JEE Chemistry 2026 Question with Solution

Given: Four alkenes I, II, III, IV are to be arranged in decreasing order of stability.

Find: The correct decreasing order of stability.

Principle: Alkene stability increases with greater alkyl substitution because of hyperconjugation. For disubstituted alkenes, trans is more stable than cis due to lower steric strain.

The stability order used is:

$$\text{tetra-substituted} > \text{tri-substituted} > \text{trans-disubstituted} > \text{cis-disubstituted}$$

From the given structures:

1. I is a tetra-substituted alkene, so it is the most stable.
2. III is a tri-substituted alkene, so it is next in stability.
3. II is a trans-disubstituted alkene, so it is more stable than the cis isomer.
4. IV is a cis-disubstituted alkene, so it is the least stable.

Therefore, the decreasing order of stability is:

$$\text{I} > \text{III} > \text{II} > \text{IV}$$

So, the correct option is B.

Given: The four alkenes are labeled I, II, III, IV.

Find: Their decreasing order of stability.

Step 1: Identify the substitution around the double bond.

• I: tetra-substituted
• III: tri-substituted
• II: trans-disubstituted
• IV: cis-disubstituted

Step 2: Compare on the basis of hyperconjugation and steric effects.

• More alkyl groups on the double bond increase stability.
• Between geometrical isomers, trans is more stable than cis.

Step 3: Arrange accordingly:

$$\text{I} > \text{III} > \text{II} > \text{IV}$$

Hence, the correct option is B, which is Option (2).