Arrange the following alkenes in decreasing order of stability.
Choose the correct answer from the options given below:

Solution

Chemistry ma’am · 4 steps

Count the alkyl groups on the double bond first, then break the tie between cis and trans by crowding.

1. Identify each alkene. I is 2,3-dimethylbut-2-ene, with four methyls on the double bond. III is 2-methylbut-2-ene, with three methyls. II and IV are both but-2-ene with two methyls. In II the methyls are on opposite sides (trans), and in IV they are on the same side (cis).
2. More substitution wins. Each alkyl group on a double-bond carbon brings α-hydrogens. These stabilise the π bond by hyperconjugation. So the more alkyl groups the double bond carries, the more stable the alkene.

   Alkene	Alkyl groups on C=C	α-H atoms	Extra factor
   I	4 (tetrasubstituted)	12	most hyperconjugation
   III	3 (trisubstituted)	9	next best
   II (trans)	2	6	methyls far apart, no crowding
   IV (cis)	2	6	methyls on the same side, steric strain

3. Cis versus trans. II and IV have the same number of α-hydrogens, so hyperconjugation cannot separate them. In the cis isomer the two methyls crowd each other on the same side of the double bond. That strain makes IV less stable than II.
4. So. The order is I > III > II > IV, which is option (c).

The idea to keep. Alkene stability rises with the number of alkyl groups on the double bond (hyperconjugation). When the counts are equal, trans beats cis because cis groups bump into each other.

Answer from the NTA final answer key. Solution by PrepWiser.