Answer:

When 4-methyl cyclopentene is treated with OsO4 followed by work-up with NaHSO3/H2O, two different diols would result. Let's break down the reaction steps and explain why.

Step 1: Reaction with OsO4
When 4-methyl cyclopentene reacts with OsO4, it undergoes a process called syn dihydroxylation. This reaction involves the addition of two hydroxyl groups (OH) across the double bond of the cyclopentene ring. The OsO4 acts as a catalyst, facilitating the addition of the hydroxyl groups.

The reaction proceeds as follows:
- The double bond in the cyclopentene ring acts as a nucleophile, attacking the electrophilic OsO4.
- The OsO4 is then reduced, resulting in the formation of an osmate ester intermediate.
- The osmate ester intermediate is then hydrolyzed, leading to the formation of two vicinal diols (1,2-diols) on the cyclopentene ring.

Step 2: Work-up with NaHSO3/H2O
After the reaction with OsO4, the resulting osmate ester intermediate is further treated with NaHSO3/H2O. This reagent is used to reduce the osmate ester back to the diol.

The reaction proceeds as follows:
- The NaHSO3 acts as a reducing agent, breaking the Os-Os bond in the osmate ester and converting it back to the diol.
- The diol is then obtained as the final product.

Conclusion:
In this reaction, when 4-methyl cyclopentene is treated with OsO4 followed by work-up with NaHSO3/H2O, two different diols are formed. This is because the OsO4 adds two hydroxyl groups across the double bond, resulting in the formation of two vicinal diols on the cyclopentene ring. The subsequent reduction of the osmate ester with NaHSO3/H2O converts the intermediate back to the diol, yielding two different diols as the final products.