
Which of the following statements regarding 1, 2-dimethyl cyclopropane (I) and 1, 2-dimethyl cyclobutane (II) are wrong?
A.Both of them show three stereoisomers
B.The cis form of both is optically inactive (meso) and the trans form of both has a pair of enantiomers
C.The cis form of both has a pair enantiomers and the trans form of both is optically inactive
D.The meso form of both is optically inactive due to presence of center of symmetry
Answer
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Hint:To answer this question, you must recall drawing stereoisomers of an organic compound. Stereoisomers are isomers that differ only in the spatial arrangement of atoms. Stereoisomers can be of two types, optically active stereoisomers and mesomeric isomers.
Complete step by step answer:
If the stereoisomers are non- superimposable mirror images of each other, they are said to be optically active. For a compound to show optical isomerism, it is essential that it does not have a plane of symmetry in its structure.
Enantiomers are a pair of molecules which are non-super imposable mirror images of each other. They possess the property of chirality. Those isomers which rotate the plane of polarized light to the left side are called as laevo rotatory and those isomers which rotate the plane of polarized light to right direction are known as dextro rotatory.
Both 2-dimethyl cyclopropane (I) and 1, 2-dimethyl cyclobutane (II) have three stereoisomers.
The cis forms of both the compounds are meso forms and thus are inactive.
The trans form of both the compounds form a pair of enantiomers each and are thus optically active. The correct options are A and B.
So, the wrong statements and the answer of the question are C and D.
Note:
Optical activity is a characteristic of chiral molecules only. We know that a chiral molecule is one that cannot be superimposed on its mirror image. Such a molecule and its non-super imposable mirror image form a pair of enantiomers. Enantiomers rotate the plane of plane polarized light to the same extent but in the opposite direction, that is, one of them is dextrorotatory and the other is laevo rotatory.
Complete step by step answer:
If the stereoisomers are non- superimposable mirror images of each other, they are said to be optically active. For a compound to show optical isomerism, it is essential that it does not have a plane of symmetry in its structure.
Enantiomers are a pair of molecules which are non-super imposable mirror images of each other. They possess the property of chirality. Those isomers which rotate the plane of polarized light to the left side are called as laevo rotatory and those isomers which rotate the plane of polarized light to right direction are known as dextro rotatory.
Both 2-dimethyl cyclopropane (I) and 1, 2-dimethyl cyclobutane (II) have three stereoisomers.
The cis forms of both the compounds are meso forms and thus are inactive.
The trans form of both the compounds form a pair of enantiomers each and are thus optically active. The correct options are A and B.
So, the wrong statements and the answer of the question are C and D.
Note:
Optical activity is a characteristic of chiral molecules only. We know that a chiral molecule is one that cannot be superimposed on its mirror image. Such a molecule and its non-super imposable mirror image form a pair of enantiomers. Enantiomers rotate the plane of plane polarized light to the same extent but in the opposite direction, that is, one of them is dextrorotatory and the other is laevo rotatory.
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