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Conformers of a meso-compound [duplicate]
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Laura Miller
Conformers of a meso-compound [duplicate]
The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:
When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.
Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2 and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.
To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).
The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:
When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.
Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2 and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.
To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).
The staggered conformation of the meso-isomer you have shown does not have a (mirror) plane of symmetry but rather a center of symmetry, which accounts for this conformations lack of optical activity. See the links in the comments above.More
I believe you misunderstood my point. The molecule is achiral and meso. The staggered conformation you have shown has a center of symmetry in the center of the C2-C3 bond. The other two staggered conformations form a racemate. Yes, the rotation by 180 degrees gives the eclipsed conformation with a plane of symmetry, no rotation and low concentration. The eclipsed conformation is a test for a meso cmpd. The other 2 eclipsed conformations are also a racemate. Your diagram should show "center of symmetry".More
The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:
When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.
Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2 and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.
To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).
The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:
When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.
Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2 and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.
To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).
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