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Will this beta-branched secondary alkyl halide undergo SN1 or SN2?
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Lauren Barcas
Will this beta-branched secondary alkyl halide undergo SN1 or SN2?
Murpy, TJ, J. Chem. Ed.2009, 86, 519-524 shows that secondary alkyl halides do not undergo SN1 reactions.
This is counter to what we have been (and still are in many cases) teaching.
Murpy, TJ, J. Chem. Ed.2009, 86, 519-524 shows that secondary alkyl halides do not undergo SN1 reactions.This is counter to what we have been (and still are in many cases) teaching.
Perhaps you could give some more details about the paper in question and ideally provide a link to it. At the moment this seems more like a comment than an answer.More
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Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.
Full mechanism:
I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.
I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.
Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.
Full mechanism:
I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted. I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.
Jerepierre, How could I have known SN1 would have won? Solvent effet and branching at beta makes the situation a bit ambiguous and when we see CN-, we are tempted to do SN2?More
You make a hydride shift from the tertiary carbon.More
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It is in $2^\circ$ position, so both, $\ce{S_{N}1}$ and $\ce{S_{N}2}$ can happen.
$\ce{Br}$ is a good leaving group probably the third best in the groups very common(depends on opinion, but it is a good one anyways though):
$\ce{OTs>I>Br>Cl>...}$ but LG accelerates both reactions
$\ce{CN-}$ is a very good nucleophile in the list of common ones(depends on opinion, but it is a good one anyways though): $\ce{PhS- > CN- > I- > ...}$, the second best so $\ce{S_{N}2}$ is favourable.
The solvent is water, which is polar protic which highly favours $\ce{S_{N}1}$
The temperature is normal so there's no elimination
Finally fourth point overthrows thirds and moreover water reduces nucleophilicity of $\ce{CN^-}$
It is in $2^\circ$ position, so both, $\ce{S_{N}1}$ and $\ce{S_{N}2}$ can happen.
$\ce{Br}$ is a good leaving group probably the third best in the groups very common(depends on opinion, but it is a good one anyways though):$\ce{OTs>I>Br>Cl>...}$ but LG accelerates both reactions
$\ce{CN-}$ is a very good nucleophile in the list of common ones(depends on opinion, but it is a good one anyways though): $\ce{PhS- > CN- > I- > ...}$, the second best so $\ce{S_{N}2}$ is favourable.
The solvent is water, which is polar protic which highly favours $\ce{S_{N}1}$
The temperature is normal so there's no elimination
Finally fourth point overthrows thirds and moreover water reduces nucleophilicity of $\ce{CN^-}$
I think the cyanide won't be a good nucleophile because it will react with the water molecules to form a hydrogen cyanide. this will make the cyanide weak so there is no way the cyanide can display the bromide. therefore a strong nucleophile is required
I think the cyanide won't be a good nucleophile because it will react with the water molecules to form a hydrogen cyanide. this will make the cyanide weak so there is no way the cyanide can display the bromide. therefore a strong nucleophile is required
Although HCN is a weak acid, in water, $\ce CN^-$ will stay as is because if it takes $\ce H^+$ from water, the net equilibrium constant of the reaction will be $\ce {K_w}/{K_a} $ = $\ce 10^{10} * 10^{-16}$ = $\ce 10^{-6}$More
Murpy, TJ, J. Chem. Ed. 2009, 86, 519-524 shows that secondary alkyl halides do not undergo SN1 reactions. This is counter to what we have been (and still are in many cases) teaching.
Murpy, TJ, J. Chem. Ed. 2009, 86, 519-524 shows that secondary alkyl halides do not undergo SN1 reactions.This is counter to what we have been (and still are in many cases) teaching.
More
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Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.
Full mechanism:
I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.
I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.
Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.
Full mechanism:
I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.
I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.
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I think the cyanide won't be a good nucleophile because it will react with the water molecules to form a hydrogen cyanide. this will make the cyanide weak so there is no way the cyanide can display the bromide. therefore a strong nucleophile is required
I think the cyanide won't be a good nucleophile because it will react with the water molecules to form a hydrogen cyanide. this will make the cyanide weak so there is no way the cyanide can display the bromide. therefore a strong nucleophile is required
More
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