2

Consider the following reaction:

enter image description here

Our interest is the second step. Notice that after protonation of oxygen, the water molecule attacks in an S$_{\text{N}}2$ fashion.

My question is, why does this reaction follow S$_{\text{N}}2$ mechanism? A protonated oxygen is an excellent leaving group. Therefore it must result in a carbocation formation. In other words, the reaction must follow S$_{\text{N}}1$ mechanism.

What am I missing here?

Source: Chapter-9, Organic chemistry by David Klein

newbie105
  • 397
  • 1
  • 13
  • 1
    "Must" is a very strong word. Why can't a good leaving group take part in an SN2 reaction? Surely you are aware that iodides react faster than chlorides in SN2 reactions, precisely because iodide is a better leaving group than chloride. So leaving group ability clearly affects both SN1 and SN2 reactions. – orthocresol Jan 07 '21 at 14:52
  • @orthocresol I understand. My intuition here was that $\ce{R-O+}$ doesnt require a water molecule to initiate the opening, thereby forming a carbocation on its own – newbie105 Jan 07 '21 at 14:55
  • So I guess this is a purely experimental determination? – newbie105 Jan 07 '21 at 14:57
  • The main difference here is that because the leaving group is tied to the electrophile, even if it does leave, it snaps right back. Also, in general, you frequently require coaxing to get an SN1 reaction to take place. In other words, you need to drive that reaction forward. The electrophile doesn't just pop off that easily. – Zhe Jan 07 '21 at 15:14
  • @Zhe Im not sure why thats the case. In my opinion the oxygen wants to get rid of the $+$ charge ASAP. Why would it join back again? Am I missing something? – newbie105 Jan 07 '21 at 15:18
  • 2
    Would you rather have that positive charge on an oxygen or a carbocation? – Zhe Jan 07 '21 at 15:22
  • 2
    The facts dictate that the mechanism is SN2. The product is a trans-diol. – user55119 Jan 07 '21 at 16:01

0 Answers0