[o0CY0o]

1.alkoxide is a stronger base than hydroxide
But how come cyclic hemiacetal is base catalyzed?

2. On my notes, there is a statement that acid chloride + grinard reagent will reduce to alcohol. But acid chloride is acidic, why it won't react with grinard reagent?

Thanks

[sjb]

1) do you mean specific base catalysis (i.e OH^-), or just generic base (B^-)?

2) Some organometallic compounds will react with acid chlorides to give ketones, which you can then reduce to the alcohol. Without knowing what you were actually taught, I'm not sure if that's what you mean, or not.

[Arctic-Nation]

While I'm unsure what is meant with the first question, the second question is faulty in that Grignard reagents don't reduce carbonyl compounds in a strict sense of the word.

I've attached an image showing what happens when a Grignard reagent reacts with an acid chloride. The most important to realize is that the first intermediary (the second structure) is not stable under the reaction conditions: it quickly collapses into the ketone by losing chloride, and a ketone, as you should know, is (also) very reactive towards strong nucleophiles such as Grignard reagents. This results in the addition of a second equivalent of Grignard reagent and the formation of an alcohol.

Also, while acid chlorides have acidic alpha hydrogens (for example, in the general molecule R-CH₂-COCl, the CH₂ hydrogens are acidic), this acidity is completely superseded by the extreme electrophilicity of the carbonyl moiety. This means (in this case) that even though Grignard reagents are very strong bases, they prefer to act as nucleophiles towards acid chlorides.
In fact, acid chlorides prefer to react as electrophiles instead of acids so much, that it takes strong non-nucleophilic bases (triethylamine is one) to deprotonate them (which results in the formation of ketenes, by loss of chloride).

[BeepoGirl]

Just wondered if this applies to all carbonyl compounds, particularly esters?

[Arctic-Nation]

The general rule is that when you have a good leaving group in the tetrahedral intermediate, this will collapse into the ketone and result in double alkylation. Acid chlorides, anhydrides and esters all have good leaving groups and will react further with a Grignard reagent. Amides usually do not, as the amide anion is a particularly bad leaving group, even under the circumstances of this reaction, but there's a reason why the use of more stable Weinreb amides is so popular.

Then there's the secondary carbonyl compound (formed by collapse of the intermediate) to consider. Usually this compound is a ketone. As ketones are less reactive than acid chlorides and anhydrides, with a careful reaction setup and choice of reagents you might be able to avoid the formation of too much alcohol. When working with esters, however, the ketone is far more reactive and will thus react much faster than the parent ester, resulting in the exclusive formation of alcohol.