[Burningkrome]

So, this is a question (and answer) from my textbook which left me with some questions. The main one is; based on the products, this reaction appears to occur via both SN2 and E2 (please correct me if I’m wrong). But it seems like it would be more likely to favor SN1/E1.

QUESTIONS: Can anyone tell me why it would be SN2/E2?

…OR IF IT IS actually SN1/E1, why there is not also some Hoffman product listed?? It seems that if a carbocation intermediate is formed, there would be just as likely a chance for the hydrogen to be pulled off the 1’carbon and the 2’carbon…?

IF IT WERE an SN2/E2, then the –OCH3 as a nucleophile would need to attack the 3’ carbon (kicking off the OSO2CH3, which I think is feasible because the Methane is a small molecule) and also the –OCH3 as a base would attack the hydrogen off the 2’ carbon making a double bond. HOWEVER, as an E2…why would the hydrogen on the 1’ carbon not be similarly attacked making the Hoffman product?

IF IT WERE an SN1/E1, then the (SN1 is obvious) but for the E1…why wouldn’t the hydrogens from BOTH the 2’carbon and 1’carbon fill in the electron gap…creating both some Zatsev and Hoffman products?
 
I have listed the pertinent information for determining SNx/Ex…listed in order of importance (according to my text).

1.   Substrate: So, to quote my text, if the carbon attached to the leaving group is 3’ you will almost ALWAYS have an SN1/E1 reaction. BUT, I continued with the other factors as a check.
2.   Strength of the Nucleophile/Base: Both seem to favor SN2/E2 (please correct anything about my assumptions here that are wrong).
3.   Leaving group: OSO2CH3 is an excellent leaving group…which should (again) favor an SN1/E1 with a carbocation.
4.   The methane solvent is protic (an acid) and is polar (holds a neg charge after losing the proton). A polar Aprotic would favor both SN1/E1 and SN2/E2…but a polar protic should actually favor an SN1/E1 (again).

Thanks for the *delete me* 

[Dan]

The base/nucleophile is methanol, not methane - methane is CH₄. You also refer to the methoxide (CH₃O^-) nucleophile - there is no methoxide, it is methanol.

2.   Strength of the Nucleophile/Base: Both seem to favor SN2/E2 (please correct anything about my assumptions here that are wrong).

Methanol is a weak nucleophile and a weak base. Review the factors that contribute to nucleophilicity. Also, you have quoted the pKb of methanol as -0.9, which is way off. The pKb of methanol is probably around 15-16 (as it is for water).

4.   The methane solvent is protic (an acid) and is polar (holds a neg charge after losing the proton). A polar Aprotic would favor both SN1/E1 and SN2/E2…but a polar protic should actually favor an SN1/E1 (again).

You have written in the text that a protic solvent favours SN1/E1 (and not SN2/E2), but you have written in the table that a protic solvent increases the rate of all the mechanisms. Which are you going with?

As for regioselectivity, is the reaction under thermodynamic or kinetic control?

[Burningkrome]

Hey...sorry for the typos I was trying to do to many things when I wrote this. See my corrections below...

- I know its methanol ;-) I Just brain-fried while typing. I don't think I mentioned methyloxide. I think I was talking about methanesulfonate.

- So the pKa of Methanol is 15.5...so doesn't that make the pKb of methanol = pKw/pKa = 14/15.5 = 0.9? Sorry about that...I just realized I had -0.9 not 0.9. Is 0.9 incorrect?

- As I understand it a polar aprotic favors both SN1/E1 AND SN2/E2 by sake of its polarity...but a protic polar solvent favors SN1/E1 because it cages the nucleophile, preventing an SN2 style from attacking the carbon. Although It would not prevent an E2 attack on the hydrogen.

Q1: Isn't methnol a polar PROtic solvent?

Lets just simplify the question for now...forget everything else :-) 

Q2: Assuming an E1 reaction is correct - after the LG (methanesulfonate) has left leaving the carbocation...why would it form more zaitsev product than hoffman product?

Q# OR...if it is an E2 reaction...why?

As for the thermodynamic/kinetic control...you know what I know. All I really have to work with is the question as shown. Since "heat" was not added in the question, I am assuming kinetic.

[Dan]

- I know its methanol ;-) I Just brain-fried while typing. I don't think I mentioned methyloxide. I think I was talking about methanesulfonate.

You wrote:

IF IT WERE an SN2/E2, then the –OCH3 as a nucleophile would need to attack the 3’ carbon (kicking off the OSO2CH3, which I think is feasible because the Methane is a small molecule) and also the –OCH3 as a base would attack the hydrogen off the 2’ carbon making a double bond.

I assumed -OCH3 was methoxide ion, ^-OCH₃

- So the pKa of Methanol is 15.5...so doesn't that make the pKb of methanol = pKw/pKa = 14/15.5 = 0.9? Sorry about that...I just realized I had -0.9 not 0.9. Is 0.9 incorrect?

No, you've got confused here. K_a is not the same as pK_a.

K_aK_b = K_w

pX = -logX, therefore -log(K_aK_b) = pK_w (= 14)

and it follows that: -logK_a -logK_b = pK_w
and finally: pK_a + pK_b = pK_w

So, the classic misunderstanding now is that if we take methanol with pK_a 15.5 then apply the equation:

15.5 + pK_b = 14

and that for methanol pK_b = -1.5

This implies that methanol is a strong base, around 10⁵ times stronger than ammonia (pK_b 4.7), which is nonsense. So what is wrong?

The equation pK_a + pK_b = pK_w relates the pK_a of an acid to the pK_b of its conjugate base.

i.e. the pK_a of methanol is 15.5, so we can say the pK_b of methoxide (not methanol) is -1.5.

The pK_a of a compound is directly related to the pK_b of it's conjugate base, and likewise the pK_b of a compound is directly related to the pK_a of its conjugate acid. But, the pK_a of a compound is not related to the pK_b of the same compound. If a compound is a weak base, it does not mean it is a strong acid and vice versa - an example is methane, it is a very weak base but also a very weak acid.

a protic polar solvent favors SN1/E1 because it cages the nucleophile, preventing an SN2 style from attacking the carbon.

Essentially yes, but a protic solvent won't prevent S_N2 it will just slow it down relative to S_N1.

My issue is that in this table:

The bottom line reads "Solvent: Polar protic = favours SN1&2/E1&2", which is a direct contradiction of what you've just told me.

Q2: Assuming an E1 reaction is correct - after the LG (methanesulfonate) has left leaving the carbocation...why would it form more zaitsev product than hoffman product?

Is there a thermodynamic difference between the Zaitsev and Hoffman products? Which is the kinetically favoured product and which is the thermodynamically favoured product?

[orgopete]

My take,

A strong acid corresponds with a weak conjugate base. HCl is a strong acid and Cl(-) is a weak base. With methanol, there are two pKa values to be concerned with, it is both a base to the conjugate acid CH3OH2(+) and an acid to the conjugate base CH3O(-). In this reaction, it is a nucleophile and a base.

Re Zaitsev products, my take. Zaitsev favor more substituted products. If you generate a carbocation, the electron best donated toward that carbocation are from substituted carbons. If the proton accompanying the electrons (hydride) is lost, it gives the most substituted product, the Zaitsev product.

You should recognize this is the same rule as governs rearrangements. If you generate a carbocation adjacent to a tertiary center, the hydride can migrate to give a more stable tertiary carbocation.

[Burningkrome]

Thanks to both of you for the help. I took OChem a few years ago, and am returning to school...so this is a lot of review and forgotten things :-) Some more questions if you;re willing.

1. CH3O- : Now I understand what you are referring to. You're saying that methanol does NOT disassociate the hydrogen to become CH3O- (methoxide) in this reaction? Is not, how does the OCH3 get attached in place of the methansulfonate? I assumed it the methanesulfonate disassociated, leaving a carbocation...allowing the CH3O- to attack. But if the hydrogen does not disassociate...then what is the mechanism?

2. I reversed my thinking on the polar protic. Polar APROtic favors SN1/E1, SN2/E2...polar protic favors SN1/E1 while slowing or preventing SN2/E2. Correct?

3.

Is there a thermodynamic difference between the Zaitsev and Hoffman products? Which is the kinetically favoured product and which is the thermodynamically favoured product?

Apparently I have no idea. I'm assuming, since the question doesn't mention heat (as it does in other questions) that this is under kinetic control...? Perhaps you can help me understand the differenc ein this case?

4. Maybe it would (again) be best to simplify. Forget the chart under the reaction picture. Let's just assume its completely wrong. How does this reaction proceed to the shown results? Is it E1/SN1 or E2/SN2. What are the mechanisms that end with both products?

Thanks again for the *delete me*

[Dan]

1. CH3O- : Now I understand what you are referring to. You're saying that methanol does NOT disassociate the hydrogen to become CH3O- (methoxide) in this reaction? Is not, how does the OCH3 get attached in place of the methansulfonate? I assumed it the methanesulfonate disassociated, leaving a carbocation...allowing the CH3O- to attack. But if the hydrogen does not disassociate...then what is the mechanism?

Methanol is the nucleophile, see for example: http://en.wikipedia.org/wiki/SN1_reaction#Mechanism

2. I reversed my thinking on the polar protic. Polar APROtic favors SN1/E1, SN2/E2...polar protic favors SN1/E1 while slowing or preventing SN2/E2. Correct?

Aprotic favours S_N2, because the nucleophile is poorly solvated and therefore more reactive.

Protic disfavours S_N2, because the nucleophile is stabilised by hydrogen bonding and therefore less nucleophilic. In addition, polar protic favours carbocation formation because it solvates the cation, but also solvates the anionic leaving group through hydrogen bonding.

4. Maybe it would (again) be best to simplify. Forget the chart under the reaction picture. Let's just assume its completely wrong. How does this reaction proceed to the shown results? Is it E1/SN1 or E2/SN2. What are the mechanisms that end with both products?

There are mistakes in the chart, but I think now you can correct them and decide on the mechanism. I can't give you the final answer because I have to follow the forum rules. You have a weak nucleophile, a weak base, a good leaving group and a polar protic solvent. S_N2 is very hindered at a tertiary centre, and you can potentially for a tertiary carbocation. Do you think S_N2/E2 or S_N1/E1 is more likely?

[Burningkrome]

Corrected chart....

[Dan]

Why is methoxide in the table? It is a strong base that would favour E2, but it is not present in the reaction.

You have listed the pK_b of MeSO₃^- (methanesulfonate) as 7. Where did you get this value from? It is way off. Since the pK_a of methanesulfonic acid is about -2, we can calculate that the pK_b of methanesulfonate is around 16.

There is another product of this reaction that I think is important when considering why the thermodynamic product of the reaction forms: What happened to the methanesulfonate (can you balance the equation)?

[Burningkrome]

Hey Dan,

I put methoxide in the table because (at the time) I still wasn't sure if the methanol's hydrogen disassociated in this reaction or not. I guess not. Can you confirm for me this is correct....methanol does not disassociate the hydrogen into CH3O- ?

So, if not...maybe this explains the third product. I initially thought that perhaps it is the Sulfonate ion that yanks the hydrogen off the methanol. But, since the oxy on the methanol should have a less stable charge (be more reactive) than the oxy on the sulfonate, I assumed no. It WOULD explain some things. I.e the sulfonate disassociates to form a carbocation...the sulfonate ion yanks the hydrogen from methanol, leaving it free to bond with the 3'carbon. Is this the mechanism? If I balance the equation... that's all I can come up with.

Finally, to be honest...I don't think I understand completely what we are discussing between kinetic control and thermodynamic control. Based on what you're saying...I don't believe I'm thinking of the right things here.

Can you explain to me or direct me to a tutorial that explains what is kenetic control and what is thermodynamic control within a reaction?

[Dan]

Can you confirm for me this is correct....methanol does not disassociate the hydrogen into CH3O- ?

Yes. I have already said the methoxide is not present (twice).

sulfonate disassociates to form a carbocation...the sulfonate ion yanks the hydrogen from methanol

No, methansulfonate is not basic enough to deprotonate methanol. The sulfonate dissociates to give the carbocation. Methanol can then add to the the carbocation directly, to form a protonated ether, which is then deprotonated by the sulfonate (or a molecule of methanol, methanol has a similar basicity to methanesulfonate). This is analogous to the mechanism here: http://en.wikipedia.org/wiki/SN1_reaction#Mechanism which shows water (not hydroxide) adding to a tertiary carbocation.

So you have identified that methanesulfonic acid MeSO₃H, a strong acid, is produced in this reaction. The word equations would be:

Starting material + methanol Ether product + methanesulfonic acid

and

Starting material Alkene + methanesulfonic acid

The point I am getting at is that since a strong acid is produced, the reactions are reversible. An alkene can be protonated by a strong acid to reform the tertiary carbocation. Since the carbocation intermediate can reform under the reaction conditions (without heat, but heat would help), we will reach an equilibrium in which the more thermodynamically stable alkene is the major product. More alkyl-substituted alkenes are more thermodynamically stable: http://en.wikipedia.org/wiki/Zaitsev's_rule#Thermodynamic_considerations