Which of the following molecules has a plane of symmetry?

• A. $2_{-\mathrm{bromopropane}}$
• B. ${\mathrm{trans}}_{1_{,2}}-{\mathrm{dibromo}}_{\mathrm{cyclohexane}}$
• C. ${\mathrm{cis}}_{1_{,2}}-{\mathrm{dibromo}}_{\mathrm{cyclohexane}}$
• D. $2_{-\mathrm{chlorobutane}}$

Answer: (C)

Verified step by step guidance

1. Identify the molecular structures of each compound mentioned: 2-bromopropane, trans-1,2-dibromocyclohexane, cis-1,2-dibromocyclohexane, and 2-chlorobutane. Visualize or draw their 3D conformations to better understand their symmetry elements.
2. Recall that a plane of symmetry (also called a mirror plane) divides a molecule into two halves that are mirror images of each other. Look for such a plane in each molecule by checking if one half can be reflected onto the other exactly.
3. For 2-bromopropane and 2-chlorobutane, consider their alkyl chain and substituent positions. Because these are open-chain molecules with substituents on a single carbon, assess whether any plane can bisect the molecule symmetrically.
4. For the cyclohexane derivatives, remember that the ring can adopt chair conformations. In the trans-1,2-dibromocyclohexane, the bromines are on opposite sides of the ring, which usually breaks symmetry. In the cis-1,2-dibromocyclohexane, the bromines are on the same side, which can create a plane of symmetry through the ring and substituents.
5. Conclude which molecule has a plane of symmetry by confirming that the cis-1,2-dibromocyclohexane can be divided into two mirror-image halves, while the others cannot due to their substituent positions and molecular geometry.