Final answer:
In the second propagation step of the free-radical bromination of 2,3-dimethylbutane, we can form two unique products with bromine attached to the secondary carbon, and two additional non-unique products with bromine attached to different primary carbon atoms. There cannot be three or four unique products due to the symmetrical nature of the molecule.
Step-by-step explanation:
The second propagation step of the free-radical bromination of 2,3-dimethylbutane involves abstracting a hydrogen atom from the organic substrate and forming a new carbon-bromine bond, along with inorganic coproducts such as hydrogen bromide (HBr).
- a. One product with bromine attached to the secondary carbon: Adding bromine to the secondary carbon (the 2-position or the 3-position of 2,3-dimethylbutane) will result in 2-bromo-2,3-dimethylbutane or 3-bromo-2,3-dimethylbutane, respectively.
- b. Two products with bromine attached to different carbon atoms: This leads to isomers where bromine can be added to primary carbon atoms at the 1-position or the 4-position, yielding 1-bromo-2,3-dimethylbutane or 4-bromo-2,3-dimethylbutane.
- c. Three products with bromine attached to different carbon atoms: Considering the symmetrical nature of 2,3-dimethylbutane, the third option would be identical to the second. There are no distinct three carbons where bromine can add to form unique products different from those mentioned in b.
- d. Four products with bromine attached to different carbon atoms: Again, due to symmetry, there cannot be four unique products. As with the previous point, there are no additional unique carbons for bromine to attach to form different products.
The examples provided in the initial question (2,2-dibromobutane, 2-chloro-2-methylpropane, etc.) suggest various types of substituted alkanes. However, for the specific reaction question regarding 2,3-dimethylbutane, we are focusing on the bromination products.