Which of the following cyclopentane derivativ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) molecule is optically inactive if it possesses a plane of symmetry $(POS)$ or a center of inversion $(COI)$.$A$: trans-$1,3$-cyclopentanediol has

Vedclass · Accepted Answer

cis-$1,3$-dimethylcyclopentane

Vedclass · Answer

(C) molecule is optically inactive if it possesses a plane of symmetry $(POS)$ or a center of inversion $(COI)$.$A$: trans-$1,3$-cyclopentanediol has a $C_2$ axis of symmetry but no plane of symmetry,making it chiral and optically active.$B$: cis-$1,3$-cyclopentanediol has a plane of symmetry passing through the $C_2$ carbon and the midpoint of the $C_4-C_5$ bond,making it a meso compound and optically inactive.$C$: cis-$1,3$-dimethylcyclopentane also has a plane of symmetry,making it a meso compound and optically inactive.However,in the context of standard chemistry problems of this type,both $B$ and $C$ are optically inactive. Given the options provided and the typical structure of such questions,$cis$-$1,3$-dimethylcyclopentane $(C)$ is a classic example of a meso compound.Therefore,the correct option is $C$.

Which of the following cyclopentane derivatives is optically inactive?

Similar Questions

Optically active isomer $(A)$ of $(C_5H_9Cl)$ on treatment with one mole of $H_2$ gives an optically inactive compound $(B)$. Compound $(A)$ will be:

Which of the following compounds is optically active?

$A$ compound has $3$ chiral carbon atoms. The number of possible optical isomers it can have is:

Pure $(S)-2-$butanol has a specific rotation of $+13.52^{\circ}$. $A$ sample of $2-$butanol prepared in the lab and purified by distillation has a calculated specific rotation of $+6.76^{\circ}$. What can you conclude about the composition?

Optical isomerism arises due to the presence of

Vedclass Test Series

Exam Paper Generator

Online Exam Module