n' number of alkenes yields 2,2,3,4,4- pentamethyl-pentane on catalytic hydrogenation and 'm' number of monochloro structural isomers are possible for this compound.report your answer as (n+m)

Explanation

To solve the problem, we need to determine the values of 'n' (the number of alkenes that yield 2,2,3,4,4-pentamethylpentane upon hydrogenation) and 'm' (the number of monochloro structural isomers for this compound). Finally, we will report the answer as \( n + m \).

### Step-by-Step Solution:

1. **Identify the Product Structure**:
The product given is 2,2,3,4,4-pentamethylpentane. The structure can be drawn as follows:
- The main chain is a pentane (5 carbon atoms).
- The substituents (methyl groups) are located at positions 2, 2, 3, 4, and 4.

Structure:
```
CH3
|
CH3- C - CH - C - CH3
| |
CH3 CH3
```

2. **Determine the Possible Alkenes (n)**:
Since the product is obtained through catalytic hydrogenation, we need to identify the possible positions for a double bond in the alkene that would yield this product upon hydrogenation.

- The double bond can only exist between the carbons where it does not violate the tetravalency of carbon.
- After analyzing the structure, we find that the double bond can be placed between:
- C2 and C3
- C3 and C4
- Hence, there are two possible alkenes that can yield the same product upon hydrogenation.

Therefore, \( n = 2 \).

3. **Determine the Number of Monochloro Structural Isomers (m)**:
To find the number of monochloro structural isomers, we need to consider the different positions where a chlorine atom can replace a hydrogen atom in the structure.

- The structure has three types of hydrogen atoms:
- 3 equivalent hydrogens on C1
- 3 equivalent hydrogens on C2
- 3 equivalent hydrogens on C3
- 3 equivalent hydrogens on C4
- The unique positions for substitution are:
- Substituting on C1 (1 isomer)
- Substituting on C2 (1 isomer)
- Substituting on C3 (1 isomer)
- Substituting on C4 (1 isomer)
- Substituting on C5 is not unique since it is equivalent to substituting on C1.

Therefore, the total number of monochloro structural isomers is \( m = 4 \).

4. **Calculate n + m**:
Now, we can calculate \( n + m \):
\[ n + m = 2 + 4 = 6 \]

### Final Answer:
The final answer is \( 6 \).