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NEET CHEMISTRYHydrocarbonsMedium

Question

When 2-Bromopentane reacts with ethanolic KOH, the main product that is formed is:

A

Cis-2-pentene

B

Trans-2-pentene

C

1-pentene

D

None of the above

Step-by-Step Solution

When 2-bromopentane is heated with an alcoholic solution of potassium hydroxide (ethanolic KOH), it undergoes dehydrohalogenation (β\beta-elimination). According to Zaitsev's (Saytzeff's) rule, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms. Thus, pent-2-ene (81%) is formed as the major product over pent-1-ene (19%) . Furthermore, pent-2-ene can exist as cis and trans isomers. The trans-isomer is more stable than the cis-isomer due to lesser steric hindrance between the alkyl groups. Hence, trans-2-pentene is the main product formed.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonsbromopentanereactsethanolicproductformed

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