TPR Orgo Solvent Question

[tuhtles]

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I didn't like this particular question from TPR online passage...

Which of the following would be the best solvent for the formation of t-butyl methyl ether with sodium t-butoxide?

A. Methanol
B. Water
C. Acetic Acid
D. Benzene

The answer was D (Benzene).

I thought all these answer choices were crappy and I would think t-butanol would be the ideal solvent. Perhaps I'm thinking way into this but how does it make any sense to have benzene as the solvent? Out of the other 3 answer choices, benzene is clearly the less crappy answer and the question particularly asked for "best solvent" but then it says "for the FORMATION". Well, wouldn't t-butoxide reside in an aqueous phase where as the methyl halide would reside in the organic phase (benzene in this case). So if these two reactants reside in different immisible layers, then wouldn't this not be the best for the reaction to create the ether?

The first 3 answer choices are essentially the same since t-butoxide would deprotonate them, forming t-butanol. And hence why the ideal solvent should be t-butanol....

Could this question just be a fluke? Or am I just over thinking it too much?

 

[doctorswole]

I didn't like this particular question from TPR online passage...

Which of the following would be the best solvent for the formation of t-butyl methyl ether with sodium t-butoxide?

A. Methanol
B. Water
C. Acetic Acid
D. Benzene

The answer was D (Benzene).

I thought all these answer choices were crappy and I would think t-butanol would be the ideal solvent. Perhaps I'm thinking way into this but how does it make any sense to have benzene as the solvent? Out of the other 3 answer choices, benzene is clearly the less crappy answer and the question particularly asked for "best solvent" but then it says "for the FORMATION". Well, wouldn't t-butoxide reside in an aqueous phase where as the methyl halide would reside in the organic phase (benzene in this case). So if these two reactants reside in different immisible layers, then wouldn't this not be the best for the reaction to create the ether?

The first 3 answer choices are essentially the same since t-butoxide would deprotonate them, forming t-butanol. And hence why the ideal solvent should be t-butanol....

Could this question just be a fluke? Or am I just over thinking it too much?

if I remember correctly- you want polar aprotic solvents for sn2 reactions and this looks like sn2 formation

benzene is the only aprotic solvent available as an answer, even though it isn't very polar

edit: and I think you are over thinking the part about the aqueous vs organic layers

 

[loltopsy]

if I remember correctly- you want polar aprotic solvents for sn2 reactions and this looks like sn2 formation

benzene is the only aprotic solvent available as an answer, even though it isn't very polar

edit: and I think you are over thinking the part about the aqueous vs organic layers

Memorizing solvents for each reaction is a poor way to go, IMO. Better to understand the logic behind it.

This reaction, judging by the product and known substrate, is an SN2 reaction between methyliodide and sodium t-butoxide.

A. Methanol
If we add methanol, the fastest thing that would get deprotonated is the methanol. Acid base reactions occur way, way, way, way faster than SN2. The formation of significant quantities of sodium methoxide would result is a lot of dimethylether production. The methoxide would SN2 onto the methyliodide.


B. Water
Same thing would happen as with methanol. We would form sodium hydroxide, which would SN2 onto methyliodide and form methanol.

C. Acetic Acid
Again, same as above. We would form sodium acetate, which would react with methyliodide giving an undesirable product.

D. Benzene
Benzene's protons are not very acidic at all! The reaction with t-butoxide wouldn't occur. This frees up t-butoxide to react with methyliodide and give desired product.



Also, I do agree with you that t-butanol would be useful reagent, but they didn't give that choice, so we don't have to consider it.

 

[Czarcasm]

Just to add on to what others contributed:

I usually answer these types of questions with this line of thinking:

First I look at the nucleophile/base: t-butoxide -- this is a strong base and a weak nucleophile (due to its bulkiness). Immediately, I know that if an acid-base reaction can occur, it WILL occur because as loltopsy mentioned, acid/base reactions occur very quickly.

However, it's likely the passage provided a non-acidic substrate (to test SN2/SN1 or E2/E1). Judging by the final product, it was likely something like methyl chloride. I immediately recognize this as our electrophile, with the chloride being the good leaving group.

Also bare in mind strong bases can undergo E2 (elimination) to form an alkene. However, because the substrate we're considering is methyl -- and because E2 requires primary, secondary, or tertiary carbons (where the LG is), no E2 reaction will occur. (You can't form a double bond with 1 carbon).

Alright. So then immediately after you might consider E1 or SN1. However, both of these competing mechanisms have a carbocation intermediate and as I'm sure you know, the more substituted intermediate, the more stable it is. Therefore, it's safe to assume there is no reaction for E1 or SN1 -- because either of these mechanisms would never occur on a methyl (unless the LG was allylic or benzylic which likely isn't the case here).

This leaves us with SN2, which is a bit unusual because we have a weak nucleophile. But how is our electrophile for SN2? SN2 mechanism prefers methyl and primary (beyond that is too hindered for back-side attack). We have a methyl electrophile, which is fantastic for SN2. Now, for SN2, what you want to keep in mind here is that an aprotic solvent is preferred. The reason being is because protic solvents (solvents capable of having O-H or N-H bonds), STABILIZE and therefore weaken the nucleophile. The nucleophile we're using for SN2 is already considered weak. Therefore, benzene is preferred for this reason.

Also bare in mind that methanol and water both can act as weak nucleophiles, so they would compete with the bulky t-butoxide (and also be deprotonated in the process). Acetic Acid is an aprotic solvent, but it can be attacked by our nucleophile because there's a partial charge on the carbon (ie. it's another good electrophile). This would also result in mixed products, which is acceptable but not the BEST option here. Benzene (also aprotic) is non-reactive for SN2. This IS the best choice for our solvent, because we yield the most product.

(Also, I just want to add one more thing: protic solvents (choice A and B) are favorable for SN1/E2 because they stabalize carbocation intermediates and are actually essential for the mechanism to occur at any favorable rate -- SN2, USUALLY has aprotic solvents, but in some cases it may be protic -- but for this situation above, based on the condition of the nucleophile, you most definitely require an aprotic solvent). Hence the reason which choice D is the correct answer here.

 

[doctorswole]

Memorizing solvents for each reaction is a poor way to go, IMO. Better to understand the logic behind it.

This reaction, judging by the product and known substrate, is an SN2 reaction between methyliodide and sodium t-butoxide.

A. Methanol
If we add methanol, the fastest thing that would get deprotonated is the methanol. Acid base reactions occur way, way, way, way faster than SN2. The formation of significant quantities of sodium methoxide would result is a lot of dimethylether production. The methoxide would SN2 onto the methyliodide.


B. Water
Same thing would happen as with methanol. We would form sodium hydroxide, which would SN2 onto methyliodide and form methanol.

C. Acetic Acid
Again, same as above. We would form sodium acetate, which would react with methyliodide giving an undesirable product.

D. Benzene
Benzene's protons are not very acidic at all! The reaction with t-butoxide wouldn't occur. This frees up t-butoxide to react with methyliodide and give desired product.



Also, I do agree with you that t-butanol would be useful reagent, but they didn't give that choice, so we don't have to consider it.

understanding that you want polar aprotic solvents for sn2 is not really memorizing something for each reaction

it is something you should know, and there is more to it than just alternate products/side reactions

you should get hindrance of the reacting species by the solvent if you use a polar protic solvent in sn2, which is why sn1 is more likely with polar protics