Which of the following biphenyls is optically active? [A] [B] [C] [D]
Hint: Here, the compound which will not have a plane of symmetry will be optically active. Due to steric hindrance caused by the ortho substituents, the two rings will not be planar and will not have a plane of symmetry and will be optically active.
Complete step by step answer: We know that compounds which contain a chiral centre show optically active. When a beam of polarised light is passed through an optically active compound, it changes the direction of the plane polarised light which shows us that the compound is optically active. We know that a biphenyl is an organic compound and it has two phenyl rings attached to each other through a single bond. To answer this question, we have to understand the factors that affect the optical activity of biphenyl. Firstly, if three or more ortho positions in the biphenyl is substituted i.e. occupied by a certain group, then the biphenyl will be resolvable i.e. optically active. The second point is there should not be any plane of symmetry and neither of the rings should have a plane of symmetry for the biphenyl compound to be optically active. And lastly, the biphenyls which are resolvable should not contain identical substitutions on the same ring. If both the rings contain similar substituents, there will be no chiral carbon. Now, let us discuss each compound in the question and check if they are optically active or not. In the last option, in the compound given to us both the substitution is on the same ring. This will cause ortho-effect I.e. the ring will move away from the plane of the first ring due to steric hindrance provided by the ortho substitutions. Here, if we pass a horizontal mirror through the middle, we will get symmetric structures. Even though they are not planar still it is optically inactive as it has a centre of symmetry. In the third option, the compound given to us has a centre of symmetry (i) which means on moving equal distances from the centre on towards the opposite sides, we will land on the symmetrical structures. Therefore, this is also optically inactive. In the first option, there is no ortho substitution. Thus, they are on the same plane and therefore it is optically inactive. And lastly, in the second option, due to the steric hindrance of the bulky ortho substituents the rings are not planar. Thus, it does not have a plane of symmetry. Therefore, it is optically active. Therefore, the correct answer is option [B] .
Note: We can also answer this question in a more general way by taking the mirror image of each of the given compounds. If the mirror images are non-superimposable i.e. if we keep the mirror image and the actual compound on top of each other, we will not get the same compound. Such isomers are called enantiomers. They rotate the plane of polarised lights equally in the opposite direction thus showing no net optical activity. However, if the mirror image obtained is superimposable i.e. we get the same compounds by placing the mirror image on top of the actual compound, such structures are diastereomers. They will show optical activity. Here, only the correct option is a diastereomer and the others are enantiomers. Therefore, it is optically active.