Select the correct answer:
a.
i. `PhOC_(2)H_(5)` ii. `Ph---O---Ph`
iii. PhI iv. `C_(2)H_(5)OC_(2)H_(5)`
b.
i. `p--NO_(2)--C_(6)H_(4)--OC_(2)H_(5)`
ii. `p--NO_(2)--C_(6)H_(4)--O--C_(6)H_(4)--NO_(2)--p`
iii. `C_(2)H_(5)--O--CH_(3)`
iv. `p--NO_(2)--C_(6)H_(5)--I`.
c.
i. `C_(6)H_(5) CH_(2)--O--Ph`
ii. `C_(6)H_(5) CH_(2)--O--C_(2)H_(5)`
iii. `Ph--O--C_(2)H_(5)`
iv. `C_(2)H_(5)--O--C_(2)H_(5)`
d.
i. `C_(6)H_(5)CH_(2)--O--C_(2)H_(5)`
ii. `C_(6)H_(5)--CH_(2)--O--Ph`
iii. `Ph--O--C_(2)H_(5)`
iv. `C_(2)H_(5)--O--C_(2)H_(5)` .
e.
i.
ii.
iii.
iv.
f.
i.
ii.
iii.

Explanation

a. iv. `C_(2)H_(5)O^((-))` acts as a base. It abstracts `H^(o+)` form phenol to form `PhO^((-))` ion.

`C_(2)H_(5)O^((-))` is a stronger nucleophile than `PhO^((-))`. Hence, the product is obtained by path II. (Acidic character: `PhOH gt C_(2)H_(5)OH`. (Basic and nuclepohilic character: `PhO^((-)) lt C_(2)H_(5)O^((-))`)
iii. `CH_(3)O^((-))` acts as a base. It abstraacts `H^(o+)` from p-nitrophenol to form `p-NO_(2)--C_(6)H_(4)O^((-))`.

`CH_(3)O^((-))` is a stronger nucleophile than `p-NO_(2)--C_(6)H_(4)O^((-))`, hence the product is obtained by path II. (Acidic character: `p-NO_(2)--C_(6)H_(4)OH gt CH_(3)OH`) (Basic and nuclepohilic character: `p-NO_(2)--C_(6)H_(4)O^((-)) lt CH_(3)O^((-))`)
c. ii. `C_(2)H_(5)O^((-))` acts as a base. It abstracts `H^(o+)` form PhOH to form `PhO^((-))` ion. `C_(2)H_(5)O^((-))`

d. i. `PhO^(-)` acts as a base and it abstracts `H^(o+)` form `C_(2)H_(5)OH` to form `C _(2)H_(5)O^(-)` . `C_(2)H_(5)O^(-)` is a stronger nucleophile than PhO^(-)` .

e. ii. `PhO^(-)` acts as a base and it abstracts `H^(o+)` form p-nitro phenol to give `p-NO_(2)C_(6)H_(5)O^(-)` . `PhO^(-)` is stronger nucleophile than `p-NO_(2)C_(6)H_(5)O^(-)`
(Acidic character: `p-NO_(2)C_(6)H_(5)OH gt PhOH`) (Basic and nucleophilic character: `p-NO_(2)C_(6)H_(5)O^(-) lt PhO^(-)`)
Moreover, strong EWG in (A) favours ArSN reaction.
f. ii. acts as a base and it abstracts `H^(o+)` from phenol to give `PhO^(-)` . `PhO^(-)` is a stronger nucleophile than `O_(2)N` . Hence, the product is like in (e).