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Explain the relative rates of RX with H2...

Explain the relative rates of `RX` with `H_2 O/EtOH` at `25^@ C` as given :
(i) `MeBr (2140)`
(ii) `MeCH_2 Br(171)`
(iii) `Me_2 CHBr (4.99)`
(iv) `Me_3CBr (1010)`
(b) Why is `EtOH` added to water ?

Text Solution

Verified by Experts

(a) In case of the first three halides by `SN^2`, the rate decreases from `(i)` to `(iii)` due to steric hindrance. `H_2O` is a nucleophile. In `(iv)`, mechanism changes from `SN^2` to `SN^1`, that is why there is a sharp increase in the reactivity.
(b) `H_2 O` is a poor solvent for `RX` and `EtOH` is added to increase the ionisation of `H_2 O`.
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