2-Bromopentane is heated with alcoholic KOH. Which is the correct statement for this reaction?

A. Only pent-2-ene is formed

B. Only pent-1-ene is formed

C. Both pent-2-ene and pent-1-ene are formed

D. Pent-2-ene will be the major product as it is more stable

Choose the correct answer from the options given below:

C, D only

The correct answer is option 4. C, D only.

When 2-bromopentane is heated with alcoholic KOH, an elimination reaction (\(E_2\) mechanism) occurs, leading to the formation of alkenes.

2-Bromopentane has the structure

During the elimination reaction, the KOH deprotonates a hydrogen atom adjacent to the carbon atom bonded to the bromine atom, resulting in the formation of a double bond and the release of \(HBr\).

Formation of pent-2-ene:

If the hydrogen atom is removed from the third carbon, the double bond forms between the second and third carbon atoms, producing pent-2-ene.

Formation of pent-1-ene:

If the hydrogen atom is removed from the carbon at the first position, the double bond forms between the first and second carbon atoms, producing pent-1-ene.

Major Product:

According to Saytzeff's rule, the more substituted alkene is generally the major product. Pent-2-ene is more substituted and therefore more stable compared to pent-1-ene. Therefore, pent-2-ene will be the major product.

Conclusion:

Both pent-2-ene and pent-1-ene are formed, but pent-2-ene will be the major product.

Correct answer: Option 4 (C and D only).

Note:

Saytzeff's Rule: Saytzeff's rule, also known as Zaitsev's rule, is a principle used in organic chemistry to predict the outcome of elimination reactions, particularly in the formation of alkenes. The rule states:"In an elimination reaction, the most substituted alkene (the one with the greatest number of alkyl groups attached to the carbon atoms of the double bond) is the major product."