Consider the following molecule(X).
The major product formed when the given molecule (X) is treated with HBr(1 eq) is :

Solution

Chemistry ma’am · 4 steps

Three double bonds, one HBr: the proton goes where it leaves the most stable, tertiary carbocation.

1. Find the alkenes. Molecule X has three C=C bonds. With only 1 equivalent of HBr, just one of them reacts. The fastest one is the bond whose protonation gives the most stable carbocation.

   Double bond in X	Substitution	Best carbocation after adding H⁺
   Top five-membered ring	disubstituted (two CH carbons)	secondary
   Six-membered ring	disubstituted (two CH carbons)	secondary
   Bottom five-membered ring, ending at the ring-fusion carbon	trisubstituted	tertiary, on the ring-fusion carbon

2. Markovnikov addition. H⁺ adds to the CH end of the trisubstituted alkene. The positive charge lands on the ring-fusion carbon, which carries three carbon groups, so it is a tertiary carbocation and the most stable one available.
3. Bromide attacks. Br⁻ bonds to this tertiary carbon. The other two double bonds stay as they were, one in the top five-membered ring and one in the six-membered ring.
4. So. The major product has Br on the tertiary ring-fusion carbon, and the bottom five-membered ring is now fully saturated. That is option (a).

The idea to keep. When a molecule has several alkenes and HBr is limited, compare the carbocations each would give. The most substituted alkene gives the most stable cation, and Br ends up on that carbon.

Answer from the NTA final answer key. Solution by PrepWiser.