(a) Ester having only one `alpha-H` does not undergo Claisen condensation `(C.C)` with `EtONa` because the ketoester formed cannot be converted to its enolate anion due to the absence of the second `alpha-H` atom. As a result, the equation does not shift to right.
However, in the presence of a very strong base such as `Ph_3 Na`, such esters undergo `C.C` to give `beta-`keto ester because a very strong base acts in two ways.
(i) It withdraws the weak acidic `alpha-H` atom irreversily from the ester (analogue `C.C` mechanim step (i) is highly reversible).
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(ii) It completely removes one of the aqueous products i.e. `C_2 H_5 OH`, so that equation shifts to the right.
Mechanism of `C.C.:`
The `beta-`keto ester having an active methylene group is acidic and reacts with `C_2 H_4 ONa` to form enolate (Step (iv)). The subsequent acidification with aq. `CH_3 COOH (1 : 1)` regenerates `beta-`ketoster.
The first three steps are in unfavourable equation state. Hence, excess of `C_2 H_5 ONa` is used to force the equation to shift in the forward direction by the formation of a resonance-stabilished enolate anion [Step(iv))]. This is substantiated by the fact that esters having only one `alpha-H` atom do not undergo `C.C` with `C_2 H_5 ONa`.
(b)
(1)
(2)
`(2)` is more resonance stabilised than `(1)` due to `(-I)` effect of `(Ph)` group. So `(C=Cl)` bond is formed and os difficult tp hydrolyse by `H_2 O`.
Hence, acetyl chloride is hydrolysed rapidly.
( c) Because their enolate anions are more stabilised by resonance, their negative charge is delocalised into two carbonyl groups.
