My organic textbook states the pathway for propyl tert-butyl ether in dilute sulfuric acid will follow E1 with a tertiary carbocation intermediate. It mentions that Sn1 will only occur with good nucleophiles.

It does not explain why however. Why is this any different than a tertiary haloalkane in water? I expect that the alcohol is a good leaving group, similar to water.

enter image description here

Vollhardt, P.; Schore, N. Molecules: Structure and Function, 7th edition; W. H. Freeman: New York, 2014​, p350

  • $\begingroup$ Tertiary carbocation? Protonate the ether oxygen, followed by splitting off the propanol together with an H-shift, and then elimination? That sounds a bit weird. You might as welldirectly attack the tertiary H-Atom in the protonated ether with H2O, making this an E2. $\endgroup$
    – Karl
    Commented Sep 9, 2018 at 21:09
  • $\begingroup$ There is no H-shift as far as I know. I have included the relevant excerpt from the book in the post now. $\endgroup$
    – ttdijkstra
    Commented Sep 9, 2018 at 23:19
  • $\begingroup$ Undoubtedly E1 over E2. $\endgroup$
    – user55119
    Commented Sep 11, 2018 at 23:19
  • $\begingroup$ This is tert-butyl ether, not isobutyl! $\endgroup$
    – Karl
    Commented Sep 12, 2018 at 19:03

1 Answer 1


It isn't necessarily the same situation as the haloalkane in water. In the sulfuric acid case, there is a large amount of acid present, and as you rightly say the alcohol and water are similar leaving groups, so the SN1 product is easily decomposed back to the tertiary carbocation. As a result, the mechanism tends towards the thermodynamic product, which apparently is the alkene (probably for entropic reasons.

enter image description here

Compare this with the alkyl halide in water. Here, the halide is a much better leaving group than water. Additionally, there is a low concentration of acid in solution (at most 1 equivalent, from the generation of HX). This means that loss of water from the SN1 product is much less favourable.

That said, I'm sure you'd get a mixture of E1 and SN1 products from the haloalkane reaction you describe, I merely wanted to point out that the ether and haloalkane cases aren't as similar as you may have presumed.

  • $\begingroup$ Is there a difference in temperature between the top and bottom mechanisms as well? (e.g. is the top one under heated conditions vs the bottom one under cold conditions) $\endgroup$
    – chemN00b
    Commented Oct 30, 2022 at 17:41

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