CHAPTER 7: Question 1

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 180^o 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.

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.

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.
 

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 methyl groups, and this is sufficient to classify the methyl groups as 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 120^o rotation around the axis indicated. Therefore, the two methyl groups are homotopic.

The three rotatable structures shown below illustrate that the methyl groups in 2-chloro-2-methylpropane can be interconverted by rotation, and show the identical nature of compounds A and B.
 

An alternative way to answer the question is to note that the compound possesses a C₂ axis as shown. A 180^o 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.

 The three rotatable structures shown below illustrate the interconvertability of the methyl groups in this compound, and the identical nature of compounds A and B.
 

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