Biphenyl, shown, rotates freely about the inter-ring bond and is achiral. Assume that rotation about the inter-ring bond is completely blocked whenever all four ortho positions carry substituents other than $H$ , so the two rings are held perpendicular. Which of the tetra-ortho-substituted biphenyls shown as options are chiral under this assumption?

AStructure P
BStructure Q
CStructure R
DStructure S

Answer: A, D

A locked perpendicular biphenyl is chiral (atropisomerism) only if neither ring has a mirror plane containing the inter-ring axis, i.e. each ring must carry two different ortho groups. P has

NO X_{2}

and

COOH

on each ring (6,6-dinitrobiphenyl-2,2-dicarboxylic acid): chiral. S has

CH X_{3}

and

Cl

on each ring: chiral. Q has two

CH X_{3}

on each ring, so the plane of either ring is a mirror plane: achiral. R has one ring with

CH X_{3}

and

NO X_{2}

but the other ring with two

CH X_{3}

; the plane of the first ring reflects the second ring onto itself, so R is achiral.

Practice Stereoisomerism More on Optical isomerism and chirality from structures

Previous questionNext question

Answer: A, D

More questions on Optical isomerism and chirality from structures