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Chapter 22: Problem 42

Cis-trans isomerism is also possible in molecules with rings. Draw the cis and trans isomers of 1,2 -dimethylcyclohexane. In Exercise 41, you drew all of the noncyclic structural and geometric isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~F}\). Now draw the cyclic structural and geometric isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{~F}\).

Short Answer

Expert verified
The cis isomer of 1,2-dimethylcyclohexane has both methyl groups on the same side of the ring, while the trans isomer has them on opposite sides. The cis isomer can be represented as: \(\require{chemfig}\chemfig{*6((-,,[:-120]Me)-(<[:-150]Me)-(-)-(-)-(-)-(-))}\) The trans isomer can be represented as: \(\chemfig{*6((-,,[:-120]Me)-(<:[:-30]Me,[:-60])-(-)-(-)-(-)-(-))}\) The cyclic structural isomer of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{F}\) can be drawn as a three-membered ring containing one fluorine atom: \(\chemfig{*3((-F)-(<[:-60]H)(<:[:60]H)-)}\) No geometric isomers are available for C4H7F as it has only one substituent (the fluorine atom).

Step by step solution

01

Understanding cis-trans isomerism

Cis-trans isomerism occurs when two atoms or groups of atoms are attached to the same carbon atom in a molecule, and there are two non-hydrogen atoms separately bonded to these carbon atoms. The isomers are classified as cis if the two non-hydrogen atoms are on the same side of the molecule, and trans if they are on opposite sides.
02

Drawing the cis isomer of 1,2-dimethylcyclohexane

To draw the cis isomer of 1,2-dimethylcyclohexane, first draw a cyclohexane ring. Now, place a methyl group (CH3) on carbon 1 and another methyl group on carbon 2. In the cis isomer, both methyl groups are on the same side of the ring (either both above or both below the plane of the ring). This is the cis isomer: \[ \text{(cis)1,2-dimethylcyclohexane: } \require{chemfig}\chemfig{*6((-,,[:-120]Me)-(<[:-150]Me)-(-)-(-)-(-)-(-))} \]
03

Drawing the trans isomer of 1,2-dimethylcyclohexane

To draw the trans isomer of 1,2-dimethylcyclohexane, follow the same method used for the cis isomer. However, this time, place one methyl group above the plane of the ring and the other one below the plane. By doing this, you will get the trans isomer: \[ \text{(trans)1,2-dimethylcyclohexane: } \chemfig{*6((-,,[:-120]Me)-(<:[:-30]Me,[:-60])-(-)-(-)-(-)-(-))} \]
04

Drawing cyclic structural isomers of C4H7F

To begin with, we are going to draw the cyclic structural isomer of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{F}\), which has one fluorine atom. Note that in our previous example, the exercise asked for non-cyclic isomers. To draw cyclic isomers, we need to have a closed ring structure. Let's start by drawing a three-membered ring containing one fluorine atom: \[ \text{Cyclic structural isomer: } \chemfig{*3((-F)-(<[:-60]H)(<:[:60]H)-)} \] This is the cyclic structural isomer of C4H7F.
05

Drawing geometric isomers of C4H7F

Since the cyclic structure we have drawn in Step 4 has only one substituent (the fluorine atom), there are no geometric isomers (cis or trans isomers) to draw. This is because geometric isomers occur when there are two non-hydrogen atoms or groups of atoms on the same carbon atom. Therefore, we have completed our task of drawing cis and trans isomers of 1,2-dimethylcyclohexane and the cyclic structural isomer of C4H7F. No geometric isomers are available for C4H7F.

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