stable carbocation is the secondary benzylic C (PhCH+CH2CH3 type is less stable vs PhCH(+)CH2CH3... actually Markovnikov places Br on C1 (benzylic) because carbocation forms there). The product is PhCHBr-CH2-CH3? No: Ph-CH=CH-CH3 + HBr -> Ph-CHBr-CH2-CH3 (Markovnikov: H adds to C2 giving benzylic carbocation at C1, then Br- attacks C1). C1 becomes a chiral center. Two stereoisomers (R and S) = 2 major products.

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