Hi William – it would be an oxime ether if there is an R group on the oxygen. Do a search for “oxime ether” and you should find lots of information.

You stated that an oxime has the C=N double bond with two organic constituents on the carbon and a hydroxyl group on the nitrogen. If the hydroxyl group were instead an ether, would it still be an oxime or would it have a different name?

I’m curious as I found a pharmaceutical compound (Roxithromycin) that wikipedia says contains an N-oxime side chain. However, every source I’ve seen on oximes says they have an R1R2C=NOH general formula, not an R1R2C=NOR3 formula.

Thanks,
William

Where is it said that they polymerize? Using MgSO4 or Na2SO4 to make an imine is pretty standard practice.

How stable aliphatic imines are if the water is absorbed by a sulfate? It’s often said they polymerize, but that’s it. No more information.

When you draw out the enamine that would form on the “most substituted” side, you’ll see that the methyl group on the 2-position has a steric clash with one of the alkyl groups on the nitrogen. It’s called “allylic strain”. Enamines tend to form on the less-substituted “alpha carbon”.

Hello James can you please throw some light on major product formed when a 2-methyl cyclohexanone reacts CH3I via enamine ? Also Explain which type of enamine formation dominate ie kinetic or thermodynamic enamine ?

Hi Alex, you have a great point, recently brought up in the comments of this post. https://www.masterorganicchemistry.com/2017/09/01/reductive-amination/
I think EB is correct in saying that the first step should be addition of the amine to the carbonyl, and that the proton source assists with departure of the resulting OH. I’m going to put fixing this in my correction queue.

This reaction mechanism has puzzled me over the last few days. The pKa for deprotonating a ketone is around -7, whereas many protonated amines of interest are around 10. If we only reason in terms of protonated ketone and amine and unprotonated ones, then we see directly that in the pH range 3 to 7 the product [protonated ketone][unprotonated base] should remain rather constant, and be extremely low.

After doing some digging, I find that some authors prefer writing a concerted mechanism, and some attack an unprotonated ketone. The latter is certainly less electrophilic. However, at pH 5, it is 10^12 times more abundant, which in free energy terms means RT ln (10) =69 kJ/mol=16.5 kcal/mol extra in favor of this pathway.

If we opt for the latter mechanism, we should expect the rate to depend on pH, but for a different reason: at low pH, there is not enough amine to attack ketone (protonated or not). At high pH, the elimination of water becomes rate-limiting (since the H2O+ moiety will spend more time as an OH moiety).

Of course, my thinking is based on pKa values in water, maybe the mechanism is quite different in organic solvents. What are your thoughts on this?

Thanks for catching that typo, Matt.