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Among the following, the optically inactive compound is:
   A)
   B)
  C)
  D)

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Last updated date: 20th Jun 2024
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Answer
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Hint:To solve this question, it is required to have knowledge about chirality of an atom. An atom is considered to be chiral if and only if all the substituent groups attached to it are different. A compound which contains at least one chiral atom will be an optically active compound and will be able to form enantiomers of itself.

Complete step by step answer:
We shall analyse each compound to find out if it has a chiral atom or not:
In option A, we know that each carbon on benzene contains one hydrogen atom and has partial double bond with two carbon atoms. Thus, they are not chiral. The nitrogen atom attached to it forms three bonds. One is with the benzene ring, one is with an ethyl group and the other is with a methyl group. So, each of the bonds are with a different group. Thus, the nitrogen atom is chiral and the compound will be optically active.
In option B, we know that each carbon on benzene contains one hydrogen atom and has partial double bond with two carbon atoms. Thus, they are not chiral. The phosphorus atom attached to it forms three bonds. One is with the benzene ring, one is with the methyl group and the last one is with an allylic group with three carbons. So, each of the bonds are with a different group. Thus, the phosphorus atom is chiral and the compound will be optically active.
In option C, the compound is an allene with each of the substituents as different halogen. Though the substituents are different, the presence of a double bond on the carbons will make them achiral atoms. Thus, the compound will be optically inactive.
In option D, we know that each carbon on benzene contains one hydrogen atom and has partial double bond with two carbon atoms. Thus, they are not chiral. The carbon attached to it has four bonds. One is with the benzene ring, one is with hydrogen, one is with hydroxide and the last one is with a carboxylic group. As all the bonds are with a different group thus, the carbon will be chiral. Thus, the compound will be optically active.
$\therefore $ The correct option is option C.

Note: All compounds form mirror images, but only the mirror images which do not superimpose with each other are called enantiomers and form optically active compounds. The images which do not superimpose with each other and are also not mirror images of each other are called diastereomers.