The first step is to draw a three dimensional representation of the molecule as shown below. The two hydrogen atoms are then separately replaced by a new substituent (X) as shown, generating structures A and B. Compounds A and B are identical since they can be interconverted by a 180o rotation around the axis indicated. Therefore, the two hydrogen atoms are homotopic.
The first step is to draw a three dimensional representation of the
molecule as shown below. The two hydrogen atoms are then separately replaced
by a new substituent (X) as shown, generating structures A and B. Compounds
A and B are non-superimposable mirror images of one another and are therefore
enantiomers. Note that in structures A and B, the carbon atom is attached
to four different substituents. Hence, the two hydrogen atoms are enantiotopic.
The first step is to draw a three dimensional representation of the molecule as shown below. The two methyl groups are then separately replaced by a new substituent (X) as shown, generating structures A and B. Compounds A and B are non-superimposable mirror images of one another and are therefore enantiomers. Note that in structures A and B, the central carbon atom is attached to four different substituents. Hence, the two methyl groups are enantiotopic.
The first step is to draw a three dimensional representation of the
molecule as shown below. The two methyl groups are then separately replaced
by a new substituent (X) as shown, generating structures A and B. Compounds
A and B are identical since they can be interconverted by a 120o
rotation around the axis indicated. Therefore, the two methyl groups are
homotopic.
An alternative way to answer the question is to note that the 3D representation
shown above possesses a C3 axis as shown. A 120o
rotation around this axis (ie carrying out the symmetry operation C31)
interconverts the locations of the two methyl groups, and this is sufficient
to classify the methyl groups as homotopic.
An alternative way to answer the question is to note that the 3D representation
shown above possesses a plane of symmetry as shown but not a proper axis
of rotation. Reflection in the plane of symmetry (ie carrying out the symmetry
operation s) interconverts the
locations of the two hydrogen atoms, and this is sufficient to classify
the hydrogen atoms as enantiotopic.
An alternative way to answer the question is to note that the compound
possesses a C2 axis as shown. A 180o rotation
around this axis (ie carrying out the symmetry operation C21)
interconverts the locations of the two hydrogen atoms, and this is sufficient
to classify the hydrogen atoms as homotopic.
An alternative way to answer the question is to note that the compound
does not possess a proper axis of rotation so the hydrogen atoms cannot
be homotopic. The molecule does possess a plane of symmetry, but reflection
in this plane does not interchange the location of the two hydrogen atoms,
so they are not enantiotopic. Hence, the two hydrogen atoms must be diastereotopic.
An alternative way to answer the question is to note that the compound
does not possess a proper axis of rotation so the phenyl groups cannot
be homotopic. The
molecule does possess a plane of symmetry, but reflection in this plane
does not interchange the location of the two phenyl groups, so they are
not enantiotopic. Hence, the two phenyl groups must be diastereotopic.
An alternative way to answer the question is to note that the compound
does not possess a proper axis of rotation so the bromine atoms cannot
be homotopic. The
molecule does possess a plane of symmetry, but reflection in this plane
does not interchange the location of the two bromine atoms, so they are
not enantiotopic. Hence, the two bromine atoms must be diastereotopic.
An alternative way to answer the question is to note that the 3D representation
shown above possesses a plane of symmetry as shown but not a proper axis
of rotation. Reflection in the plane of symmetry (ie carrying out the symmetry
operation s) interconverts the
locations of the two carboethoxy groups, and this is sufficient to classify
the carboethoxy groups as enantiotopic.
An alternative way to answer the question is to note that the 3D representation
shown above possesses a plane of symmetry as shown but not a proper axis
of rotation. Reflection in the plane of symmetry (ie carrying out the symmetry
operation s) interconverts the
locations of the two hydrogen atoms, and this is sufficient to classify
the hydrogen atoms as enantiotopic.
An alternative way to answer the question is to note that the compound
does not possess a proper or improper axis of rotation so the methyl groups
cannot be homotopic or enantiotopic. Hence, the two methyl groups must
be diastereotopic.
An alternative way to answer the question is to note that the 3D representation
shown above possesses a plane of symmetry (horizontally along the C=C bond).
Reflection in the plane of symmetry (ie carrying out the symmetry operation
s) interconverts the locations
of the two hydrogen atoms, and this is sufficient to classify the hydrogen
atoms as enantiotopic.
An alternative way to answer the question is to note that the compound
does not possess a proper or improper axis of rotation so the methyl groups
cannot be homotopic or enantiotopic. Hence, the two methyl groups must
be diastereotopic.
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