Question

(a)- Define racemic mixture.
(b)- Give IUPAC name of $C{{H}_{3}}-C{{H}_{2}}-CH(C{{H}_{3}})-CHO$.

Answer 1

Hint: A racemic mixture is defined on the basis of enantiomers. In IUPAC name of $C{{H}_{3}}-C{{H}_{2}}-CH(C{{H}_{3}})-CHO$, the numbering of carbon starts from the aldehyde group because it is a functional group.

Complete step by step answer:
(a)- The optical activity of the compound is the ability of the compound to rotate the plane of polarized light. And optically active compounds are those which rotate the plane of polarized light.
So those compounds which have a carbon atom on which the attached groups are different can rotate the plane of polarized light. This means that the compound must be asymmetric. If the compound is symmetric then it will not show optical activity or it is an optically inactive compound. The carbon which has all 4 groups attached differently is termed as chiral carbon atom in that molecule.
So, the optically active compounds which are non-superimposable mirror images of each other are called enantiomers and the phenomenon is called enantiomerism.
An example of an optically active compound is bromochloroiodomethane. It is asymmetric and has a non-superimposable mirror image.

So a mixture containing an equal amount of two enantiomers is known as a racemic mixture or racemic modification.

(b)- For the IUPAC name of $C{{H}_{3}}-C{{H}_{2}}-CH(C{{H}_{3}})-CHO$:
The numbering will start from the aldehyde group because it is a functional group. So, the longest chain is of four carbon atoms, and the second carbon atom has a methyl group. So the name will be:

2-Methylbutan-1-al or 2-Methylbutanal.

Note: It must be noted that a racemic mixture is always optically inactive because the rotation of one molecule is exactly opposite of the other molecule, hence cancel each other and remain optically inactive. The numbering of the chain must be done in such a way that the functional group and substituents attached get the lowest possible numbering.