3
$\begingroup$

I was asked to compare the reaction between sodium hydride and water and sodium borohydride and water. I know that the former acts more violently than the latter, but apparently, I must use a curly arrow mechanism to show this.

I think that the HOMO for the latter reaction is the oxygen lone pairs, and the LUMO is the B-H bond, but I'm having difficulty moving on from there.

What attacks what first?

$\endgroup$
  • 4
    $\begingroup$ I don’t see much point in a curly-arrow mechanism for the reaction of $\ce{NaBH4}$ with water. And I don’t see the point of doing it with $\ce{NaH}$, either. In one case, you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’ $\endgroup$ – Jan Oct 2 '15 at 22:17
5
$\begingroup$

Their mechanisms are easily searchable online.

This is from 2009 from art-xy.com:

Notice that the reaction would be different in the presence of metal catalysts:

$$\ce{NaBH4 + 2H2O ->[metal catalyst] NaBO2 + 4H2}$$

This is from 2007 from chembook.co.uk:


Which reaction is more violent is already explained by Jan:

In one case [$\ce{NaH}$], you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case [$\ce{NaBH4}$] you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’

Therefore the former is more violent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.