Which one of the following compounds is optic | vedclass

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

Question

(A) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).$1$. $2-$Chloropentane

Vedclass · Accepted Answer

$2-$Chloropentane

Vedclass · Answer

(A) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).$1$. $2-$Chloropentane: $CH_3-CHCl-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is a chiral center. Thus,it is optically active.$2$. $3-$Chloropentane: $CH_3-CH_2-CHCl-CH_2-CH_3$. The $C-3$ atom is bonded to two identical ethyl groups $(-CH_2CH_3)$,so it is achiral.$3$. $2-$Chloropropane: $CH_3-CHCl-CH_3$. The $C-2$ atom is bonded to two identical methyl groups $(-CH_3)$,so it is achiral.$4$. $2-$Chloro$-2-$methylbutane: $CH_3-CCl(CH_3)-CH_2-CH_3$. The $C-2$ atom is bonded to two identical methyl groups $(-CH_3)$,so it is achiral.Therefore,$2-$Chloropentane is the optically active compound.

Which one of the following compounds is optically active?

Similar Questions

$d-$tartaric acid and $l-$tartaric acid are

Explain the following terms:
$(i)$ Plane polarised light
$(ii)$ Optical activity
$(iii)$ Chirality
$(iv)$ Chiral centre (atom)
$(v)$ Enantiomers
$(vi)$ Diastereomers
$(vii)$ Meso compounds
$(viii)$ Plane of symmetry and Achiral molecule
$(ix)$ Centre of symmetry

Which of the following signs indicates that the sugar is actually 'dextrorotatory'?

Which of the following is expected to be optically active?

What is the number of chiral carbon atoms present in $3,4-$Dibromohexane?

Vedclass Test Series

Exam Paper Generator

Online Exam Module