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Why do tosylation and mesylation of alcohols follow different mechanisms?
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May Chen
Why do tosylation and mesylation of alcohols follow different mechanisms?
The only real difference between an alkanesulfonyl chlorides and arylsulfonyl chlorides is the $\alpha$-hydrogen. In a mechanistic study of the hydrolysis of methanesulfonyl chloride, the reaction was found to have a very small secondary kinetic isotope effect for $\mathrm{pH}<6.7$, indicating no bond breakage to the isotopically labelled hydrogen and thereby an $\mathrm{S_N2}$ reaction mechanism.
For $\mathrm{pH}>6.7$ there is a large primary kinetic isotope effect, indicating bond breakage to the isotopically labelled hydrogen, corresponding to the formation of the sulfene intermediate.$^{[1]}$ This shows that though methanesulfonyl chloride isn't incapable of direct attack by alcohols, in the presence of basic amines, which serve the additional purpose of driving the reaction to completion by accepting the alcohol proton, it is more quickly deprotonated to form the sulfene.
While pyridine isn't that great of a base ($\mathrm{p}K_\mathrm{b}=5.21$) and won't form the sulfene to any great extent, triethylamine ($\mathrm{p}K_\mathrm{b}=11.01$) and trimethylamine ($\mathrm{p}K_\mathrm{b}=9.81$) are much stronger bases and will deprotonate methanesulfonyl chloride to give the sulfene to a greater extent. The mechanism is just as dependent on the base as it is the sulfonating agent, so all these considerations must be taken into account.
The only real difference between an alkanesulfonyl chlorides and arylsulfonyl chlorides is the $\alpha$-hydrogen. In a mechanistic study of the hydrolysis of methanesulfonyl chloride, the reaction was found to have a very small secondary kinetic isotope effect for $\mathrm{pH}<6.7$, indicating no bond breakage to the isotopically labelled hydrogen and thereby an $\mathrm{S_N2}$ reaction mechanism.
For $\mathrm{pH}>6.7$ there is a large primary kinetic isotope effect, indicating bond breakage to the isotopically labelled hydrogen, corresponding to the formation of the sulfene intermediate.$^{[1]}$ This shows that though methanesulfonyl chloride isn't incapable of direct attack by alcohols, in the presence of basic amines, which serve the additional purpose of driving the reaction to completion by accepting the alcohol proton, it is more quickly deprotonated to form the sulfene.
While pyridine isn't that great of a base ($\mathrm{p}K_\mathrm{b}=5.21$) and won't form the sulfene to any great extent, triethylamine ($\mathrm{p}K_\mathrm{b}=11.01$) and trimethylamine ($\mathrm{p}K_\mathrm{b}=9.81$) are much stronger bases and will deprotonate methanesulfonyl chloride to give the sulfene to a greater extent. The mechanism is just as dependent on the base as it is the sulfonating agent, so all these considerations must be taken into account.
I finally got around to reading this paper that has already been cited multiple times. Thank you for not saying that "the pKa of MsCl is 6.7" - that is inaccurate and quite illogical, too (a typical sulfone has pKa ~25-30). I guess the bottom line is that a similar mechanism with TsCl would operate More
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The pKa of Mesyl Chloride can be looked at from the following work by J. F. King, J. Y. L. Lam, and S. Skonieczny in their paper - J. Am. Chem. Soc. 1992, 114, 1743 - 1749.
They have shown that in pH range 6.7-11.8, rate-determining attack by hydroxide anion forms sulfene which is quickly trapped by water. So the Pka of Mesyl Chloride can be said to be 6.7; this is less than the Pka of Et3N at 10.75. So Mesyl chloride will be forming the sulfene intermediate if treated with Et3N. And the SN2 attack by Oxygen happens in both cases, I don't know what you wanted to ask in your first question. Your mechanisms themselves show the attack in both cases by oxygen on the sulfur atom. I hope I am correct in my analogy. Apologies if I am wrong somewhere.
The pKa of Mesyl Chloride can be looked at from the following work by J. F. King, J. Y. L. Lam, and S. Skonieczny in their paper - J. Am. Chem. Soc. 1992, 114, 1743 - 1749.
They have shown that in pH range 6.7-11.8, rate-determining attack by hydroxide anion forms sulfene which is quickly trapped by water. So the Pka of Mesyl Chloride can be said to be 6.7; this is less than the Pka of Et3N at 10.75. So Mesyl chloride will be forming the sulfene intermediate if treated with Et3N. And the SN2 attack by Oxygen happens in both cases, I don't know what you wanted to ask in your first question. Your mechanisms themselves show the attack in both cases by oxygen on the sulfur atom. I hope I am correct in my analogy. Apologies if I am wrong somewhere.
1 & 2. An acid-base reaction can only occur on MsCl, since as you point out it is the only one with acidic alphaprotons. The protons are indeed quite acidic since the negative charge is stabilized by the two oxygens. (pKa ~6.7, thanks @Mrityunjay Gupta)
Et3N is quite basic (pKa ~11), and acid-base reactions will always occur first since they are much faster than other reactions. I'm quite sure the alcohol also attacks the sulfur on MsCl, but the a/b-reaction is just faster. The elimination of Cl- on MsCl follows an E1Cb-mechanism.
Source: Clayden Organic chem. 2nd ed pp. 403-404, J. Am. Chem. Soc., 1992, 114 (5), pp 1743–1749 (pKa of MsCl)
1 & 2. An acid-base reaction can only occur on MsCl, since as you point out it is the only one with acidic alphaprotons. The protons are indeed quite acidic since the negative charge is stabilized by the two oxygens. (pKa ~6.7, thanks @Mrityunjay Gupta)
Et3N is quite basic (pKa ~11), and acid-base reactions will always occur first since they are much faster than other reactions. I'm quite sure the alcohol also attacks the sulfur on MsCl, but the a/b-reaction is just faster. The elimination of Cl- on MsCl follows an E1Cb-mechanism.
Source: Clayden Organic chem. 2nd ed pp. 403-404, J. Am. Chem. Soc., 1992, 114 (5), pp 1743–1749 (pKa of MsCl)
The only real difference between an alkanesulfonyl chlorides and arylsulfonyl chlorides is the $\alpha$-hydrogen. In a mechanistic study of the hydrolysis of methanesulfonyl chloride, the reaction was found to have a very small secondary kinetic isotope effect for $\mathrm{pH}<6.7$, indicating no bond breakage to the isotopically labelled hydrogen and thereby an $\mathrm{S_N2}$ reaction mechanism.
For $\mathrm{pH}>6.7$ there is a large primary kinetic isotope effect, indicating bond breakage to the isotopically labelled hydrogen, corresponding to the formation of the sulfene intermediate.$^{[1]}$ This shows that though methanesulfonyl chloride isn't incapable of direct attack by alcohols, in the presence of basic amines, which serve the additional purpose of driving the reaction to completion by accepting the alcohol proton, it is more quickly deprotonated to form the sulfene.
While pyridine isn't that great of a base ($\mathrm{p}K_\mathrm{b}=5.21$) and won't form the sulfene to any great extent, triethylamine ($\mathrm{p}K_\mathrm{b}=11.01$) and trimethylamine ($\mathrm{p}K_\mathrm{b}=9.81$) are much stronger bases and will deprotonate methanesulfonyl chloride to give the sulfene to a greater extent. The mechanism is just as dependent on the base as it is the sulfonating agent, so all these considerations must be taken into account.
$^{[1]}$ King, J. F.; Lam, J. Y. L.; Skonieczny, S. Journal of the American Chemical Society 1992, 114 (5), 1743–1749.
The only real difference between an alkanesulfonyl chlorides and arylsulfonyl chlorides is the $\alpha$-hydrogen. In a mechanistic study of the hydrolysis of methanesulfonyl chloride, the reaction was found to have a very small secondary kinetic isotope effect for $\mathrm{pH}<6.7$, indicating no bond breakage to the isotopically labelled hydrogen and thereby an $\mathrm{S_N2}$ reaction mechanism.
For $\mathrm{pH}>6.7$ there is a large primary kinetic isotope effect, indicating bond breakage to the isotopically labelled hydrogen, corresponding to the formation of the sulfene intermediate.$^{[1]}$ This shows that though methanesulfonyl chloride isn't incapable of direct attack by alcohols, in the presence of basic amines, which serve the additional purpose of driving the reaction to completion by accepting the alcohol proton, it is more quickly deprotonated to form the sulfene.
While pyridine isn't that great of a base ($\mathrm{p}K_\mathrm{b}=5.21$) and won't form the sulfene to any great extent, triethylamine ($\mathrm{p}K_\mathrm{b}=11.01$) and trimethylamine ($\mathrm{p}K_\mathrm{b}=9.81$) are much stronger bases and will deprotonate methanesulfonyl chloride to give the sulfene to a greater extent. The mechanism is just as dependent on the base as it is the sulfonating agent, so all these considerations must be taken into account.
$^{[1]}$ King, J. F.; Lam, J. Y. L.; Skonieczny, S. Journal of the American Chemical Society 1992, 114 (5), 1743–1749.
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The pKa of Mesyl Chloride can be looked at from the following work by J. F. King, J. Y. L. Lam, and S. Skonieczny in their paper - J. Am. Chem. Soc. 1992, 114, 1743 - 1749.
They have shown that in pH range 6.7-11.8, rate-determining attack by hydroxide anion forms sulfene which is quickly trapped by water. So the Pka of Mesyl Chloride can be said to be 6.7; this is less than the Pka of Et3N at 10.75. So Mesyl chloride will be forming the sulfene intermediate if treated with Et3N. And the SN2 attack by Oxygen happens in both cases, I don't know what you wanted to ask in your first question. Your mechanisms themselves show the attack in both cases by oxygen on the sulfur atom. I hope I am correct in my analogy. Apologies if I am wrong somewhere.
The pKa of Mesyl Chloride can be looked at from the following work by J. F. King, J. Y. L. Lam, and S. Skonieczny in their paper - J. Am. Chem. Soc. 1992, 114, 1743 - 1749.
They have shown that in pH range 6.7-11.8, rate-determining attack by hydroxide anion forms sulfene which is quickly trapped by water. So the Pka of Mesyl Chloride can be said to be 6.7; this is less than the Pka of Et3N at 10.75. So Mesyl chloride will be forming the sulfene intermediate if treated with Et3N. And the SN2 attack by Oxygen happens in both cases, I don't know what you wanted to ask in your first question. Your mechanisms themselves show the attack in both cases by oxygen on the sulfur atom. I hope I am correct in my analogy. Apologies if I am wrong somewhere.
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1 & 2. An acid-base reaction can only occur on MsCl, since as you point out it is the only one with acidic alphaprotons. The protons are indeed quite acidic since the negative charge is stabilized by the two oxygens. (pKa ~6.7, thanks @Mrityunjay Gupta)
Et3N is quite basic (pKa ~11), and acid-base reactions will always occur first since they are much faster than other reactions. I'm quite sure the alcohol also attacks the sulfur on MsCl, but the a/b-reaction is just faster. The elimination of Cl- on MsCl follows an E1Cb-mechanism.
Source: Clayden Organic chem. 2nd ed pp. 403-404, J. Am. Chem. Soc., 1992, 114 (5), pp 1743–1749 (pKa of MsCl)
1 & 2. An acid-base reaction can only occur on MsCl, since as you point out it is the only one with acidic alphaprotons. The protons are indeed quite acidic since the negative charge is stabilized by the two oxygens. (pKa ~6.7, thanks @Mrityunjay Gupta)
Et3N is quite basic (pKa ~11), and acid-base reactions will always occur first since they are much faster than other reactions. I'm quite sure the alcohol also attacks the sulfur on MsCl, but the a/b-reaction is just faster. The elimination of Cl- on MsCl follows an E1Cb-mechanism.
Source: Clayden Organic chem. 2nd ed pp. 403-404, J. Am. Chem. Soc., 1992, 114 (5), pp 1743–1749 (pKa of MsCl)
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