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So while studying about Wolff-Kishner reduction reaction to form alkanes from carbonyl groups. I read that it is not suitable for base sensitive groups as the environment is highly basic and as an example the following reaction was given:

                                              Reduction of 4-Chloro cyclohexanone

Now the desired product that should've formed is

                                                          cyclohexyl chloride

But this doesn't happen as -Cl is a base sensitive group and so reacts with KOH. But how can a halogen be base sensitive when it already has 7 e- and is highly electronegative? How is it possible for it to react with OH-?

Also, I don't understand the reaction mechanism of this reaction so can someone please explain how this takes place?

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  • $\begingroup$ Have you studied the E2 mechanism? en.wikipedia.org/wiki/Elimination_reaction $\endgroup$ – Jabbamanga Jan 7 at 19:10
  • $\begingroup$ I only have a vague idea about it and haven't yet studied it. So can you please elaborate how it is relevant here and what is the reaction mechanism of this reaction, perhaps in an answer instead of a comment? $\endgroup$ – A. Joshi Jan 7 at 19:19
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    $\begingroup$ The chloride can also displace directly with hydrazine $\endgroup$ – Waylander Jan 7 at 22:10
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You've probably realised already that a possible side reaction is the SN2 nucleophilic substitution of Cl by HO-, which would look like this: enter image description here

Evidently, this is not the product we actually get. Hydroxide ions are considered 'hard' nucleophiles, as they have a high negative charge density, and they tend to react best with 'hard' electrophiles with high positive charge density. The electrophilic carbon here is reasonably 'soft' whilst the proton adjacent to it is considered 'hard' due to its small size (you can learn more about this here). As a result, hydroxide ions like to act as bases instead of nucleophiles, and the mechanism favoured is not the SN2 shown above, but in fact the competing E2 elimination shown below.

enter image description here

This proton may not look particularly acidic, but formation of a new alkene alongside loss of a reasonably stable chloride leaving group drives the deprotonation.

Another contributing factor is that whilst the substitution has a reasonably small change in entropy, the elimination results in a greater number of species than in the reactants, so has a large positive entropy. This is relevant because Wolff-Kishner reduction is usually done at a reasonably high temperature, and entropy becomes a greater influence on the spontaneity of a mechanism as the temperature increases. This further favours E2 over SN2.

The chloride is considered a base-sensitive group for the above reasons, and this second pathway is inevitable alongside a Wolff-Kishner reduction in a highly basic solution.

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I think the reaction as you told will be very inefficient. To explain why, I quote the paragraph from page-905 of David R Klein's Organic Chemistry (*):

"The pH of the solution is an important consideration during imine formation, with the reaction being greatest when the pH is around (4.5). If the pH is too high (i.e: no acid catalyst is used) , the carbinolamine is not protonated, so the reaction occurs more slowly. If the pH is too low, most of the amine molecules will be protonated to give ammonium ions, which are not nucleophilic. Under these conditions, step 2 of the mechanism occurs too slowly. As a result, care must be taken to ensure optimal pH of the solution during imine formation"

Though the above paragraph concerns imine formation, it applys equally well to hydrazone formation which occurs as an intermediate in the standard scheme of wolf Kishner reduction of ketones.

The mechanism of the reduction can be found here

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  • $\begingroup$ Formation of hydrazones are irreversible, unlike imines. That is one reason why they are used as derivatizing agents (2,4-DNP's, semicarbazones, etc.) The Wolff-Kishner is conducted under basic conditions, not pH 4.5. $\endgroup$ – user55119 Jan 7 at 22:45
  • $\begingroup$ Hmm but you had written in this answer that it's irreversible. Correct me if I misread anything. $\endgroup$ – Buraian Jan 8 at 7:55

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