Comments on: Elimination Reactions Are Favored By Heat

By: Kashyap Patel

Kashyap Patel — Tue, 02 Jan 2024 21:34:31 +0000

There is an issue with the article. Temperature changes don’t alter the activation energy (Ea), because temperature itself doesn’t nothing to stabilize the transition state. The Ea remains the same, but higher temperature allows for the products to have enough kinetic energy to reach the transition state. So the ultimate reason why the elimination product is favoured at higher temperatures is because it leads to an increase in entropy. At lower temperatures the products have a lower probability of overcoming the Ea for the elimination reaction and so the substitution reaction will dominate.

By: Kirti sharma

Kirti sharma — Wed, 09 Nov 2022 08:07:02 +0000

Crystal clear explanation thankyou sir by heart

By: komal

komal — Thu, 10 Mar 2022 04:06:08 +0000

Hi, I don’t understand the unit ‘kcal/g-mol’ mentioned in the second research paper, could you explain that?

By: James Ashenhurst

James Ashenhurst — Sat, 24 Aug 2019 04:06:04 +0000

In reply to Matt. I removed a figure that was in there previously.

By: James Ashenhurst

James Ashenhurst — Sat, 24 Aug 2019 04:04:15 +0000

In reply to zach. I think you'd find that reaction to be very slow at that temperature. If confined to NaOH as nucleophile, a better way to try to get the substitution would be to try to use phase transfer catalysis (18-crown-6) with NaOH in a polar aprotic solvent at somewhat low temperature.

By: James Ashenhurst

James Ashenhurst — Thu, 15 Aug 2019 18:00:33 +0000

In reply to Sabrina jones. Awesome - thanks for letting me know Sabrina!

By: Sabrina jones

Sabrina jones — Fri, 06 Apr 2018 05:50:37 +0000

Thank you so much for these articles – they have been so helpful, clear, and easy to understand as I’ve studied for my midterm!!

By: zach

zach — Sat, 21 Oct 2017 01:08:36 +0000

So, for example could you take 2-bromo-3-ethylpentane and treat it with NaOH (at say -78degC) to get a majority of the 2-ol product as opposed to the -ene?

By: Idan

Idan — Thu, 09 Jun 2016 21:16:30 +0000

In reply to Idan.

*Both meaning both reactions
**What I’m trying to ask is if we should only consider the activation energy of the transition state (as opposed to the products), why does the article mainly covers the ΔG of the products?
Would it be right to claim that the same applies for the transition state? (the transition state of an elimination reaction also includes more species than the competing SN reaction)

By: Idan

Idan — Thu, 09 Jun 2016 20:30:35 +0000

I get the part about elimination being more entropically favorable due to the increase in product species amount, but if you say that we should only consider the ΔG of the transition state don’t they both form only one “transition species”? Why should this entropy reasoning apply to the transition state?