[pocopo]

I don't recall ever seeing that base before and I do not know how it would integrate itself with those two reactants.  Could someone help get me started?

Thanks in advance

[discodermolide]

The base deprotonates the alpha protons of the ketone, this anion attacks the imine carbon.

[pocopo]

Thanks, alright so I had drawn out the mechanism.  I have the it so the base attacked the alpha carbon and had attached itself at that point.  By doing so a double bond formed between the alpha carbon and the O gets a negative charge.  It then attacks the imine carbon and attaches as that point.  The double bond breaks and the N gets a negative charge.  Does the negative charge move down to form a double bond between N and Ph2? 

[discodermolide]

Probably not it may pick up a proton and form the NH.

[pocopo]

There is no water present in the first step so I don't think that will happen.  I was thinking the N- might attack the double bonded O.  Since intramolecular reactions are usually what your supposed to look for first.  Therefore a ring would form.  Does this sound right?

[discodermolide]

OK did not realise there was no work up. It attacks the carbonyl to the ketone and eliminates ^-CCl₃

[pocopo]

In the product the CCl3 was replaced by an OH group.  If the CCl3 leaves how does the OH group attach?  Wouldn't it substitute somehow?

[Dan]

I disagree with that elimination. I think the carbon in the CCl₃ must be incorporated in the product.

You don't need a workup to protonate the N^- in the first step - you protonated the phenoxide catalyst in the first step (to form a phenol). Once you form the N^-, you can deprotonate the phenol, regenerating your phenoxide catalyst and forming an NH.

Hint for the OP: That borohydride reagent is a common alternative to sodium cyanoborohydride.

[pocopo]

Hi Dan,
I had spoken to my professor and was told that it was correct that CCl3 was not a leaving group.  I was also told that what I had said about the N- attacking the double bonded O was incorrect.  Since that large base is a catalyst it must be regenerated.  Therefore it comes in and attacks the N-, which re-protonates the base (reforming the base).  Then from working on the problem with my prof I had found that LiAl(O+Bu)3H comes in and reduces the ketone.  He had told me to think about the O- bonded to the CCl3 and what might happen.  I don't see how a -OH can be subbed in for the CCl. : /

[Dan]

Then from working on the problem with my prof I had found that LiAl(O+Bu)3H comes in and reduces the ketone.

Ok, I think I know what is happening now. I initially suspected reductive amination in the reducing step, but was not happy about the ring strain in the imine intermediate. Now I see an alternative that is more likely.

Hint: KOH will deprotonate the alcohol; Cl^- is a good leaving group.