[Il Divo]

Just a bit of confusion regarding a reaction between Nitriles with Alkyl Groups supplied by Grignard Reagents.

1) What's stopping the Grignard Reagent from R-Mg-X from supplying a second R group to the Carbon-Nitrogen bond, similar to what happens via addition of LiAlH4? In the former case, we generate an imine, but with the other case we get a primary amine. I'm just having trouble understanding why a very strong base/nucleophile from Grignard reagents won't add again, but H: anion will to generate Nitrogen with a -2 charge.

2) With the Grignard reagent, we generate an imine intermediate and hydroxide to protonate the Nitrogen. From here, everywhere I look mentions hydrolysis with mild acid. But wouldn't it be possible to perform hydrolysis using the OH already generated by imine formation?

The Carbon-Nitrogen bond in the imine will still be fairly electrophilic, making it a great target for attack by hydroxide. By manipulating equilibrium, is it still possible to produce the Ketone using basic conditions?

[Archer]

1) The Grignard forms an N-Mg-X bond which prevents further reaction.

see http://www.organic-chemistry.org/namedreactions/grignard-reaction.shtm for a mechanism.

RE LAH the nitrogen co-ordinates with the Al but not in the same way so a second hydride can attack producing N^- which has Li⁺ as the counter ion.

2) When ever I have done these reactions the work up is acidic, usually saturated ammonium chloride. This is sufficiently acidic to hydrolyse the imine in a number of cases, even if not completely. I have, on occasion, had to add a little HCl to finish of the difficult ones.

[opsomath]

I would add to Archer's excellent reply this:

What you get when a Grignard adds to a nitrile is essentially a negative ion of the form

R₂C=N^-

except it has a magnesium counterion. Now a Grignard is also very similar to a carbanion (a carbon with a negative charge). So, it will be very difficult for the Grignard to react with this thing that is already negatively charged and electron-rich.

In LAH, all the bonds are more covalent (both the N-Al bonds and the Al-H bonds) and this is not the case.

[Archer]

What you get when a Grignard adds to a nitrile is essentially a negative ion of the form

R₂C=N^-

except it has a magnesium counterion. Now a Grignard is also very similar to a carbanion (a carbon with a negative charge). So, it will be very difficult for the Grignard to react with this thing that is already negatively charged and electron-rich.

In LAH, all the bonds are more covalent (both the N-Al bonds and the Al-H bonds) and this is not the case.

This is much better than how I explained it 

And @Il Divo, very good question, I never really thought about this so my initial thought was "because it does" so thanks for making me think about something I take for granted.

[Il Divo]

Nah, you both did a fantastic job. I appreciate the help.

But just to be clear, from a theory stand point: what is stopping hydrolysis of Imines via base vs. acid? Hydrolysis of Esters for example can occur via both mechanisms, either by making the carbonyl more electrophilic or using base as a strong nucleophile.

That's what makes this odd to me as addition of water to that negative ion results in formation of OH anion, which itself seems like could lead to hydrolysis without further addition of acid.

[opsomath]

You are asking good questions, Divo. The only place I am finding basic hydrolysis of imines is when the parent amine is conjugated to something else, which makes sense in that basic hydrolysis has to go through the R2N- intermediate, so anything that stabilizes that will make base hydrolysis easier. By contrast, acid workup only needs to form R₂NH₂⁺, which is easy with any reasonably strong acid. So it's all about choosing a hydrolysis route that works with your structures. No doubt if you had an imine structure which was very acid-sensitive but not base-sensitive at all (an acetal-protected thing, for instance) you could hydrolyze it with strong NaOH solution or something even if the nitrogen was just on an alkyl skeleton and not conjugated to anything.

[orgopete]

I foresee two technical reasons acid is used. First, the principle of elimination of the weakest base from a tetrahedral intermediate should rule. If water were to add, the intermediate would possess two alkyl groups, an OH, and an NH2 (after abstraction of a proton from water which generates  hydroxide). As leaving groups, HO(-) is the weakest base and thus the best leaving group. This will regenerate the imine (after deprotonation by hydroxide).

This principle similarly applies to hydrolysis of amides. Addition of hydroxide to an amide (and protonation-deprotonation), gives a tetrahedral intermediate with an NH2, an OH, an O(-), and a carbon. Hydroxide is the weakest base so the hydrolysis fails. If the base concentration is sufficiently high, then a second deprotonation of the intermediate can occur to give a doubly charged intermediate. Now, NH2(-) is the weakest base, and a carboxylate is formed.

The second reason is hydroxides of magnesium are milky and difficult to work with. Acidification gives a soluble magnesium salt.

[opsomath]

The second reason is hydroxides of magnesium are milky and difficult to work with. Acidification gives a soluble magnesium salt.

I strongly suspect that this is a big part of it. Practical considerations are every bit as important as theoretical ones in determining what standard practice is. Also, gooey oxide byproducts suck.

[Il Divo]

Thank you all again for the help. Your replies were exactly what I needed!