The compound `C_8H_9Cl(A)` on treatments with KCN followed by hydrolysis gives `C_9H_10O_2(B)`.Ammonium salt of B on dry distillation yields C.Which reacts with alkaline solution of bromine to gives `C_8H_11N`.(D),Another compound E `(C_6H_10O)` is obtained by the action of nitrous acid on D. or by the action of aquouspotash on A,E on oxidation gives `F(C_8H_10O)` Which gives the inner anhydride G on heating. The compound D is reaction with `CHCl_3+NaOH` gives a compound H.
The compound A on reaction with AgCN gives :

Options

A. Compound H
B. Compound B
C. Compound G
D. Compound D

Explanation

To solve the problem step-by-step, we will analyze the reactions and compounds mentioned in the question systematically.

### Step 1: Identify Compound A
The compound `C8H9Cl (A)` reacts with `KCN` followed by hydrolysis to yield `C9H10O2 (B)`.
- The presence of chlorine suggests that `A` could be a benzyl chloride derivative. A plausible structure for `A` could be `C6H5CH2Cl` (benzyl chloride) or a similar compound.

### Step 2: Reaction of A with KCN and Hydrolysis
When `A` reacts with `KCN`, the chlorine is replaced by a cyano group, forming an intermediate which upon hydrolysis gives `B (C9H10O2)`.
- The hydrolysis of the cyano group leads to the formation of a carboxylic acid. Thus, `B` could be `C6H5CH2COOH` (benzoic acid derivative).

### Step 3: Formation of Compound C
The ammonium salt of `B` undergoes dry distillation to yield `C`.
- The dry distillation of an ammonium salt of a carboxylic acid typically yields an amine. Thus, `C` could be `C6H5CH2NH2` (benzylamine).

### Step 4: Reaction of C with Alkaline Bromine
`C` reacts with an alkaline solution of bromine to give `C8H11N (D)`.
- This reaction is likely a bromination of the amine, resulting in a secondary amine. Thus, `D` could be `C6H5CH(CH3)NH2` (a substituted amine).

### Step 5: Formation of Compound E
Compound `E (C6H10O)` is obtained by the action of nitrous acid on `D` or by the action of aqueous potash on `A`.
- The reaction of a primary amine with nitrous acid typically yields an alcohol. Thus, `E` could be `C6H5C(OH)(CH3)`.

### Step 6: Oxidation of E to Form F
`E` on oxidation gives `F (C8H10O)`.
- The oxidation of an alcohol typically results in a ketone or aldehyde. Thus, `F` could be `C6H5C(=O)CH3`.

### Step 7: Heating F to Form G
`F` gives an inner anhydride `G` on heating.
- The heating of a compound with two carboxylic acid groups or derivatives can lead to the formation of an anhydride. Thus, `G` could be an anhydride derived from `F`.

### Step 8: Reaction of D with CHCl3 + NaOH
The compound `D` reacts with `CHCl3 + NaOH` to give compound `H`.
- This is a classic carbylamine reaction, where an amine reacts with chloroform and a base to form isocyanides. Thus, `H` could be `C6H5C(CH3)N=C=O`.

### Step 9: Reaction of A with AgCN
The compound `A` on reaction with `AgCN` gives:
- The reaction of an alkyl halide with `AgCN` typically yields an isocyanide. Thus, `A` gives the same isocyanide as `H`.

### Final Summary of Compounds
- **A**: C8H9Cl (Benzyl chloride)
- **B**: C9H10O2 (Benzoic acid derivative)
- **C**: C8H11N (Benzylamine)
- **D**: C8H11N (Substituted amine)
- **E**: C6H10O (Alcohol)
- **F**: C8H10O (Ketone/Aldehyde)
- **G**: Inner anhydride
- **H**: Isocyanide