Question

The optical rotation of a racemic mixture of 2-methyl 1-butanol is:

• A. +11.5
• B. +5.75
• C. +2.87
• D. 0 (zero)

Hint:

Racemic mixtures are optically inactive because the equal and opposite rotations of enantiomers cancel each other.

Answer 1

The Correct Option is D

A racemic mixture contains equal amounts of both enantiomers (optical isomers) of a chiral compound. Since one enantiomer rotates plane-polarized light to the right (dextrorotatory) and the other to the left (levorotatory) by equal amounts, their rotations cancel each other out. Hence, the net optical rotation of a racemic mixture is zero.

Answer 2

The optical rotation of a racemic mixture of 2-methyl-1-butanol is:

Step 1: Understand what a racemic mixture is:
A racemic mixture (also called a racemate) consists of equal amounts of two enantiomers — mirror image isomers of a chiral compound. These enantiomers rotate plane-polarized light in equal magnitude but opposite directions. One rotates it to the right (dextrorotatory, +), and the other to the left (levorotatory, –).

Step 2: Structure of 2-methyl-1-butanol:
2-methyl-1-butanol has a chiral center at the 2nd carbon. Therefore, it exists as two enantiomers that are non-superimposable mirror images of each other.

Step 3: Optical rotation of a racemic mixture:
When these two enantiomers are present in equal quantities, their optical activities cancel each other out. The positive rotation of one is exactly neutralized by the negative rotation of the other.

Result:
The net optical rotation of the racemic mixture becomes zero, making the mixture optically inactive even though each enantiomer is optically active.

Therefore, the optical rotation of a racemic mixture of 2-methyl-1-butanol is:
\[ \boxed{0} \]