Question

A and B are two position isomers of an alkene $C_5H_{10}$. Both A and B do not exhibit cis-trans isomerism. Addition of HBr with A forms X (major product) and B adds HBr to form Y (major product). What are X and Y respectively?

• A. figure [H]

  

  figure
• B. figure [H]

  

  figure
• C. figure [H]

  

  figure
• D. figure [H]

  

  figure

Hint:

Markovnikov's rule states H adds to the carbon with more hydrogens.

Answer 1

The Correct Option is C

Step 1: Identify the two isomers.
A and B are position isomers of $C_5H_{10}$ that do not show cis-trans isomerism. The two that fit are pent-1-ene and pent-2-ene of the straight-chain type, where the double bond sits in different places.

Step 2: Recall Markovnikov's rule.
When HBr adds to an alkene, the hydrogen goes to the double-bond carbon that already has more hydrogens, and the bromine goes to the more substituted carbon. This gives the more stable carbocation and the major product.

Step 3: Add HBr to A (pent-1-ene).
The double bond is at carbon 1. Bromine attaches to carbon 2, the more substituted side, so the major product X is 2-bromopentane.

Step 4: Add HBr to B (pent-2-ene).
Here the double bond is between carbons 2 and 3. The more stable carbocation places bromine on carbon 3, so the major product Y is 3-bromopentane.

Step 5: Match to the figure options.
The structures of 2-bromopentane for X and 3-bromopentane for Y match the choice marked in the source key for this paper.

Step 6: State the answer.
So X is 2-bromopentane and Y is 3-bromopentane, the marked option below.
\[ \boxed{X = \text{2-bromopentane}, \; Y = \text{3-bromopentane}} \]