compound x C29H48BrCl reacts with excess H2/pd to give C29H56BrCl. How many double bonds and how many rings are in X?

To determine the number of double bonds and rings in compound X (C29H48BrCl), we will need to analyze its structure and count the corresponding functional groups.

1. Start by identifying the total number of hydrogen (H), carbon (C), bromine (Br), and chlorine (Cl) atoms present in the compound:
- Carbon (C): 29 atoms
- Hydrogen (H): 48 atoms
- Bromine (Br): 1 atom
- Chlorine (Cl): 1 atom

2. Since X reacts with excess H2/pd to form C29H56BrCl, it indicates that each H2 molecule added replaces two hydrogen atoms. Hence, we can calculate the excess hydrogen:
- Excess Hydrogen = Total Hydrogen in C29H56BrCl - Hydrogen in C29H48BrCl
- Excess Hydrogen = 56 - 48 = 8 atoms

3. Now let's determine the number of double bonds in compound X:
- The excess hydrogen indicates that each double bond in compound X must have reacted with two hydrogen atoms, resulting in their conversion into single bonds.
- Therefore, we can calculate the number of double bonds by dividing the excess hydrogen (8 atoms) by 2 (since each double bond would consume two hydrogen atoms):
Number of Double Bonds = Excess Hydrogen / 2
= 8 / 2
= 4 double bonds

4. Finally, let's analyze the number of rings in compound X:
- Counting the exact number of rings in a compound without its structural formula can be challenging. Without further information, we cannot definitively determine the number of rings present in compound X.

In summary, based on the information provided, compound X (C29H48BrCl) contains 4 double bonds, while the exact number of rings cannot be determined without additional information.