Write bond line structural formulas for two constitutionally isomeric primary alkyl bromides with the formula C4H9Br, a secondary alkyl bromide, and a tertiary alkyl bromide with the same formula.

To address your question, let’s first clarify what we mean by constitutionally isomeric alkyl bromides. These are compounds that have the same molecular formula, C4H9Br in this case, but differ in the connectivity of their atoms.

Primary Alkyl Bromides

1. **1-Bromobutane**: This is the simplest primary alkyl bromide. The bond line structural formula can be represented as:

       Br  
        |  
H3C – CH2 – CH2 – CH3

2. **2-Bromobutane**: This is another primary alkyl bromide, where the bromine is bonded to the second carbon. The bond line representation is:

       Br   
         |  
H3C – CH – CH2 – CH3
        |  
        H

Secondary Alkyl Bromide

1. **3-Bromobutane**: As a secondary alkyl bromide, the bromine atom is attached to the carbon that is connected to two other carbon atoms. The structural formula is represented as:

        Br  
          |  
H3C – CH – CH2 – CH3
          |  
          H

Tertiary Alkyl Bromide

1. **tert-Butyl bromide (1-Bromo-2-methylpropane)**: In this case, the bromine is attached to a carbon atom that is bonded to three other carbon atoms. The bond line structural formula can be shown as:

          Br  
          |  
         C  
       / |  
    H3C   C−H  
       \
       CH3

In summary, we’ve represented the bond line structural formulas for two constitutional isomeric primary alkyl bromides, one secondary alkyl bromide, and one tertiary alkyl bromide, all matching the formula C4H9Br. Understanding these structures is crucial for studying organic chemistry and illustrates how the arrangement of atoms can lead to different chemical properties.