When learning about E1 we were taught that three things are important factors to judge if the reaction is E1 or E2

  1. Substrate ( 3,2,1 favours E2 but only 3 favours E1 )
  2. Base ( Stronger Base favours E2 , weaker Base disfavours E2 hence suggests E1 ) But E1 requires a base however weak
  3. Leaving group ( E1 is more sensitive to LG because it initiates E1 ) Bad LG disfavours both , excellent LG favours E1 more

This was from David M Kleins Book " Organic Chemistry I as a Second Language " But in Section 11.6 he says this Markonikov addition of OH and H

fact, the reverse path (starting from the alcohol and ending with the alkene) is a reaction that we have already studied. It is just an E1 reaction

Then he goes on to show this picture Equilibrium between addition of OH and H markonikov and Elimination of OH

Initial question : How do you have an elimination reaction without a base, that too in the presence of an acid ? (Although I understand that E1 's rate doesnt depend on a Base but it is still needed) Answer : there is a Base ( H20) which makes it look very much like E1 .

Edit : it behaves very much like an E1 but with an extra step

Step 0: converts the bad LG to an excellent LG

Step 1: LG leaves

Step 2 : Base abstracts proton to make alkene

New question: Why does water behave like a base in an acidic mixture? And why does it not react with the H+ already in the solution ?

  • $\begingroup$ you are dehydrating an alcohol.. why not? Also, no, base is not needed for all E1 reactions.. $\endgroup$ Sep 2, 2020 at 16:40
  • $\begingroup$ then what are the necessary conditions for E1 reaction ? $\endgroup$ Sep 2, 2020 at 17:03
  • $\begingroup$ so first of all, concentrated H2SO4 also has some amounts of water in it, so you have a base to work with. I think, "Why does water behave like a base in an acidic mixture? And why does it not react with the H+ already in the solution?" is indeed possible, but this reaction is in equilibrium forming some definite amounts of both. So, water is anyway abstract a proton. $\endgroup$
    – Floatoss
    Jan 6, 2021 at 12:39


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