Which of the following reaction is correct?

Solution

Chemistry ma’am · 3 steps

Check each product against the real mechanism: only HI on 1-methylcyclohexene follows its rule correctly.

1. Method. For each option, ask what the reagent really does. Then check that the carbon count and the position of the new group in the given product match.

   Option	What really happens	Correct product	Verdict
   (a) amide with Br₂ and NaOH	Hofmann bromamide degradation removes one carbon	C₂H₅NH₂, not C₂H₅CH₂NH₂	wrong
   (b) cyclohexene with Br₂, heat or UV	Free-radical substitution at the allylic carbon, not addition	3-bromocyclohexene	wrong
   (c) 1-methylcyclohexene with HI	Markovnikov addition through a tertiary carbocation	1-iodo-1-methylcyclohexane	correct
   (d) CH₃CH₂CH₂NH₂ with HNO₂	The -NH₂ group is replaced by -OH and N₂ is released. No carbon is lost	CH₃CH₂CH₂OH, with N₂ and H₂O	wrong

2. Why (c) works. H⁺ adds to the ring CH carbon of the double bond. This leaves the positive charge on the carbon that carries CH₃, which makes a stable tertiary carbocation. I⁻ then attacks that carbon, so I and CH₃ end up on the same carbon.
3. So. Only reaction (c) shows the correct product, so the answer is (c).

The idea to keep. In HX addition, H goes to the double-bond carbon with more hydrogens, so X lands where the most stable carbocation forms. Peroxide anti-Markovnikov addition works only with HBr, never with HI.

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