Question

An alkene X (\( C_4H_8 \)) does not exhibit cis-trans isomerism. Reaction of X with \( Br_2 \) in the presence of UV light gave Y. What is Y?

• A.
• B. }
• C.
• D.

Hint:

For reactions involving bromine and alkenes, remember that in the presence of UV light, radical substitution occurs at the allylic position rather than simple electrophilic addition across the double bond.

Answer 1

The Correct Option is C

Step 1: Identifying the given alkene The molecular formula of the alkene is \( C_4H_8 \). Since it does not exhibit cis-trans isomerism, it must be a symmetrical alkene where free rotation is possible. The most likely candidate is 2-methylpropene (isobutene): \[ CH_2 = C(CH_3)CH_3 \] Step 2: Reaction with Bromine in UV light
- When an alkene reacts with \( Br_2 \) in the presence of UV light, a radical substitution reaction occurs instead of the usual electrophilic addition.
- The allylic hydrogen (hydrogen on the carbon next to the double bond) is replaced by bromine due to radical stability.
- In isobutene, the most stable radical forms at the allylic position, leading to the formation of 3-Bromo-2-methylpropene as the major product.
Step 3: Examining the given options
- Option 1: Incorrect, as it shows a different bromo-derivative that is unlikely in radical bromination.
- Option 2: Incorrect, as it shows an extended conjugated system that does not form under these conditions.
- Option 3: Correct, as it represents 3-Bromo-2-methylpropene, the expected product.
- Option 4: Incorrect, as it depicts an incorrect bromination position.
Step 4: Conclusion
Since 3-Bromo-2-methylpropene is the correct major product, the answer is Option (3).