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It is achiral. The cyclopropane ring is planar. The substituted C will be $\mathrm{sp^3}$ hybridised (tetrahedral). If you consider two of the C's in one plane and the H and Br in the other plane, these two planes will make 90° angle between them and one can draw a mirror plane along the plane containing H and Br. Thus, it will be achiral.


Here is a 3-D conformer from PubChem As you can clearly see, a plane of symmetry can be sent along the black line perpendicular to the plane of the screen. Hence, the molecule is achiral. If you take a mirror image, you can ultimately super-impose it again on the parent form Here is an illustrative 3D image(courtesy of andselisk) which clearly shows the ...


On rotating the compound B by 180°, we get the two -OH groups on the left side and the two H atoms on the right. In other words, B would then be an enantiomer of A. Note: on rotating a Fischer projection by 180°, the compound remains the same; it's just that all the atoms have to be rotated, i.e. the atoms on C3 in this case would go to C2 and vice versa.


Biphenyl 3 is the only optically active compound here. These stereoisomers are due to the hindered rotation about the 1,1'-single bond of the compound (Ref.1). Biphenyl 2 is noty optically active, because partially allowed rotation about the 1,1'-single bond of the compound (rotation is only partially restricted). To illustrate this phenomenon, I depicted ...


(3) is not chiral, so maybe it is (2).

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