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Chemical Properties of Phenols

A, a colourle...

Question

(A), a colourless solution, gives white precipitate (B) with NaOH solution but the precipitate dissolves in excess of NaOH forming (C). (C) does not give any precipitate with $H_{2} S$ but on boiling with $N H_{4} C l$ , white precipitate (B) appears. (A) also gives yellow precipitate with $A g N O_{3}$ . (A), (B) and (C) are as follows:
(A) : $A l B r_{3}$ (B) : $A l (O H)_{3}$ and (C) : $N a A l O_{2}$
If true enter 1, if false enter 0.

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Solution

(A) is $A l B r_{3}$ , (B) is $A l (O H)_{3}$ and (C) is $N a A l O_{2}$ .
Aluminum bromide (A) reacts with sodium hydroxide to give a white precipitate of aluminum hydroxide.
$A l B r_{3} + 3 N a O H \to A l (O H)_{3} ↓ w h i t e p p t + 3 N a B r$
The precipitate dissolves in excess of NaOH to form sodium meta aluminate which is (C).
$A l (O H)_{3} + N a O H \to N a [A l O_{2}] s o d i u m m e t a a l u m i n a t e + 2 H_{2} O$
Sodium meta aluminate does not give precipitate with hydrogen sulphide.
When sodium metaaluminate is boiled with ammonium chloride, the precipitate of aluminum hydroxide is formed.
Aluminum bromide reacts with silver nitrate to form yellow precipitate of silver bromide.
$A l B r_{3} + 3 A g N O_{3} \to 3 A g B r y e l l o w p r e c i p i t a t e + A l (N O_{3})_{3}$

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