Consider the following molecule(X). [figure] The major product formed when the given molecule (X) is treated with HBr (1 eq) is :
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Answer
(1) [figure]
Solution
X has three C=C bonds; 1 equivalent of HBr adds to only one of them by electrophilic addition (H+ first, then Br-). The C=C giving the most stable carbocation reacts.
• C=C in the six-membered ring: protonation gives a secondary carbocation.
• C=C in the upper-left five-membered ring: protonation gives a secondary carbocation.
• C=C in the lower five-membered ring is trisubstituted (one carbon is the ring-fusion carbon). H+ adds to the CH carbon (Markovnikov's rule), giving a tertiary carbocation on the ring-fusion carbon (stabilised by +I effect and hyperconjugation of three alkyl groups).
3° carbocation > 2° carbocation, HBr adds to this double bond.
Br- attacks the tertiary carbocation Br on the ring-fusion carbon; the other two C=C bonds remain unchanged.
Answer: (1)