[Rutherford]

I have to determine from the picture what is the correct mechanism for the decomposition of methyl-ester if the following facts are known:
1)rate of the reaction depends on the concentration of both the ester and I^-;
2)the corresponding ethyl-ester decomposes 10 times slower that the methyl-ester
3)the rate depends on the slowest step of the mechanism
There were more options but I left with those two in the picture, don't know hot to choose from them what is the correct one, by the facts that are given.

[fledarmus]

Make two models, the methyl ester and the ethyl ester, and then answer these questions:

1) if the rate depends on both the concentration of the iodine and the concentration of the ester, and the rate depends on the slowest step of the reaction, which step in these two mechanisms is rate determining?

2) For your two rate determining steps, move your iodine ion into your model in the direction which it would need to move. You know the ethyl ester reacts 10x slower than the methyl ester -  In which molecule would this more likely be the case?

To answer question 2, it might help to actually assign a name to the type of reaction that is occurring, and to the type of group it is occurring on.

[Rutherford]

In mechanism A, it is the second step, in mechanism B it is the first step.
The directions of I^- movements are assigned.
"You know the ethyl ester reacts 10x slower than the methyl ester -  In which molecule would this more likely be the case"- I don't understand this, how the other molecule can help here, could you explain it more clearly?

[fledarmus]

What is the mechanism of the second step in mechanism A? This is a very well known reaction, probably one of the first you learned in Org 1.

Did you make the models I suggested? Did you try moving the I^- ion the way it would need to move to react, and see whether the two models were different in any respect?

[Rutherford]

It is nucleophilic substitution. For the 1st mechanism, I^- can easier attack the partial positive C atom (positive because O pushes the electrons towards itself) in methyl ester than in ethyl ester, because in ethyl ester the C atom is less positive (the methyl group pushes its electrons toward the C atom in CH₂), so the first mechanism is much slower for ethyl-ester than methyl-ester, while for the second mechanism there wouldn't be any big difference in the rate. The answer is A (which is correct). Correct me if I am somewhere wrong in my conclusions.

[fledarmus]

I agree with your answer, but not your argument. Look closely at mechanism A - this mechanism involves an S_N2 substitution reaction, using I^- as the nucleophile and the carboxylate anion as the leaving group. If it was merely a question of stabilizing a positive charge, the primary carbon of the methyl ester would be less stable than the secondary carbon of an ethyl ester. Stabilizing a positive charge would argue an S_N1 mechanism, and ethyl cations being more stable than methyl cations, you would expect the ethyl group to react faster, not 10x slower. However, sterically, an S_N2 reaction is much faster at a primary center than at a secondary center. The presence of an extra methyl group partially blocks the backside attack required by an S_N2 reaction , and I^- is a very large atom to try to push in.

The reaction at the carbonyl center in mechanism B is an addition reaction, and sterically would really only be affected by atoms directly attached to the carbonyl carbon, unless there was some sort of ring system to pull groups around. The atoms attached to the carbonyl are identical for both the ethyl and the methyl ester (they both have a secondary carbon on one side and an oxygen on the other) so you would expect very little difference in reaction rate from changing a methyl ester to an ethyl ester - the change is too far away from the reactive center.

[Rutherford]

Okay, so it is about space availability rather than the charge. Thanks for the help.