(a) Draw the monomer 1,2 -dibromoethene. (b) Draw the polymer formed from this monomer, making it at least three monomer units long.

To draw 1,2 -dibromoethene, first, recognize that it is an ethene derivative. Ethene has the molecular formula C2H4, and its structure contains a carbon-carbon double bond. In 1,2 -dibromoethene, two hydrogen atoms are replaced by two bromine atoms. The "1,2" in the name indicates that the bromine atoms are attached to the first and second carbons, resulting in the molecular formula C2H2Br2. The structure of the compound is as follows: \(Br-C=C-Br\)

A polymer is a large molecule formed by joining multiple monomer units through a process called polymerization. For 1,2 -dibromoethene, this process involves breaking the double bond between the carbon atoms in the monomer unit and joining the monomers through single bonds. To draw the polymer, follow these steps: 1. Begin by drawing the initial 1,2 -dibromoethene monomer with a broken double bond, so it will appear as: \(Br-C-C-Br\). 2. Now, draw another 1,2 -dibromoethene monomer, such that the carbon with a free bond on the first monomer can form a single bond with the carbon with a free bond on the second monomer. 3. Add a third monomer to the chain, again linking the carbon with a free bond to a carbon with a free bond on the third monomer. 4. Finally, put parentheses around the entire structure, with a subscript 'n' outside the parentheses to indicate that this is a chain of monomers that can be repeated n times. The final polymer chain with at least three monomer units of 1,2 -dibromoethene will look like this: \([Br-C-C-Br]_1-[Br-C-C-Br]_2-[Br-C-C-Br]_3 \cdots\) or simply \([Br-C-C-Br]_n\)