Comments on: Deciding SN1/SN2/E1/E2 – The Solvent

By: James Ashenhurst

James Ashenhurst — Wed, 15 Feb 2023 02:38:39 +0000

In reply to Aayushi. Usually in the Williamson the substrate is a primary alkyl halide. When the substrate is secondary, elimination will compete.

By: Aayushi

Aayushi — Tue, 14 Feb 2023 18:00:37 +0000

Relating to the last sentence I said… Alcohol Is a polar protic solvent… (but it’s less polar than water… and the textbook kept saying that a less polar solvent favours elimination…)
My last sentence more fittingly refers to the Williamson synthesis.
Sorry for the several comments, I am trawling this site for my chemistry final in 2 weeks.

By: Aayushi

Aayushi — Sun, 12 Feb 2023 14:05:09 +0000

Oh a couple of other people have asked about the alcoholic and aqueous KOH already-
I’ve found the source with the explanation:
https://flic.kr/p/2ogxqFf
The book says that alkoxide ion prefers to be a base rather than a nucleophile for steric reasons, as opposed to OH- which can go in for an SN2 easier.
The textbook is Pradeep’s New Course Chemistry for Class 12
So yeah, exception to polar protic solvents favouring E2 over SN2.

By: Aayushi

Aayushi — Sun, 12 Feb 2023 13:17:39 +0000

The second reaction tripped me up because ethoxide ion in ethanol is the usual reagent for the Williamson ether synthesis, which is an SN2 reaction.
(Not on topic, but my textbook also had something confusing about whether it’s better to have a tertiary alkyl halide substrate or a tertiary alkoxide nucleophile for that SN2 reaction…I may ask about it on the comments section of that page)
But as you say it’s not that the SN2 reaction won’t occur at all, it’s that the E2 reaction will occur in greater proportion, I guess. The Williamson page said how we get alkoxide ion from alcohol in situ with Na or NaH, and if the alkoxide ion is given beforehand, we could use a polar aprotic solvent to prevent elimination from competing.
Also, in my course we hear that aqueous KOH leads to an SN2 reaction while alcoholic KOH leads to an E2 reaction. I heard the explanation is that OH- is a good nucleophile for an SN2 reaction, and when there’s alcohol and KOH the OH- can deprotonate the alcohol to get alkoxide ion, which is a stronger base than a nucleophile, so takes a proton for an E2 reaction. Does this hold up?

By: James Ashenhurst

James Ashenhurst — Wed, 20 Jul 2022 18:14:56 +0000

In reply to J. Fluoride ion (not HF) is perfectly capable of acting as a nucleophile - e.g. TBAF (tetrabutylammonium fluoride)

By: J

Wed, 20 Jul 2022 04:09:59 +0000

F is a very weak nucleophile because it would rather take a H+ and leave than attack an electrophilic center. This article is incorrect in this regard. This is a special exception.

By: DM

DM — Mon, 02 May 2022 19:41:43 +0000

Can one claim that the polarity of the solvent affects also E1/SN1 ratio?
Both pathways form the same carbocation as the product of the rate-determining step. But in the second step, the solvent can interact with the (usually not charged) base-nucleophile. The higher the solvent’s polarity, the stronger the interaction (that lowers the nucleophilicity more than basicity).
Are these interactions negligible here? Does it have empirical support?
Generally, what effect the solvent has in E1/SN1 reactions?

By: James Ashenhurst

James Ashenhurst — Tue, 26 Oct 2021 19:18:23 +0000

In reply to Parr. CH3O(-) is a strong nucleophile, and therefore it will be either SN2 or E2.

By: Parr

Parr — Sat, 23 Oct 2021 11:15:17 +0000

A little doubt, if solvent consists of Ch3O- ions and Ch3Oh the reaction will be Sn1 or Sn2?

By: James Ashenhurst

James Ashenhurst — Fri, 10 Apr 2020 16:28:18 +0000

In reply to Gon. Primary or secondary alkyl halide? Because if it's primary, it should have been clear from the start that it's an SN2. The real question is what happens with secondary alkyl halides. There is vast inconsistency in how instructors handle this, so I strongly suggest SN2 for polar aprotic solvents, and E2 for polar protic.