The major product obtained in the following reaction is
Choose the correct answer from the options given below
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When applying Markovnikov's rule to highly substituted alkenes, the halogen atom always ends up attached to the more highly substituted carbon of the original double bond because it hosts the more stable carbocation.
Step 1: Understanding the Question:
The problem presents an addition reaction of a hydrohalic acid (HI) across the double bond of an unsymmetrical alkene (3-methylhex-3-ene) and requires us to predict the major product using IUPAC nomenclature guidelines.
Step 2: Key Formula or Approach:
The addition of hydrogen halides to unsymmetrical alkenes proceeds via electrophilic addition following
Markovnikov's rule. The electrophile ($\text{H}^+$) adds to the double-bonded carbon to form the more stable carbocation intermediate (preferring tertiary over secondary), followed by nucleophilic attack by the halide ion ($\text{I}^-$).
Step 3: Detailed Explanation:
The starting alkene 3-methylhex-3-ene features a double bond between carbon-3 and carbon-4:
$$\text{CH}_3-\text{CH}_2-\text{C(CH}_3\text{)=CH}-\text{CH}_2-\text{CH}_3$$
Carbon-3 is attached to a methyl group and an ethyl group, making it a highly substituted olefinic carbon.
Carbon-4 is attached to a single hydrogen atom and an ethyl group.
When $\text{H}^+$ adds to carbon-4, a highly stable $3^\circ$ carbocation is formed at carbon-3. Then, the iodide nucleophile ($\text{I}^-$) attacks this tertiary carbocation.
The resulting structure is:
$$\text{CH}_3-\text{CH}_2-\text{C(I)(CH}_3\text{)}-\text{CH}_2-\text{CH}_2-\text{CH}_3$$
Numbering the longest continuous six-carbon chain from the left side gives a methyl group and an iodo group both located at position 3. Therefore, the systematic IUPAC name is
3-iodo-3-methylhexane.
Step 4: Final Answer:
The major product is 3-Iodo-3-methylhexane, which corresponds to option (C).
