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An organic compound having a carbon attached to four different groups is optically active. But, is the opposite also true? That is, do all optically active organic compounds have chiral carbons? Not necessarily. Presence or absence of chiral centre is not the sufficient criterion for optical activity. The ultimate criterion is the presence or absence of either plane or centre of symmetry. Two compounds which are non-superimposable mirror images of each other are called enantiomers. If a compound contains more than one chiral carbon, new words are required to describe the relationship between various stereoisomers of the compounds. Those words are diastereomers and mesomers. 

Optically active compounds among the following is :


A
CH3CH2CH(Cl)CH2CH3
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B
CH3CH2CD(Cl)CH2CH3
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C
CH3CH2CH(D)CH2CH3
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D
CH3CH2CH(D)CH3
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Solution

The correct option is D $$CH_{3}-CH_{2}-CH(D)-CH_{3}$$
x= optical centre

Thus the compound in the option D has asymmetric carbon atom as all the four valencies of carbon indicated by 'x' is satisfied by four different group. 

Thus the compound in option D is optically active.

60503_25939_ans_67d654c7c7154e5eb64b6a9915884cb7.png

Chemistry

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