Rotation of groups about double bonds are usually restricted due to large torsional strain and it is due to this reason that cis- and trans-isomers have their independent existence. However, certain structural modifications in the compound makes the rotation of groups about double bond possible and hence, for such compound, cis- and trans-isomer exist together in equilibrium.

Identify which of the following compounds has such structural feature and its cis and trans isomers exist together in equilibrium:

(A) $\ce{CH3-CH=CH-CHO}$
(B) $\ce{CH3-CH=CH-CH3}$
(C) $\ce{cyclohexyl-CH=CH-cyclohexyl}$
(D) $\ce{CH3-CH=NH}$

Could someone explain the answer step-by-step and the theory behind it?

I think that rotation of groups about a double bond can be possible due to conjugation of double bonds, so the answer should be (A). But I am afraid I haven't completely understood the logic behind it.

  • 3
    $\begingroup$ Have you considered keto-enol tautomers? $\endgroup$ – Karsten Theis Mar 31 at 11:14
  • $\begingroup$ I would suggest D as the H atom can easily tunnel from one side on the N atom to the other. Tunnelling cannot happen in any of the other molecules as the groups are too massive. $\endgroup$ – porphyrin Apr 1 at 14:46
  • $\begingroup$ See also chemistry.stackexchange.com/questions/60800/… $\endgroup$ – Buck Thorn Apr 12 at 14:18

Option (D) is the answer. This may be predicted by redistribution of hybrid orbital theory or simply "Bent's Rule".

Clearly, during movement of the lone pair from one sp2 hybrid orbital to the other(to interconnect between the cis and trans isomers), there has to be an intermidiate transition state, during flipping when the lone pair resides in an unhybridized p-orbital. However the activation energy to achieve this intermidiate transition state is low for option (D).

Between N and H (as in the question), the electronegativity of N is higher than H. This pulls the shared electron pair in the N-H bond towards the N atom. In order to stabilize the high electron density, there is a redistribution of hybridization observed. The s character of the sp2 hybrid orbital of N (forming bond with the H ) increases ( eg from sp2 to sp1.8). This causes a corresponding decrease of s character of the other sp2 hybrid orbitals of N, such as the one containing the lone pair. The decrease in s character is associated with a a corresponding increase in p character of the lone pair containing orbital( eg from sp2 to sp2.3). Since it is easier to achieve transition from sp2.3 to unhybridized p orbital (having 100% p character), in the transition state it is easier to mediate between geometric isomer, possibly even existing in equilibrium.

For further reading see this answer:

Why are oxime geometrical isomers stable?


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