Hydrolysis of C$_6$H$_5$COO$^{18}$CH$_3$ with aqueous NaOH produces what final product.

(a) CH$_3$$^{18}$OH

(b) CH$_3$CO$^{18}$ONa

(c) CH$_3$C$^{18}$OONa

(d) CH$_3$$^{18}$ONa

(e) CH$_3$OH

The correct answer is (a), which is puzzling to me. I think it should be (d), as when the radiolabeled methoxy leaving group is pushed out by the reformation of the carbonyl, it will be negatively charged. The only cation that is floating around is Na$^+$.

Where, oh were, did this magical proton appear in basic conditions?

EDIT: Since I think my comments will get longer I'll just post my question here. I agree that the methoxy is in equilibrium with water:

CH$_3$O$^-$ + H$_2$O $\rightleftharpoons$ CH$_3$OH + OH$^-$

But the pK$_a$ of methanol is 15.54, and the pK$_a$ of water is 15.74. Thus this equilibrium will favor the methoxy/water side. Thus, the slight majority of product should be a methoxy salt.


You are surely right in that the labeled oxygen will stay with the alcohol and not with the acid. Now is the time to recall what do you know about the comparative acidity of alcohol and... wait, what does the word "aqueous" stand for, really?

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    $\begingroup$ Ah, I see. The labeled methoxy will be protonated by the water in the system. $\endgroup$ – coloratura Sep 1 '15 at 22:37
  • $\begingroup$ But, the equilibrium CH$_3$O$^-$ + H$_2$O $\rightleftharpoons$ CH$_3$OH + OH$^-$ should favor the left hand side, as methoxy is the weaker base. So, wouldn't the majority of the hydrolysis product be in methoxy form? $\endgroup$ – coloratura Sep 1 '15 at 22:43
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    $\begingroup$ You are overthinking it; your first answer was right. $\endgroup$ – Ivan Neretin Sep 1 '15 at 22:57
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    $\begingroup$ It would be more proper to take proton from transient acid. Also methoxide is stronger base. $\endgroup$ – Mithoron Sep 1 '15 at 23:25
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    $\begingroup$ The product of the hydrolysis is benzoic acid: $$\ce{C6H5COOH}$ which is promptly deprotonated. As the reaction proceeds, it gets less basic. $\endgroup$ – Ben Norris Sep 2 '15 at 2:14

To understand why the answer a is correct, we need to understand two things.

  1. As Ivan has pointed out water is a stronger acid than methanol, so a solution of sodium methoxide and water will be mostly converted into sodium hydroxide and methanol.

  2. The which of the single bonds CH3-O and C(=O)-O will break during the reaction, I am sure that the CH3-O bond will stay intact.

Lastly why do you assume that oxygen-18 is radioactive,

Oxygen has three stable isotopes, 16, 17 and 18.

Oxygen-14 has a half life of about 1 minute while oxygen-15 has a half life of 2 minutes. I think that it would be very hard to make an organic compound with a radioactive oxygen lable and then to do a chemistry experiment on it. I think that making an ester and doing the experiment within 20 minutes of getting the oxygen-15 enriched water from a cyclotron would be the test of all tests in terms of working very quickly.

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