Question

The major product formed in the following reaction is : $CH _{3} CH = CHCH \left( CH _{3}\right)_{2} \stackrel{ HBr }{\longrightarrow}$

• A. $CH _{3} CH _{2} CH _{2} C ( Br )\left( CH _{3}\right)_{2}$
• B. $\operatorname{Br}\left( CH _{2}\right)_{3} CH \left( CH _{3}\right)_{2}$
• C. $CH _{3} CH _{2} CH ( Br ) CH \left( CH _{3}\right)_{2}$
• D. $CH _{3} CH ( Br ) CH _{2} CH \left( CH _{3}\right)_{2}$

Answer 1

The Correct Option is A

The reaction given is the addition of hydrogen bromide (HBr) to an alkene, specifically, $CH_3CH=CHCH(CH_3)_2$. This reaction follows Markovnikov's rule, which states that in the addition of HX to an alkene, the hydrogen (H) attaches to the carbon with the most hydrogen atoms already present, and the halide (X) attaches to the carbon with fewer hydrogen atoms.

1. Identify the double bond in the alkene: The double bond is between the second carbon (C-2) and the third carbon (C-3) in the chain.
2. Apply Markovnikov's Rule: In this case, the H from HBr will attach to the C-2 atom (as it has more hydrogen compared to C-3), and the Br will attach to the C-3 atom.
3. The resulting major product will thus have the H added to C-2, converting the double bond to a single bond, and Br attached to C-3.

The structure changes from $CH_3CH=CHCH(CH_3)_2$ to $CH_3CH_2CH_2C(Br)(CH_3)_2$.

This matches the following option: $CH_3CH_2CH_2C(Br)(CH_3)_2$.

Therefore, the major product formed in this reaction is indeed $CH_3CH_2CH_2C(Br)(CH_3)_2$, as per Markovnikov's rule. The other options would represent products formed through alternative pathways or incorrect application of the rule.