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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: scientific on March 29, 2017, 05:55:57 PM

Title: which is the best way to prepare (CH3)3COCH3?
Post by: scientific on March 29, 2017, 05:55:57 PM
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  :rarrow:
(b) (CH3)3COH + CH3I  in H2SO4 :rarrow:
(c) (CH3)3CH + CH3OH in H2SO4  :rarrow:
(d) (CH3)3COK + CH3OH :rarrow:
(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?
Title: Re: which is the best way to prepare (CH3)3COCH3?
Post by: Dan on March 30, 2017, 01:48:07 AM
Quote from: scientific on March 29, 2017, 05:55:57 PM
(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?
Title: Re: which is the best way to prepare (CH3)3COCH3?
Post by: scientific on March 30, 2017, 02:39:19 PM
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?
Title: Re: which is the best way to prepare (CH3)3COCH3?
Post by: wildfyr on March 30, 2017, 02:50:56 PM
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.
Title: Re: which is the best way to prepare (CH3)3COCH3?
Post by: Babcock_Hall on March 30, 2017, 05:21:17 PM
@OP, I would also consider the possibility of a competing side-reaction with respect to (a).
Title: Re: which is the best way to prepare (CH3)3COCH3?
Post by: organic92 on April 15, 2017, 06:25:30 PM
Obviously, (e) (CH3)3COK + CH3I

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