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Topic: Product?  (Read 4719 times)

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Offline camptzak

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Product?
« on: August 09, 2012, 12:13:46 AM »
An aldehyde functionality reacts with a primary amine in the presence of sodium cyanide and BH3

I think that the primary amine forms an Imine by reaction with the aldehyde and releases H20

The sodium cyanide could then attack the BH3 and create a hydride nucleophile, which would in turn attack the lumo of the Imine.

The nitrogen would have the negative charge but would be stabilized by the BH2CN.
quench with HCl and you will have NaCl, BH2CN, and the reductive amination is complete.
is this correct?

Is this correct?
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Offline PhDoc

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Re: Product?
« Reply #1 on: August 09, 2012, 12:20:57 AM »
You're really close. The NaCN (Lewis base) and BH3 (Lewis acid) form NaCNBH3, which is a reducing agent of choice for protonated imines. You need a proton source in your "soup" to produce a solution of pH 3.

What happens if you take your aldehyde and amine, dissolve them in EtOH, and then add Na(OAc)3BH ?
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Offline orgopete

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Re: Product?
« Reply #2 on: August 09, 2012, 12:39:07 AM »
Sort of, but the nitrogen will not have a negative charge.

Let's back up a minute. The electrons of borohydride are very easily protonated. If CN replaces one of the hydrides, the electrons are less easily protonated, that is, cyanoborohydride is now more acid stable. If an amine and an aldehyde react, an imine does form. If the pH is low enough, it will become protonated. What might happen next?
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Offline camptzak

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Re: Product?
« Reply #3 on: August 09, 2012, 12:40:04 AM »
You're really close. The NaCN (Lewis base) and BH3 (Lewis acid) form NaCNBH3, which is a reducing agent of choice for protonated imines. You need a proton source in your "soup" to produce a solution of pH 3.

What happens if you take your aldehyde and amine, dissolve them in EtOH, and then add Na(OAc)3BH ?

I think the aldehyde and amine would react to form an Imine. The Na(OAc)3BH would generate a hydride nucleophile and attack the Imine.

The negatively charged nitrogen would remove a proton from the EtOH. Reducing the amine completley.

The EtO- would stabilize on the B(OAc)3

is that right?
"Chance favors the prepared mind"
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Offline camptzak

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Re: Product?
« Reply #4 on: August 09, 2012, 12:45:00 AM »
Sort of, but the nitrogen will not have a negative charge.

Let's back up a minute. The electrons of borohydride are very easily protonated. If CN replaces one of the hydrides, the electrons are less easily protonated, that is, cyanoborohydride is now more acid stable. If an amine and an aldehyde react, an imine does form. If the pH is low enough, it will become protonated. What might happen next?

If the Imine becomes protonated it will be very susceptible to attacks from nucleophiles to form the amine. Maybe another borohydride will attack to form the amine?

And you are saying that the cyano groups prevent a proton from snatching the hydride and creating hydrogen gas? that makes sense
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Offline orgopete

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Reductive amination reaction with cyano or triacetoxyborohydride
« Reply #5 on: August 09, 2012, 12:51:22 AM »

I think the aldehyde and amine would react to form an Imine. The Na(OAc)3BH would generate a hydride nucleophile and attack the Imine.

The negatively charged nitrogen would remove a proton from the EtOH. Reducing the amine completley.

The EtO- would stabilize on the B(OAc)3

is that right?

No, if the pH is not low enough, the borohydride competes with the amine to reduce the aldehyde and you get a lot of alcohol as well. You need to reduce the pH. If you do so, you will protonate the imine. It will now be more reactive. All you need is a reducing agent that can survive low pH conditions. Remember, you are trying to get the carbon to accept the electrons of boron, but a proton can do the same thing. You need a less reactive or more acid stable borohydride.
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Offline camptzak

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Re: Reductive amination reaction with cyano or triacetoxyborohydride
« Reply #6 on: August 09, 2012, 12:59:29 AM »

I think the aldehyde and amine would react to form an Imine. The Na(OAc)3BH would generate a hydride nucleophile and attack the Imine.

The negatively charged nitrogen would remove a proton from the EtOH. Reducing the amine completley.

The EtO- would stabilize on the B(OAc)3

is that right?

No, if the pH is not low enough, the borohydride competes with the amine to reduce the aldehyde and you get a lot of alcohol as well. You need to reduce the pH. If you do so, you will protonate the imine. It will now be more reactive. All you need is a reducing agent that can survive low pH conditions. Remember, you are trying to get the carbon to accept the electrons of boron, but a proton can do the same thing. You need a less reactive or more acid stable borohydride.

If you have a low pH how would that prevent formation of the alcohol? could the hydride still attack the aldehyde in the presence of acid? or is the oxygen not basic enough to be protonated?
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Offline orgopete

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Re: Product?
« Reply #7 on: August 09, 2012, 01:46:59 PM »
Okay, it been awhile. This was originally published by Borch in the 1970s. As I recall, at lower pH, the reduction of ketones and aldehydes slowed. If an imine formed it too was not reduced as one might expect. It should be easier for a carbonyl group to be reduced by dumping electrons onto the oxygen than onto a nitrogen. However, if an imine became protonated, the iminium salt would reduce faster. Therefore, this combination was effective for reductive amination as it selectively reduced iminium salts in the presence of aldehydes and ketones.

See http://www.springerlink.com/content/j73w3197510t7710/
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