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

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?

## 1 Answer

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):

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

• OP has used oven-dried glassware and I assumed OP did the reaction under $\ce{N2}$ atmosphere. Hence, I think $\ce{HCl}$ was released by the reaction between $\ce{Me3SiCl}$ and $\ce{Si-OH}$ residue on glassware surface. Would you agree? – Mathew Mahindaratne Apr 26 '19 at 22:23
• OP did not specify the reaction run under N2. I'm more inclined to think that the addition or possibly the weighing out of the TMSCl may have been done in air. OP comment? – Waylander Apr 26 '19 at 22:54
• @Waylander I did it under N2. However, I still could see some white fume. Maybe because it's not possible to completely eliminate water in the air? – chemrese May 2 '19 at 12:36
• possibly you are getting the TMSiCl reacting with Et3N vapour over the reaction mixture as you have excess Et3N – Waylander May 2 '19 at 12:48
• I carefully conducted every step under nitrogen but the reaction ultimately failed (twice already) so I am now doing with LiHMDS... – chemrese May 2 '19 at 15:02