Question

(a) Explain how the sterochemistry of `SN^1` and `SN^2` differs.
(b) Explain the formation of `60 %` inverted and `40 %` racemic product from a typical `SN^1` reaction.
(c) Account for the following in terms of the rate of solvent in the above problem.
(i) The reaction is first order.
(ii) Under what conditions the rate expression is more than first order ?
(iii) What is the order and molecularity ?

Answer 1

(a) `SN^2` gives complete inversion of the configuration of the attacked `C`. If a free `R^+` were to form in the `SN^1` reaction, it would be achiral because it would possess a plane of symmetry incorporating its three `sigma` -bonds Its `p` atomic orbital could be attacked equally well at either of its faces, giving a completely racemic product.
(b) The actual mechanism does not involve a free `R^(oplus)` As `X` leaves and the bond angles open up from `109^@` to `120^@`, there is room for the solvent molecule (`HS` : Polar protic solvent) to approach from the rear. The anion is also solvated by H-bonding as it leaves `R`

This `T.S` passes onto a di-solvated intimate ion-pair intermediate :
`(HS ....R^+ X^(Ө) ...HS : )`
If the solvation bond to `C` get stronger and `X^(Ө)` leaves completely, an inverted product is obtained.
If H-bond is stronger, it will remove `X^(Ө)` and hence `R^+` is formed which gives racemic product.
(c) (i) Althrough the solvent molecules are present in the rate determing `T.S`., they to not appear in rate expression. In such cases, the reaction in pseudo first order. This shows that solvent molecules intervence in the rate determing `T.S`.
(ii) When an inert solvent (e.g., benzene) is used with a small amount of nucleophilic protic solvent `(MeOH)`, then rate expression is :
`R = K[RX][MeOH]^2` (Third order)
(iii) It will be termolecular, but it is rare. Solvating molecule that helps to remove `X^(Ө)` will appear first, followed by back-side attacking molecule Each solvent is biomolecular.