Comparing heat of hydrogenation among 1‐methylcycloprop‐1‐ene, 3‐methylcycloprop‐1‐ene and methylenecyclopropane

In the given problem, we have three compounds and have been asked to compare their heat of hydrogenation.

Starting Premise

We start from the fact that the heat of hydrogenation of a compound is inversely proportional to the stability of the double bond in the system. Using this we can say that the opposite of the order of stability would be the answer

Comparing stability of the compounds

Here, the three compounds given contain three-member rings. Cyclopropane rings with all carbons being $ce{sp^3}$ have a very high ring strain since the angle between the carbons are $mathrm{60 ^circ}$ when the expected $ce{C-C}$ bond angle is $mathrm{109.28^circ}$.

Therefore, the higher the total strain, the higher the heat of hydrogenation, since the the final product is same in all cases. Finding the total strain in the three cases using WebMO, the total strain energy is as follows.

$$ small begin{array} {lcc} hline text{Compound} &text{Total strain energy(kcal/mol)} \ hline text{methylenecyclopropane} &text{224.836} \ hline text{3-methylcyclopropene} &text{389.564} \ hline text{1-methylcyclopropene} &text{395.751} \ hline text{1-methylcyclopropane} &text{90.490} \ hline end{array} $$

Now, from this table, we can see that methylenecyclopropane (2) has the highest stability, then $ce{3-methylcyclopropene}$ (1) and finally $ce{1-methylcyclopropene}$ (3)

Edit: After Rahul Verma's comment, taking hyperconjugation(due to similar total strain) for (3) and (1), due to better hyperconjugation on (3), (3) would be more stable.

Therefore the order of stability would be (2) > (3) > (1)

Order of Heat of Hydrogenation

With the above data, we can say that the order of Heat of Hydrogenation would be (1) > (3) > (2)

Source:Relation between total strain energy and stability