I got to know about what neighboring group participation is, but does it always occur if there are lone pair donating groups in beta or delta positions. Or does it only occur in some kind of medium or is there any specific condition that is required?

For example, I have attached a picture below. I was solving some problems on this topic and came across this one.


The product I expected an epoxide(diazotization followed by attack of lone pair of oxygen) ,

enter image description here

But the answer given was a Ring contraction followed by formation of an aldehydic group,

enter image description here

So, to be clear, Why didn't Neighboring group participation take place, or in this case why an epoxide wasn't formed?

The mechanism I predicted: enter image description here

  • $\begingroup$ Can you draw out the mechanism for the epoxide formation? $\endgroup$
    – orthocresol
    Jun 29 '21 at 15:12
  • $\begingroup$ @orthocresol I have added the mechanism I predicted that would take place. Can you please help me out? $\endgroup$
    – Srini
    Jun 29 '21 at 15:28
  • 4
    $\begingroup$ The Tiffeneau rearrangement has been addressed here: chemistry.stackexchange.com/questions/116689/… Remember that the t-butyl group anchors the chair conformation. $\endgroup$
    – user55119
    Jun 29 '21 at 15:42
  • 6
    $\begingroup$ Good work with the mechanism. Now, you've drawn what's basically an SN2 reaction. You remember that the nucleophile must approach from a certain angle in an SN2 reaction...? $\endgroup$
    – orthocresol
    Jun 29 '21 at 15:55
  • 2
    $\begingroup$ You are right in saying that it must approach from the rear. Only that in your case, the $\ce{OH}$ group is not rear to $\ce{NH2}$ - both of them are faced to the right side. Since $\ce{NH2}$ is to pointing left, the attacking/migrating group from adjacent carbon must point to the right. The right pointing group is the alkyl part, which migrates causing ring contraction. It's always the opposite-side-pointing that matters - not up/down or cis/trans for the cyclohexane substituents. $\endgroup$
    – TRC
    Jun 30 '21 at 4:02

In the given compound the $\ce{-OH}$ group and the $\ce{-NH2}$ and later $\ce{N2^+}$ group are not in anti position with respect to each other. Equatorial positions in cyclohexane are not anti with respect to each other.

Below I have drawn newmann projection of the reactant which should help you visualise that a carbon (encircled) is instead in anti-relation to the leaving group, instead of $\ce{-OH}$. The cyclohexane cannot flip here either since $\ce{t-Bu}$ has locked the conformation would cause heavy steric repulsion in axial location making the ring unstable if it were to flip.

enter image description here

To show neighbouring group participation it is important that the lone pair donor be anti with respect to the leaving group. Hence here this is not availiable. Instead the anti-carbon attacks (undergoing ring contraction) because the resulting carbocation is stabilized by resonance with $\ce{-OH}$ as shown by this mechanism

Footnote:- The reaction does not involve hydride transfer to give ($\ce{t-Bu}$) substituted cyclohexanone as a major product because here $S_n2$ mechanism is faster than $S_n1$ (for most cases of secondary carbon in polar solvents and we can't have direct attack of hydrogen on $\ce{N2+}$ because it is not anti to leaving group) and there is not a great deal of difference in stability of 5 and 6 membered rings. However it still does occur to some extent.


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