[scientific]

I came across this question and I think I was able to cut out 3/5 of the incorrect answers, but please let me know if my reasoning is also incorrect.


Which of the following is the best method for preparing (CH₃)₃COCH₃?
(a) NaOCH₃ + (CH₃)₃CCl 
(b) (CH₃)₃COH + CH₃I  in H₂SO₄
(c) (CH₃)₃CH + CH₃OH in H₂SO₄ 
(d) (CH₃)₃COK + CH₃OH
(e) (CH₃)₃COK + CH₃I


I think (b) and (c) are unlikely because the acid is too strong and would likely protonate something before the two other compounds even came into contact.
(d) is unlikely because -OH is a terrible leaving group, and even if (CH₃)₃CO^- attacked the C on the methanol, I don't see it leaving.

I'm stuck between (a) and (e). Now, I would choose (a) normally because in (e) CH₃I just seems too unlikely to me to create a 1° anion, although I know iodine is a very good leaving group. However, my answer key says the answer is (e). so i'm a little confused. is it entirely because iodine is a very good LG? Or do sterics play a very important role here?

[Dan]

(e) CH₃I just seems too unlikely to me to create a 1° anion, although I know iodine is a very good leaving group.

What mechanism did you envisage involving a 1° anion?

[scientific]

It wasn't so much a true anion formation, since it should be SN2. But I kind of don't understand how (e) is still better than (a)--is the sterics of (CH₃)₃CCl really so severe that it's unlikely to occur?

[wildfyr]

There are a 3 factors I see in comparing these two:

1: Which counter ion leads to a better nucelophile: K⁺ or Na⁺?
2: Which is a better leaving group Cl^- or I^-
3: Which is more favorable, a primary alkoxy attacking a tertiary carbocation during an SN1, or a tertiary alkoxide attacking a primary alkyl halide during SN2.

In my mind, number 3 has the most influence.

[Babcock_Hall]

@OP, I would also consider the possibility of a competing side-reaction with respect to (a).

[organic92]

Obviously, (e) (CH3)3COK + CH3I

it's an SN2 reaction where the tert-butanol K alcoxide reacts with methyl iodide.