[omegasynthesis999]

Why do complex hydrazones such as phenylhydrazine (H2NNHC6H5) react with a regular aldehydes/ketones to form an imine product? Can't they technically react to form the enamine product? Why does it seem the imine is preferred over the enamine?

[macman104]

Imine and Enamine are tautomers of each other.  Also, the C=N bond is stronger than the C=C bond, just like the C=O bond is preferred over the C=C bond in the keto/enol tautomer.

[omegasynthesis999]

Attached is what I am talking about. So are you saying that both products are possible (the ones I have drawn)?

Also you raise an interesting idea about imine/enamine tautomerization. I am familiar with enol/keto tautomerization but how exactly would the imine/enamine tautomerize [ie. what conditions would you use (basic/acidic) and mechanistically speaking how would things attach/detach]?  For example, how could you tautomerize the attached imine/enamine combo?

[macman104]

I do not believe your "enamine" product will form.  The "imine" product can tautomerize to an enamine form.  If remove a hydrogen from the on of the methyls and form a double bond and then protonate the nitrogen, you will get an enamine.  While your two products are examples of enamine and imine, they are not tautomers of eachother, but they each have a corresponding tautomer.  I don't have access to drawing software right now, but if you need more info, I can try and draw something up later.

[omegasynthesis999]

I can see why they wouldn't be tautomers. But in terms of the enamine and imine I have drawn, is it not possible to form the enamine or rather will the enamine not form too much because it is an unfavored product?

For example, aldehydes are more reactive than ketones. By analogy, are primary amines better nucleophiles than secondary amines?

Thanks.

[spirochete]

I think I see the question you are asking.   As you've shown, there are two reasonable looking products that could form from the reaction your considering.  And in the course of the reaction, both products probably form transiently.  But enamine/imine formations are reactions which are very easily reversible.  That is, the energy of activation for the reverse set of reactions is relatively low.  This means that the thermodynamic product will be favored.  Only the primary amine forms a stable, uncharged product with a C=N pi bond, which is stronger than a C=C pi bond.  This makes the imine more stable.

Another example of this concept is the addition of HBr to an alkene with multiple pi bonds.  But in that case it's easier to control the course of the reaction by altering temperature.  Enamine/ime formation is reversable at a lower temperature.

We can also consider the nucleophility of each nitrogen in the reactant, but this is probably not as important to the course of the reaction because the product is thermodynamically controlled.  The phenyl group next to the nitrogen is mildly electron withdrawing by resonance, making it less nucleophilic than primary amine.  Also, it is more stearically hindered.  If we had a regular alkyl group (slightly electron donating) that might make a secondary nitrogen more nucleophilic,but I believe the imine product would still form because of the reversability of the reaction.

[omegasynthesis999]

Your ideas about reversibility of enamine/imine formation make sense. Furthermore it would make sense for the thermodynamic product to form as you said.  Thanks for the insights.

[spirochete]

One minor correction to the last paragraph: it's not reasonable to say that alkyl groups would increase net nucleophility by donating electron density.  Stearic hindrance is definitely the stronger effect, which decreases nucleophility of course.