Show the product, including stereochemistry, of the reaction of the epoxide below with HBr. The image illustrates the molecular structure of camphor. It is depicted using a ball-and-stick model, where atoms are represented as spheres and bonds as sticks connecting them.### Detailed Description:1. **Carbon Atoms (C):** - Represented by grey spheres. - Form the backbone of the camphor molecule, creating a ring structure at the core.2. **Hydrogen Atoms (H):** - Shown as smaller white spheres. - Attached to the carbon atoms, filling the valency requirements in the molecule.3. **Oxygen Atom (O):** - Highlighted in red, indicating its significance. - Single oxygen atom bonded to the carbon framework, contributing to the ketone group in camphor.### Explanation:Camphor is a terpenoid compound with a notable ketone functional group, which is the oxygen atom double-bonded to a carbon within the ring structure. It has a bicyclic structure, typical of many natural terpenes. The arrangement showcases the three-dimensional shape that is crucial for its biochemical properties and aromatic characteristics.

By HBr addition epoxide ring opened Nucleophile addition is opposit plane and more substituted carbon. Given epoxide is (Structure) : Mechanism : 1) Protonation of epoxide oxygen : 2) Nu- Br- attacks :In the next step Br- nucleophile attacks on opposite plane and more hindered carbon, so that epoxide ring is opened. Stereochemistry : Nu- attacks on more hindered (i.e. more substituted carbon and attacking of Nu- to opposite plane Since it is cyclic form hence above the plane is blocked, further attcaking must be below the plane. (i.e. trans addition )