FW: MT 2 Practice Exam V3, #28

[Zellmer, Robert]

I received question from someone about #28 on Practice Exam V3.
The answer given is correct.  The explanation is not.  The explanation given could
apply to either of the hydrogens marked as I or II.  So what is the correct explanation?

The H atom marked II is more acidic because of the F atom (an electron withdrawing
atom) on the alpha carbon atom to the left of the C=O carbon atom with the H atom
marked II.  That's the C atom with the H atom labeled with Roman numeral III.  That's
the C atom highlighted with the red/purple.  The F atom on this carbon draws electron
density away from the -OH bond on the C to the right.  That makes the H atom labeled
II more acidic than the one labeled I left.

[A picture containing text, clock, watch  Description automatically generated]

The -CO2H group on the left has an alpha C atom to the right of it.  However,
that alpha carbon atom has 2 H atoms on it, no electron withdrawing groups.

Remember, electron withdrawing groups on an alpha carbon will make the H
atom of the -OH group of the -CO2H group more acidic.  The more electronegative
the electron withdrawing group the greater the affect (stronger acid).  The more
electron withdrawing groups on an alpha carbon the greater the affect (stronger
acid).

The explanation given in the solutions is true for why -OH hydrogen atom of a
-CO2H functional group is more acidic than the -OH hydrogen in alcohols.
Ethanol, CH3CH2OH, has a Ka ~ 10^-16. Acetic acid, CH3CO2H, has a Ka of
1.8 x 10^-5.  Alcohols don't behave as acids in water.

Look at my notes about Carboxylic acids.  All of this is discussed in the notes
and lectures.

Dr. Zellmer


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