Conformational Analysis of Disubstituted Cyclohexane | Stereochemistry | Organic Chemistry
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- Опубліковано 13 лис 2018
- In case of disubstituted cyclohexane, there are three possibilities that the Group -R is attached to 1 and 2 positions, 1 and 3 position or 1 and 4 position.
In the case of 1, 2 disubstituted cyclohexane if both groups are axial or both groups are equatorial in these cases R groups are in opposition to each other and therefore it is trans isomer. But if one group is axial and other is equatorial then it is Cis isomer.
Cis isomer doesn’t have any element of symmetry and therefore cannot be superimposed on its mirror image. However, when the chair conformer flips to give the other chair conformer, the axial groups change to equatorial and vice versa.
Further, as the groups are the same, the strain must also be the same in both conformations and their amount must also be equal and hence a cis-isomer of 1, 2 disubstituted cyclohexane, in which the two groups are identical will be optically inactive.
The trans configuration can exist in two different conformations 1a, 2a, and 1e, 2e of which the equatorial conformation will be more stable than the axial conformation. Also, this equatorial trans-isomer will be more stable than the cis isomer as both substituent groups are in equatorial positions.
Further, as neither conformation has an element of symmetry, and also they cannot be converted into their mirror images by flipping, therefore, the trans-1, 2-disubstituted cyclohexane exists as pairs of enantiomers.
In the case of 1, 3 disubstituted cyclohexane if both groups are axial or both groups are equatorial in these cases R groups are in opposition to each other and therefore it is trans isomer. But if one group is axial and other is equatorial then it is Cis isomer.
In trans isomer of 1,3 disubstituted cyclohexane, one substituent is always equatorial and the other axial. The two trans-conformers are identical and do not have any plane of symmetry. They are non-superimposable and thus compounds are resolvable.
In the cis-conformers either both substituents are axial or both are equatorial. In both cases, they have a plane of symmetry passing through carbon atoms 2 and 5. The cis-e, e-form is more stable than the cis-a, a-form because in cis a-a case there is a steric interaction between the two axial groups. The cis-e, e-form is also more stable than the trans-isomer by about 1.8 kcal/mole.
In this case in cis conformer, one of the groups is axial and the other at the equatorial position (a-e). In the trans conformer, both groups are at axial (a-a) or both are equatorial (e-e).
The cis forms are identical and do not have a plane of symmetry. Hence they are optically active.
The trans isomers have a plane of symmetry and hence optically inactive. The trans isomer which has both groups in equatorial position is the most stable isomer of all.
In general, it has been observed that the compounds tend to take up chair conformation wherever possible.
the chair conformation with a maximum number of equatorial substituents will be most preferred conformation.
However, these observations hold good only if other internal forces due to dipole interactions or hydrogen bonding are absent. When these are present they have also to be taken into consideration while deciding which conformer is more stable.
The energy barriers between the different conformations are too small to prevent interconversions.
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3:53 the cis 1,4 disubstituted cyclo hexane too have plane of symmetry [(plane passing through C1-C4) cuts the groups and cyclo hexane into 2 equal halves] therefore it is optically "inactive".
Maybe he refer to the case that two substituent are different.
Both two substituent are named R but it probably means just general substituents like R1 and R2.
I have literally enjoyed every bit of this video lecture
Plz tell me how to calculate energy in 1,2disubstituted cyclohexane
How is cis isomer of 1,2 di-sub cyclohexane optically inactive ? The mirror images are not superimposable even after flipping one of the mirror images , someone plz clarify this thing !
Nice video sir
Thank you sir
Sir , at ( 1,2 position as cis form ) it does fullfill the conditions of being optically active (i.e. 1. No symmetry, and mirror image non superimposable) than why is it optically inactive.
Sir plz tell me how calculate energy in 1,2 substituted cyclohexane
Sir pls tell bout mono.substituted cyclo hexane and conformation of Ethane
👌👌
But how to determine whether the compound has plane of symmetry or not.
Bro, draw a hypothetical mirror by yourself at possible configurations ,and if at any configuration you find that half the molecule is mirror image of the other then the molecule has plane of symmetry.
Love from pakistan.....thanks for superb explanation
Thanks and welcome
Hello sir can you please make a video lecture on 5th semester chemistry of kurukshetra university
Plz send me the syllabus at cepekmedia@gmail.com
1,4-opticaly inactive
Love from Pakistan 🇵🇰🇵🇰🇵🇰
Thanks @Rizwan..