[H]Organic Chemistry

kyzers0ze

Singapore1073 Posts

October 05 2010 05:55 GMT

This is my first homework help blog.

Q: Sn2 reaction of Dibromomethylbenzene C6H5CHBr2 with NaOH yields benzaldehyde rather than dihydroxymethylbenzene C6H5CH(OH)2.Why is this so.

I was thinking after the 1st Sn2, a sort of E1 reaction would occur due to the presence of the electronegative OH causing the Br to fall off, but instead of forming an alkene, the alcohol group becomes an aldehyde.

Anyone else have any reasoning?

Speedbump

New Zealand338 Posts

October 05 2010 07:07 GMT

I believe that this would be due to the basic environment. After the Sn2 reactions are completed (forming dihydroxymethylbenzene), a third hydroxyl anion will force the E2 reaction to occur, forming benzaldehyde.

It has been quite a while since I've done this though. (2-3 years) If anyone comes up with a better explanation, I'd go with them. Hope I was some help.

bbq ftw

United States139 Posts

October 05 2010 07:12 GMT

You initially get the monohydroxyl via Sn2/Sn1, that's right. IIRC benzyl halides can go through either substitution mechanism.

So you have a benzyl carbon w/ hydroxy + bromo group on it.

deprotonate the hydroxyl (remember its a bit acidic due to electron withdrawing power of the benzene ring-- pKa ~ 15 for benzyl alcohol and pKa ~ 15.7 for water), kick out the bromine (push the electrons from the alkoxide to the benzyl carbon), voila. Conjugated system provides a bit of driving force for this reaction too.

As to why you don't get the dihydroxyl--you might have some but the equilibrium process will also take it to the aldehyde (you can deprotonate the dihydroxyl and expel a hydroxide ion, giving the aldehyde). Again, yielding a stable conjugated system.

tl;dr yes, you're right.

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