[doof]

I'm just a little confused about the mechanisms behind the Wolff-Kishner Method
I get what the hydrazine and KOH do in the process
I'm just wondering if the ethylene glycol used in the reaction plays any part in the change in chemistry of the product or is it just an innocent bystander?
is it broken down into something else or do i still have the same amount at the end of the reaction as i do at the start. If so, would it be removed with diethyl ether extraction?

[dunno260]

The ethylene glycol serves a couple of purposes.    First its a high boiling solvent which allows the reaction to be run at a higher temperature.  It also helps to stabilize the transition state in the reaction.  It can also serve as a proton donor to the carbanion formed during the reaction (depending on if there is water around, the pKa of an alochol and water is similar though water is slightly more acidic.  Mechanism here http://www.organic-chemistry.org/namedreactions/wolff-kishner-reduction.shtm

Assuming your workup for the reaction is aqueous, you would recover the same amount of ethylene glycol as you started out with as any alkoxide that is formed will be reprotonated during the workup.  Ethylene glycol is very polar so if you do an extraction with ether, you would expect most of the ethylene glycol to go into the aqueous layer though some will go with the ether too. 

[doof]

Cool, thanks for all that
though now i have one more question

would the reaction still work if it was performed in water rather than an alcohol? though if it did the efficiency would probably be lower, is that right?

[dunno260]

It should work, though its going to be slower as you aren't funneling your intermediates to the product by removing water so your intermediates are reversible.  In addition, the reason alcohol solvents such as ethylene glycol and related high boiling point solvents are used is that it allows for an increase in temperature to increase the rate of the reaction.  You could do this in water using a pressure vessel of some sort (or in a low boiling point solvent such as methanol, ethanol, or isopropanol.  The reaction isn't reversible though once it reaches the final product and that helps.

[doof]

Ah ok, so working in an alcohol is a better option then. cool thanks

actually after looking more closely at the mechanisms involved another thought has occurred to me,
the second part of the process involves cleavage of a C-N bond, however the molecule i'm treating has an NH2 group elsewhere in the molecule. I dont want this to be removed. Will it be removed in this process or is the mechanism of this method more dependent on the negative charge place on the C when the double bond is between the two nitrogen atoms rather than the between nitrogen and carbon. is it also dependent on there being two nitrogens being attached in the process and those nitrogens being able to be removed as N2 gas?

[doof]

Actually i have another additional question, can NaOH be used in place of KOH in this method?

[cpncoop]

Hi,

Any strong base will work for this type of reduction (CsOH, NaOH, LiOH, KOH).  Additionally, you can run the reaction in the presence of water (I usually use hydrazine hydrate anyways), but as mentioned, it is slightly slower.  Good luck.

[orgopete]

I believe KOH is used because less glass is dissolved than with NaOH. If you use NaOH, you can note a significant etching of the glass vessel.