Q15-15-31E - Organic Chemistry (Mcmurry)

Question

Cyclopropane is highly reactive because of its large amount of angle strain. Methylcyclopropenone, although even more strained than cyclopropanone, is nevertheless quite stable and can even be distilled. Explain, taking the polarity of the carbonyl group into account.

Cyclopropanone                                           Methylcyclopropenone

Step-by-Step Solution

Answer

Methylcyclopropenone                                       2 pi electrons

                                                                                More stable

                                                                           The polarity of the carbonyl group

Stability of methylcyclopropenone

The stability of methyl cyclopropenone can be explained by Huckel’s rule using the polarity of its carbonyl group. Huckel’s rule is concerned with compounds containing one planar ring in which each atom has a p orbital as in benzene. His calculations show that planar monocyclic rings containing $\left(4n+2\right)\mathit{\pi }$ electrons, where n=0,1,2,3,….., and so on, have closed shells of delocalized electrons like benzene and should have substantial resonance energies. In other words, planar monocyclic rings with 2,6,10,14,…., delocalized electrons should be aromatic.

Aromatic compounds

Aromatic compounds have double bonds. They are in a conjugated system. The polarity of the carbonyl group in mehtylcyclopropenone is as shown:

Methylcyclopropenone                                       2 pi electrons

                                                                                     More stable

                                                 The polarity of the carbonyl group

Due to the polarity of the carbonyl group mehtylcyclopropenone exists in the above two forms. The second form, in which cyclopropenyl cation is highly stable due to it obeys Huckel’s rule $\left(4n+2\right)\pi$electrons.