[rleung]

Hey again,

I have a quick question about E1 reactions.  In my book (Wade), when it lists examples, the base that transfers electrons to a H in order to deprotonate the molecule is always the base listed in the chemical equation, but in the solutions to some of the end-of-chapter problems, it sometimes uses a Br- base to deprotonate the molecule instead.  For example:

H2C-CH-CH3 ---> Br- + H2C-CH-CH3 ---> H2C=CH-CH3
        |                             |
        H                            H

In the solutions manual, the Br- that is a product of the first, slow reaction is used to deprotonate the molecule; the solvent is water, however.  How do I know that Br- will be used as the base and not the solvents, such as HOH or HOCH3-?  The book always just uses the solvent for deprotonation.

Thanks.

Ryan

[alphahydroxy]

errmm... not quite sure about your structures/reagents there...any chance you could re-draw them using isis/chemdraw? or otherwise make it clearer?

kinda looks like you're deprotonating an alkane to give an alkene... not the most favourable process i've encountered!


maybe i'm just being fick tho?

[rleung]

Hmm, I will work on illustrating it.  I actually do not have ChemDraw, but I will try to illustrate it using Paintbrush on Windows.

Ryan

[alphahydroxy]

chemdraw is only a free download away....

so is isis/draw, come to think of it

[rleung]

Alright, I finally got this worked out.  

CH3-CHBr-CH3 is reacted with water/acetone and heat.  Below is the solutional manual's answer for the E mechanism.  It is E1.  I guess my question is why wouldn't acetone deprotonate the  CH3-C(+)H-CH3 in the second step instead of Br-?  My own proposed mechanism is in the next post.  Thanks so much.

Ryan

[rleung]

Here is my own proposed mechanism.

[rleung]

Whoops, forgot to show the formation of the double bond in my proposed mechanism above.  Here it is again.

Ryan

[rleung]

Hey,

I noticed that you said ChemDraw was a free program, but I can't seem to find where you can download the full version for free.  I googled it and only found trial versions that have a time limit.  Is there a place where it is free?  Thanks.

Ryan

[movies]

The way you have that drawn in the last picture looks like an E2 mechanism to me.  Also your structure for acetone is wrong.

It'd actually more likely be water that did the deprotonation; water is a lot more basic than acetone (pK_as of about -1.74 and -7, respectively).

[lemonoman]

I always thought ChemDraw was a paid-for program...but's free to use at places like Unviersities and stuff...

[Mitch]

but's free to use at places like Unviersities and stuff...

Its not free at all.

[alphahydroxy]

it's free if you put some effort in...

[rleung]

Hmm, then how do I know Br- does the deprotonation and not water?  Is it because Br- is a better base than water is?

Ryan

[movies]

Hmm, then how do I know Br- does the deprotonation and not water?  Is it because Br- is a better base than water is?

Ryan

No, because water is a better base than Br^-, if there is water in the pot it will do the deprotonation, most likely.  Bromide would only deprotonate under dry conditions when there was no other good base around.  It looks nice to draw it that way because you end up with all neutral species, but it's probably not the case in aqueous solution because in water salts tend to be more stable as solvated ions than as complexes.

That said, it's not wholely unreasonable to draw Br^- as the base since it is a very polarizable ion.  Just remember that in reality, even if Br^- does the deprotonation, it will rapidly dissociate since HBr is a very strong acid.

[movies]

it's free if you put some effort in...

Please don't encourage software piracy on the forums.

For all interested, Cambridge Software offers a free two-week trial of ChemDraw.