Explain the following :
(a) Esters can be prepared by the reaction of `RCOOH` and `R'X` in basic medium. Why is this method not suitable for the preparation of `t-butyl` ethanoate ?
(b) How is `t-butyl` ethanoate prepared ?
( c) Explain the rate esterification of the following acids with `MeOH`.

(d) Why cannot `HCl` be used for the conversion of `RCOOH` to `RCOCl` ?
( e) Why can esters be prepared more efficiently by the squence : `RCOOH rarr RCOCl rarr RCOOR'`, rather than `RCOOH rarr RCOOR'` ?
(f) Why are acyl azides less reactive than `ROCOCl` but a little more reactive than anhydride ?
`K_a` of `N_3 H = 2.6 xx 10^-5` and `K_a` of `CH_3 COOH = 1.8 xx 10^-5`.

Explanation

(a) (i)
(ii)
(iii)
The carboxylate anion `(I)` is sufficiently basic and it would bring about the elimination in `3^@` alkyl halides.
(b) It can be prepared by the Markovnikov's addition of `CH_3 COOH` with isobutene.

( c) More the `EDG`, lesser favoured is the reaction (esterification). More the `EWG`, more favoured is the reaction.
`ED` power `[(+I)` effect] increases from `(I)` to `(IV)`. So the rate of esterification of `(I) gt (II) gt (III) gt (IV)`.
Alternatively,as the size of substituents on the `alpha-C` atom increases, the tetrahedrally bonded intermediate becomes more sterically hindered. The greater the crowding, the more sterically hindered. The greater the crowding, the larger is `Delta H` of the activation of the transition state and the slower is the reaction.
(d) The reaction of `RCOOH` with `HCl` to give `RCOCl` and `H_2 O` is a reversible reaction. `RCOCl` and `H_2 O` react rapidly to give starting material. `H_2 O` cannot be removed as it is formed as in the case of ester formation because `HCl (g)` would excape.
`RCOOH + HCl hArr RCoCl + H_2 O`.
( e) Reaction of `RCOOH` with `R'OH` is slow, and the equilibrium must be shifted to the right by removing `H_2 O`.
`RCOOH + R'OH hArr RCOOR' + H_2 O`
`RCOCl` is easily prepared under non-equilibrium conditions if `SOCl_2` is used. The exothermic reaction of `RCOCl` with an alcohol completes by evolving `HX(g)`.
`RCOCl + R' OH rarr RCOOR' + HCl, Delta H =` negative.
(f) `N_3 H` is slightly more acidic than `CH_3 COOH. N_3^(Ө)` is slightly less basic than `CH_3 COO^(Ө)` and , therefore, `N_3^(Ө)` is a slightly better leaving group. `RCON_3` is thus less reactive than `RCOCl` but a little more reactive than the anhydride.