Question

draw the skeletal structure of (r)-3-bromo-5,5-dimethylcyclohex-1-ene. use a dash or wedge bond to indicate stereochemistry of substituents on asymmetric centers, where applicable.

Answer 1

To draw the structure of (R)-3-bromo-5,5-dimethylcyclohex-1-ene, begin with a hexagonal cyclohexane ring with a double bond, then add two methyl groups at the fifth carbon and a bromine atom at the third carbon, using a wedge bond to demonstrate the (R)-configuration stereochemistry.

Step-by-step explanation:

The question requires drawing the skeletal structure of (R)-3-bromo-5,5-dimethylcyclohex-1-ene, indicating the stereochemistry using dash or wedge bonds. In organic chemistry, a skeletal structure is a simplified way of representing a molecule without showing all hydrogen atoms. It focuses on the carbon skeleton and heteroatoms (atoms other than hydrogen and carbon), making it easier to visualize complex structures.

Steps to Draw the Structure

1. Start with drawing the cyclohex-1-ene as a six-carbon ring with one double bond. This will be represented as a hexagon with one side drawn as a double line, indicating the presence of a double bond.
2. Add two methyl groups (–CH3) at the fifth carbon. The methyl groups will be represented as lines (bonds) with 'CH3' at the ends.
3. The bromine atom (Br) will be attached to the third carbon of the cyclohexene ring. To indicate the (R) configuration, and considering that it is the only stereocenter in this molecule, a wedge bond will be used to show that the bromine atom is coming out of the plane of the page towards the viewer.
4. Review the final structure to ensure that each carbon atom in the ring has four bonds (including the implicit hydrogen atoms not shown in a skeletal structure), and double-check that the stereochemistry at the third carbon is correctly portrayed by the wedge bond.

It's important to understand that in skeletal structures, carbon atoms are represented at the vertex (corner) of each line intersection or at the end of a line, and hydrogen atoms bonded to carbons are not drawn. The assumption is that each carbon has enough hydrogen atoms to complete its tetravalency, unless indicated otherwise by the presence of functional groups or heteroatoms.