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

### Problem

Which of the following orders is correct for heat of hydrogenation of these compounds?

A) 1 > 3 > 2
B) 3 > 2 > 1
C) 3 > 1 > 2
D) 2 > 1 > 3

D) 2 > 1 > 3

### My Attempt

Heat of hydrogenation is inversely proportional to stability. Stability is directly proportional to the number of α-hydrogen. The heat of hydrogenation is inversely proportional to the number of α-hydrogens:

1: one α-hydrogen;
2: four α-hydrogens;
3: five α-hydrogens.

Also, here in the 3, the intermediate formed is more stable due to dancing resonance. So, the order of heat of hydrogenation must be 1 > 2 > 3, which contradicts the answer.

Edit:The given solution to this question contidicts our argument

• Consider the angles of the double bonds. Stability is in this case principally determined by ring strain. – Waylander Jul 15 '20 at 10:00
• can you please elaborate it as an answer @Waylander? – Anonymous Jul 15 '20 at 10:01
• why is the order heat of hydrogenation is failing when we compare with the $alpha$ hydrogens? @Waylander – Anonymous Jul 15 '20 at 10:03
• The number of alpha hydrogens is only a determining factor when considering unstrained alkenes or alkenes of identical strain. – Waylander Jul 15 '20 at 10:15
• I think the given answer is incorrect, as 2 is the most stable! Also, can you explain how do you see dancing resonance in 3? – Rahul Verma Jul 15 '20 at 13:40

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)

• I guess you forgot about hyperconjugation. If you'll take account of both the factors, then answer should be 1 > 3 > 2 (as I got). Btw, you can take ring-strain of 1 & 3 to be nearly same. – Rahul Verma Jul 15 '20 at 13:39