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 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?
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Markovnikov's rule states H adds to the carbon with more hydrogens.
Updated On: Jun 10, 2026
- figure [H]
- figure [H]
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Markovnikov's rule for HBr addition.
Step 2: Analysis
A = Pent-1-ene (forms 2-bromopentane). B = Pent-2-ene (forms 3-bromopentane). Both are position isomers and lack stereocenters at the site of double bond to allow standard cis-trans isomerism given the conditions.
Step 3: Conclusion
X is 2-bromopentane and Y is 3-bromopentane.
Final Answer: (C)
Markovnikov's rule for HBr addition.
Step 2: Analysis
A = Pent-1-ene (forms 2-bromopentane). B = Pent-2-ene (forms 3-bromopentane). Both are position isomers and lack stereocenters at the site of double bond to allow standard cis-trans isomerism given the conditions.
Step 3: Conclusion
X is 2-bromopentane and Y is 3-bromopentane.
Final Answer: (C)
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