Q40. An alkyl halide with molecular formula C₆H₁₃Br on dehydrohalogenation gave two isomeric alkenes

Q: What is the correct answer for CUET UG Chemistry Q40?

The correct answer is d. Let's work backward from the ozonolysis products. The four compounds (CH₃COCH₃, CH₃CHO, CH₃CH₂CHO, (CH₃)₂CHCHO) are: Acetone, Acetaldehyde, Propanal, 2-Methylpropanal. These products indicate the structures of alkenes A and B:- Acetone + Propanal comes from 2-Methylpent-2-ene (CH₃-C(CH₃)=CH-CH₂-CH₃).- Acetaldehyde + 2-Methylpropanal comes from 2-Methylpent-1-ene (CH₃-CH(CH₃)-CH=CH₂).Now, we ne

Q: Q40. An alkyl halide with molecular formula C₆H₁₃Br on dehydrohalogenation gave two isomeric alkenes A and B with molecular formula C₆H₁₂. On reductive ozonolysis, A and B gave four compounds…

The correct answer is d. Let's work backward from the ozonolysis products. The four compounds (CH₃COCH₃, CH₃CHO, CH₃CH₂CHO, (CH₃)₂CHCHO) are: Acetone, Acetaldehyde, Propanal, 2-Methylpropanal. These products indicate the structures of alkenes A and B:- Acetone + Propanal comes from 2-Methylpent-2-ene (CH₃-C(CH₃)=CH-CH₂-CH₃).- Acetaldehyde + 2-Methylpropanal comes from 2-Methylpent-1-ene (CH₃-CH(CH₃)-CH=CH₂).Now, we ne

Q: What are the options for this CUET UG Chemistry question?

A. 2-Bromohexane · B. 2,2-Dimethyl-1-bromobutane · C. 2-Bromo-2,3-dimethylbutane · D. 3-Bromo-2-methylpentane

Question

Q40. An alkyl halide with molecular formula C₆H₁₃Br on dehydrohalogenation gave two isomeric alkenes A and B with molecular formula C₆H₁₂. On reductive ozonolysis, A and B gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH₂CHO and (CH₃)₂CHCHO. The alkyl halide is:

FirstInTest · Accepted Answer

Let's work backward from the ozonolysis products. The four compounds (CH₃COCH₃, CH₃CHO, CH₃CH₂CHO, (CH₃)₂CHCHO) are: Acetone, Acetaldehyde, Propanal, 2-Methylpropanal. These products indicate the structures of alkenes A and B:-   Acetone + Propanal comes from 2-Methylpent-2-ene (CH₃-C(CH₃)=CH-CH₂-CH₃).-   Acetaldehyde + 2-Methylpropanal comes from 2-Methylpent-1-ene (CH₃-CH(CH₃)-CH=CH₂).Now, we need an alkyl halide C₆H₁₃Br that yields both 2-Methylpent-2-ene and 2-Methylpent-1-ene upon dehydrohalogenation.(d) 3-Bromo-2-methylpentane (CH₃-CH(CH₃)-CHBr-CH₂-CH₃) on dehydrohalogenation can yield 2-Methylpent-2-ene (by elimination of H from C1 or C5) and 2-Methylpent-1-ene (by elimination of H from C5), and 3-Methylpent-2-ene (by elimination of H from C4). If we assume the four ozonolysis products are derived from *two* alkenes, and considering Saytzeff and Hofmann elimination rules to guide product formation, 3-Bromo-2-methylpentane fits the requirement to produce substituted alkenes which upon ozonolysis give the mentioned products.

Answer

2-Bromohexane

Answer

2,2-Dimethyl-1-bromobutane

Answer

2-Bromo-2,3-dimethylbutane

Answer

3-Bromo-2-methylpentane

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