Optical isomerism Questions in English

(A) compound is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1-$chloropentane: $CH_2(Cl)-CH_2-CH_2-CH_2-CH_3$. The carbon atom bonded to chlorine is attached to two identical hydrogen atoms,so it is achiral.
$2-$chloropentane: $CH_3-CH(Cl)-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H, -Cl, -CH_3, \text{ and } -CH_2CH_2CH_3$. Since all four groups are different,it is chiral.
$1-$chloro$-2-$methylpentane: $CH_2(Cl)-CH(CH_3)-CH_2-CH_2-CH_3$. The $C-2$ atom is bonded to $-H, -CH_3, -CH_2Cl, \text{ and } -CH_2CH_2CH_3$. Since all four groups are different,it is chiral.
$3-$chloro$-2-$methylpentane: $CH_3-CH(CH_3)-CH(Cl)-CH_2-CH_3$. The $C-3$ atom is bonded to $-H, -Cl, -CH(CH_3)_2, \text{ and } -CH_2CH_3$. Since all four groups are different,it is chiral.
Therefore,$1-$chloropentane is the only achiral compound.

(B) The correct answer is $(B)$ $CH_3-CH(Cl)-COOH$.
Optical activity is shown by molecules that possess a chiral center,which is a carbon atom bonded to four different groups.
In $CH_3-CH(Cl)-COOH$,the central carbon atom is bonded to $-H$,$-CH_3$,$-Cl$,and $-COOH$ groups.
Since all four groups are different,the carbon is chiral,making the molecule optically active.

(D) An enantiomorph (or enantiomer) exists if a molecule contains at least one chiral center (a carbon atom bonded to four different groups).
In $CH_3-CH_2-CH^*(NH_2)-CH_3$,the carbon atom marked with an asterisk $(*)$ is bonded to four different groups: $-H$,$-NH_2$,$-CH_3$,and $-CH_2CH_3$.
Since it possesses a chiral center,it can exist as a pair of non-superimposable mirror images,known as enantiomorphs.

(C) The condition for a compound to exist as enantiomers is the presence of at least one chiral center (an asymmetric carbon atom bonded to four different groups).
$A$. $CH_3CH(OH)CO_2H$ has a chiral center at the $C_2$ position.
$B$. $CH_3CH_2CH(CH_3)CH_2OH$ has a chiral center at the $C_3$ position.
$C$. $C_6H_5CH_2CH_3$ (ethylbenzene) does not have any chiral center,as the $CH_2$ group is bonded to two identical hydrogen atoms.
$D$. $C_6H_5CHClCH_3$ has a chiral center at the $C_1$ position.
Therefore,$C_6H_5CH_2CH_3$ cannot exist as enantiomers.

(C) Lactic acid $(CH_3CH(OH)COOH)$ contains a chiral carbon atom,which is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-COOH$.
Because of the presence of this chiral center,lactic acid exists as two non-superimposable mirror images (enantiomers).
Therefore,it exhibits optical isomerism.

(B) An optically active compound must contain at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $2-$chlorobutane $(CH_3-CH(Cl)-CH_2-CH_3)$,the second carbon atom is bonded to a hydrogen atom,a chlorine atom,a methyl group,and an ethyl group.
Since all four groups are different,$2-$chlorobutane is chiral and thus optically active.
The other options ($n-$propanol,$n-$butanol,and $4-$hydroxyheptane) do not contain a chiral carbon atom.

(A) compound exhibits optical isomerism if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $CH_3-CHCl-CH_2-CH_3$ ($2$-chlorobutane),the $C-2$ atom is bonded to $-H$,$-Cl$,$-CH_3$,and $-C_2H_5$ groups.
Since all four groups are different,the $C-2$ atom is chiral,and thus the compound shows optical isomerism.

(C) An object is considered 'achiral' if it is superimposable on its mirror image.
$1$. The letters $P$ and $F$ are chiral because their mirror images are non-superimposable on the original letters.
$2$. $A$ pair of hands is chiral because the left hand cannot be superimposed on the right hand.
$3$. $A$ ball is achiral because it is symmetrical and its mirror image is superimposable on the original object.
Therefore,the correct option is $(C)$.

(A) The process of separating a racemic mixture into its individual $d$ and $l$ enantiomers is known as resolution.

(B) Lactic acid is $CH_3-CH(OH)-COOH$.
It contains only one chiral center.
Hence,two optical isomers (enantiomers) are possible.
The number of optical isomers is given by the formula $2^n$,where $n$ is the number of chiral carbon atoms.
For lactic acid,$n = 1$,so the number of optical isomers = $2^1 = 2$.

(D) An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
In $CH_3-CH(OH)-CH_2Br$,the second carbon atom is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2Br$.
Therefore,it is an asymmetric carbon atom.

(D) The correct answer is $(D)$ $CH_3-CH(OH)-CH_2-CH_3$.
$A$ chiral structure is one in which a central carbon atom is bonded to $4$ different atoms or groups.
In $CH_3-CH(OH)-CH_2-CH_3$,the second carbon atom is bonded to $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$ groups.
Since all $4$ groups are different,this carbon is a chiral center.

(A) An organic compound exhibits optical isomerism if it contains a chiral center.
$A$ carbon atom is considered chiral (or asymmetric) when it is bonded to four different groups or atoms.
Therefore,the correct condition is that all four groups linked to the carbon atom must be different.

(B) . Polarimeter is an instrument used for measuring the optical rotation.
It consists of two Nicol prisms,one called the polarizer (near the light source) and the other called the analyser (near the eye).
In between the polarizer and analyser,a glass tube containing the solution of an optically active compound is placed.

(C) Plane-polarized light consists of light waves in which the vibrations occur in a single plane.
When this light passes through certain compounds,they rotate the plane of the plane-polarized light by a specific angle.
This property,where a compound rotates the plane of plane-polarized light,is known as $Optical \ activity$.

(A) Meso-tartaric acid contains a plane of symmetry which divides the molecule into two identical halves.
The optical rotation caused by one half is cancelled by the other half within the same molecule.
This is known as internal compensation or molecular symmetry.

(B) The structure of $2-$hydroxy$-2-$methylbutanoic acid is $CH_3-CH_2-C(OH)(CH_3)-COOH$.
In this molecule,the carbon atom at position $2$ is bonded to four different groups: $-OH$,$-CH_3$,$-COOH$,and $-CH_2CH_3$.
Since there is $1$ chiral center,the number of stereoisomers is given by $2^n$,where $n$ is the number of chiral centers.
Here,$n = 1$,so the number of stereoisomers is $2^1 = 2$.

(D) Optical isomers are non-superimposable mirror images of each other. They possess identical physical and chemical properties in an achiral environment. However,they differ in their interaction with plane-polarized light. Optical isomers rotate the plane of plane-polarized light in opposite directions (one rotates it clockwise,the other counter-clockwise). Polarimetry is the experimental technique used to measure this rotation,thereby allowing us to distinguish between optical isomers.

(B) The given compound is $(CH_3)_2C=CH-CH(CH_3)-COOH$.
$1$. Geometrical isomerism: The double bond is $(CH_3)_2C=CH-$. Since one of the carbons of the double bond (the left one) is attached to two identical groups (two $CH_3$ groups),it cannot exhibit geometrical isomerism.
$2$. Optical isomerism: The molecule contains a chiral carbon atom at the $CH(CH_3)$ position (the carbon attached to $H$,$CH_3$,$COOH$,and the $-CH=C(CH_3)_2$ group). Since all four groups attached to this carbon are different,it is a chiral center.
Therefore,the compound exhibits optical isomerism.

(D) . Diastereoisomers are stereoisomers that are not mirror images of each other.
Optical isomers that lack a mirror-image relationship are classified as diastereoisomers.

(D) An asymmetric centre (chiral centre) is a carbon atom bonded to four different groups.
In $Lactic$ $acid$ $(CH_3-CH(OH)-COOH)$,the central carbon atom is bonded to a $-H$,$-OH$,$-CH_3$,and $-COOH$ group.
Since all four groups are different,it contains an asymmetric centre.

(C) chiral object is one that is non-superimposable on its mirror image.
$A$. $A$ shoe is chiral because its mirror image cannot be superimposed on the original.
$B$. $A$ screw is chiral because it has a specific handedness (right-handed or left-handed).
$C$. $A$ screwdriver,specifically a standard one with a symmetrical handle and shaft,possesses a plane of symmetry and is superimposable on its mirror image,making it achiral.
Therefore,a screwdriver cannot be used to exemplify a chiral structure.

(B) molecule is optically active if it is chiral,meaning it contains at least one carbon atom bonded to four different groups (a chiral center).
In $C_2H_5CH(CH_3)C_3H_7$,the central carbon atom is bonded to four different groups: $-H$,$-CH_3$,$-C_2H_5$,and $-C_3H_7$.
Since all four substituents are different,the molecule is chiral and therefore optically active.

(D) The compound $CH_3CH(Br)CH(Br)COOH$ contains $n = 2$ chiral carbon atoms.
Since the molecule is unsymmetrical,the number of stereoisomers is given by the formula $2^n$.
Here,$n = 2$,so the number of stereoisomers = $2^2 = 4$.
These $4$ stereoisomers consist of $2$ pairs of enantiomers.

(A) compound is chiral if it contains at least one chiral center (a carbon atom bonded to four different groups).
$(A)$ $DCH_2-CH_2-CH_2-Cl$: This molecule has no chiral center because the carbon atoms are bonded to identical hydrogen atoms or are not bonded to four distinct groups.
$(B)$ $CH_3-CH_2-CHD-Cl$: The carbon atom attached to $D$ and $Cl$ is bonded to $H$,$D$,$Cl$,and an ethyl group $(-CH_2CH_3)$. Since these four groups are different,it is chiral.
$(C)$ $CH_3-CHD-CH_2-CH_2-Cl$: The carbon atom attached to $D$ is bonded to $H$,$D$,$CH_3$,and $-CH_2CH_2Cl$. Since these four groups are different,it is chiral.
$(D)$ $CH_2D-CHCl-CH_3$: The carbon atom attached to $Cl$ is bonded to $H$,$Cl$,$CH_2D$,and $CH_3$. Since these four groups are different,it is chiral.
Therefore,the correct option is $A$.

(D) . $CHClBrI$
$A$ chiral molecule is one that contains at least one chiral center,which is a carbon atom bonded to four different atoms or groups.
In $CHClBrI$,the central carbon atom is bonded to four distinct groups: $H$,$Cl$,$Br$,and $I$.
Since all four substituents are different,the molecule is chiral and lacks a plane of symmetry.

(B) The correct answer is $(B)$.
$2-$butanol $(CH_3-CH(OH)-CH_2-CH_3)$ is an optically active compound because it contains a chiral (asymmetric) carbon atom,which is bonded to four different groups: a hydrogen atom,a hydroxyl group,a methyl group,and an ethyl group.
$1-$butanol and $3-$butanol do not possess a chiral center,and $4-$heptanol is a symmetric molecule.

(A) $d-$tartaric acid and $l-$tartaric acid are non-superimposable mirror images of each other.
An enantiomer is one of two stereoisomers that are mirror images of each other and are non-superimposable (not identical),much as one's left and right hands are the same except for being reversed along one axis (the hands cannot be made to appear identical simply by reorientation).
Since $d-$tartaric acid and $l-$tartaric acid are non-superimposable mirror images,they are classified as enantiomers.

(A) Optical isomerism arises due to the presence of an asymmetric carbon atom (chiral center). For a molecule to be optically active,it must be chiral,meaning it lacks any internal symmetry elements such as a plane of symmetry,a center of symmetry,or an alternating axis of symmetry.

(B) The bromination of propionic acid $(CH_3CH_2COOH)$ at the $\alpha$-carbon produces $2$-bromopropionic acid $(CH_3CH(Br)COOH)$.
In $2$-bromopropionic acid,the carbon atom attached to the bromine atom is a chiral center because it is bonded to four different groups: $-H$,$-CH_3$,$-Br$,and $-COOH$.
Due to the presence of this chiral center,$2$-bromopropionic acid exists as two non-superimposable mirror images,which are known as enantiomers.
Enantiomers are a type of optical isomer.
Therefore,the pair of $2$-bromopropionic acids formed are optical isomers.

(A) Optical isomerism is exhibited by compounds that possess a chiral (asymmetric) carbon atom.
In lactic acid $(CH_3CH(OH)COOH)$,the central carbon atom is bonded to four different groups: a methyl group $(-CH_3)$,a hydroxyl group $(-OH)$,a carboxylic acid group $(-COOH)$,and a hydrogen atom $(-H)$.
Because this central carbon atom is asymmetric,it creates a chiral center,which is responsible for the optical activity of the molecule.

(D) Stereoisomers that are not mirror images of each other are known as diastereomers.
$1$. Enantiomers are stereoisomers that are non-superimposable mirror images of each other.
$2$. Diastereomers are stereoisomers that are not mirror images of each other.
$3$. Meso compounds (or mesomers) are achiral molecules that contain multiple stereocenters but are optically inactive due to an internal plane of symmetry.
$4$. Tautomers are structural isomers that exist in dynamic equilibrium with each other.

(D) compound is optically active if it lacks a plane of symmetry or center of symmetry,making it chiral.
$CH_2Cl_2$ and $CHCl_3$ are achiral due to the presence of planes of symmetry.
Meso-tartaric acid is achiral because it contains an internal plane of symmetry,making it optically inactive.
Glyceraldehyde $(CH_2(OH)CH(OH)CHO)$ contains a chiral carbon atom (the central carbon is bonded to four different groups: $-H, -OH, -CHO, -CH_2OH$),which makes it optically active.

(A) Enantiomers have identical physical properties (except for the direction of rotation of plane-polarized light) and identical chemical properties in an achiral environment. However,they exhibit different chemical properties when interacting with other chiral molecules and often show different biological activities. Therefore,the statement that they have the same physical properties is incorrect.

(A) $Meso$-tartaric acid contains two chiral carbon atoms.
Due to the presence of a plane of symmetry within the molecule,the optical rotation caused by one half of the molecule is cancelled by the other half.
This phenomenon is known as internal compensation.
Therefore,$meso$-tartaric acid is optically inactive.

(A) An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
In $CH_3CHClCOOH$,the central carbon is attached to four different groups: $-H$,$-Cl$,$-CH_3$,and $-COOH$.
Therefore,it is an asymmetric carbon atom.

(B) $ (b) $ $A$ racemic mixture is an equimolar mixture of two enantiomers,which are non-superimposable mirror images of each other.
Because the rotation caused by one enantiomer is cancelled by the other,the resulting mixture is optically inactive.

(B) chiral compound is one that contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $1-$chloro$-2-$methylbutane $(CH_3-CH_2-CH(CH_3)-CH_2Cl)$,the carbon atom at position $2$ is bonded to four different groups: $-H$,$-CH_3$,$-CH_2CH_3$,and $-CH_2Cl$.
Therefore,it is a chiral molecule.

(B) disymmetric object is defined as an object that lacks any element of symmetry (like a plane of symmetry or center of inversion). Such objects are non-superimposable on their mirror images and are referred to as chiral. Therefore,the correct option is $(B)$.

(D) chiral compound is one that contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $2,3,4-$trimethyl hexane,the carbon atom at position $3$ is bonded to a hydrogen atom,a methyl group,an ethyl group,and a $sec-$butyl group.
Since all four groups attached to the $C-3$ atom are different,it is a chiral center.
Therefore,$2,3,4-$trimethyl hexane is a chiral compound.

(B) The number of optical isomers for a compound with $n$ dissimilar asymmetric carbon atoms is given by the formula $2^n$.
Given $n = 2$,the number of optical isomers $= 2^2 = 4$.

(B) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $2-$chlorobutane $(CH_3-CHCl-CH_2-CH_3)$,the second carbon is bonded to a hydrogen atom,a chlorine atom,a methyl group,and an ethyl group.
Since all four groups are different,$2-$chlorobutane is chiral and thus optically active.
$3-$chloropentane and $2-$chloropropane do not contain any chiral carbon atoms.

(D) An enantiomerically pure acid (let it be $R-COOH$) reacts with a racemic mixture of an alcohol (let it be $R'-OH$ and $S'-OH$).
The reaction produces two esters: $R-COOR'$ and $R-COOS'$.
Since $R'$ and $S'$ are enantiomers,the resulting esters $R-COOR'$ and $R-COOS'$ are diastereomers.
Therefore,the product is a mixture of diastereomers.

(B) compound exhibits optical isomerism if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $CH_3CHClBr$,the central carbon atom is bonded to four different groups: $-H$,$-CH_3$,$-Cl$,and $-Br$.
Therefore,$CH_3CHClBr$ is a chiral molecule and exhibits optical isomerism.

(D) $2-$methylbutanoic acid $(CH_3-CH_2-CH(CH_3)-COOH)$ contains a chiral carbon atom at the $C-2$ position.
This carbon is bonded to four different groups: $-H$,$-CH_3$,$-CH_2CH_3$,and $-COOH$.
Due to the presence of this chiral center,the molecule is optically active and exhibits optical isomerism,which is a type of stereoisomerism.

(A) To determine the chirality ($R/S$ configuration) of the chiral center,we assign priorities to the groups attached to the central carbon atom based on the Cahn-Ingold-Prelog $(CIP)$ sequence rules based on atomic number:
$1$. $-Br$ (atomic number $35$) is priority $1$.
$2$. $-Cl$ (atomic number $17$) is priority $2$.
$3$. $-CH_3$ (atomic number $6$ for $C$) is priority $3$.
$4$. $-H$ (atomic number $1$) is priority $4$.
Since the lowest priority group $(-H)$ is on a dashed bond (pointing away from the viewer),we look at the sequence $1$ $\rightarrow 2$ $\rightarrow 3$.
The sequence $Br$ $\rightarrow Cl$ $\rightarrow CH_3$ follows a clockwise direction.
Therefore,the configuration is $R$.

(B) Optical isomerism is exhibited by molecules that possess a chiral center,which is a carbon atom bonded to four different groups.
In the molecule $HC \equiv C - CH(Cl) - CH_3$,the central carbon atom is bonded to four distinct groups: a hydrogen atom $(-H)$,a chlorine atom $(-Cl)$,a methyl group $(-CH_3)$,and an ethynyl group $(-C \equiv CH)$.
Since all four groups attached to this carbon are different,the molecule is chiral and will exhibit optical isomerism.

(B) $CH_3-CH(OH)-COOH$ (Lactic acid) contains a chiral carbon atom (asymmetric carbon) attached to four different groups: $-H$,$-OH$,$-CH_3$,and $-COOH$.
Therefore,it exhibits optical isomerism.
It does not have a double bond with restricted rotation,so it does not show geometrical isomerism.

(A) . Chirality in a carbon compound arises due to the $sp^3$ hybridized,tetrahedral geometry of the carbon atom,which allows for the attachment of four different groups to a single carbon center,creating a non-superimposable mirror image.