Q.22 Which one among the following structures is the most stable conformer of
(Z)–pent–2–ene?
Introduction
Understanding the most stable conformer of (Z)-pent-2-ene is essential for mastering alkene stereochemistry and conformational analysis in organic chemistry.
The question asks which Newman-like representation around the C2–C3 double-bond region best minimizes steric repulsions while still representing the Z configuration
(higher‑priority groups on the same side).
Step 1: What is (Z)-pent-2-ene?
(Z)-pent-2-ene has the structure CH3–CH=CH–CH2–CH3, with the higher‑priority groups (the two ethyl/methyl‑containing substituents)
on the same side of the C=C bond.
Around the C=C bond, rotation is restricted, but limited twisting of the σ bonds at C1–C2 and C3–C4 allows different conformers that change relative H/H and
CH3/CH3 distances without breaking the double bond.
The most stable conformer will minimize steric repulsion, especially between the two bulky CH3/CH2 groups and between any syn‑periplanar hydrogens.
Analysis of each option
For all options, assume the C=C bond is drawn horizontally and the “up–down” orientation shows which groups are on the same side of the double bond (Z configuration).
Option (A)
In (A), both CH3 groups are drawn cis and very close, almost eclipsed across the C=C.
This arrangement creates strong steric crowding between the two alkyl groups and also places several H atoms in close proximity, leading to high torsional and steric strain.
Therefore, option (A) represents a high‑energy conformer and is not the most stable.
Option (B)
In (B), the CH3 groups are still on the same side (Z), but one methyl is placed almost syn to a hydrogen on the opposite carbon, and the second methyl is close to the other hydrogen.
Steric interactions are somewhat relieved compared with (A), yet there is still substantial crowding because the substituents are not arranged in a staggered-like fashion around the C2–C3 framework.
Option (B) is more stable than (A) but still not the lowest‑energy conformer.
Option (C)
In (C), one CH3 and one H are drawn on the same side, placing a methyl group opposite a hydrogen in a way that increases CH3–H and H–H proximity.
The drawing also partially distorts the Z requirement, effectively pushing the higher‑priority groups toward a pseudo‑E‑like orientation to reduce repulsion, which does not correctly depict the Z stereochemistry.
Because it compromises the Z configuration and keeps significant steric contacts, option (C) is incorrect for the most stable Z conformer.
Option (D) – Correct answer
In (D), the two CH3-bearing substituents remain on the same side of the double bond (correct Z configuration), but they are oriented so that each methyl is roughly anti to a hydrogen on the other carbon.
This arrangement spreads bulky groups as far apart as is possible while preserving the Z relationship, minimizing both CH3–CH3 and H–H repulsions and closely resembling a staggered‑like pattern around the double-bond environment.
Hence, option (D) gives the most stable conformer of (Z)-pent-2-ene.
Key points for exams
- Always maintain the correct Z/E relationship when comparing conformers.
- Among all valid Z conformers, the most stable is the one that maximizes distance between bulky CH3/alkyl groups.
- Minimize eclipsing or syn‑periplanar interactions to lower steric and torsional strain.
- For this question, only option (D) satisfies both the Z configuration and lowest steric strain, so it is chosen as the most stable conformer.
