Question

Which of the following compounds can in principle be isolated in optically active form at room temperature?

a. cis-1,4-dimethylcyclohexane
b.trans-1,3-dimethylcyclohexane
c.1,1-dimethylcyclohexane
d.cis-1-ethyl-3-methylcyclohexane

Answer 1

Concepts and reason

Chiral compounds are optically active and achiral compounds are optically inactive.

The optically compound and its mirror image are not superimposable to each other. It is possible if the compound does not possess any plane of symmetry.

Fundamentals

Plane of symmetry is an imaginary plane through which reflection takes place.

The compound which can rotate plane polarized light in clockwise or anti-clockwise direction is known as optically active compound.

The compounds which have plane of symmetry are as follows:

The compounds cis-1,4-dimethylcyclohexane and 1,1-dimethylcyclohexane are achiral compounds and trans-1,3-dimethylcyclohexane and cis-1-ethyl-3-methylcyclohexane are chiral compounds.

Therefore, cis-1,4-dimethylcyclohexane and 1,1-dimethylcyclohexane are optically inactive and trans-1,3-dimethylcyclohexane and cis-1-ethyl-3-methylcyclohexane are optically active.

Hence, trans-1,3-dimethylcyclohexane and cis-1-ethyl-3-methylcyclohexane can be isolated in optically active form.

Ans:

The compounds which can be isolated in optically active form at room temperature are as follows:

Answer 2

Ansi, . . . 7. The optically inactive molecules will be . 11-dimethyleyelo hexane and cis. 1; 4 - dimethylcyclohexane as shown as Мен , Me L / Me 1 Explanation - A plane of Symmetry is an imaginary plane which, when passes through a molecule, divided it into two similar parts, – Here, the above two molecules Contain a plane of symmetry and hince they arie optically inactive, cs Scanned with CamScanner . m er

Ansi - The optically active molecule will be Cis-l-ethyl-3-methyleyclohexane which shown as follows. Explamalioner ₂. A plane of Symmetry is an imaginary plane which; when passes through a molecule, divided it into two similar parts, . tere, the above molecule does not contain a plane of symmetry and hence it is optically active. .. Scanned with CamScanner

7. The optically active molecule will be troms 1,3- dimethylcyclohexane which is Shown as follows. - Me. - Explanations À plane of symmetry is an imaginary plane which; when passes through a molecule, divided it into two similar parls. . Here the above molecule does not contain a plane a plane of symmetry and hence it is optically active. Therefore the answer is, eis-l-ethyl-3-methylcyclohexane - Cs Srans 11, 3-dimethyl cyclohexane -