Explain optical activity of 2-chlorobutane.

Answer: 2-Chlorobutane, CH₃–C*H(Cl)–CH₂–CH₃, has a chiral carbon (C-2) bonded to four different groups: –H, –Cl, –CH₃ and –C₂H₅. Hence the molecule has no plane of symmetry and its mirror image is not superimposable on it.

[Figure described: two tetrahedral structures drawn as mirror images, each with C at the centre carrying H, Cl, CH₃ and C₂H₅; the mirror between them.]

So 2-chlorobutane exists as a pair of enantiomers. One rotates plane-polarised light to the right (dextrorotatory, (+)) and the other to the left by the same angle (laevorotatory, (–)). An equimolar mixture of the two is a racemic mixture (±), which is optically inactive due to external compensation.