Question #4341e

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Answer 1 · 2016-09-15T21:53:24

Here's how I do it.

Step 1. Draw the bond-line structure.

Bond-line

We see that there are two chiral carbons (those bearing the $"Cl"$ atoms) — they are attached to four different groups.

The maximum number of stereoisomers is $2^n$ , where $n$ is the number of chiral centres.

There is no internal plane of symmetry, so the number of stereoisomers is $2^2 = 4$ .

Step 2. Convert the structure to a wedge-dash formula.

Wedge-Wedge

Here the $"C-Cl"$ bonds are both wedges.

The four possible combinations are: wedge-wedge, dash-dash, wedge-dash, and dash-wedge.

Step 3. Draw the structures for the three remaining combinations..

Dash-Dash

Wedge-Dash

Dash-Wedge