Stereoisomerism in dimethylcyclohexane

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1,1-dimethylcyclohexane has no stereoisomers due to the absence of stereocenters, making it achiral. In contrast, 1,2- and 1,3-dimethylcyclohexane can exist as cis and trans isomers, with the cis forms being meso due to their planes of symmetry, while the trans forms are chiral and have enantiomers. The discussion emphasizes that even molecules with stereocenters can be achiral if they possess symmetry. The concept of time average symmetry in cyclohexane chairs is introduced, indicating that these conformations can also be treated as achiral. Overall, the analysis confirms the stereochemical properties of dimethylcyclohexane derivatives based on their structural characteristics.
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Homework Statement



1,1-dimethylcyclohexane No stereoisomers
1,2-dimethylcyclohexane (Cis,Trans) Do 1,2- , 1,3- exhibit enantiomerism?
1,3-dimethylcyclohexane (Cis,Trans)
1,4-dimethylcyclohexane (Cis-Achiral,Trans-Achiral)

Please justify your answer.

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The Attempt at a Solution



All the fields were blank. I got some of them. Please verify whether they are correct or not.
 
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Here are the steps to solving these problems:

1) Are there any stereocenters? These are atoms connected to 4 different "things." No stereocenters means the molecule is achiral. This means it has no enantiomer, and it's mirror image is the same molecule. It also means the molecule has a plane of symmetry.

2) Is there just one stereocenter? The molecule must be chiral, ie it does have an enantiomer.

3) Are there multiple stereocenters? This is when you have to look for symmetry. Can you cut the molecule in half, along any plane, and end up with the exact mirror reflection on either side? In this case the molecule is still achiral even though it has stereocenters. These kind of molecules are called "meso"

Now I'll answer your questions:

1) Correct. There are no stereocenters here. There are multiple planes of symmetry.

2)+3) These can be addressed together. Both Cis isomers are meso. They have two stereocenters but also possesses plains of symmetry.

Both Trans isomers are chiral and thus exist as two unique enantiomers. Note that both molecules possesses a special kind of symmetry called rotational symmetry which often fools people into thinking they're achiral. Rotational symmetry means you can spin the molecule around a certain number of degrees and end up with the same molecule. Rotational symmetry does NOT necessarily make a molecule achiral.

4) Correct on both counts. The most obvious reason is they both have planes of symmetry. It may or may not be obvious that they both lack stereocenters too.
 
Thanks Spirochete. Your explanation was useful.

Did you answer 2),3) while referring to the chair conformer of dimethylcyclohexane. Your answers strongly indicate that you have considered dimethylcyclohexane as a molecule with a planar ring. For 1),4), it doesn't make any difference. We can end up with the right answer even if we take dimethylcyclohexane's ring as a plane.
 
I believe the definition of achiral is that molecule has a plane of symmetry in any conformation. Although cyclohexane chairs are never flat, that is one possible conformation.

Also achiral cyclohexane chairs spend exactly equal amounts of time in the two conformations. So you can think of them as having time average symmetry. This concept of time average symmetry will be useful later when you learn about NMR (nuclear magnetic resonance imaging).
 
Assuming dimethylcyclohexane exists always as a chair, what would the answer for 2),3) be?
If you encounter any molecule whose mirror is nothing but the molecule which results after flipping(which you should assume to be an instantaneous process), then treat it as achiral. Though this might not be the actual definition of an achiral molecule, this, most probably, is the answer which was expected.
 
An achiral chair molecule spends 50% of time in conformation "A" and 50% of the time in exactly opposite conformation "B" so it has time average symmetry. The flip happens almost infinitely fast compared to the time spent in each conformation so you're correct that for most purposes we can assume exactly 50% of the time is spent in each "perfect" chair conformation.

Given this fact, none of my answers have changed.
 
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