Final answer:
To draw a diene's structure that reacts with one equivalent of HBr, you would typically draw a compound like 1,3-butadiene and add HBr across one of the double bonds following Markovnikov's rule to get 3-bromo-1-butene.
Step-by-step explanation:
The student is asking to draw the structure of the diene that reacts with one equivalent of HBr. A diene is an organic compound containing two double bonds. When reacting with HBr, a typical diene such as 1,3-butadiene would add HBr across one of its double bonds. The product would be a substituted alkene, with one less double bond than the original diene.
For this exercise, let's consider the reaction of 1,3-butadiene with HBr. Since we are adding one equivalent of HBr, only one of the double bonds will react. Here is how you would draw it step-by-step:
- Draw the backbone of 1,3-butadiene, which consists of four carbon atoms in a row with double bonds between carbon 1 and 2, and carbon 3 and 4.
- Add hydrogen atoms to give each carbon atom four bonds in total, remembering that the double bonds count as two of those bonds.
- Now add one equivalent of HBr across one of these double bonds. Follow the rule of Markovnikov's addition where H adds to the carbon with more hydrogen atoms, and Br adds to the carbon with fewer hydrogen atoms.
- After reacting with HBr, the 1,3-butadiene becomes 3-bromo-1-butene.
The final structure will have a three-carbon chain with one double bond and one bromine atom attached to the third carbon atom.