Determine Correct Stability Order of Given Alkenes

To determine the correct stability order of the given alkenes, we analyze their hyperconjugative stability, which is directly proportional to the number of $\alpha$-hydrogens ($\alpha\text{-H}$) attached to the $sp^2$-hybridized carbon atoms of the double bond.

1. Structure and $\alpha$-hydrogens of each alkene:

   • C. 2,3-Dimethylbut-2-ene:
     $\text{Structure: } (CH_3)_2C=C(CH_3)_2$
     The $sp^2$-hybridized carbon atoms are bonded to four methyl groups.
     $\text{Number of } \alpha\text{-H} = 3 + 3 + 3 + 3 = 12\ \alpha\text{-H}$

   • A. 2-Methylbut-2-ene:
     $\text{Structure: } (CH_3)_2C=CH-CH_3$
     The $sp^2$-hybridized carbon atoms are bonded to three methyl groups.
     $\text{Number of } \alpha\text{-H} = 3 + 3 + 3 = 9\ \alpha\text{-H}$

   • B. cis-But-2-ene:
     $\text{Structure: } CH_3-CH=CH-CH_3$
     The $sp^2$-hybridized carbon atoms are bonded to two methyl groups.
     $\text{Number of } \alpha\text{-H} = 3 + 3 = 6\ \alpha\text{-H}$

   • D. Prop-1-ene:
     $\text{Structure: } CH_3-CH=CH_2$
     The $sp^2$-hybridized carbon atoms are bonded to one methyl group.
     $\text{Number of } \alpha\text{-H} = 3\ \alpha\text{-H}$

2. Comparing Stability:

   Since the stability of an alkene increases as the number of hyperconjugative structures (and hence the number of $\alpha$-hydrogens) increases: $\text{Stability } \propto \text{Number of } \alpha\text{-hydrogens}$

   Comparing the number of $\alpha$-hydrogens for each compound: $\text{C } (12\ \alpha\text{-H}) > \text{A } (9\ \alpha\text{-H}) > \text{B } (6\ \alpha\text{-H}) > \text{D } (3\ \alpha\text{-H})$

Thus, the correct stability order is $\text{C} > \text{A} > \text{B} > \text{D}$.

This corresponds to Option A.