# Reaction of acyl chlorides with diazomethane and silver(I) oxide

$$\ce{R-COOH ->[SOCl2,CH2N2][Ag2O/CH3OH] X}$$

While solving Carboxylic derivatives recently, I came across this. so following the reagents one by one-

$$\ce{SOCl2}$$ will form acid chloride.

$$\ce{CH2N2}$$ (diazomethane) on the other hand reacts with carboxylic acid to give a methyl ester. but we are currently on acid chloride, this part is a little confusing to me.

$$\ce{Ag2O}$$ in aqueous/alcoholic medium should give ($$\ce{CH3O-}$$) ions.

How to proceed? What is wrong with my reaction mechanism?

Your first reaction is correct. You do get an acid chloride. The reaction would proceed with the first step being what you have proposed:

$$\ce{R-COOH ->[SOCl2] R-COCl}$$

Now, the next step is a reaction known as Arndt-Eistert Reaction:

$$\ce{R-COCl ->[CH2N2][Ag2O/CH3OH] R-CH2-COOCH3}$$

The mechanism for this reaction is as follows:

Note: Mechanism is from Organic-Chemistry.org

• What this mechanism, which is correct, does not tell you is that a minimum of two equivalents of CH2N2 is required and the acyl chloride should be added -- slowly --so that the second equivalent of CH2N2 consumes HCl as CH3Cl and N2 at the second step. Inverse addition, allows for chloride formed in structure 2 to displace N2 and form the alpha-chloroketone. Jul 27 '20 at 22:07
• Is the slow addition to prevent the formation of the alpha chloroketone and allow the reaction of Cl- with CH2N2. just want to make sure if I've got it right.. Jul 28 '20 at 7:09
• The slow addition is to avoid a rush of N2 gas and to insure an excess of CH2N2 is present. More than 2 equivalents is advisable to insure an excess of CH2N2 at the end of the addition. The excess CH2N2 is what deprotonates your second structure. If it weren't present, then the chloride can displace N2 in the second structure by an SN2 mechanism. From a practical standpoint, I would guess that you have never prepared diazomethane. Its dangerous stuff! So the less it is handled the better. Invariably, one would add the acyl halide to the CH2N2. Jul 28 '20 at 11:24
• (continued) The excess CH2N2 essentially scavenges the HCl produced in the reaction. Diazomethane is a stronger base than the diazoketone so it, CH2N2, gets the proton. I know that your structures are from your link but numbered structures are a big help in discussions. – user55119 5 mins ago Jul 28 '20 at 11:38