Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Synthesisgirl on November 12, 2011, 11:43:42 PM
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Hi.
I was trying this problem I was given, but I am not sure it is the correct answer. Well honestly, I think it's wrong, but it's the only way I can think of it.
The picture is below. The series of reactions I would perform would be:
1)LiAlH(OtBu)3 ---> To turn the carbonyl into an aldehyde because if I used LiAlH4 I would over-reduce the compound into an alcohol
2) H3O+ ---> To protonate the Oxygen that has left the carbonyl
3) N(CH3)2 with NaBH3CN ----> To produce an imine then the NaBH3CN will reduce the compound into the amine.
The reason I am unsure about these series of reactions is because before I would be able to add my amine, there would be an intramolecular reaction between the alcohol and the aldehyde to go back to the original lactone as intramolecular reactions are faster than intermolecular reactions.
Also, I thought of simply just adding the N(CH3)2 with NaBH3CN as my one and only step but I don't think it works because I won't be able to form an imine, and without imine formation the NaBH3CN cannot reduce it into the amine I want.
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You're on the right track and you correctly sniffed out the dilemma with going through the aldehyde (although you wouldn't get the lactone back, it'd be a lactol and the structure would shift between open and closed forms like a sugar).
There's a more direct way to do this. Try thinking of the lactone as an ester. How would you convert an ester to an amine?
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1)LiAlH(OtBu)3 ---> To turn the carbonyl into an aldehyde because if I used LiAlH4 I would over-reduce the compound into an alcohol
Note that LiAl(OtBu)3H is a nucleophilic hydride source, consequently it will react with an aldehyde much faster than it will with a lactone (same relative chemoselectivity as LiAlH4, just weaker). At low temp the lactol will hopefully be stable enough to limit further reduction, but a much more reliable option is to use an electrophilic reducing agent. I'd go for DIBAL, which reduces esters and lactones faster than aldehydes (opposite chemoselectivity compared to LiAlH4).
If it was me, I would probably go for (i) DIBAL; (ii) HNMe2, NaBH3CN
You can do this without going through the aldehyde, it's still 2 steps though - I think that's what Doc Oc is getting at.
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Excess dimethylamine, potassium carbonate, reflux?
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Open the lactone with the dimethylamine (if necessary add a cat amount of NaCN) then reduce the amide with BH3.DMS at reflux in THF. The entire thing could probably be done in THF.
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then reduce the amide with BH3.DMS at reflux in THF.
Is this compatible with the alkene? I'd have chosen LiAlH4 to avoid potential problems with hydroboration.
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So I guess I can just add dimethylamine and then reduce it with NaBH3CN :)
I've attached my proposed mechanism to show the spot where I was a little shady before (The ring closing up again).
Thanks guys :)
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So I guess I can just add dimethylamine and then reduce it with NaBH3CN :)
I've attached my proposed mechanism to show the spot where I was a little shady before (The ring closing up again).
Thanks guys :)
I don't think the amide will reduce with cyanoborohydride. You need something stronger.
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I don't think the amide will reduce with cyanoborohydride. You need something stronger.
Well, if I use something stronger, won't the alkene be reduced as well?
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I don't think the amide will reduce with cyanoborohydride. You need something stronger.
Well, if I use something stronger, won't the alkene be reduced as well?
No
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The route you've proposed is the one I was thinking, although you would probably need an acid catalyst to get it going. Discodermolide is right, NaCNBH3 will not reduce an amide to an amine. There is a way to reduce amides to amines and it won't affect the alkene (hint: the reagent has been mentioned in this thread)
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I ended up using:
1) HN(CH3)2, LiAlH4
Was this correct. Can the dimethylamine actually open up the lactone?!
I checked the solutions my professor had made up, and his steps were:
1) LiAlH4
2) PCC
3) HN(CH3)2, NaCNBH4
But wouldn't this be impossible considering as soon as he generates the ketone with PCC there will be an intramolecular reaction with the alcohol attacking the ketone (regenerationg of lactone/lactol) and the amine won't be formed because intramolecular reactions are faster than intermolecular reactions.
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Those are the reagents I was thinking.
Your prof's answer is correct, although it's funny that someone mentioned that route here and tried to avoid it because it's a bit inelegant (at least in my opinion).
A) The product of the PCC oxidation is an aldehyde, not a ketone.
B) Lactone and lactol are not the same thing. I may be nitpicking a bit here, but one of those things will react with the amine as suggested and the other wouldn't, so nomenclature is important.
C) Remember that lactols (or any hemiacetal for that matter) can reversibly open and close. So while the alcohol would attack the aldehyde and form a closed ring structure, it very easily opens to allow the amine to react with the aldehyde.
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Those are the reagents I was thinking.
Your prof's answer is correct, although it's funny that someone mentioned that route here and tried to avoid it because it's a bit inelegant (at least in my opinion).
A) The product of the PCC oxidation is an aldehyde, not a ketone.
B) Lactone and lactol are not the same thing. I may be nitpicking a bit here, but one of those things will react with the amine as suggested and the other wouldn't, so nomenclature is important.
C) Remember that lactols (or any hemiacetal for that matter) can reversibly open and close. So while the alcohol would attack the aldehyde and form a closed ring structure, it very easily opens to allow the amine to react with the aldehyde.
Cool! Yes, very unaesthetic :P
Do you think I would be able to argue with my prof that 1) HN(CH3)2, LiAlH4 is correct if he marks it incorrect?
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1) If we're worried about the hemiacetal, why shouldn't we be as or even more worried about the hemiaminal?
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Do you think I would be able to argue with my prof that 1) HN(CH3)2, LiAlH4 is correct if he marks it incorrect?
No, it's two steps: 1) HNMe2; 2) LiAlH4
Another good alternative is: 1) DIBAL; 2) HNMe2, NaBH3CN.
In the latter approach, DIBAL achieves in one step what LiAlH4/PCC achieves in two. That will probably be easier to argue with the prof, but I'd expect either of those routes to work, and they are certainly both better than the 3 step procedure.