Question

Draw a stereoisomer of trans-1,2-dibromocyclopentane. (Note that the question asks for a different stereoisomer of the named compound and not the named compound itself.) Use the wedge/hash bond tools to indicate stereochemistry where it exists.

In cases where there is more than one answer, just draw one.

Answer 1

A stereoisomer of trans-1,2-dibromocyclopentane is the cis-1,2-dibromocyclopentane, with both bromine atoms on the same side of the cyclopentane ring, visualized by using wedges to indicate the orientation towards the viewer.

Step-by-step explanation:

To draw a stereoisomer of trans-1,2-dibromocyclopentane, we must change the arrangement of the substituents around one or more of the carbons in the cyclopentane ring. Given that the trans stereoisomer means that the two bromine atoms are on opposite sides of the ring, a different stereoisomer would have both bromine atoms on the same side of the ring, which would be the cis isomer.

You would represent the two bromine atoms on adjacent carbon atoms both projecting either above or below the plane of the ring using wedged and dashed lines to indicate three-dimensional orientation. A wedge indicates a bond coming out of the page towards the viewer, while a dash indicates a bond going into the page away from the viewer.

Unfortunately, without visual tools, I cannot provide an actual drawing here. However, imagine the cyclopentane ring as a pentagon with carbon atoms at the corners. Two adjacent carbon atoms would have bromine substituents. If both bromines are drawn with wedges (pointing towards you), it will represent a cis-1,2-dibromocyclopentane, the stereoisomer we wanted to demonstrate.