Optical isomerism Questions in English

(C) $2,3-$dichlorobutane has two chiral carbon atoms at positions $2$ and $3$.
It can exist in different stereoisomeric forms,including enantiomers and a meso form.
Since it possesses chiral centers and can exhibit non-superimposable mirror images as well as a meso form,it primarily exhibits optical isomerism.

(C) racemic mixture is an equimolar mixture of two enantiomers.
Since the two enantiomers rotate the plane of plane-polarized light in opposite directions by the same magnitude,their effects cancel each other out.
Therefore,the net optical rotation of a racemic mixture is $0$ (zero).

(D) Optical isomerism is exhibited by molecules that possess a chiral center (a carbon atom bonded to four different groups).
In the given options,let us analyze the structure of $CH_2=CHCH(OH)CH_3$:
This molecule is $but-3-en-2-ol$.
The carbon atom at position $2$ is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH=CH_2$.
Since this carbon is chiral,the molecule exhibits optical isomerism.
Other options like $CH_3COCH_2CH_3$ (butanone),$CH_3CH_2CH_2CHO$ (butanal),and $(CH_3)_2CHCHO$ ($2$-methylpropanal) do not contain a chiral carbon atom.

(B) The given compound is $4-methylpent-3-enoic acid$ with the structure $(CH_3)_2C=CH-CH(CH_3)COOH$.
$1$. Geometrical Isomerism: The double bond is between $C_3$ and $C_4$. At $C_4$,there are two identical methyl groups attached,so it does not show geometrical isomerism.
$2$. Optical Isomerism: There is a chiral carbon atom at the $C_2$ position (the carbon attached to $-H$,$-CH_3$,$-COOH$,and the $-CH=C(CH_3)_2$ group). Since it has a chiral center,it exhibits optical isomerism.
$3$. Stereoisomerism: Since optical isomerism is a type of stereoisomerism,the compound exhibits stereoisomerism.

(D) For a molecule to exhibit optical isomerism,it must contain at least one chiral center (an asymmetric carbon atom bonded to four different groups).
Let us analyze the structures:
$A$: $CH_3-C(CH_3)=CH-CH_2-CH_3$ ($2-$Methyl$-2-$pentene) - No chiral center.
$B$: $CH_3-CH=C(CH_3)-CH_2-CH_3$ ($3-$Methyl$-2-$pentene) - No chiral center.
$C$: $CH_2=CH-CH_2-CH(CH_3)-CH_3$ ($4-$Methyl$-1-$pentene) - No chiral center.
$D$: $CH_2=CH-CH(CH_3)-CH_2-CH_3$ ($3-$Methyl$-1-$pentene) - The carbon at position $3$ is bonded to a hydrogen atom,a methyl group,an ethyl group,and a vinyl group $(-CH=CH_2)$. Since all four groups are different,it is a chiral center.
Therefore,$3-$Methyl$-1-$pentene exhibits optical isomerism.

(D) The molecular formula $C_4H_8$ corresponds to butene isomers.
$1$-butene and $2$-butene exhibit chain and positional isomerism.
$2$-butene exhibits geometrical isomerism.
However,none of the isomers of $C_4H_8$ contain a chiral carbon atom,therefore they do not exhibit optical isomerism.

(D) The given molecule is $CH_3-CH(OH)-CH(OH)-CH(OH)-CH_3$.
It has $n = 3$ chiral centers.
The molecule is unsymmetrical.
The number of optically active isomers $(a)$ is given by $a = 2^{n-1} = 2^{3-1} = 2^2 = 4$.
The number of meso isomers $(m)$ is given by $m = 2^{(n-2)/2} = 2^{(3-1)/2} = 2^1 = 2$.
Wait,for an unsymmetrical molecule with $n$ chiral centers,the total number of stereoisomers is $2^n = 2^3 = 8$.
However,the question asks for optical isomers.
For $CH_3-CH(OH)-CH(OH)-CH(OH)-CH_3$,the number of stereoisomers is $2^n = 2^3 = 8$.
These consist of $4$ pairs of enantiomers (optically active).
Thus,the total number of optical isomers is $8$.

(D) The molecule $CH_3CH(Br)CH(Br)COOH$ contains $2$ chiral carbon atoms $(n = 2)$.
Since the molecule is unsymmetrical,the number of optical isomers is $2^n = 2^2 = 4$.
These $4$ optical isomers consist of $2$ pairs of enantiomers.

(B) racemic mixture is an equimolar mixture of two enantiomers. For a compound to exist as a racemic mixture,it must possess at least one chiral center (an asymmetric carbon atom bonded to four different groups).
$1$. $CH_3CH_2CH(OH)CH_2CH_3$ (Pentan$-3-$ol): The central carbon is bonded to two identical ethyl groups $(-CH_2CH_3)$,so it is achiral.
$2$. $CH_3CH(OH)COOH$ (Lactic acid): The central carbon is bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$. Since all four groups are different,it is chiral and can exist as a racemic mixture.
$3$. $CH_3CH_2NHCH_2CH_3$ (Diethylamine): The nitrogen is bonded to two identical ethyl groups,making it achiral.
$4$. $CH_3CH_2CHDCl$ ($1$-chloro$-1-$deuteropropane): While this is chiral,lactic acid is the standard textbook example for a simple racemic mixture.

(C) The given molecule is $CH_3-CH(OH)-CH(OH)-CH_3$,which is $2,3-butanediol$.
It has $n = 2$ chiral centers.
The molecule is symmetric because the groups attached to both chiral centers are identical.
For a symmetric molecule with an even number of chiral centers $(n=2)$:
Number of optically active isomers $(a)$ = $2^{n-1} = 2^{2-1} = 2^1 = 2$.
Number of meso isomers $(m)$ = $2^{(n/2)-1} = 2^{(2/2)-1} = 2^0 = 1$.
Total number of stereoisomers = $a + m = 2 + 1 = 3$.

(D) When a molecule has $n$ asymmetric carbon atoms and can be divided into two identical halves,it possesses a plane of symmetry.
For such molecules,the number of optically active isomers (enantiomeric pairs) is $2^{(n/2-1)}$.
The number of meso forms is $2^{(n/2-1)}$.
The total number of stereoisomers is the sum of optically active forms and meso forms,which is $2^{(n-1)} + 2^{(n/2-1)}$.

(D) molecule or object is chiral if it is non-superimposable on its mirror image.
Among the given letters,the letter $P$ does not have a plane of symmetry,making its mirror image non-superimposable on itself.
Therefore,$P$ is chiral.

(C) chiral molecule is one that contains a chiral center,which is a carbon atom bonded to four different groups.
In $CH_3CH(OH)CH_2CH_3$ (butan$-2-$ol),the second carbon atom is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$.
Therefore,it is a chiral molecule.

(B) Butane-$2,3$-diol has two chiral centers at $C-2$ and $C-3$.
The structure is $CH_3-CH(OH)-CH(OH)-CH_3$.
It exists in three stereoisomeric forms:
$1$. $(2R, 3R)$-butane-$2,3$-diol (optically active)
$2$. $(2S, 3S)$-butane-$2,3$-diol (optically active)
$3$. $(2R, 3S)$-butane-$2,3$-diol (meso form,optically inactive due to internal compensation).
Thus,there are $2$ optically active stereoisomers.

(C) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In $CH_3CHClCOOH$,the central carbon atom is bonded to a $-CH_3$ group,a $-Cl$ atom,a $-COOH$ group,and an $-H$ atom.
Since all four groups are different,this carbon is chiral,making the molecule optically active.

(B) The correct answer is $(b)$.
Butan$-2-$ol,represented as $CH_3-CH(OH)-CH_2-CH_3$,exhibits optical isomerism.
This is because the carbon atom at the $C-2$ position is a chiral center,meaning it is bonded to four different groups: a hydrogen atom $(-H)$,a hydroxyl group $(-OH)$,a methyl group $(-CH_3)$,and an ethyl group $(-CH_2CH_3)$.

(C) molecule is optically active if it contains a chiral carbon atom (a carbon atom bonded to four different groups).
In $3-$methylheptane,the carbon at position $3$ is bonded to a hydrogen atom,a methyl group,an ethyl group,and a butyl group.
Since all four groups are different,the carbon at position $3$ is chiral.
Therefore,$3-$methylheptane is optically active.

(B) $2,3-$dichlorobutane has two chiral carbon atoms at positions $2$ and $3$.
It exists in three stereoisomeric forms: a pair of enantiomers and a meso compound.
Since it exhibits enantiomerism and meso-form formation,it is a classic example of optical isomerism.
Therefore,the correct option is $B$.

(B) To determine the $R$-configuration,we assign priorities to the groups attached to the chiral carbon based on the Cahn-Ingold-Prelog $(CIP)$ rules: $Cl(1) > C_2H_5(2) > CH_3(3) > H(4)$.
In the Fischer projection,the configuration is $R$ if the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,provided the lowest priority group $(H)$ is on a vertical bond.
Analyzing the structures,the configuration $R$ corresponds to the arrangement where the priority sequence is clockwise. Based on the provided image,the third structure shows the $R$-configuration where $Cl$ is on the left,$C_2H_5$ is on top,$CH_3$ is on the right,and $H$ is at the bottom. This corresponds to the structure $C_2H_5 - C(Cl)(CH_3) - H$ (Option $B$).

(A) Optical isomerism requires the presence of at least one chiral carbon atom (a carbon atom bonded to four different groups).
Maleic acid $(HOOC-CH=CH-COOH)$ contains $sp^2$ hybridized carbons in a double bond and lacks a chiral carbon atom,so it does not exhibit optical isomerism.
Tartaric acid,lactic acid,and $\alpha$-amino acids all contain at least one chiral carbon atom (indicated by $*$ in the structure),allowing them to exhibit optical isomerism.

(B) The molecule $CH_3CH(OH)COOH$ (lactic acid) contains a chiral carbon atom bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-COOH$.
Stereoisomers that are non-superimposable mirror images of each other are known as enantiomers.
Since the two structures of lactic acid are non-superimposable mirror images and are both optically active,they are classified as enantiomers.

(C) compound that does not rotate plane polarised light is optically inactive.
If a compound is thought to be chiral but shows no optical rotation,it could be a racemic mixture (an equimolar mixture of two enantiomers where the rotation of one is cancelled by the other,i.e.,externally compensated) or it could be a meso compound (which contains chiral centers but is achiral due to an internal plane of symmetry,i.e.,internally compensated).
Therefore,the observation of no rotation does not definitively prove the compound is meso; it could also be a racemic mixture.

(D) molecule is chiral if it contains at least one chiral carbon atom,which is a carbon atom bonded to four different groups.
$1$. $2-$Hydroxypropanoic acid: $CH_3-CH(OH)-COOH$. The $C-2$ atom is bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$. It is chiral.
$2$. $2-$Butanol: $CH_3-CH(OH)-CH_2-CH_3$. The $C-2$ atom is bonded to $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$. It is chiral.
$3$. $2,3-$Dibromopentane: $CH_3-CH(Br)-CH(Br)-CH_2-CH_3$. Both $C-2$ and $C-3$ are bonded to four different groups. It is chiral.
$4$. $3-$Bromopentane: $CH_3-CH_2-CH(Br)-CH_2-CH_3$. The $C-3$ atom is bonded to $-H$,$-Br$,and two identical $-CH_2CH_3$ groups. Since it is bonded to two identical groups,it is achiral.
Therefore,$3-$Bromopentane is not chiral.

(B) molecule is expected to rotate the plane of plane-polarised light if it is optically active.
Optically active molecules must be chiral,meaning they lack a plane of symmetry or center of inversion.
$1$. Glycine $(H_2N-CH_2-COOH)$ has a plane of symmetry and is achiral.
$2$. Glyceraldehyde $(HOCH_2-CH(OH)-CHO)$ has a chiral carbon atom bonded to four different groups $(-H, -OH, -CHO, -CH_2OH)$ and is optically active.
$3$. Butane-$2$-thiol $(CH_3-CH(SH)-CH_2-CH_3)$ has a chiral carbon atom bonded to four different groups $(-H, -SH, -CH_3, -CH_2CH_3)$ and is optically active.
$4$. $1,2$-Diphenylethane-$1,2$-diamine can exist as a meso compound (optically inactive) or as chiral enantiomers depending on the configuration.
In the context of standard chemistry questions,both Glyceraldehyde and Butane-$2$-thiol are classic examples of chiral molecules. However,Glyceraldehyde is the most fundamental example of a chiral molecule in carbohydrates. Given the options,both $B$ and $C$ are chiral. Assuming a single choice is required,Glyceraldehyde is the standard textbook example.

(B) To determine the absolute configuration,we assign priorities to the groups attached to each chiral center using the Cahn-Ingold-Prelog $(CIP)$ rules.
For the first chiral center (left): The groups are $-OH$ $(1)$,$-COOH$ $(2)$,$-CH(OH)COOH$ $(3)$,and $-H$ $(4)$. Since $-H$ is on a dashed bond (pointing away),the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,which corresponds to the $R$ configuration.
For the second chiral center (right): The groups are $-OH$ $(1)$,$-COOH$ $(2)$,$-CH(OH)COOH$ $(3)$,and $-H$ $(4)$. Since $-H$ is on a dashed bond (pointing away),the sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,which corresponds to the $R$ configuration.
Thus,the absolute configuration is $R, R$.

(B) compound is optically active if it is chiral,meaning it lacks a plane of symmetry,center of inversion,or an improper axis of rotation.
$A$: Trans$-1,3-$dichlorocyclobutane has a plane of symmetry,so it is achiral.
$B$: $(2S, 3S)-2,3$-dichlorobutane is a chiral molecule because it lacks any internal plane of symmetry or center of inversion. Therefore,it is optically active and will rotate plane-polarized light.
$C$: $1-$bromo$-2-$fluoro$-3-$chlorobenzene is a planar aromatic molecule,which is achiral.
$D$: $CH_2ClF$ is achiral because it has a plane of symmetry (the plane containing the $C, Cl, F$ atoms).

(B) Assigning $R/S$ configurations using Cahn-Ingold-Prelog priority rules:
For $(A)$: Priority is $-OH (1) > -CHO (2) > -CH_2OH (3) > -H (4)$. With $-H$ on the horizontal axis,the clockwise arrangement $1$ $\rightarrow 2$ $\rightarrow 3$ corresponds to $S$ configuration.
For $(B)$: Priority is $-OH (1) > -CHO (2) > -CH_2OH (3) > -H (4)$. With $-H$ on the vertical axis,the clockwise arrangement $1$ $\rightarrow 2$ $\rightarrow 3$ corresponds to $R$ configuration.
For $(C)$: Priority is $-OH (1) > -CHO (2) > -CH_2OH (3) > -H (4)$. With $-H$ on the horizontal axis,the counter-clockwise arrangement $1$ $\rightarrow 2$ $\rightarrow 3$ corresponds to $R$ configuration.
Comparing configurations: $(A)$ is $S$,$(B)$ is $R$,$(C)$ is $R$. Therefore,$(B)$ and $(C)$ are identical,while $(A)$ and $(B)$ are enantiomers,and $(A)$ and $(C)$ are enantiomers. The statement '$A$ and $C$ are identical' is incorrect.

(A) chiral centre is a carbon atom that is bonded to four different groups.
In the given Newman projection,we can convert it into a Fischer projection to identify the chiral centres.
The structure has two main carbon atoms in the backbone and a side chain containing a chiral carbon.
$1$. The carbon atom attached to $-COOH$,$-H$,$-Cl$,and the rest of the chain is chiral.
$2$. The carbon atom attached to $-CHO$,$-Cl$,$-H$,and the rest of the chain is chiral.
$3$. The carbon atom in the side chain $-CH(OH)D$ is attached to $-H$,$-OH$,$-D$,and the main chain,making it chiral.
Thus,there are $3$ chiral centres in the molecule.

(D) $1$. The starting material is a chiral ketone with an alkyne group: $CH_3-CD(CH_3)-C \equiv C-CO-CH_3$. The chiral center at the $CD(CH_3)$ position is fixed as $(R)$ or $(S)$.
$2$. Reaction $A$ involves the reduction of the alkyne to a $cis$-alkene. This creates a new double bond,but since the geometry is fixed as $cis$,no new stereocenter is formed at the alkene carbons.
$3$. Compound $C$ is the $cis$-alkene: $CH_3-CD(CH_3)-CH=CH-CO-CH_3$. It has one chiral center.
$4$. The final step is the addition of $Br_2/CCl_4$ to the $cis$-alkene. This is an anti-addition reaction across the double bond.
$5$. The addition of $Br_2$ to the $cis$-alkene creates two new chiral centers at the carbons where $Br$ atoms are added.
$6$. Since the molecule already has one chiral center,and two new ones are formed,the total number of stereoisomers is $2^n$,where $n$ is the number of chiral centers. Here,$n=3$. Total stereoisomers = $2^3 = 8$.

(A) The structure of $2,3-$dibromobutane-$1,4-$dioic acid is $HOOC-CH(Br)-CH(Br)-COOH$.
This molecule contains two chiral centers at $C2$ and $C3$.
Since the molecule is symmetric (the two ends are identical),we can determine the number of stereoisomers using the formula for symmetric molecules.
For a symmetric molecule with an even number of chiral centers $(n=2)$,the number of optically active isomers (enantiomers) is $2^{(n-1)} = 2^{(2-1)} = 2^1 = 2$.
The two optically active isomers are the $(2R, 3R)$ and $(2S, 3S)$ forms.
The other isomer is the meso form $(2R, 3S)$,which is optically inactive due to an internal plane of symmetry.
Therefore,there are $2$ optically active isomers.

(A) Enantiomers are non-superimposable mirror images of each other.
They possess identical physical properties,such as boiling point,melting point,and solubility in a given solvent,in an achiral environment like methanol.
Therefore,the pair consisting of enantiomers will exhibit identical solubilities.

(D) For a molecule to be optically active,it must lack a plane of symmetry $(P.O.S.)$ and a center of symmetry $(C.O.S.)$.
$A$: This is a substituted allene with an odd number of double bonds. The terminal groups are in the same plane,and it possesses a plane of symmetry. Thus,it is optically inactive.
$B$: This is a bicyclic compound with a plane of symmetry passing through the molecule. Thus,it is optically inactive.
$C$: This is $cis-1,3-dimethylcyclopentane$. It has a plane of symmetry passing through the $C_2$ and $C_5$ atoms. Thus,it is optically inactive.
$D$: This is a tartaric acid derivative. In the given configuration,there is no plane of symmetry or center of symmetry. Therefore,it is chiral and optically active.

(C) Lactic acid $(CH_3-CH(OH)-COOH)$ on heating undergoes intermolecular esterification to form a cyclic diester known as lactide.
The lactide molecule contains two chiral centers.
The possible stereoisomers are $(R,R)$,$(S,S)$,and $(R,S)$ (meso form).
Therefore,a total of $3$ stereoisomeric products are formed.

(A) The formula for specific rotation $[\alpha]$ is given by: $[\alpha] = \frac{\alpha}{l \times c}$
Where:
$\alpha$ is the observed rotation = $+10^{\circ}$
$l$ is the path length in $dm$ = $1 \ dm$
$c$ is the concentration in $g/mL$ = $\frac{1 \ g}{10 \ mL} = 0.1 \ g/mL$
Substituting the values:
$[\alpha] = \frac{10}{1 \times 0.1} = \frac{10}{0.1} = +100^{\circ}$
Therefore,the specific rotation is $+100^{\circ}$.

(B) To determine the configuration,we assign priorities to the groups attached to the chiral carbon atom using the Cahn-Ingold-Prelog $(CIP)$ rules based on atomic number:
$1$. $-OH$ (Oxygen,$Z=8$)
$2$. $-Cl$ (Chlorine,$Z=17$)
$3$. $-NH_2$ (Nitrogen,$Z=7$)
$4$. $-H$ (Hydrogen,$Z=1$)
Wait,re-evaluating priorities:
$1$. $-Cl$ $(Z=17)$
$2$. $-OH$ $(Z=8)$
$3$. $-NH_2$ $(Z=7)$
$4$. $-H$ $(Z=1)$
In the given structure,the lowest priority group $(-H)$ is on a dashed bond (pointing away).
Tracing the path from priority $1$ $\rightarrow 2$ $\rightarrow 3$:
$Cl$ $\rightarrow OH$ $\rightarrow NH_2$
This path is counter-clockwise.
Therefore,the configuration is $S$.

(C) The compound $1-$bromo$-3-$chlorocyclobutane has two chiral centers at positions $1$ and $3$ of the cyclobutane ring.
Since the ring is symmetric,we must consider the geometric configurations (cis and trans) and the optical activity.
For the $cis-$isomer,the molecule has a plane of symmetry,making it achiral (meso form).
For the $trans-$isomer,the molecule exists as a pair of enantiomers because it lacks a plane of symmetry.
Thus,the total number of stereoisomers is $1$ (cis-meso) $+ 2$ (trans-enantiomers) $= 3$ stereoisomers.

(D) chiral carbon atom is a carbon atom that is bonded to four different groups.
Let us analyze the structure:
$1$. The carbon atom attached to the $H$ atom is bonded to an ethyl group,a $D$ atom,the rest of the chain,and an $H$ atom. Since all four groups are different,this is a chiral center.
$2$. The carbon atom attached to the $D$ atom is bonded to the previous chiral carbon,a $D$ atom,the carbon with the $OH$ group,and an $H$ atom (implied). This is a chiral center.
$3$. The carbon atom attached to the $OH$ group is bonded to the carbon with the $D$ atom,an $OH$ group,an $H$ atom (implied),and the rest of the chain. This is a chiral center.
$4$. The carbon atom attached to the $Me$ group is bonded to the rest of the chain,an $H$ atom (implied),a methyl group,and another methyl group (the terminal $Me$ and the branch $Me$). Since it is bonded to two identical methyl groups,it is not a chiral center.
Thus,there are $3$ chiral carbon atoms in the given compound.

(A) molecule is optically active if it contains at least one chiral center (a carbon atom bonded to four different groups).
$A$. Lactic acid $(CH_3-CH(OH)-COOH)$ has a chiral carbon atom bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$ groups. Thus,it is optically active.
$B$. Propane $(CH_3-CH_2-CH_3)$ has no chiral center.
$C$. Cyclohexane is a symmetric molecule and does not possess a chiral center.
Therefore,the correct option is $A$.

(C) The molecule $2$-hydroxycyclopentanethiol has two chiral centers at the $C_1$ and $C_2$ positions of the cyclopentane ring.
Since the two chiral centers are attached to different groups (one to $-OH$ and one to $-SH$),the molecule does not possess a plane of symmetry in any configuration.
The number of stereoisomers for a molecule with $n$ chiral centers and no internal symmetry is given by $2^n$.
Here,$n = 2$,so the number of stereoisomers is $2^2 = 4$.
These consist of two pairs of enantiomers: $(1R, 2R)$,$(1S, 2S)$,$(1R, 2S)$,and $(1S, 2R)$.

(C) Enantiomers are non-superimposable mirror images of each other.
In option $C$,the two structures are mirror images of each other and are non-superimposable because the central carbon is chiral (bonded to four different groups: $-OCH_3, -Cl, -D, -H$).
Therefore,the pair in option $C$ represents enantiomers.

(C) To determine the configuration of the chiral centers,we use the Cahn-Ingold-Prelog $(CIP)$ priority rules.
For the chiral center at $C-2$ (the upper one attached to $COOH$,$CH_3$,$H$,and the rest of the chain):
$1$. Priority $1$: $-COOH$ $(-C(=O)OH)$
$2$. Priority $2$: $-CH(OH)CH_2CH_2OH$ (the lower chiral center group)
$3$. Priority $3$: $-CH_3$
$4$. Priority $4$: $-H$
Since the lowest priority group $(-H)$ is on a horizontal bond,we reverse the result. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise $(R)$,so reversing it gives $S$. Thus,$C-2$ is $2S$.
For the chiral center at $C-3$ (the lower one attached to $OH$,$H$,$CH_2CH_2OH$,and the upper group):
$1$. Priority $1$: $-OH$
$2$. Priority $2$: $-CH(CH_3)COOH$ (the upper chiral center group)
$3$. Priority $3$: $-CH_2CH_2OH$
$4$. Priority $4$: $-H$
Since the lowest priority group $(-H)$ is on a horizontal bond,we reverse the result. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise $(S)$,so reversing it gives $R$. Thus,$C-3$ is $3R$.
Therefore,the configuration is $2S, 3R$.

(A) The given structures are Fischer projections of polyhydroxy compounds. Let us analyze the configuration at each chiral center.
For the first molecule (left): The chiral centers from top to bottom have configurations (assuming standard priority rules): $C_2$ ($OH$ on right),$C_3$ ($OH$ on right),$C_4$ ($OH$ on left).
For the second molecule (right): The chiral centers from top to bottom have configurations: $C_2$ ($OH$ on left),$C_3$ ($OH$ on left),$C_4$ ($OH$ on right).
Since the configurations at all chiral centers are inverted,the two molecules are non-superimposable mirror images of each other.
Therefore,they are enantiomers.

(D) The reaction of $CH_3-CH=CH-COOH$ (crotonic acid) with $Br_2$ is an electrophilic addition reaction.
This reaction involves the anti-addition of bromine across the double bond.
The product formed is $CH_3-CH(Br)-CH(Br)-COOH$ ($2$,$3$-dibromobutanoic acid).
This molecule contains two chiral centers at $C_2$ and $C_3$.
For a molecule with two chiral centers where the groups attached to the chiral carbons are different,the number of stereoisomers is $2^n$,where $n$ is the number of chiral centers.
Here,$n = 2$,so the number of stereoisomers is $2^2 = 4$.

(A, C) First,determine the configuration of each chiral center using the Cahn-Ingold-Prelog $(CIP)$ priority rules: $1: -OH$,$2: -C_2H_5$,$3: -CH_3$,$4: -H$.
For compound $A$: The $-H$ is on a horizontal bond. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is clockwise,so the configuration is $S$.
For compound $B$: The $-H$ is on a horizontal bond. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise,so the configuration is $R$.
For compound $C$: The $-H$ is on a horizontal bond. The sequence $1$ $\rightarrow 2$ $\rightarrow 3$ is counter-clockwise,so the configuration is $R$.
Comparing the configurations: $A$ $(S)$ and $B$ $(R)$ are enantiomers. $B$ $(R)$ and $C$ $(R)$ are identical (they are the same molecule). $A$ $(S)$ and $C$ $(R)$ are enantiomers.
Therefore,the correct statements are that $A$ and $B$ are enantiomers,and $A$ and $C$ are enantiomers.

(B) The given compound is $CH_3-CH=CH-CH(OH)-CH(OH)-CH_3$.
It contains $3$ stereogenic centers:
$1.$ One carbon-carbon double bond $(CH=CH)$ which exhibits geometrical isomerism $(cis/trans)$.
$2.$ Two chiral carbon atoms $(CH(OH))$ which exhibit optical isomerism.
Since the molecule is unsymmetrical,the total number of stereoisomers is calculated using the formula $2^n$,where $n$ is the total number of stereogenic centers.
Here,$n = 3$.
Therefore,the total number of stereoisomers $= 2^3 = 8$.

(B) molecule is optically active if it lacks a plane of symmetry,center of inversion,or improper axis of rotation (i.e.,it is chiral).
$A$: The structure is $2,3$-butanediol in a Fischer projection. It has a plane of symmetry passing through the center,making it a meso compound (optically inactive).
$B$: The structure is $2$-chlorobutane. The central carbon is bonded to four different groups $(-H, -Cl, -CH_3, -CH_2CH_3)$,making it a chiral center. It lacks any plane of symmetry and is therefore optically active.
$C$: The structure is $trans$-$1,2$-cyclopentanediol. It has a $C_2$ axis of symmetry but no plane of symmetry or center of inversion,making it chiral and optically active. However,looking at the provided image,it represents $trans$-$1,2$-cyclopentanediol,which is chiral.
$D$: The structure is $cis$-$1,3$-dimethylcyclopentane. It has a plane of symmetry passing through the $C_2$ carbon,making it a meso compound (optically inactive).
Comparing the options,$B$ is a classic example of a chiral molecule with a single stereocenter.

(C) The compound $CH_3-CH(OH)-CH(Cl)-CH_2-COOH$ contains two chiral carbon atoms ($C2$ and $C3$).
Since the molecule is unsymmetrical,the number of optical isomers is calculated using the formula $2^n$,where $n$ is the number of chiral centers.
Here,$n = 2$,so the number of optical isomers $= 2^2 = 4$.

(A) The given structures are Fischer projections of tartaric acid derivatives.
Let us assign the configuration to the chiral centers.
In the first structure,the top chiral carbon has $-OH$ on the right and the bottom chiral carbon has $-OH$ on the right.
In the second structure,the top chiral carbon has $-OH$ on the left and the bottom chiral carbon has $-OH$ on the left.
These two structures are non-superimposable mirror images of each other.
Therefore,they are enantiomers.

(A) compound is optically active if it lacks a plane of symmetry and contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$A$. $CH_3-CH(OH)-COOH$ (Lactic acid) has a chiral carbon atom bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$. It is optically active.
$B$. The structure is $2,3$-butanediol. The meso form has a plane of symmetry and is optically inactive.
$C$. The structure is $1,2$-dichlorocyclobutane. The cis-isomer has a plane of symmetry and is optically inactive.
$D$. $CH_3-CH=CH-CH_3$ (But-$2$-ene) is a planar molecule and does not contain a chiral carbon atom; it is optically inactive.
Therefore,the correct answer is $A$.