Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: geoffn09 on July 09, 2008, 10:49:13 PM
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I believe the equation is balanced in a 1:1 mol ratio. I took the 0.1 g of benzil and converted it to mols of benzil. I then took that number and multiplied it times the g/mol amount for NaBH4 to get the mass needed.
Is that at all correct? I've been stuck on this for too long..
Thanks,
Geoff
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Sounds OK, if you will show the numbers it will be even possible to check.
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I disagree. If you consider the mechanism of the reduction you'll see that you only require one equivalent of hydride for this reduction. NaBH4 has four potential hydrides, so technically you require 0.25 equivalents of NaBH4, which is one equivalent of hydride.
In practice however, an excess is often required as the reduction is usually carried out in water, methanol or ethanol - these solvents also react with NaBH4 so a certain percentage of the hydride is destroyed.
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Thank you for your clarification. I was referring just to the approach to the calculations, ignoring specific reaction and its stoichiometry.
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The mechanism in the lab shows that the hydride creates two enantiomers, (+) and (-) benzoin in a racemic mixture. So would that mean it requires 0.5 of the NaBH4?
If thats wrong and it is 0.25 hydride would I just multiply 0.25 by the mol value of the 0.1 g of benzil?
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The mechanism in the lab shows that the hydride creates two enantiomers, (+) and (-) benzoin in a racemic mixture. So would that mean it requires 0.5 of the NaBH4?
I don't understand how you have come to this conclusion...
To clarify: One "hydride ion" reacts with one molecule of benzil to produce one molecule of benzol. The molecule of benzol formrd could be the (+) enantiomer or the (-) enantiomer - there is a 50/50 chance in this case.
So we want each molecule of benzil to react with one "hydride ion". Sodium borohydride therefore can theoretically react in this way four times (it is NaBH4, so for each molecule (or mole) of benzil we require 0.25 molecules (or moles) of sodium borohydride to carry out this reaction. Does that make sense?
If thats wrong and it is 0.25 hydride would I just multiply 0.25 by the mol value of the 0.1 g of benzil?
yes
Just out of interest, which solvent are you using?
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Ethyl acetate
thanks for all the help
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So we want each molecule of benzil to react with one "hydride ion". Sodium borohydride therefore can theoretically react in this way four times (it is NaBH4, so for each molecule (or mole) of benzil we require 0.25 molecules (or moles) of sodium borohydride to carry out this reaction. Does that make sense?
It doesn't work quite that straightforward. March gives mechanism for the lithium and boron hydrides in which there's no actual hydride H- moiety that reacts but rather AlH4- or BH4- complexes, respectively. Further, it's questioned whether the once reacted complex can reduce another carbonyl.
March 6th edition, p. 1804-1805
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It doesn't work quite that straightforward. March gives mechanism for the lithium and boron hydrides in which there's no actual hydride H- moiety that reacts but rather AlH4- or BH4- complexes, respectively. Further, it's questioned whether the once reacted complex can reduce another carbonyl.
March 6th edition, p. 1804-1805
I realise this, I put "hydride ion" in inverted commas to show I didn't actually think it was real, but it was sufficient for ths discussion.
As for further reduction by a partially reacted complex, that's interesting. I don't have March's 6th Ed, but I'll have a look in my 5th Ed later, does March give a reference?. From personal experience there is certainly evidence that borohydride can reduce at least more than once - I have reduced alsdoses using 0.5 eq NaBH4 in methanol and obtained the corresponding polyols in >90% yeild. In our lab we regularly reduce lactones to diols using less than 1 eq LiBH4 in essentially quantitative yeild.
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Hi Dan, here are March's references with regards of the mechanism&species:
LiAlH4
Haubenstock, H.; Eliel, E.L. J. Am. Chem. Soc. 1962, 84, 2363;
Malmvik, A.; Obenius, U.; Henriksson, U. J. Chem. Soc. Perkin Trans. 2 1986, 1899, 1905.
NaBH4
Malmvik, A.; Obenius, U.; Henriksson, U. J. Org. Chem. 1988, 53, 221.
Have a nice weekend!
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NaBH4
Malmvik, A.; Obenius, U.; Henriksson, U. J. Org. Chem. 1988, 53, 221.
Have a nice weekend!
Thanks, that was an interesting read. I'm not sure this is directly applicable to reductions, as it is a study of the acid-base reaction of sodium borohydride with carboxylic acids. The finding that intermediate borohydride complexes are less basic than borohydride itsself is fair enough - and the fact that the complex of the type (RCO2)3BH- is not basic at RT is not overly suprising - however I am not convinced that this means complexes of the (RCO2)3BH- type are not nucleophilic hydride donors.
Na(AcO)3BH is a commercially available (mild) reducing agent, and is indeed stable in an acidic medium, but certainly has the potential to reduce aldehydes and ketones, I was playing with it a few months back and remembered this paper:
Tet lett, 1983, 24(40), 4287 (http://www.sciencedirect.com/science?_ob=ArticleListURL&_method=list&_ArticleListID=765661441&_sort=d&view=c&_acct=C000010360&_version=1&_urlVersion=0&_userid=126524&md5=2b514992a5bcef74fb1b7717109155fb)
Aside from that, the main focus of the paper on the borohydride complexes is the disproportionation of said complexes - this is an interesting observation and a nice little nugget of information I can impress the ladies with later, but the timescale of this reaction is pretty far outside the usual frame of a borohydride reduction, and in any case it would not affect the ultimate question of the thread (provided we are happy with (RO)3BH- being a reducing agent, incidentally Na(MeO)3BH is another commercially available reducing agent). Anyway, I'm going to leave it there for now before I take it too far, nice paper, thanks and have a good one yourself.