Optical isomerism Questions in English

(B) Optical activity is shown by molecules that contain at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $CH_3-CH_2-CH(OH)-CH_3$ (Butan-$2$-ol),the second carbon atom is chiral as it is attached to four different groups: $-H$,$-OH$,$-CH_3$,and $-C_2H_5$.

(D) compound is optically active if it lacks both a plane of symmetry and a center of symmetry.
Option $D$ represents a chiral molecule (a substituted tetrahydropyran derivative) that does not possess any plane of symmetry or center of symmetry,making it optically active.
The other options ($A$,$B$,and $C$) possess planes of symmetry,rendering them optically inactive.

(D) For an allene $(>C=C=C < )$ to exhibit optical isomerism,the substituents on each terminal carbon atom must be different.
In the given options:
$A$) $Cl-CH=C=CH-Cl$: The terminal carbons have substituents $(H, Cl)$ and $(H, Cl)$. This molecule has a plane of symmetry if the groups are in the same plane,but allenes are non-planar. However,if the terminal groups are identical ($Cl, H$ on both sides),it can show geometrical isomerism,but it is achiral due to a plane of symmetry in the planar conformation or center of symmetry. Wait,for allenes $abC=C=Cab$,if $a \neq b$,it is optically active. Here,$a=Cl, b=H$. Both sides have $Cl$ and $H$. This molecule is achiral.
$B$) $Br-CH=C=CH-Br$: Similar to $A$,this is achiral.
$C$) $CH_3-CH=C=CH-CH_3$: Similar to $A$ and $B$,this is achiral.
$D$) $CH_3-CH=C=CH-Cl$: Here,one terminal carbon has $(CH_3, H)$ and the other has $(Cl, H)$. Since the groups on each terminal carbon are different ($CH_3 \neq H$ and $Cl \neq H$),this molecule is chiral and will show optical isomerism.
Re-evaluating the question: The question asks which will $NOT$ show optical isomerism. Options $A, B,$ and $C$ all have identical groups on the terminal carbons,making them achiral. This suggests a potential error in the question or options. Assuming the question intended to ask which one $DOES$ show optical isomerism,the answer is $D$.

(C) Optical activity requires a molecule to be chiral,meaning it lacks a plane of symmetry,center of inversion,or improper axis of rotation.
$A$. $Cl-CH=C=CH-Cl$ is an allene with different groups on each terminal carbon,making it chiral and optically active.
$B$. $Br-CH=C=CH-Br$ is an allene with different groups on each terminal carbon,making it chiral and optically active.
$C$. Meso-tartaric acid has a plane of symmetry passing through the molecule,making it achiral and optically inactive.
$D$. The Newman projection shows a molecule that lacks a plane of symmetry and is chiral,thus it is optically active.
Therefore,the correct answer is $C$.

(C) compound is optically active if it lacks a plane of symmetry and a center of inversion (i.e.,it is chiral).
$(I)$ $1,1$-dimethyl-$4$-chlorocyclohexane: This molecule has a plane of symmetry passing through the $C1$ and $C4$ atoms. Therefore,it is achiral and optically inactive.
$(II)$ $4$-chlorocyclohex-$1$-ene: The $C4$ atom is bonded to four different groups ($-H$,$-Cl$,and two different paths around the ring). Thus,$C4$ is a chiral center. The molecule lacks a plane of symmetry and is optically active.
$(III)$ Penta-$2,3$-diene: This is an allene with an even number of double bonds. The terminal carbons are substituted such that the molecule lacks a plane of symmetry and a center of inversion. It is axially chiral and optically active.
Therefore,compounds $(II)$ and $(III)$ are optically active.

(C) chiral center is a carbon atom that is bonded to four different groups.
In the given structure,we analyze each carbon atom:
$1$. The two carbons in the cyclohexane ring attached to the side chain are chiral because they are bonded to different groups (the ring path clockwise vs. counter-clockwise,the side chain,and the hydrogen atom).
$2$. The carbon atom attached to the chlorine atom is chiral as it is bonded to a hydrogen,a chlorine,the cyclohexane ring,and the carbon with the bromine.
$3$. The carbon atom attached to the bromine atom is chiral as it is bonded to a hydrogen,a bromine,a methyl group,and the carbon with the chlorine.
Thus,there are $4$ chiral centers in the compound.

(D) An optically inactive compound is one that does not rotate the plane of polarized light.
$A$: $A$ meso-compound contains chiral centers but possesses an internal plane of symmetry,making the molecule achiral and optically inactive.
$B$: cis$-1,2-$cyclohexanediol has a plane of symmetry passing through the molecule,making it achiral and optically inactive.
$C$: $A$ chiral molecule with a single stereocenter is always optically active.
Since both $A$ and $B$ are optically inactive,the option 'All' is incorrect. However,based on the provided options,if the question implies identifying which of the given structures are inactive,both $A$ and $B$ qualify. Given the standard nature of such questions,if the options are meant to be specific structures,we evaluate them individually. Since $A$ and $B$ are both inactive,the most appropriate answer is that the question might be flawed or intended to select multiple. Assuming the question asks for an example,both $A$ and $B$ are correct.

(C) molecule is optically active if it lacks a plane of symmetry,a center of symmetry,and an alternating axis of symmetry.
$1$. $CH_2=C=CH_2$ (allene) is achiral because it has a plane of symmetry.
$2$. $4,4'$-difluorobiphenyl is achiral due to the presence of a plane of symmetry.
$3$. $CH_3-CH=C=CH-CH_3$ (penta$-2,3-$diene) is chiral because the two terminal carbons have different groups attached,and the molecule lacks a plane of symmetry.
$4$. The structure shown in option $D$ is $2,3$-dihydroxybutane. Depending on the configuration (e.g.,the $(2R, 3R)$ or $(2S, 3S)$ isomers),it can be optically active. However,in the context of comparing these specific choices,the allene derivative in $C$ is a classic example of axial chirality,while the structure in $D$ is a meso compound if it has an internal plane of symmetry. Given the specific Fischer projection provided,it represents a chiral molecule (specifically,the $(2R, 3R)$ or $(2S, 3S)$ form). Both $C$ and $D$ are optically active. Typically,in such multiple-choice questions,the most distinct example of chirality is sought. $C$ is definitely chiral. $D$ as drawn is also chiral.

(D) $1$. $2,2'-dibromo-6,6'-diiodobiphenyl$ is optically active due to restricted rotation around the single bond connecting the two benzene rings (atropisomerism) caused by bulky ortho-substituents.
$2$. $Butan-2-ol$ $(CH_3CH(OH)CH_2CH_3)$ contains a chiral carbon atom bonded to four different groups,making it optically active.
$3$. $N,N-methylethylisopropylammonium \ hydroxide$ contains a chiral nitrogen atom (quaternary ammonium salt) bonded to four different groups,making it optically active.
$4$. Since all three compounds are optically active,the correct option is $D$.

(A) To determine if two Fischer projections are enantiomers,we assign the $R/S$ configuration to the chiral center in each structure.
For structure $I$,the priority order is: $-OH (1) > -CH=CH_2 (2) > -CH_3 (3) > -H (4)$. With $-H$ on a horizontal bond,the clockwise direction corresponds to the $S$ configuration.
For structure $II$,the priority order is: $-OH (1) > -CH=CH_2 (2) > -CH_3 (3) > -H (4)$. With $-H$ on a horizontal bond,the counter-clockwise direction corresponds to the $R$ configuration.
Since $I$ and $II$ are non-superimposable mirror images of each other,they are enantiomers.
Therefore,the correct combination is $I$ and $II$.

(B) Dichloroethene $(C_2H_2Cl_2)$ exists in three isomeric forms: $1,1$-dichloroethene,$cis$-$1,2$-dichloroethene,and $trans$-$1,2$-dichloroethene.
$1,2$-dichloroethene exhibits geometrical isomerism due to restricted rotation around the $C=C$ double bond.
However,all isomers of dichloroethene are planar molecules.
Since they possess a plane of symmetry,they are achiral and cannot exhibit optical isomerism.

(A) substance that can rotate the plane of plane-polarized light is known as an $Optically \ active$ substance.
This property is known as optical activity,which arises due to the presence of a chiral center in the molecule.

(A) For a molecule containing $n$ chiral centers (asymmetric carbon atoms) where the molecule is not symmetrical,the total number of stereoisomers is given by the formula $2^n$.

(C) The given compound is $CH_3CH(OH)COOH$ (lactic acid).
In this molecule,the central carbon atom is bonded to four different groups: a hydrogen atom $(-H)$,a hydroxyl group $(-OH)$,a methyl group $(-CH_3)$,and a carboxylic acid group $(-COOH)$.
Since the central carbon is a chiral center,the molecule can exist as two non-superimposable mirror images.
Therefore,it exhibits optical isomerism.

(D) For a compound to exhibit optical isomerism,it must contain at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1, 1-$ Dichloropropane: $CH_3-CH_2-CHCl_2$ (No chiral carbon)
$2, 2-$ Dichloropropane: $CH_3-CCl_2-CH_3$ (No chiral carbon)
$1, 3-$ Dibromopropane: $Br-CH_2-CH_2-CH_2-Br$ (No chiral carbon)
$1, 2-$ Dichloropropane: $CH_3-CHCl-CH_2Cl$. Here,the $C-2$ atom is bonded to a methyl group $(-CH_3)$,a hydrogen atom $(-H)$,a chlorine atom $(-Cl)$,and a chloromethyl group $(-CH_2Cl)$. Since all four groups are different,$C-2$ is a chiral center.
Therefore,$1, 2-$ Dichloropropane exhibits optical isomerism.

(B) Enantiomers are non-superimposable mirror images of each other.
They possess identical physical properties (like melting point,boiling point,refractive index) except for the direction in which they rotate plane-polarized light.
They exhibit identical chemical properties in an achiral environment,but they react differently with other optically active substances (chiral reagents).
Crucially,enantiomers often exhibit different biological properties because biological systems (enzymes,receptors) are chiral and interact differently with specific enantiomers.
Therefore,the statement that they have identical biological properties is incorrect.

(C) Isomers that are non-superimposable mirror images of each other are known as $Enantiomers$.
$Diastereomers$ are stereoisomers that are not mirror images of each other.
$Metamers$ are isomers resulting from the unequal distribution of carbon atoms on either side of a functional group.
$Mesomers$ (or $Meso$ compounds) are optically inactive compounds that possess chiral centers but have an internal plane of symmetry.

(C) meso compound is an optically inactive molecule that contains two or more stereocenters but also has an internal plane of symmetry.
In $CH_3-CH(OH)-CH(OH)-CH_3$ ($2$,$3$-butanediol),the molecule can exist in a meso form where the two chiral carbons have opposite configurations ($R,S$ configuration),leading to an internal plane of symmetry.
Thus,the correct option is $C$.

(D) The molecular formula $C_7H_{16}$ corresponds to an alkane $(C_nH_{2n+2})$.
To find optically active isomers,we look for chiral centers in the branched isomers of heptane.
The isomers of heptane are:
$1$. $n$-Heptane (achiral)
$2$. $2$-Methylhexane: $CH_3-CH(CH_3)-CH_2-CH_2-CH_2-CH_3$. The $C_2$ carbon is chiral. It has $2$ enantiomers.
$3$. $3$-Methylhexane: $CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$. The $C_3$ carbon is chiral. It has $2$ enantiomers.
$4$. $3$-Ethylpentane (achiral)
$5$. $2,2$-Dimethylpentane (achiral)
$6$. $2,3$-Dimethylpentane: $CH_3-CH(CH_3)-CH(CH_3)-CH_2-CH_3$. The $C_3$ carbon is chiral. It has $2$ enantiomers.
$7$. $2,4$-Dimethylpentane (achiral)
$8$. $3,3$-Dimethylpentane (achiral)
$9$. $2,2,3$-Trimethylbutane (achiral)
Total optically active isomers = $2$ (from $2$-methylhexane) + $2$ (from $3$-methylhexane) + $2$ (from $2,3$-dimethylpentane) = $6$.

(B) For a compound to exhibit optical isomerism,it must contain at least one chiral carbon atom (a carbon atom bonded to four different groups).
In Butan-$2$-ol $(CH_3-CH(OH)-CH_2-CH_3)$,the second carbon atom is bonded to a hydrogen atom,a hydroxyl group $(-OH)$,a methyl group $(-CH_3)$,and an ethyl group $(-CH_2CH_3)$.
Since all four groups attached to the $C-2$ atom are different,it is a chiral center.
Therefore,Butan-$2$-ol exhibits optical isomerism.

(C) The process of converting one enantiomer (e.g.,$d$-form) into a mixture containing both enantiomers ($d$ and $l$ forms) in equal proportions is known as $Racemisation$.
If the question refers specifically to the inversion of configuration at a chiral center during a reaction (like $S_N2$),it is called $Walden \text{ } Inversion$.
However,the conversion of a pure optical isomer into a racemic mixture is defined as $Racemisation$.

(C) Tartaric acid exists in different stereoisomeric forms.
$m-$ tartaric acid (meso-tartaric acid) is optically inactive due to an internal plane of symmetry.
$dl-$ tartaric acid (racemic mixture) is also optically inactive because the rotation caused by the $d-$ form is cancelled by the $l-$ form.
$d-$ tartaric acid (dextrorotatory) is optically active as it rotates plane-polarized light to the right.
Therefore,the correct answer is $d-$ tartaric acid.

(A) molecule is defined as chiral if it lacks a plane of symmetry or a center of inversion,making it non-superimposable on its mirror image. This property is the fundamental requirement for optical activity.

(D) Optical isomerism is exhibited by compounds that possess a chiral center (an asymmetric carbon atom bonded to four different groups).
$A$: Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ has chiral carbons and can exhibit optical isomerism.
$B$: Lactic acid $(CH_3-CH(OH)-COOH)$ has a chiral carbon atom at the $C-2$ position.
$C$: $2$-methylbutanoic acid $(HOOC-CH_2-CH(CH_3)-C_2H_5)$ has a chiral carbon atom at the $C-3$ position.
$D$: Propene $(CH_3-CH=CH_2)$ does not have any chiral carbon atom because the $sp^2$ hybridized carbons are bonded to identical hydrogen atoms,and the $sp^3$ carbon is bonded to three identical hydrogen atoms. Thus,it does not exhibit optical isomerism.

(B) In $1, 2-$dichlorocyclohexane,the carbon atoms at positions $1$ and $2$ are bonded to four different groups: a hydrogen atom,a chlorine atom,the rest of the cyclohexane ring,and the adjacent carbon atom.
Since each of these two carbons is bonded to four distinct groups,they are chiral centers.
Therefore,there are $2$ chiral carbons in $1, 2-$dichlorocyclohexane.

(D) Meso-tartaric acid contains two chiral carbon atoms but possesses a plane of symmetry within the molecule.
Due to this plane of symmetry,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.

(B) molecule is optically active if it contains at least one chiral center (an asymmetric carbon atom) and lacks a plane of symmetry or center of symmetry.
$1$. Succinic acid $(HOOC-CH_2-CH_2-COOH)$ has no chiral center.
$2$. Lactic acid $(CH_3-CH(OH)-COOH)$ has one chiral carbon atom attached to four different groups $(-H, -OH, -CH_3, -COOH)$,making it optically active.
$3$. Meso tartaric acid has chiral centers but possesses a plane of symmetry,making it optically inactive (meso compound).
$4$. Chloro acetic acid $(ClCH_2-COOH)$ has no chiral center.
Therefore,Lactic acid is the correct answer.

(C) Optical activity refers to the ability of a chiral molecule to rotate the plane of plane-polarized light. This rotation is measured using an instrument called a $Polarimeter$.

(D) The structure of $2, 3-$dibromobutanoic acid is $CH_3-CH(Br)-CH(Br)-COOH$.
This molecule contains two chiral centers at $C_2$ and $C_3$.
The number of stereoisomers is given by $2^n$,where $n$ is the number of chiral centers.
Here,$n = 2$,so the total number of stereoisomers is $2^2 = 4$.
These four stereoisomers consist of two pairs of enantiomers,which are optically active.
Therefore,the number of optical isomers is $4$.

(C) For a compound with $n$ different chiral centers,the total number of stereoisomers is given by the formula $2^n$.
Given that the compound has $n = 3$ different chiral centers.
Therefore,the number of stereoisomers = $2^3 = 8$.

(C) molecule is achiral if it possesses a plane of symmetry or a center of inversion.
$(a)$ $2$-bromocyclohexanol has chiral centers and no plane of symmetry,so it is chiral.
$(b)$ $2$-chlorohexane has a chiral carbon at the $C2$ position,so it is chiral.
$(c)$ $1,4$-dimethylcyclohexane (specifically the cis-isomer) has a plane of symmetry passing through $C1$ and $C4$,making it achiral.
$(d)$ $1$-chloro-$2$-deuterocyclohexane has chiral centers and no plane of symmetry,so it is chiral.
Therefore,the correct option is $C$.

(A) Tartaric acid is $HOOC-CH(OH)-CH(OH)-COOH$.
It contains two chiral carbon atoms.
The total number of stereoisomers is given by $2^n$ where $n$ is the number of chiral centers,but since the molecule is symmetric,we use the formula for meso compounds.
For tartaric acid,there are two optically active enantiomers ($d$ and $l$ forms) and one optically inactive meso form.
Thus,the number of optically active isomers is $2$.

(D) molecule is optically active if it contains at least one chiral center (a carbon atom bonded to four different groups).
$1$. Alanine $(CH_3CH(NH_2)COOH)$ has a chiral carbon.
$2$. Lactic acid $(CH_3CH(OH)COOH)$ has a chiral carbon.
$3$. Malic acid $(HOOCCH_2CH(OH)COOH)$ has a chiral carbon.
$4$. $2, 2-$Dimethylbutanoic acid $(CH_3CH_2C(CH_3)_2COOH)$ has no chiral carbon because the $C-2$ atom is bonded to two identical methyl groups. Therefore,it is not optically active.

(C) molecule is achiral if it possesses a plane of symmetry or a center of symmetry,or if it lacks a stereocenter.
$A$: $CH_3CH(Br)CH=CH_2$ has a chiral carbon at the second position (bonded to $-H$,$-CH_3$,$-Br$,and $-CH=CH_2$),so it is chiral.
$B$: The structure shows a carbon bonded to four different groups ($-OH$,$-H$,$-CH_3$,and $-CH_2CH_2CH_3$),so it is chiral.
$C$: The structure represents a meso compound (specifically a form of tartaric acid derivative). It contains a plane of symmetry passing through the center of the molecule,making it achiral.
$D$: The structure shows a carbon bonded to four different groups ($-H$,$-CH_3$,$-OH$,and $-Br$),so it is chiral.
Therefore,the achiral compound is $C$.

(B) Non-superimposable mirror images are known as enantiomers. Enantiomers are a type of stereoisomer that are mirror images of each other but cannot be superimposed on one another.

(D) The given compound is $CH_3-CHBr-CHBr-COOH$.
It contains two chiral carbon atoms at positions $C_2$ and $C_3$.
The molecule is unsymmetrical because the groups attached to the chiral carbons are different ($CH_3$ and $COOH$).
For an unsymmetrical molecule with $n$ chiral centers,the number of stereoisomers is given by $2^n$.
Here,$n = 2$,so the number of stereoisomers = $2^2 = 4$.

(C) The given compound is $CH_3-CH(COOH)-CH(COOH)-CH_2CH_3$.
It contains two chiral carbon atoms (marked with *): $CH_3-CH^*(COOH)-CH^*(COOH)-CH_2CH_3$.
Since the two chiral carbons are attached to different groups (one is attached to $-CH_3$ and $-H$,the other to $-CH_2CH_3$ and $-H$),the molecule is asymmetric.
The number of optical isomers is given by the formula $2^n$,where $n$ is the number of chiral centers.
Here,$n = 2$,so the number of optical isomers = $2^2 = 4$.

(D) stereogenic $C-$ atom (chiral center) is a carbon atom bonded to four different groups.
$1$. In $1,2-$ dibromobutane $(CH_3-CH_2-CH(Br)-CH_2Br)$,the $C-2$ atom is bonded to $-H, -Br, -CH_2CH_3,$ and $-CH_2Br$. Thus,it is chiral.
$2$. In $1-$ bromo $-2-$ methylcyclopentane,the $C-1$ and $C-2$ atoms are bonded to four different groups,making them chiral centers.
$3$. In $1,1,2-$ tribromopropane $(CH_3-CH(Br)-CHBr_2)$,the $C-2$ atom is bonded to $-H, -Br, -CH_3,$ and $-CHBr_2$. Thus,it is chiral.
Since all the given molecules contain at least one stereogenic carbon atom,the correct answer is $D$.

(B) Optical activity is shown by molecules containing at least one chiral carbon atom (a carbon atom attached to four different groups).
In $CH_3-CH_2-CH(OH)-CH_3$ $(\text{Butan-2-ol})$,the second carbon atom is chiral because it is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$.
Thus,it is optically active.

(A) Enantiomerism is exhibited by molecules that are chiral,meaning they possess at least one chiral carbon atom (a carbon atom bonded to four different groups).
$1.$ Diphenyl methanol: The central carbon is bonded to one $H$ atom,one $OH$ group,and two identical phenyl groups. Since it has two identical groups,it is achiral and cannot exhibit enantiomerism.
$2.$ $1-$Bromo$-2-$chlorobutane: The carbon at position $2$ is bonded to $H$,$Cl$,$CH_3$,and $CH_2Br$. All four groups are different,so it is chiral.
$3.$ $2-$Butanol: The carbon at position $2$ is bonded to $H$,$OH$,$CH_3$,and $CH_2CH_3$. All four groups are different,so it is chiral.
$4.$ Tartaric acid: It contains two chiral carbon atoms and can exist as enantiomers (e.g.,$(+)-$ and $(-)-$tartaric acid).
Therefore,Diphenyl methanol cannot exhibit enantiomerism.

(B) meso compound is an optically inactive molecule that contains chiral centers but also possesses an internal plane of symmetry or a center of symmetry.
For a molecule to exhibit a meso form,it must have at least two identical chiral centers and a symmetric structure.
In $CH_3CH(OH)CH(OH)CH_3$ (butane$-2,3-$diol),the two central carbons are chiral and identical. In the eclipsed conformation,it can have a plane of symmetry,making it a meso compound.

(A) molecule is chiral if it has a chiral center (a carbon atom bonded to four different groups). Otherwise,it is achiral.
$(i)$ $(i)$ $CH_3CH(Br)OH$: The central carbon is bonded to $-H$,$-CH_3$,$-Br$,and $-OH$. Since all four groups are different,it is chiral.
(ii) $CH_3CH(Br)_2$: The central carbon is bonded to two identical $-Br$ atoms. Thus,it is achiral.
(ii) $(i)$ $CH_3CH(OH)CH_2CH_2CH_3$: The carbon bonded to $-OH$ is attached to $-H$,$-CH_3$,$-OH$,and $-CH_2CH_2CH_3$. All four groups are different,so it is chiral.
(ii) $CH_3CH_2CH(OH)CH_2CH_3$: The carbon bonded to $-OH$ is attached to $-H$,$-OH$,and two identical $-CH_2CH_3$ groups. Since two groups are identical,it is achiral.
(iii) $(i)$ $CH_3CH(Br)CH_2CH_3$: The carbon bonded to $-Br$ is attached to $-H$,$-CH_3$,$-Br$,and $-CH_2CH_3$. All four groups are different,so it is chiral.
(ii) $CH_3CH_2CH_2CH_2Br$: The carbon bonded to $-Br$ is attached to two $-H$ atoms. Since two groups are identical,it is achiral.

(B) chiral molecule is defined as a molecule that is non-superimposable on its mirror image.
This property is known as chirality.
Such molecules lack any internal plane of symmetry or center of inversion,which makes them optically active.

(N/A) Configuration: The spatial arrangement of functional groups (atoms) around a carbon atom is called the spatial configuration of that carbon.
Below are two spatial structures of the same compound $CHClBrI$.
The arrangement of groups (atoms) attached to the carbon in structure $(A)$ and structure $(B)$ is not the same. These structures $(A)$ and $(B)$ are mirror images of each other,which is called the reflection of a configuration.
The reflection of $(A)$ is $(B)$ and the reflection of $(B)$ is $(A)$. The configuration of carbon in structure $(A)$ is the reflection of the configuration of carbon in structure $(B)$. If $(A)$ and $(B)$ are different,then the optical activity of $(A)$ and $(B)$ is opposite,and their chirality is also opposite.

(N/A) Optical activity: When ordinary light is passed through a Nicol prism,it produces plane-polarized light. When this plane-polarized light is passed through a solution of a compound in a polarimeter,if the compound rotates the plane of polarized light,it is called an optically active compound. They are denoted as $(+)$ or $(-)$.
Polarimeter: It is an instrument used to measure optical activity. The angle of rotation of plane-polarized light is measured using this instrument.
$Dextrorotatory$ $(d)$ or $(+)$ isomer: $A$ compound that rotates the plane of polarized light to the right,i.e.,in the clockwise direction,is called a dextrorotatory or $d$-form. This rotation is indicated by a $(+)$ sign before the angle of rotation.

(N/A) $(i)$ Laevorotatory isomer:
$A$ compound is called laevorotatory $(l)$ or $(-)$ if its solution rotates the plane of plane-polarized light to the left (anticlockwise direction). The $l$-rotation is indicated by placing a $(-)$ sign before the value of the angle of rotation.
$(ii)$ Optical isomers:
The $(+)$ and $(-)$ isomers of a compound are known as optical isomers,and this phenomenon is called optical isomerism.

(N/A) Louis Pasteur laid the foundation of modern stereochemistry in $1848$.
$(i)$ He observed that crystals of certain compounds exist as mirror images.
$(ii)$ He demonstrated that aqueous solutions of equal concentration of these two types of crystals show equal but opposite optical rotation.
$(iii)$ He believed that the difference in optical activity is related to the three-dimensional arrangement (configuration) of atoms in the crystals.
Note: Configuration refers to the three-dimensional arrangement of atoms around the carbon atom.