Question

Consider 1-bromo-2-methylpropane and draw the following. (a) The staggered conformation(s) of lowest energy (b) The staggered conformation(s) of highest energy

Short answer

Question: Identify the lowest and highest energy staggered conformations of 1-bromo-2-methylpropane. Answer: The lowest energy staggered conformation has the hydrogen atom opposite to the bromine atom, while the highest energy staggered conformation has the ethyl group opposite to the bromine atom.

Step-by-step solution

Draw the molecule 1-bromo-2-methylpropane

First, draw the main structure of 1-bromo-2-methylpropane with a central carbon bonded to a bromine atom, a methyl group, a hydrogen atom, and an ethyl group (C2H5).

Draw the staggered conformations

To draw the staggered conformations, imagine the molecule as if you were looking down the C-Br bond axis. You need to rotate the molecule so that the groups on the front and back carbons are maximally separated (180° dihedral angle) to achieve the staggered conformations. Draw two different staggered conformations: one with the hydrogen atom (H) opposite to the bromine atom (Br) and another with the ethyl group (C2H5) opposite to the bromine atom (Br).

Identify the lowest and highest energy staggered conformations

Staggered conformations with minimal steric hindrance (i.e., the groups are farther apart) have lower energy, whereas staggered conformations with more steric hindrance have higher energy. (a) In the staggered conformation with the hydrogen atom opposite to the bromine atom, there is less steric hindrance because hydrogen is smaller in size compared to the ethyl group. Therefore, this conformation represents the staggered conformation of lowest energy for 1-bromo-2-methylpropane. (b) In the staggered conformation with the ethyl group opposite to the bromine atom, there is more steric hindrance due to the larger size of the ethyl group. Thus, this conformation represents the staggered conformation of highest energy for 1-bromo-2-methylpropane.