Any chemists here

TiAg335i · 11-28-2008, 05:41 PM

I know this is gen. chem stuff but I need to know:

Please balance the following:

4 cyclohexanone + sodium borohydride + 4CH3OH --> 4 cyclohexanol

The product end of the reaction seems to be missing reagents, otherwise this is impossible to balance. Anybody know the missing reagents?

Schmed · 11-28-2008, 06:59 PM

You are obviously missing Na and B in the product side.

mpimping · 11-28-2008, 07:09 PM

Has anybody noticed how bad some of them have been lately?

Kev · 11-28-2008, 07:41 PM

Quote:

Originally Posted by TiAg335i

I know this is gen. chem stuff but I need to know:

Please balance the following:

4 cyclohexanone + sodium borohydride + 4CH3OH --> 4 cyclohexanol

The product end of the reaction seems to be missing reagents, otherwise this is impossible to balance. Anybody know the missing reagents?

Hmm........ I haven't done O-Chem for a long long time, but here's my take. Sodium Borotetrahydride is the H donor in this case and the methanol acts as a nucleophilic catalyst (it will be regenerated at the end of the reaction).

4 cyclohexanone + 4NaBH4 -- CH3OH / CH3O- --> 4 cyclohexanol + 4Na+ + 4BH3

or

4 cyclohexanone + 4NaBH4 + 4CH3OH --> 4 cyclohexanol + 4Na+ + 4BH3 +4CH3OH

TiAg335i · 11-28-2008, 07:54 PM

Quote:

Originally Posted by Kev

Hmm........ I haven't done O-Chem for a long long time, but here's my take. Sodium Borotetrahydride is the H donor in this case and the methanol acts as a nucleophilic catalyst (it will be regenerated at the end of the reaction).

4 cyclohexanone + 4NaBH4 -- CH3OH / CH3O- --> 4 cyclohexanol + 4Na+ + 4BH3

or

4 cyclohexanone + 4NaBH4 + 4CH3OH --> 4 cyclohexanol + 4Na+ + 4BH3 +4CH3OH

Hmm..thanks but I am not sure if that's correct. Sodium Borohydride is able to reduce 4 moles of cyclohexanone, and thus that ratio cannot possible be right. I don't know if a car forum is the right place to look for this information, haha. But I appreciate your help.

Mods feel free to delete/close this thread.

Kev · 11-28-2008, 08:08 PM

Quote:

Originally Posted by TiAg335i

Hmm..thanks but I am not sure if that's correct. Sodium Borohydride is able to reduce 4 moles of cyclohexanone, and thus that ratio cannot possible be right. I don't know if a car forum is the right place to look for this information, haha. But I appreciate your help.

Mods feel free to delete/close this thread.

I have not touched O-Chem since MCAT, which was at least 5 years ago, so I'm a little rusty with it. I looked it up again and indeed 1 NaBH4 can reduce 4 cyclohexanone, so the correct formula would be:

4 cyclohexanone + NaBH4 + 4CH3OH --> 4 cyclohexanol + 4 NaCH3O + B

vachss · 11-28-2008, 08:46 PM

Quote:

Originally Posted by Kev

I have not touched O-Chem since MCAT, which was at least 5 years ago, so I'm a little rusty with it. I looked it up again and indeed 1 NaBH4 can reduce 4 cyclohexanone, so the correct formula would be:

4 cyclohexanone + NaBH4 + 4CH3OH --> 4 cyclohexanol + 4 NaCH3O + B

Umm... where'd the extra 3 Na's come from on the product side? You're right that the Methanol and the NaBH4 each donate 4 hydrogens to make up the 8 required to transform 4 Cyclohexanone to 4 Cyclohexanol, but I don't know what becomes of the NaB(4+) and the 4 CH3O(-). Somehow I don't think that NaB(OCH3)4 is a stable product (though it does balance the equation if used on the RHS).

Sadly I last took OChem during the Carter administration I'm reduced to Googling for hints rather than having any really solid knowledge here.

Kev · 11-28-2008, 08:47 PM

Quote:

Originally Posted by vachss

Umm... where'd the extra 3 Na's come from on the product side? You're right that the Ethanol and the NaBH4 each donate 4 hydrogens to make up the 8 required to transform 4 Cyclohexanone to 4 Cyclohexanol, but I don't know what becomes of the NaB(4+) and the 4 CH3O(-). Somehow I don't think that NaB(OCH3)4 is a stable product (though it does balance the equation if used on the RHS).

Sadly I last took OChem during the Carter administration I'm reduced to Googling for hints rather having any really solid knowledge here.

Yeah, you are right, stupid me......

Back to the drawing board again.

I still think that NaBH4 can only donate one equivalent of hydride, compared to 4 H- in LiAlH4.

DevJ · 11-28-2008, 08:50 PM

hey dude when do you need the answer by?

Kev · 11-28-2008, 08:57 PM

Quote:

Originally Posted by vachss

Umm... where'd the extra 3 Na's come from on the product side? You're right that the Methanol and the NaBH4 each donate 4 hydrogens to make up the 8 required to transform 4 Cyclohexanone to 4 Cyclohexanol, but I don't know what becomes of the NaB(4+) and the 4 CH3O(-). Somehow I don't think that NaB(OCH3)4 is a stable product (though it does balance the equation if used on the RHS).

Sadly I last took OChem during the Carter administration I'm reduced to Googling for hints rather than having any really solid knowledge here.

I think the product that you wrote is correct. The product is sodium tetramethylborate

4 cyclohexanone + NaBH4 + 4MeOH ---> 4 cyclohexanol + NaB(OMe)4

xxForgedxx · 11-28-2008, 09:11 PM

No youre doing it wrong:

Chill Glass + 2 shots white grape juice + 2 shots Ciroc Vodka + 3 Grapes = Grape Martini

TiAg335i · 11-28-2008, 09:22 PM

Quote:

Originally Posted by Kev

Yeah, you are right, stupid me......

Back to the drawing board again.

I still think that NaBH4 can only donate one equivalent of hydride, compared to 4 H- in LiAlH4.

All 4 hydrogens on NaBH4 can EQUALLY act as nucleophiles which can attack the electrophilic carbonyl carbon on cyclohexanone. Therefore, one mole of NaBH4 can react with 4 moles of ketone. I'm going to try to work up an answer now.

TiAg335i · 11-28-2008, 09:24 PM

4 cyclohexanone + NaBH4 + 4MeOH ---> 4 cyclohexanol + 4NaB(OMe)4

Still unbalanced....

TiAg335i · 11-28-2008, 09:42 PM

This is what I propose:

4 cyclohexanone + NaBH4 + 4CH3OH --> 4 cyclohexanol + B(OCH3)4 + Na+

What say you?

Kev · 11-28-2008, 10:08 PM

Quote:

Originally Posted by TiAg335i

4 cyclohexanone + NaBH4 + 4MeOH ---> 4 cyclohexanol + 4NaB(OMe)4

Still unbalanced....

You have an extra 4 in the product side. It should read:

4 cyclohexanone + NaBH4 + 4MeOH ---> 4 cyclohexanol + NaB(OMe)4

Quote:

Originally Posted by TiAg335i

This is what I propose:

4 cyclohexanone + NaBH4 + 4CH3OH --> 4 cyclohexanol + B(OCH3)4 + Na+

What say you?

It's the same as my equation, except you neglected to put a negative charge on the tetramethylborate, so your equation is unbalanced.

TiAg335i · 11-28-2008, 10:34 PM

Ok cool. Glad we got it worked out. Thanks for your help.

Kev · 11-28-2008, 10:48 PM

Quote:

Originally Posted by TiAg335i

Ok cool. Glad we got it worked out. Thanks for your help.

No problem. I was surprised how much I actually forgot in a matter of 8 years...... I took O-Chem in Fall '99 and Spring '00.

11-28-2008, 05:41 PM	#1
TiAg335i Trance Life 333 Rep 4,706 Posts Drives: 2007 E92 335i Join Date: Jan 2007 Location: Long Island, NY iTrader: (4)	Any chemists here I know this is gen. chem stuff but I need to know: Please balance the following: 4 cyclohexanone + sodium borohydride + 4CH3OH --> 4 cyclohexanol The product end of the reaction seems to be missing reagents, otherwise this is impossible to balance. Anybody know the missing reagents?
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11-28-2008, 06:59 PM	#2
Schmed Private First Class 21 Rep 191 Posts Drives: E30M3, E70X5, F25X3 Join Date: Jan 2008 Location: Rocky Mountains iTrader: (1)	You are obviously missing Na and B in the product side. __________________ 2013 X3 3.5 \| SAP \| DHP \| NAV \| CWP \| AW \| Mojave
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11-28-2008, 07:09 PM	#3
mpimping Banned 867 Rep 336 Posts Drives: gti Join Date: Sep 2006 Location: ny,ny iTrader: (0)	TV commercials Has anybody noticed how bad some of them have been lately?
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11-28-2008, 08:50 PM	#9
DevJ Colonel 156 Rep 2,736 Posts Drives: R8 V10+/YMB M4 Join Date: Nov 2007 Location: North Texas iTrader: (1)	hey dude when do you need the answer by? __________________ ‘23 RRS First Edition '22 X7 M50i '18 R8 V10+ http://www.bimmerpost.com/forums/attachment.php?attachmentid=133578&d=1204654487[/IMG]
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11-28-2008, 09:11 PM	#11
xxForgedxx Lieutenant 620 Rep 516 Posts Drives: Slow/Fat Join Date: Aug 2008 Location: USA iTrader: (1)	No youre doing it wrong: Chill Glass + 2 shots white grape juice + 2 shots Ciroc Vodka + 3 Grapes = Grape Martini
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11-28-2008, 09:24 PM	#13
TiAg335i Trance Life 333 Rep 4,706 Posts Drives: 2007 E92 335i Join Date: Jan 2007 Location: Long Island, NY iTrader: (4)	4 cyclohexanone + NaBH4 + 4MeOH ---> 4 cyclohexanol + 4NaB(OMe)4 Still unbalanced....
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11-28-2008, 09:42 PM	#14
TiAg335i Trance Life 333 Rep 4,706 Posts Drives: 2007 E92 335i Join Date: Jan 2007 Location: Long Island, NY iTrader: (4)	This is what I propose: 4 cyclohexanone + NaBH4 + 4CH3OH --> 4 cyclohexanol + B(OCH3)4 + Na+ What say you?
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11-28-2008, 10:34 PM	#16
TiAg335i Trance Life 333 Rep 4,706 Posts Drives: 2007 E92 335i Join Date: Jan 2007 Location: Long Island, NY iTrader: (4)	Ok cool. Glad we got it worked out. Thanks for your help.
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