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Offline 113zami

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equilibrium shifts
« on: July 30, 2008, 06:21:57 PM »
Which statement CORRECTLY explains the following acid-base reaction at equilibrium?

CH3H2C:- + HOC(CH3)3 <---> CH3CH3 + -:Ö:C(CH3)3


A. Weaker acid, HOC(CH3)3, and weaker base, CH3H2C:-, will be present in abundance.
B. Stronger acid, HOC(CH3)3, and stronger base, CH3H2C:-, will be present in abundance.
C. Weaker acid, CH3CH3, and stronger base, -:Ö:C(CH3)3, will be present in abundance.
D. Stronger acid, CH3CH3, and weaker base, -:Ö:C(CH3)3, will be present in abundance.
E. Weaker acid, CH3CH3, and weaker base, -:Ö:C(CH3)3, will be present in abundance

explanation:
Correct Answer: E
By nature, weak acids and weak bases are weakly dissociated to their respective conjugate bases and conjugate acids. Equilibrium position staying far right will only begin shifting to the left when stronger acids or stronger bases are added to the solution to selectively deplete CH3H2C:- or HOC(CH3)3 respectively. {Note: Reaction in the question involves an organometallic compound (Grignard Reagent); and it generally makes sense to consider the strong base, -:Ö:C(CH3)3, a weaker one when compared to CH3H2C:-}


I tried to understand the blue part but I just don't get it, so equilibrium shifts from right to left when you add strong acid, but if you're adding them to CH3H2C:- or HOC(CH3)3 to deplete them then the equilibrium should shift to the right (not left as they say) , am I correct??

please help

Offline azmanam

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Re: equilibrium shifts
« Reply #1 on: July 31, 2008, 11:34:22 AM »
I think deplete should be Replete (or replenish).  Perhaps a typo? 

But you can't add BOTH the stronger acid and stronger base to move the equlibrium to the left.  Adding an acid will reprotonate the alkoxide to regenerate the alcohol (but will not regenerate the Grignard reagent).  Adding a really strong base might deprotonate the alkane, but will not regenerate the alcohol.

Basically, this is another wording of the mnemonic 'Weaker Acid Wins.'  Look on both sides of the equilibrium arrow for the 2 acids in the reaction.  They are HOC(CH3)3 and CH3CH3.  The pka of the alcohol is around 15-18 and the pka of the alkane is around 40-45.  The side of the equilibrium with the weaker acid (the higher pka) will be the favored side of the equilibrium.  Thus the side with the alkane will be favored here.

If you disrupt the equilibrium by adding a different acid or base, the position of the equilibrium might change.  (actually, that's a poor description.  When you add a new reagent, you'll have to write out a NEW equilibrium reaction).  If you add a stronger acid than the protonated alcohol on the left, the new equilibrium is:

ROH   +   A-    --><--   HA   + RO-

(where HA is the strong acid you added, and A- is its conjugate base).  Now the weaker acid is on the left, and the equilibrium will adjust as such.  Note the Grignard is not pictured in this new equilibrium.  It will not participate here and will remain as the alkane.

You can do the same analysis if you add a strong (really strong) base.
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Offline 113zami

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Re: equilibrium shifts
« Reply #2 on: August 09, 2008, 08:31:13 AM »

If you disrupt the equilibrium by adding a different acid or base, the position of the equilibrium might change.  (actually, that's a poor description.  When you add a new reagent, you'll have to write out a NEW equilibrium reaction).  If you add a stronger acid than the protonated alcohol  on the left , the new equilibrium is:

ROH   +   A-    --><--   HA   + RO-

(where HA is the strong acid you added, and A- is its conjugate base).


if HA is the strong acid you're adding and you're adding it to the ROH on the left, shouldn't the A- in your equation be HA? then you'll have RO- and H2O in the products??

Offline azmanam

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Re: equilibrium shifts
« Reply #3 on: August 09, 2008, 06:48:22 PM »
For acid/base, it doesn't really matter which side has which acid since it's an equilibrium.  I put HA on the right in order to keep ROH on the left, as it is in your original example.  My point was that if you add a new reagent, you establish a new equilibrium.  Since the new equilibrium still involves the alcohol, I kept the alcohol on the same side of the arrow for reference.

I didn't say I added a stronger acid to the alcohol, I said I added a stronger acid than the alcohol, and it was added to the already established equilibrium, which at that point contained the alkane and the alkoxide.
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