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Why can NCl3 be hydrolyzed but CCl4 cannot?
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Matthew Dubeau
Why can NCl3 be hydrolyzed but CCl4 cannot?
There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave.
$\ce{NCl3}$
The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{SiCl4}$ and $\ce{CCl4}$. Since the oxygen in water usually attacks nucleophilicly, the attack happens on the chlorine; it can proceed nicely with an $\mathrm{S_N2}$ mechanism (the rear side of all three chlorines are pointing into the solution). Note that another water molecule or the same will likely protonate the nitrogen either before or concertedly, so the leaving group won’t be $\ce{NCl2-}$ but rather $\ce{NHCl2}$. The other reaction product is $\ce{HOCl}$.
Other concerted mechanisms including only one water molecule might be possible, but sound less likely.
$\ce{SiCl4}$
Silicon atoms are rather large compared to both carbon and nitrogen: They have an additional electron shell. (Layman’s simplification, but it works.) As such, even in crowded cases like tri-iso-propylsilyl chloride $\ce{((CH3)2CH)3-SiCl}$ with three bulky iso-propyl groups, the silicon atom can be rear-side attacked by water (or catalysts such as 4-(N,N-dimethylamino)pyridine) liberating a chloride atom in a traditional $\mathrm{S_N2}$ mechanism.
No d-orbital contribution is required to explain it. In fact, silicon does not have ‘low-lying’ or ‘accessable’ d-orbitals; that is an old, superceded explanation.
The hydrolysis products of $\ce{SiCl4}$ are $\ce{Si(OH)4}$ (which will likely condensate to some polysilicate or $\ce{SiO2}$) and $\ce{HCl}$.
$\ce{CCl4}$
Technically, this could also proceed via an $\mathrm{S_N2}$ mechanism. However, carbon is much smaller and the chlorine atoms much larger in relation, so the rear-side attack is disfavoured sterically.
Instead, this molecule has access to the $\mathrm{S_N1}$ pathway, which can be described as follows:
And so on, displacing the remaining chlorine atoms, ultimately liberating $\ce{H2CO3}$. (After the second chlorine displacement, phosgene will form which can be attacked directly by water with the $\ce{C=O}$ bond acting as an electrophile.)
Note that here again we gain $\ce{HCl}$, an indirect proof of chlorine being the more electronegative element.
Conclusion
The three molecules we considered are so different in their steric and electronic properties, that we need to consider three different mechanisms that lead us to two different sets of products (depending on which side gains the hydrogens and which formally gains hydroxides). Nowhere do we need d-orbitals.
There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave.
$\ce{NCl3}$
The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{SiCl4}$ and $\ce{CCl4}$. Since the oxygen in water usually attacks nucleophilicly, the attack happens on the chlorine; it can proceed nicely with an $\mathrm{S_N2}$ mechanism (the rear side of all three chlorines are pointing into the solution). Note that another water molecule or the same will likely protonate the nitrogen either before or concertedly, so the leaving group won’t be $\ce{NCl2-}$ but rather $\ce{NHCl2}$. The other reaction product is $\ce{HOCl}$.
Other concerted mechanisms including only one water molecule might be possible, but sound less likely.
$\ce{SiCl4}$
Silicon atoms are rather large compared to both carbon and nitrogen: They have an additional electron shell. (Layman’s simplification, but it works.) As such, even in crowded cases like tri-iso-propylsilyl chloride $\ce{((CH3)2CH)3-SiCl}$ with three bulky iso-propyl groups, the silicon atom can be rear-side attacked by water (or catalysts such as 4-(N,N-dimethylamino)pyridine) liberating a chloride atom in a traditional $\mathrm{S_N2}$ mechanism.
No d-orbital contribution is required to explain it. In fact, silicon does not have ‘low-lying’ or ‘accessable’ d-orbitals; that is an old, superceded explanation.
The hydrolysis products of $\ce{SiCl4}$ are $\ce{Si(OH)4}$ (which will likely condensate to some polysilicate or $\ce{SiO2}$) and $\ce{HCl}$.
$\ce{CCl4}$
Technically, this could also proceed via an $\mathrm{S_N2}$ mechanism. However, carbon is much smaller and the chlorine atoms much larger in relation, so the rear-side attack is disfavoured sterically.
Instead, this molecule has access to the $\mathrm{S_N1}$ pathway, which can be described as follows:
And so on, displacing the remaining chlorine atoms, ultimately liberating $\ce{H2CO3}$. (After the second chlorine displacement, phosgene will form which can be attacked directly by water with the $\ce{C=O}$ bond acting as an electrophile.)
Note that here again we gain $\ce{HCl}$, an indirect proof of chlorine being the more electronegative element.
Conclusion
The three molecules we considered are so different in their steric and electronic properties, that we need to consider three different mechanisms that lead us to two different sets of products (depending on which side gains the hydrogens and which formally gains hydroxides). Nowhere do we need d-orbitals.
There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave.
$\ce{NCl3}$
The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{SiCl4}$ and $\ce{CCl4}$. Since the oxygen in water usually attacks nucleophilicly, the attack happens on the chlorine; it can proceed nicely with an $\mathrm{S_N2}$ mechanism (the rear side of all three chlorines are pointing into the solution). Note that another water molecule or the same will likely protonate the nitrogen either before or concertedly, so the leaving group won’t be $\ce{NCl2-}$ but rather $\ce{NHCl2}$. The other reaction product is $\ce{HOCl}$.
Other concerted mechanisms including only one water molecule might be possible, but sound less likely.
$\ce{SiCl4}$
Silicon atoms are rather large compared to both carbon and nitrogen: They have an additional electron shell. (Layman’s simplification, but it works.) As such, even in crowded cases like tri-iso-propylsilyl chloride $\ce{((CH3)2CH)3-SiCl}$ with three bulky iso-propyl groups, the silicon atom can be rear-side attacked by water (or catalysts such as 4-(N,N-dimethylamino)pyridine) liberating a chloride atom in a traditional $\mathrm{S_N2}$ mechanism.
No d-orbital contribution is required to explain it. In fact, silicon does not have ‘low-lying’ or ‘accessable’ d-orbitals; that is an old, superceded explanation.
The hydrolysis products of $\ce{SiCl4}$ are $\ce{Si(OH)4}$ (which will likely condensate to some polysilicate or $\ce{SiO2}$) and $\ce{HCl}$.
$\ce{CCl4}$
Technically, this could also proceed via an $\mathrm{S_N2}$ mechanism. However, carbon is much smaller and the chlorine atoms much larger in relation, so the rear-side attack is disfavoured sterically.
Instead, this molecule has access to the $\mathrm{S_N1}$ pathway, which can be described as follows:
$$\ce{CCl4 -> CCl3+ + Cl-} \\ \ce{CCl3+ + H2O -> Cl3C-OH2+} \\ \ce{Cl3C-OH2+ + Cl- -> CCl3-OH + HCl}$$
And so on, displacing the remaining chlorine atoms, ultimately liberating $\ce{H2CO3}$. (After the second chlorine displacement, phosgene will form which can be attacked directly by water with the $\ce{C=O}$ bond acting as an electrophile.)
Note that here again we gain $\ce{HCl}$, an indirect proof of chlorine being the more electronegative element.
Conclusion
The three molecules we considered are so different in their steric and electronic properties, that we need to consider three different mechanisms that lead us to two different sets of products (depending on which side gains the hydrogens and which formally gains hydroxides). Nowhere do we need d-orbitals.
There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave.
$\ce{NCl3}$
The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{SiCl4}$ and $\ce{CCl4}$. Since the oxygen in water usually attacks nucleophilicly, the attack happens on the chlorine; it can proceed nicely with an $\mathrm{S_N2}$ mechanism (the rear side of all three chlorines are pointing into the solution). Note that another water molecule or the same will likely protonate the nitrogen either before or concertedly, so the leaving group won’t be $\ce{NCl2-}$ but rather $\ce{NHCl2}$. The other reaction product is $\ce{HOCl}$.
Other concerted mechanisms including only one water molecule might be possible, but sound less likely.
$\ce{SiCl4}$
Silicon atoms are rather large compared to both carbon and nitrogen: They have an additional electron shell. (Layman’s simplification, but it works.) As such, even in crowded cases like tri-iso-propylsilyl chloride $\ce{((CH3)2CH)3-SiCl}$ with three bulky iso-propyl groups, the silicon atom can be rear-side attacked by water (or catalysts such as 4-(N,N-dimethylamino)pyridine) liberating a chloride atom in a traditional $\mathrm{S_N2}$ mechanism.
No d-orbital contribution is required to explain it. In fact, silicon does not have ‘low-lying’ or ‘accessable’ d-orbitals; that is an old, superceded explanation.
The hydrolysis products of $\ce{SiCl4}$ are $\ce{Si(OH)4}$ (which will likely condensate to some polysilicate or $\ce{SiO2}$) and $\ce{HCl}$.
$\ce{CCl4}$
Technically, this could also proceed via an $\mathrm{S_N2}$ mechanism. However, carbon is much smaller and the chlorine atoms much larger in relation, so the rear-side attack is disfavoured sterically.
Instead, this molecule has access to the $\mathrm{S_N1}$ pathway, which can be described as follows:
$$\ce{CCl4 -> CCl3+ + Cl-} \\\ce{CCl3+ + H2O -> Cl3C-OH2+} \\\ce{Cl3C-OH2+ + Cl- -> CCl3-OH + HCl}$$
And so on, displacing the remaining chlorine atoms, ultimately liberating $\ce{H2CO3}$. (After the second chlorine displacement, phosgene will form which can be attacked directly by water with the $\ce{C=O}$ bond acting as an electrophile.)
Note that here again we gain $\ce{HCl}$, an indirect proof of chlorine being the more electronegative element.
Conclusion
The three molecules we considered are so different in their steric and electronic properties, that we need to consider three different mechanisms that lead us to two different sets of products (depending on which side gains the hydrogens and which formally gains hydroxides). Nowhere do we need d-orbitals.
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$\ce{CCl4}$ can be hydrolysed under pressure at elevated temperature. Seems the reaction proceeds via $S_N1$ mechanism. See the source.
$\ce{CCl4}$ can be hydrolysed under pressure at elevated temperature. Seems the reaction proceeds via $S_N1$ mechanism. See the source.
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