Consider the following sequence of reactions.

$\underset{{\mathrm{NaHSO}}_3}{\overset{{\mathrm{OsO}}_4}{\to }}$A$\underset{2.<mspace\; width="1em"></mspace>\mathrm{NaOH},<mspace\; width="1em"></mspace>\mathrm{heat}}{\overset{1.<mspace\; width="1em"></mspace>{\mathrm{HIO}}_4}{\to }}$B. The final product (B) is:

1. 

2. 

3. 

4. 

1. ${\mathrm{Cl}}^{\oplus }$

2. Cl^-

3. Cl

4. Cl²⁺

The major product expected from the mono-bromination of phenyl benzoate is:

1.
2.
3.
4.

(A) C₈H₁₀ $\stackrel{{\mathrm{KMnO}}_4}{\to }$ (B)  C₈H₆O₄$\underset{\mathrm{Fe}}{\overset{{\mathrm{Br}}_2}{\to }}$  C₈H₅BrO₄  (C)  (one-product only)

The structure of A in the above-mentioned reaction is-

1.		2.
3.		4.

$\mathrm{Ph}-{\mathrm{CO}}_2\mathrm{H}\stackrel{{\mathrm{SOCl}}_2}{\to }\mathrm{A}\stackrel{{\mathrm{Me}}_2\mathrm{NH}}{\to }\mathrm{B}\stackrel{\left(\mathrm{c}\right)}{\to }\mathrm{Ph}-\mathrm{CHO}$, unknown reagent (c) is:

1. LiALH₄

2. NaBH₄

3. LiAlH₄(t-BuO)₃

4. PCC/CH₂CL₂

The product (B) in the above-mentioned reaction is:

1.		2.
3.		4.

In the above reaction o/p ratio will be highest when:-
(1) R=-CH₃
(2) R=-CH₂-CH₃
(3) R=-CHMe₂
(4) R=-CMe₃

$\stackrel{{\mathrm{Cl}}_2}{\to }$ (A), Product (A) of this reaction is:

(a) 

(b) 

(c) 

(d) 

 $\underset{{\mathrm{H}}_2{\mathrm{SO}}_4}{\overset{{\mathrm{HNO}}_3}{\to }}\left(\mathrm{A}\right)\underset{\mathrm{HCl}}{\overset{\mathrm{Zn}\left(\mathrm{Hg}\right)}{\to }}\left(\mathrm{B}\right),$Product (B) of this reaction is:

(a) 

(b) 

(c) 

(d) 

p-Toluidine reacts with benzene diazonium chloride to form compound, which on boiling with aq. H₂SO₄ give .......... products:
(1) 3
(2) 2
(3) 4
(4) 5