When I initially learned about substitution reactions, there would be times where a poor leaving group had to be changed into a better one in order to proceed with a substitution reaction. For instance, changing a poor leaving group (-OH) to a better one, such as water: enter image description here

After the "good" leaving group is made, it is displaced via an SN2 reaction. The displacement of the water required a nucleophile. However, I recently came across this mechanism (dehydration): enter image description here

In this case, the water leaves by itself. My question is:

If I ever come across an instance where water (a good leaving group), must leave, how should I assess whether it can leave BY ITSELF or not?

  • 1
    $\begingroup$ Generally, you look at the cationic intermediate to see if it is stabilized or not. For example, creating a cation from the primary alcohol is too disfavored. Secondary cations on the other hand, are reasonably attainable though some forcing conditions may be required. $\endgroup$
    – Zhe
    Commented Nov 27, 2018 at 14:22
  • $\begingroup$ That does make sense since a primary carbocation is far too unstable. So would it be alright to just displace water if it results in a secondary or tertiary carbocation? $\endgroup$
    – LamGyro
    Commented Nov 27, 2018 at 14:55
  • $\begingroup$ Sn1 reactions of secondary and tertiary alcohols are standard fare in a beginning organic chemistry course. $\endgroup$
    – Zhe
    Commented Nov 27, 2018 at 18:44

1 Answer 1


In that kind of reactions, the full explanation would require advanced quantic calculation to determine the energy of all the present entities.

In chemistry, we usually prefer giving a half-quantitative explanation to explain the formation of the observed products.

In your example A —> B + C + D, A is a secondary alcohol and it CAN be transformed to a carbocation leading to B... but you can see it is far from perfect since B represents only 50% of the yield. The result would be much more clear-cut with a primary alcohol (unstable carbocation) or a tertiary one (very stable carbocation). I think you are expected to explain all that.

  • $\begingroup$ Dehydration represents 100% of the yield as shown in the figure. 3 possible isomers result from the secondary cation; this is not an indication of how stable or otherwise the secondary cation is. $\endgroup$
    – Waylander
    Commented Nov 27, 2018 at 14:46
  • $\begingroup$ @Waylander Yes you are right. I think I was too focused on the questions I had at the time and I will probably change my answer $\endgroup$
    – SteffX
    Commented Nov 27, 2018 at 14:54

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