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Topic: How can Br- deprotonate the intermediate in SN1 since it's such a weak base?
Emily_6585 Posted: Tuesday, August 17, 2021 7:41:16 PM
I'm just reviewing OChem 4 where we talked about Sn1 reactions occurring in 3 steps when we have a neutral nucleophile.

In the example Br was the leaving group, producing Br- which subsequently deprotonated the R-OH2+ (water was the electrophile) to produce a neutral product.

I don't understand how this could happen since Br- (as the conjugate base of a strong acid) is such a weak base.

Would it be more plausible to say that the water solvent would deprotonate the intermediate?

What if the solvent/neutral electrophile was an alcohol? Could the alcohol act as a base and deprotonate the intermediate?

Thanks!
Topic: Is "ether" the same thing as "diethyl ether"?
Emily_6585 Posted: Monday, August 09, 2021 10:03:28 PM
I understand an ether is a functional group, but "ether" seems to be used colloquially by the classroom companion to refer to a solvent (ie. the MCAT Tip on page 167 says "If you see ether underneath a reaction arrow think Sn2 because [it's] polar aprotic".

I'm confused because I thought diethyl ether was a non-polar solvent.

Does "ether" under a reaction arrow mean "any solvent with an ether functional group" or is it a short-hand way to refer to diethyl ether (and thus I'm wrong in thinking diethyl ether is non-polar).

Thank you!
Topic: Relationship between Ka and Kb of the exact same molecule?
Emily_6585 Posted: Friday, July 16, 2021 5:24:31 PM
Thanks for your answer, can I just confirm what you're saying?


Are you saying I can solve for the Ka of HPO4-- by doing 14/[Kb of HPO4--]?

I thought to use that equation I had to use the Kb of the conjugate base, not the Kb of the same molecule?


Perhaps I should have said stronger acid instead of strong. My question is referring to molecules that can act both as an acid and a base (ie. HPO4-- can act as an acid when it donates a proton to become PO4(3-) but could also act as a base by accepting a proton to become H2PO4-).
Topic: Relationship between Ka and Kb of the exact same molecule?
Emily_6585 Posted: Tuesday, July 13, 2021 10:07:18 PM
I know that the Ka (of an acid) times the Kb (of its conjugate base) is equal to Kw.

ie. (Ka of H2PO4-) x (Kb of HPO4--) = 1 x 10^-14 @ room temp


But is there a way to calculate the Ka of a molecule acting as an acid and from the Kb of that exact same molecule if it was to act as a base?

ie. The Ka for HPO4-- + H20 -> Po4(3-) + H3O+ from the Kb of HPO4-- + H2O -> H2PO4- + OH-


Additionally, would it be true to say that as a molecule's Ka increases its Kb decreases (because if something is a strong acid it's a weak base)? Or can you only say that if something's a strong acid then its conjugate base is weak base?
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