Question

2-Bromopentane reacts with alcoholic KOH to form major product as:

• A. Pent-1-ene
• B. Pent-2-ene
• C. Pent-3-ene
• D. Pent-1-ene and Pent-2-ene

Hint:

With small bases like \(\text{alc. KOH}\), \(\text{OH}^-\), or \(\text{OEt}^-\), Zaitsev's rule gives the more substituted alkene as major.
Bulky hindered bases like potassium tert-butoxide give the Hofmann (less substituted) alkene as major.

Answer 1

The Correct Option is B

Concept:
When haloalkanes containing \(\beta\)-hydrogen atoms are heated with a strong base such as alcoholic potassium hydroxide (\(\text{alc. KOH}\)), they undergo a \(\beta\)-elimination reaction (dehydrohalogenation) to produce alkenes.
The regiochemical outcome of this elimination is governed by Zaitsev's (Saytzeff's) rule.

Step 1: Identifying the Substrate and Reaction Centers:
The chemical structure of 2-bromopentane is given by:
\[ \overset{1}{\text{C}}\text{H}_3-\overset{2}{\text{C}}\text{H}(\text{Br})-\overset{3}{\text{C}}\text{H}_2-\overset{4}{\text{C}}\text{H}_2-\overset{5}{\text{C}}\text{H}_3 \] The bromine atom is bonded to the \(\alpha\)-carbon (C2).
There are two non-equivalent adjacent carbons bearing \(\beta\)-hydrogens:
1. The primary \(\beta_1\)-carbon at C1 containing three hydrogen atoms.
2. The secondary \(\beta_2\)-carbon at C3 containing two hydrogen atoms.

Step 2: Analysis of Competing Elimination Pathways:
Elimination of hydrogen from the C1 carbon and bromine from C2 yields pent-1-ene:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}=\text{CH}_2 \] This alkene possesses only one alkyl substituent attached to the double bond (a monosubstituted alkene).
Alternatively, elimination of hydrogen from the C3 carbon and bromine from C2 yields pent-2-ene:
\[ \text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_3 \] This alkene possesses two alkyl substituents attached to the double bond (a disubstituted alkene).

Step 3: Application of Zaitsev's Rule:
According to Zaitsev's rule, the predominant product in a dehydrohalogenation elimination is the more highly substituted alkene because it has greater hyperconjugative stabilization.
Pent-2-ene contains five hyperconjugative \(\alpha\)-hydrogens, making it thermodynamically significantly more stable than pent-1-ene, which contains only two hyperconjugative \(\alpha\)-hydrogens.
Consequently, pent-2-ene forms as the major product (approximately \(81\%\)), whereas pent-1-ene forms as the minor product.
Final Answer:
The major product formed is pent-2-ene, which corresponds to option (B).