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How do I arrive at the α-trimethylsilyloxy ketone from an unmodified ketone using TMSCl, triethylamine and mCPBA?
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Lizzy Due?as Gaytan
How do I arrive at the α-trimethylsilyloxy ketone from an unmodified ketone using TMSCl, triethylamine and mCPBA?
In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.
In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:
γ-deprotonation to the methyl group, creating an exocyclic double bond
γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system
α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system
By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.
If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.
The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.
Scheme 1: mechanism of the Rubottom oxidation.[1]
Reference:
[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.
In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.
In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:
γ-deprotonation to the methyl group, creating an exocyclic double bond
γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system
α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system
By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.
If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.
The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.
Scheme 1: mechanism of the Rubottom oxidation.[1]
Reference:
[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.
In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.
In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:
γ-deprotonation to the methyl group, creating an exocyclic double bond
γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system
α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system
By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.
If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.
The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.
Scheme 1: mechanism of the Rubottom oxidation.[1]
Reference:
[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.
In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.
In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:
γ-deprotonation to the methyl group, creating an exocyclic double bond
γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system
α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system
By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.
If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.
The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.
Scheme 1: mechanism of the Rubottom oxidation.[1]
Reference:
[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.
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