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Topic: NaBH4 and H2SO4  (Read 29237 times)

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Offline Aznhmonglor

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NaBH4 and H2SO4
« on: February 25, 2009, 07:42:33 PM »


Can anybody explain this reaction to me.

I know the NaBH4 makes the double bond O's turn into OH's but what does the H2SO4 do?
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Offline macman104

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Re: NaBH4 and H2SO4
« Reply #1 on: February 25, 2009, 07:46:06 PM »
The NaBH4 does not turn ALL C=O bonds into alcohols.  It will not reduce the acid.  Can you draw the product of the first step?  And then we can talk about the H2SO4.
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Offline Aznhmonglor

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Re: NaBH4 and H2SO4
« Reply #2 on: February 25, 2009, 07:55:29 PM »


I know its going to look like this when NaBH4 is put into the reaction. The double bonds of O go away and H's take place. The blue H's where I boxed them came from the NaBH4. Correct?
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Offline macman104

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Re: NaBH4 and H2SO4
« Reply #3 on: February 25, 2009, 07:57:22 PM »
No, did you read my post?  Do you know what groups are affected by NaBH4?  Look in your book.
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Offline Aznhmonglor

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Re: NaBH4 and H2SO4
« Reply #4 on: February 25, 2009, 08:05:00 PM »
Yes I did read your post. Kind of confused but took another shoot at it. Only affects aldehydes so does this look correct?

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Offline macman104

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Re: NaBH4 and H2SO4
« Reply #5 on: February 25, 2009, 08:06:42 PM »
Yes.  Now, what do you know about sulfuric acid.  What reactions do you know about that involve a strong acid?  Do you see any that apply here?
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Offline Aznhmonglor

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Re: NaBH4 and H2SO4
« Reply #6 on: February 25, 2009, 08:14:41 PM »
I know that H2SO4 can turn single bond O's back into double bonds. Not sure? was that even a valid sentence? lol


Looking at it now, it seems that it can form a ring with the loss of water??
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Offline expert

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Re: NaBH4 and H2SO4
« Reply #7 on: February 26, 2009, 12:34:19 AM »
Aznhmonglor, actually you maybe right on the reduction of both ketone and COOH functions. If it was only NaBH4, then only Ketone would be affected. Strong acid turns NaBH4 into borane - extremely strong reducing agent. So, your assumption on reduction of both functions is correct.
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Offline azmanam

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Re: NaBH4 and H2SO4
« Reply #8 on: February 26, 2009, 06:56:29 AM »
Quote
Strong acid turns NaBH4 into borane - extremely strong reducing agent. So, your assumption on reduction of both functions is correct.

the acid and borohydride are not mixed in the same reaction.  the borohydride reduction occurs, the product is isolated.  then acid is added and a second reaction occurs.  borane is not generated under the reaction conditions.

see here for more information on why borohydride does not reduce carboxylic acids.

http://www.chemicalforums.com/index.php?topic=27725.msg105062#msg105062
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Offline gericox

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Re: NaBH4 and H2SO4
« Reply #9 on: February 26, 2009, 11:20:10 AM »
It should give a lactone (cyclic ester) if H2SO4 is used at catalytic amount.
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Offline macman104

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Re: NaBH4 and H2SO4
« Reply #10 on: February 26, 2009, 11:59:01 AM »
Quote from: gericox on February 26, 2009, 11:20:10 AM
It should give a lactone (cyclic ester) if H2SO4 is used at catalytic amount.
That is what I am considering gericox, I'm just not sure how advanced the student's class is.  Could they just want a dehydration product?
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Offline Aznhmonglor

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Re: NaBH4 and H2SO4
« Reply #11 on: February 26, 2009, 02:58:02 PM »
I guess you guys didn't read my last post lol. I said that does it form a ring with the loss of water which is what you guys are talking about now a lactone. Now what would it look like?


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Offline gericox

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Re: NaBH4 and H2SO4
« Reply #12 on: February 27, 2009, 04:26:03 AM »
Take a book and learn organic chemistry.....find yourself....it will be more valuable for you.
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Offline CopperSmurf

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Re: NaBH4 and H2SO4
« Reply #13 on: February 27, 2009, 03:20:22 PM »
I think it does form a ring. The lone pair electrons on OH attacks the C=O and makes the OH on the COOH leave (as water), but this assumes that the H2SO4 is concentrated. Just be careful on how you draw this.
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Offline samarthj

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Re: NaBH4 and H2SO4
« Reply #14 on: November 06, 2010, 06:45:44 AM »
hey guys it will form delta lactone  :)
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