A and B in the following reactions are `:`
`RCOR' underset( KCN)overset(HCN)(rarr) A overset( B ) (rarr) R.R' ( C ) ( OH ) (CH_(2)NH_(2))`

Options

A. `A=R.R'(C ) ( CN ) ( OH), B = L i[AlH_(4) ]`
B. `A=R.R'(C ) ( COOH ) ( OH ) , B = NH_(3)`
C. `A=R.R'(C ) ( C N ) ( OH ) , B = H _(3) O ^(+)`
D. `A= R.R' CH_(2) CN, B= NaOH`

Answer

To solve the question regarding the reactions involving A and B, we need to analyze the given reactions step by step.

### Step 1: Identify the Initial Compound
The initial compound is given as \( R-C(=O)-R' \). This structure represents a ketone, specifically an unsymmetrical ketone since R and R' are different groups.

**Hint:** Remember that ketones have the general structure \( R-C(=O)-R' \), where the carbonyl carbon is bonded to two alkyl groups.

### Step 2: Reaction with KCN and HCN
The ketone undergoes a nucleophilic addition reaction with potassium cyanide (KCN) and hydrogen cyanide (HCN). In this reaction, the cyanide ion (CN⁻) acts as a nucleophile and attacks the electrophilic carbon of the carbonyl group.

**Hint:** Nucleophilic addition involves a nucleophile attacking an electrophilic carbon, leading to the formation of a tetrahedral intermediate.

### Step 3: Formation of Compound A
After the nucleophilic attack, we form a tetrahedral intermediate. This intermediate can be represented as \( R-C(OH)(R')-C \) (where C is the cyanide). Upon protonation, we get compound A, which is \( R-C(OH)(R')-CN \).

**Hint:** The addition of water or H⁺ to the tetrahedral intermediate leads to the formation of an alcohol (or hydroxyl compound).

### Step 4: Reaction of Compound A
Next, compound A undergoes a further reaction. The cyanide group (CN) in compound A is converted into an amine group (–CH₂NH₂) through reduction. This reduction is typically achieved using lithium aluminum hydride (LiAlH₄), a strong reducing agent.

**Hint:** Reducing agents like LiAlH₄ can convert nitriles (CN) into primary amines (–CH₂NH₂).

### Step 5: Identify Compound B
After the reduction of compound A, we obtain compound B, which is \( R-C(OH)(R')-CH₂NH₂ \). This indicates that the cyanide group has been successfully reduced to an amine.

**Hint:** The product of the reduction will have the same carbon skeleton as the starting ketone but will now have an amine group instead of the cyanide group.

### Conclusion
To summarize:
- **Compound A** is \( R-C(OH)(R')-CN \) (the product of the nucleophilic addition).
- **Compound B** is \( R-C(OH)(R')-CH₂NH₂ \) (the product of the reduction of A).

### Final Answer:
- A = \( R-C(OH)(R')-CN \)
- B = \( LiAlH₄ \) (lithium aluminum hydride)