Role of NaI in Formation of Silyl Enol Ethers using Chloro(trimethyl)silane

Question

I am forming silyl enol ethers using Chloro(trimethyl)silane (TMSCl), and below is an experimental protocol I am following

Silyl Enol Ether Synthesis

1. In an oven-dried 100 mL three-necked flask, Et3N (37.2 mmol) and carbonyls (30.0 mmol) were added. The mixture was stirred and heated to 30~35 ºC, and TMSCl (37.2 mmol) was added dropwise over 15 minutes. The mixture was stirred at the same temperature for 0.5 hour.
2. A solution of NaI (37.2 mmol) in acetonitrile (35 mL) was added dropwise to the mixture over 0.5 hour and stirred for 1 hour. The mixture was then heated to 40-45 ºC and stirred at this temperature for 2 hours.
3. The mixture was then cooled to room temperature and 30 mL of cold pentane was added. The mixture was poured into 30 mL of ice-cold water and extracted with pentane (15 mL × 2). The combined organic phases were dried over anhydrous Na2SO4, filtered and concentrated in vacuo. After distillation under reduced pressure, the products were obtained as colorless oils and stored at 0 ºC under an atmosphere of nitrogen for use.

I don't understand the mechanistic background of Step 2. What does NaI do?

Side question: when adding TMSCl in Step 1, I saw some white fume. I am guessing TMSCl was reacting with moisture in the air but what would be the product of H2O+ TMSCl? Is it HCl and TMS-OH?

Answer 1

The use of $\ce{NaI}$ in $\ce{MeCN}$ (a solvent in which $\ce{NaI}$ is quite soluble) is forming $\ce{Me3SiI}$ from the $\ce{Me3SiCl}$. The reason for doing this is $\ce{Me3SiI}$ is significantly more reactive with the triethylammonium enolate and will increase the yield of the TMS-enolate.

You are correct in thinking that the white fume you see when adding the $\ce{Me3SiCl}$ is $\ce{HCl}$. The product of aqueous hydrolysis of $\ce{Me3SiCl}$ is $\ce{HCl}$ plus hexamethyldisiloxane (Wikipedia):

$$\ce{2Me3SiCl + 2H2O -> 2HCl + Me3SiOSiMe3}$$