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Reductive amination of (protected) 3-Oxopropanenitrile: Will azetidin-2-ol be formed via intramolecular ring closure?
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Mohammad Sayed Immam
Reductive amination of (protected) 3-Oxopropanenitrile: Will azetidin-2-ol be formed via intramolecular ring closure?
Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.
But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:
The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.
In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):
But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.
Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.
But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:
The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.
In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):
But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.
I agree for product 1. For product 2, if you go for the intramolecular route then the product of a reductive amination should be azetidine. The 2nd step of your mechanism in the bottom picture is not correct (give it another check). However I am not convinced it will go intramolecularly because of the ring strain of the intermediate. Maybe at best you can get a mixture of the azetidine with the open chain product and possibly oligo/polymerisation products, a mess. You need to check the literature for a precedent.
In respect to the question about MeOH: Methanol should not compete with the amine, as the amine is a lot more nucleophilic. In fact MeOH is a common solvent or cosolvent for reductive aminations (or alkylations).
I agree for product 1. For product 2, if you go for the intramolecular route then the product of a reductive amination should be azetidine. The 2nd step of your mechanism in the bottom picture is not correct (give it another check). However I am not convinced it will go intramolecularly because of the ring strain of the intermediate. Maybe at best you can get a mixture of the azetidine with the open chain product and possibly oligo/polymerisation products, a mess. You need to check the literature for a precedent.
In respect to the question about MeOH: Methanol should not compete with the amine, as the amine is a lot more nucleophilic. In fact MeOH is a common solvent or cosolvent for reductive aminations (or alkylations).
Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.
But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:
The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.
In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):
But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.
Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.
But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:
The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.
In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):
But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.
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I agree for product 1. For product 2, if you go for the intramolecular route then the product of a reductive amination should be azetidine. The 2nd step of your mechanism in the bottom picture is not correct (give it another check). However I am not convinced it will go intramolecularly because of the ring strain of the intermediate. Maybe at best you can get a mixture of the azetidine with the open chain product and possibly oligo/polymerisation products, a mess. You need to check the literature for a precedent.
In respect to the question about MeOH: Methanol should not compete with the amine, as the amine is a lot more nucleophilic. In fact MeOH is a common solvent or cosolvent for reductive aminations (or alkylations).
I agree for product 1. For product 2, if you go for the intramolecular route then the product of a reductive amination should be azetidine. The 2nd step of your mechanism in the bottom picture is not correct (give it another check). However I am not convinced it will go intramolecularly because of the ring strain of the intermediate. Maybe at best you can get a mixture of the azetidine with the open chain product and possibly oligo/polymerisation products, a mess. You need to check the literature for a precedent.
In respect to the question about MeOH: Methanol should not compete with the amine, as the amine is a lot more nucleophilic. In fact MeOH is a common solvent or cosolvent for reductive aminations (or alkylations).
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