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In the following question, which I tried to solve, I became confused by which way I should proceed. Here is my thought process:

  1. SN1: If Cl detaches and forms a secondary carbocation, then there's no nearby groups to stabilize it or make a shift to turn it into a tertiary cation.
  2. SN2: I can't make out how exactly SN2 would occur
  3. E1: I can't make out how exactly E1 would occur
  4. E2: There's no stereospecific data given in the question whether the methyl group and the hydrogen lie in one plane or not, so I couldn't make out a way to do anti-elimination.

The Answer is given : E1

enter image description here


so @benzene , This is the E1 product as per the mechanism , Am i right ?enter image description here

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E1 is the correct answer. There are a couple of factors that lead me to this conclusion. First of all, heat favors elimination over substitution, meaning that this reaction will either be E1 or E2. Now that we know that this is an elimination reaction, it is just a matter of deciding if it is E1 or E2. E2 requires a strong base to occur and since $\ce{H2O}$ is a week base, we know that this reaction must follow an E1 mechanism. While E1 is favored by a tertiary carbocation intermediate, both E1 and E2 can occur with a secondary carbocation.

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  • $\begingroup$ Funny that your account name is Benzene but your account picture is a tesseract. $\endgroup$ – Aaron John Sabu May 14 '17 at 11:30

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