# Which elimination is this?

This is the question: $$\ce{Cl-CH2-CH2(Cl)2->[alc KOH][Δ]?}$$ Now, I got the answer right but I think that the reaction should go by the $$\ce{E1}$$ mechanism because of the high temperature/heat, however, my teacher says that the mechanism will be $$\ce{E2}$$. I don't understand why it will be like that. Everyone seemed to agree, so I didn't have the guts to ask in class. Please tell me why it's $$\ce{E2}$$? And then, what's the use of high temperature/heat? It was there to make ionization easy, right?

• I don't think high temperature facilitates E1, I also think it should go by E2... It's better to ask your teacher such conceptual doubts. Don't be shy in such cases.... – Zenix Apr 23 '20 at 20:23
• I guess this link about alpha elimination will be useful, it woyuld be similar to E1cb I dare say – Yusuf Hasan Apr 23 '20 at 20:26
• @Zenix E1 is generally more favorable under heating just because you need to promote dissociation of the electrophile. That doesn't mean of course that if you need heating, it's an E1... – Zhe Apr 23 '20 at 21:18
• @Zhe won't any elimination reaction be more favourable at high temperature, why specifically E1? I read your answer's last para... And first para.... I guess for E1 to occur, dominating factor will be stability of carbocation, which isn't stable (due to reasons you mentioned) – Zenix Apr 23 '20 at 22:38
• @Zenix If that was your claim, I don't really have any complaints. Frequently E2 reactions are spontaneous enough that no heating is required. In that case, if you heat it, yes, it will go faster, but you're also more likely to have less selective reactions or side reactions. – Zhe Apr 23 '20 at 22:56

An E1 mechanism is highly unlikely. I don't know what level you're at, but one could invoke the Hammond Postulate in this case, and look at the high energy cation intermediate. There are three chloride leaving groups that could generate one of two intermediates. Both of these intermediates are primary cations. A chloride substituent isn't likely to be able to stabilize this much via resonance but more likely destabilize due to inductive effects.

The other potential mechanisms to consider (that do not involve a cation intermediate) are E2 and E1cb. E1cb might work for the first step depending on where the anion is formed, especially if the inductive effect of the chlorines is enough to help stabilize the carbanion. E2 just requires an anti-periplanar arrangement, which isn't hard here with free rotation.

The reason I'd favor E2 and the reason the reaction likely requires heat is the fact that hydroxide is just an OK base. Add to that the fact that it's in a polar protic solvent that might further dampen basicity. If you tried this reaction with an amide base (e.g., LDA) in toluene, I would guess no heating is necessary.

Heating is generally helpful for E1 because the dissociation of the leaving group is slow. However, the need to heat a reaction is by no means indicative that it is E1 (it might not even be an elimination!). Any number of factors could conspire to make the reaction slow, in which case, heating might help make the reaction proceed at a faster, more practical rate.

• Ok, thanks. But i still don't understand "it's in a polar protic solvent that might further dampen basicity". I understand that polar protic solvent will dampen nucleophilicity but why basic strength ? Being in polar protic solvent(alcohol here) should increase basicity because $\ce{C2H5O-}$ ion is more basic than $\ce{OH-}$ , you can verify. Being a strong base is the reason why elimination occurs instead of substitution. Correct me if i'm wrong. – Physicsa Apr 23 '20 at 22:49
• No, I'm saying that hydroxide would be a stronger base in something like DMSO which is aprotic. The hydrogen bonding in a protic solvent will make basicity weaker. It doesn't matter if you're looking at an alkoxide or hydroxide. – Zhe Apr 23 '20 at 22:55
• Cf. the pKa values for water in water and in DMSO: studocu.com/en-ca/document/university-of-victoria/… – Zhe Apr 23 '20 at 22:58
• can you please provide a reason why hydrogen bonding would decrease basicity. I just can't understand why. – Physicsa Apr 23 '20 at 23:16
• The same reason hydroxide is more basic in a nonpolar solvent. Generally, bases have high energy lone pairs; that's what makes them a base. In our case, the base is also negatively charged. Any stabilization of the base, for example, via charge-dipole interactions in a polar solvent or hydrogen bonding in a protic solvent, make it less basic. – Zhe Apr 23 '20 at 23:24