Molecular Weight Analysis for HBr Addition

Steps:

• Calculate mass increase: Product M=164, reactant M=84, so addition of 80 (H=1 + Br=79).
• Identify reaction type: HBr adds to unsaturated compounds like alkenes (C_nH_{2n}) without mass loss.
• Confirm reactant formula: M=84 matches C6H12 (72+12), an alkene.
• Verify product: C6H12 + HBr → C6H13Br (72+13+79=164).

Why D is correct:

• D is cyclohexene (C6H12), which undergoes electrophilic addition with HBr per Markovnikov's rule, yielding bromocyclohexane (M=164).

Why the others are wrong:

• A is hexane (C6H14, M=86), saturated alkane that does not react with HBr via addition.
• B is 1,3-cyclohexadiene (C6H8, M=80), diene that adds two HBr molecules, exceeding M=164.
• C is benzene (C6H6, M=78), aromatic that resists addition, preferring substitution.

Final answer: D