When ortho dibromobenzene is subjected to mononitration X number of product are formed and when meta dibromobenzene is subjected to mononitration, Y number of products are formed. Report answer as XY.

Answer

To solve the problem of how many products are formed when ortho-dibromobenzene and meta-dibromobenzene are subjected to mononitration, we can break it down into a step-by-step analysis.

### Step 1: Understand the Structures
1. **Ortho-dibromobenzene** has two bromine substituents at the ortho positions (1,2-positions) of the benzene ring.
2. **Meta-dibromobenzene** has two bromine substituents at the meta positions (1,3-positions) of the benzene ring.

### Step 2: Mononitration of Ortho-dibromobenzene
1. When ortho-dibromobenzene is subjected to nitration (using a nitrating agent like HNO3 and H2SO4), the nitro group (NO2) can be introduced at different positions.
2. The possible positions for the nitro group in ortho-dibromobenzene are:
- At the position where one bromine is located (ortho to one bromine).
- At the position where the other bromine is located (ortho to the second bromine).
- At the position that is para to one of the bromines (which is also ortho to the other bromine).
3. Therefore, the possible products formed are:
- Nitro at position 3 (ortho to one bromine)
- Nitro at position 4 (para to one bromine)
- Nitro at position 5 (ortho to the other bromine)
4. Thus, **X = 2** distinct products.

### Step 3: Mononitration of Meta-dibromobenzene
1. When meta-dibromobenzene is subjected to nitration, the nitro group can also be introduced at different positions.
2. The possible positions for the nitro group in meta-dibromobenzene are:
- At the position that is ortho to one bromine (position 2).
- At the position that is ortho to the other bromine (position 6).
- The nitro group cannot be placed at the meta position relative to either bromine because that position is already occupied by a bromine.
3. Therefore, the possible products formed are:
- Nitro at position 2 (ortho to one bromine)
- Nitro at position 6 (ortho to the other bromine)
4. Thus, **Y = 2** distinct products.

### Step 4: Combine the Results
1. We have determined that for ortho-dibromobenzene, there are 2 products (X = 2).
2. For meta-dibromobenzene, there are also 2 products (Y = 2).
3. Therefore, the final answer expressed as XY is **22**.

### Final Answer
**22**