[Rutherford]

I got three now, but as I need only short answers I thought I post them all here. The questions are:
A)Why do we make an alkali hydrolysis and then add H₃O⁺ instead of an acid hydrolysis?
B)Is it common for an ester+Grinard to react in molar ratios 1:2? On what basis does the last conversion take place (the one with NH₃)?
C)Why did a Cl that is attached to the carbonyl group leave the molecule? Why not the one attached to -CH₂?

[discodermolide]

The first reaction, using OH- to hydrolyse the lactone stops the ring from re-forming.
The methyl ester is a good leaving group, perhaps it leaves first to give the ketone which reacts with the vinyl Grignard. The reaction with ammonia is a nucleophillic displacement on one of the Br atoms, it them attacks the other one closing the ring.
That Cl is an acid chloride and many many times more reactive than an alkyl halide.

[Rutherford]

"The first reaction, using OH^- to hydrolyse the lactone stops the ring from re-forming."
Didn't know that. In organics there are to many things to be memorized !

"The methyl ester is a good leaving group, perhaps it leaves first to give the ketone which reacts with the vinyl Grignard. The reaction with ammonia is a nucleophillic displacement on one of the Br atoms, it them attacks the other one closing the ring."
Yes, the mechanism looks like that. Is it common for esters to behave this way with Grignard (so I can know for similar problems)?
For ammonia, I suppose that it is S_N2 then.

"That Cl is an acid chloride and many many times more reactive than an alkyl halide."
Why? Shouldn't it harder form Cl- ion than the other Cl atom because of the inductive effect?

[discodermolide]

Esters react with Grignards to give initially ketones.
As I said it is an acid chloride. If you want to think of electronic effects, the carbonyl group makes the C-Cl bond quite weak and hence reactive. It is a better leaving group than the alkyl halide. Look up the reactions of acid chlorides.

[Sircodekill]

"The methyl ester is a good leaving group, perhaps it leaves first to give the ketone which reacts with the vinyl Grignard. The reaction with ammonia is a nucleophillic displacement on one of the Br atoms, it them attacks the other one closing the ring."
Yes, the mechanism looks like that. Is it common for esters to behave this way with Grignard (so I can know for similar problems)?
For ammonia, I suppose that it is S_N2 then.

Grignard reactive reduces your functional group. You have an ester that is reduced to alcohol. The methoxy leaves to form methanol. I suppose you do it at low pH (not indicated).

"That Cl is an acid chloride and many many times more reactive than an alkyl halide."
Why? Shouldn't it harder form Cl- ion than the other Cl atom because of the inductive effect?

Well you can take that bond COCl as somekind ionic. Acid chloride is used with this objective, to easily free the carbonyl which is a great electrophile

[Rutherford]

So for B) first I get vinyl-4-chlorophenyl-ketone and then if Grignard is in excess I get the other product. I will have to study the mechanism now.

I oversaw the electronegativity of chlorine. As it is little lower than the one of oxygen and the oxygen is more distant, the bonding pair of electrons of CO-Cl won't be so much attracted to the oxygen.

I am very thankful to both of you.

[discodermolide]

The mechanism of the Grignard reaction is a simple nucleophillic attack on the ester carbonyl. No mystery there.

[NL-bucky]

So for B) first I get vinyl-4-chlorophenyl-ketone and then if Grignard is in excess I get the other product. I will have to study the mechanism now.

No, actually you will only find the divinylated alcohol and the starting material. Can you think of why this is?

[Rutherford]

No. Could you explain what you mean?

The mechanism of the Grignard reaction is a simple nucleophillic attack on the ester carbonyl. No mystery there.

This one seems more complicated, as a hemiketal is made in a step. Then the metoxy group leaves the molecule.

[NL-bucky]

Let's imagine the following thought experiment. We take 2 molecules of ester, and 2 molecules of Grignard. The first Grignard attacks at the ester, leaving us with 1 ester molecule, 1 ketone molecule and 1 Grignard. The second Grignard will now attack the best electrophile, which is the ketone. Thus we end up with 1 ester molecule and 1 alcohol. A classic case of: "the product is more reactive than the starting material"

Of course, in reality the reaction is pretty irreversible so concentrations come into play, but you will not find the statistical mixture you might expect.

A similar phenomenon arises in Friedel-Crafts alkylations, which is why Friedel-Crafts acylations are preferred in syntheses.

[discodermolide]

In the Grignard case there are no reagent equivalents quoted. I assume the Grignard is in excess. Therefore the first attack gives the ketone by attack at the ester carbonyl and leaving of methoxide. The Grignard then attacks the ketone to give the alcohol.

[Rutherford]

I see why it is a better electrophile. So the ketone can be called an intermediate whose concentration changes rapidly?

There is a mechanism in the answer, but I think that it is misleading as the arrows aren't drawn properly.

[discodermolide]

The ketone is an intermediate.
The mechanism is simple attack of the C=C^- at the eater with elimination of methoxy, then a second attack of C=C^- at the ketone carbonyl to give the alcohol.

[Rutherford]

I see. I should study those mechanism more deeply because it is easier to guess the products if you practice them a lot. One last question about this reaction: why is the methoxy group leaving so easily?

[discodermolide]

Because it is a good leaving group in the case of methyl esters. It is reactive towards most nucleophiles.
There is not much to study herein terms of mechanism. Just a point of recognizing what is going on.