Meso compund, i am totally lost

[Saitama]

Homework Statement

I have been currently studying optical activity and i am stuck at this point. The book mentions that the compounds (2,3-Dichlorobutane) shown below are super imposable and its a meso compound:

But if we mentally try to keep one structure on the another, the two Cl and hydrogen atoms do not overlap and the compound should not be super imposable, i can't understand what's going on here.

Any help is greatly appreciated!

Homework Equations

The Attempt at a Solution

 

[AGNuke]

That compound is meso structure because there is plane of symmetry present in the compound, i.e. the compound is achiral. That is imagine a line between the two H-C-Cl bonds.
But if we flip H and Cl on one C, the new compound will become chiral. That's why the compound is meso.

 

[Saitama]

That compound is meso structure because there is plane of symmetry present in the compound, i.e. the compound is achiral. That is imagine a line between the two H-C-Cl bonds.
But if we flip H and Cl on one C, the new compound will become chiral. That's why the compound is meso.

I know there is a plane of symmetry but i want to see if its a meso structure on the basis of the previous knowledge like if it is super imposable or not. You see that the mirror images do not overlap each other completely.

 

[AGNuke]

Oh. Just rotate the second compound by 180° along the axis normal to your computer screen. Doing so DO NOT change the configuration of chiral Carbons, meaning the compounds remains the same in every aspect.

Int this case, you will get the first one. So, they overlap. You just need to rearrange the compound to see that.

 

[Saitama]

In this case, you will get the first one. So, they overlap. You just need to rearrange the compound to see that.

I still don't get it. If i rotate the first structure as you said by 180, i get the second structure. :(

 

[AGNuke]

I still don't get it. If i rotate the first structure as you said by 180, i get the second structure. :(

...just add vice-versa in my statement.

Since the relation is between them, it is applicable if we tweak either of them.

If you rotate the first one, you get the second. Rotate the second, you get first. That's why they are in meso configuration.

 

[Saitama]

...just add vice-versa in my statement.

Since the relation is between them, it is applicable if we tweak either of them.

If you rotate the first one, you get the second. Rotate the second, you get first. That's why they are in meso configuration.

I think i get it. Do this method always work for identifying a meso compound? I thought i would ask this because plane of symmetry also fails sometimes in identifying a meso compound but i have never seen such a case. In my book its written that "If we do not see a plane of symmetry, however, this does not mecessarily mean that the molecule is chiral." Do you know examples for this statement? That would be really helpful.

Thank you for all the help AGNuke!

 

[AGNuke]

I take you are preparing for JEE. Under the syllabus, to determine the chirality of a compound, we need to check three types of symmetry, that is, the plane of symmetry, the (alternating) axis of symmetry and the centre of symmetry.

While the above three can be checked in Wedge-Dash representation, Fisher Projection can only be checked for the plane and the axis of symmetry.

In your question, the molecule had a plane of symmetry and the two can be related by 2 degree axial symmetry.

 

[Saitama]

Looking at your profile, seems like you are also preparing for JEE. Don't take me as your competitor, i am only a shy kid trying to learn.

...the (alternating) axis of symmetry and the centre of symmetry.

I have read about it once but haven't ever used it or never came across the questions which involves these other two symmetry.

Thank you for all the help! It cleared my doubts.

 

[Harrisonized]

I'm not sure why nobody has mentioned this yet, but the situation becomes clear when you assign R and S configurations to each stereocenter. A stereocenter is defined as the position of an atom with four unique substituents. Also, lone pairs of electrons are substituents, so don't skip over an atom on which you only see three atomic substituents. This usually happens when nitrogen is locked in a cyclic structure.

What you should do in all optical-activity-type questions is first find and label all the stereocenters as R or S. If such a designation can be given to an atom (for example, carbon), that atom is chiral. For this reason, stereocenters are more commonly called chiral centers. Read the rest of the post after you have learned how to make these designations.

First, assume both molecules are different and nonsuperimposable. One is S,R and the other is R,S. As you should already know, when there are chiral centers in a molecule, the molecule should rotate plane-polarized light.

There is a special case when symmetry ruins this property: when the two molecule are actually the same. Important cases like this deserve a name, a meso compound. Meso compounds do not optically rotate light. For your problem, it's really obvious that the molecules are actually the same, because they're drawn as Fischer projections. In a Fischer projection, the side groups are out of the plane, and the "backbone" is "into" the plane. Rotating a Fischer projection does not change the fact that the side groups are still out of the plane. You should really look up Fischer projections to see why. You can see some good examples if you search Google Images.

It's more difficult to determine whether a molecule is a meso compound if it's not drawn as a Fischer projection. In the case that a molecule is drawn as a dash-wedge structure, one should attempt to recognize if it is a meso compound by exploiting symmetry instead of trying to superimpose images. However, exercise caution. As you have said yourself, planes of symmetry don't always denote meso compounds. You have to first find out if there are chiral centers in the molecule at all. I'm guessing what your book says is another caution: planes of symmetry can be difficult to find.