Optical isomerism Questions in English

(C) The number of optical isomers for a compound containing $n$ asymmetric carbon atoms (chiral centers) is given by the formula $2^n$,provided that the molecule does not possess any internal plane of symmetry (i.e.,it is not a meso compound).

(D) An asymmetric carbon atom (chiral center) is a carbon atom bonded to four different groups.
The structure of $\alpha$-hydroxy propanoic acid is $CH_3-CH(OH)-COOH$.
The central carbon atom is bonded to $-H$,$-OH$,$-CH_3$,and $-COOH$ groups.
Since all four groups are different,the central carbon is asymmetric.
Thus,$\alpha$-hydroxy propanoic acid has $1$ asymmetric carbon atom.

(C) compound is optically active if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In butan$-2-$ol $(CH_3-CH_2-CH(OH)-CH_3)$,the second carbon atom is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$.
Therefore,butan$-2-$ol is optically active.

(C) Optical isomerism is exhibited by compounds that contain at least one chiral carbon atom (a carbon atom bonded to four different groups).
$A$. $3-$Iodohexane: $CH_3-CH_2-CH(I)-CH_2-CH_2-CH_3$. The $C3$ atom is bonded to $-H, -I, -CH_2CH_3, -CH_2CH_2CH_3$. It is chiral.
$B$. $2-$Iodopentane: $CH_3-CH(I)-CH_2-CH_2-CH_3$. The $C2$ atom is bonded to $-H, -I, -CH_3, -CH_2CH_2CH_3$. It is chiral.
$C$. $2-$Iodo$-2-$methylbutane: $CH_3-C(I)(CH_3)-CH_2-CH_3$. The $C2$ atom is bonded to two identical $-CH_3$ groups. It is achiral.
$D$. $2-$Iodo$-3-$methylbutane: $CH_3-CH(I)-CH(CH_3)_2$. The $C2$ atom is bonded to $-H, -I, -CH_3, -CH(CH_3)_2$. It is chiral.
Therefore,$2-$Iodo$-2-$methylbutane does not exhibit optical isomerism.

(B) The molecular formula $C_4H_9Br$ corresponds to butyl bromide isomers.
Among the structural isomers,$2$-bromobutane $(CH_3-CH_2-CH(Br)-CH_3)$ contains a chiral carbon atom (the $C_2$ carbon is bonded to four different groups: $-H$,$-CH_3$,$-CH_2CH_3$,and $-Br$).
$A$ molecule with one chiral center exists as a pair of enantiomers (two optically active isomers: $d$- and $l$-forms).
Therefore,there are $2$ optically active isomers possible for this compound.

(B) polarimeter is an instrument used to measure the angle of rotation caused by passing polarized light through an optically active substance.
This rotation determines whether a compound is dextrorotatory ($d$ or $+$) or levorotatory ($l$ or $-$).
Therefore,it is used to determine the $d$ and $l$ configuration (optical rotation) of compounds.

(D) compound is optically inactive if it lacks a chiral center or possesses a plane of symmetry (meso compound).
$1$. $3-$chlorobut$-1-$ene: The $C3$ atom is bonded to a hydrogen,a chlorine,a methyl group,and a vinyl group. It is chiral.
$2$. $2,3-$dichlorobutane: This can exist as a meso form (achiral) or chiral enantiomers. However,$2,2-$dichloropentane is definitively achiral.
$3$. $2-$hydroxypropanoic acid (lactic acid): The $C2$ atom is bonded to a hydrogen,a hydroxyl group,a methyl group,and a carboxyl group. It is chiral.
$4$. $2,2-$dichloropentane: The $C2$ atom is bonded to two identical chlorine atoms. Since it lacks a chiral center,it is optically inactive.

(D) Enantiomers,such as $(+)$ butan-$2$-ol and $(-)$ butan-$2$-ol,are non-superimposable mirror images of each other.
They possess identical physical properties,including boiling points,melting points,and refractive indices,in an achiral environment.
They differ only in their interaction with plane-polarized light and their reactivity with other chiral substances.
Therefore,$(+)$ butan-$2$-ol and $(-)$ butan-$2$-ol have the same boiling point.

(D) The given compound is $CH_{3}CHBrCHBrCOOH$. It contains two chiral carbon atoms $(n=2)$.
Since the molecule is unsymmetrical (the two ends are different: $-CH_{3}$ and $-COOH$),it cannot be divided into two equal halves.
For an unsymmetrical molecule,the number of optical isomers is given by $2^{n}$,where $n$ is the number of chiral centers.
Here,$n=2$,so the number of optical isomers $= 2^{2} = 4$.
These $4$ isomers consist of $2$ pairs of enantiomers,and there are no meso forms $(m=0)$.
Thus,the total number of optical isomers is $4$.

(C) The correct answer is $C$. Meso compounds are those compounds whose molecules are superimposable on their mirror images despite the presence of asymmetric carbon atoms.
This is due to internal compensation.
The two halves of the molecule rotate the plane of polarised light in opposite directions and hence cancel the effect of each other,making the molecule optically inactive.
Thus,the optical inactivity in meso compounds is due to the presence of a plane of symmetry or molecular symmetry,which leads to internal compensation.

(A) The molecule $1-$bromo$-2-$methylcyclobutane contains two chiral centers at $C1$ and $C2$.
Since the ring is not symmetric,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 $4$ stereoisomers consist of two pairs of enantiomers (cis-isomer pair and trans-isomer pair),all of which are optically active.
Therefore,the maximum number of possible optical isomers is $4$.

(B) The structure of butan-$2,3$-diol is $CH_3-CH(OH)-CH(OH)-CH_3$.
It has two chiral centers $(n=2)$.
Since the molecule is symmetrical,the number of optically active stereoisomers is given by $2^{n-1}$.
Here,$n=2$,so the number of optically active isomers $= 2^{2-1} = 2^{1} = 2$.
These two optically active isomers are the $(2R, 3R)$ and $(2S, 3S)$ forms. The $(2R, 3S)$ form is a meso compound and is optically inactive.

(C) The structure of $2,3,4$-trichloropentane is $CH_3-CHCl-CHCl-CHCl-CH_3$.
In this molecule,the carbon atoms at positions $2$ and $4$ are bonded to four different groups: a hydrogen atom $(-H)$,a chlorine atom $(-Cl)$,a methyl group $(-CH_3)$,and a $-CHCl-CHCl_2$ group (for $C-2$) or a $-CHCl-CH_3$ group (for $C-4$).
The carbon atom at position $3$ is not chiral because it is bonded to two identical $-CHCl-CH_3$ groups.
Therefore,there are $2$ chiral carbon atoms present in $2,3,4$-trichloropentane.

(B) An optically inactive compound is one that does not contain any chiral centers (asymmetric carbon atoms).
$2-$Bromopropanal: $CH_3-CH(Br)-CHO$ (The $C-2$ carbon is chiral).
$3-$Bromopropanal: $Br-CH_2-CH_2-CHO$ (No chiral carbon exists).
$3-$Bromo-$2-$iodopropanal: $Br-CH_2-CH(I)-CHO$ (The $C-2$ carbon is chiral).
$2-$Bromo-$3-$iodopropanal: $CH_2(I)-CH(Br)-CHO$ (The $C-2$ carbon is chiral).
Therefore,$3-$Bromopropanal is the optically inactive compound.

(C) molecule is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
In molecule $X$,the central carbon is bonded to $-H$,$-CH_3$,$-CH_2CH_2CH_3$,and $-CH_2Br$. Since all four groups are different,the carbon is chiral,making molecule $X$ chiral.
In molecule $Y$,the central carbon is bonded to $-H$,$-Br$,and two identical $-CH_2CH_3$ (ethyl) groups. Since two groups are identical,the carbon is not chiral,making molecule $Y$ achiral.
Therefore,$X$ is chiral and $Y$ is achiral.

(C) The structure of $2-$Bromo$-3-$chlorobutane is $CH_3-CH(Br)-CH(Cl)-CH_3$.
This molecule contains two chiral centers (at $C-2$ and $C-3$).
Since the molecule does not have a plane of symmetry (the groups attached to the chiral centers are different),the number of optical isomers is given by $2^n$,where $n$ is the number of chiral centers.
Here,$n = 2$.
Therefore,the number of optical isomers $= 2^2 = 4$.
The four isomers are $(2R, 3R)$,$(2S, 3S)$,$(2R, 3S)$,and $(2S, 3R)$.

(A) molecule is chiral if it has a carbon atom bonded to four different groups.
$(i)$ The central carbon is bonded to $H$,$C_2H_5$,$CH_3$,and $Y$. All four groups are different,so it is chiral.
(ii) The central carbon is bonded to $H$,$CH_3$,$Y$,and $Y$. Since two groups $(Y)$ are the same,it is achiral.
(iii) The central carbon is bonded to $H$,$C_2H_5$,$X$,and $Y$. All four groups are different,so it is chiral.
(iv) The central carbon is bonded to $H$,$Y$,and two identical alkyl chains. Since two groups are the same,it is achiral.
Therefore,molecules $(i)$ and $(iii)$ are chiral.

(B) carbon atom bonded to four different groups or atoms is called a chiral carbon.
$A$ chiral molecule exists in two stereoisomers that are non-superimposable mirror images of each other,called enantiomers.
$1$. $CH_3-CH(OH)-CH_3$ (Isopropyl alcohol): The central carbon is bonded to two identical methyl groups,so it is achiral.
$2$. $CH_3-CH(OH)-CH_2-CH_2-CH_3$ ($2-$Pentanol): The second carbon is bonded to four different groups ($-H$,$-OH$,$-CH_3$,and $-CH_2CH_2CH_3$),making it a chiral center. Thus,it is a chiral molecule.
$3$. $BrCH_2-CH_2-CH=CH_2$ ($1-$Bromo$-3-$butene): No carbon atom is bonded to four different groups,so it is achiral.
$4$. $(CH_3)_2CH-CH_2OH$ (Isobutyl alcohol): The carbon attached to the hydroxyl group is bonded to two identical hydrogen atoms,so it is achiral.

(A) molecule is chiral if it is non-superimposable on its mirror image. This typically occurs when a carbon atom is bonded to four different groups,known as a chiral center.
Analyzing the structures:
$(I)$ $CH_3-CHCl_2$: The central carbon is bonded to two identical $Cl$ atoms. Achiral.
$(II)$ $CH_3-CHCl-CH_3$: The central carbon is bonded to two identical $CH_3$ groups. Achiral.
$(III)$ $1-bromo-1-methylcyclohexane$: The carbon at position $1$ is bonded to two identical $CH_2$ groups of the ring. Achiral.
$(IV)$ The carbon bonded to $I$ and $CH_3$ is attached to four different groups ($I$,$CH_3$,and two different paths around the ring). Chiral.
$(V)$ $CH_3-CH(CH_3)-C_2H_5$: The central carbon is bonded to two identical $CH_3$ groups. Achiral.
$(VI)$ $CH_3-CH(Cl)-C_2H_5$: The central carbon is bonded to four different groups ($H$,$CH_3$,$Cl$,$C_2H_5$). Chiral.
Thus,$(IV)$ and $(VI)$ are chiral.

(A) Enantiomers are stereoisomers that are non-superimposable mirror images of each other.
They possess identical physical properties such as melting point,boiling point,density,and refractive index in an achiral environment.
However,they differ in their interaction with plane-polarized light.
One enantiomer rotates the plane of plane-polarized light to the right (dextrorotatory,denoted by $+$),while the other rotates it to the left (levorotatory,denoted by $-$) to the same extent.
Therefore,they differ in the sign of their specific rotation.

(B) compound exhibits enantiomerism if it contains at least one chiral center (an asymmetric carbon atom,which is bonded to four different groups).
In $CH_3-CH(Br)-CH_2-CH_3$ ($2$-bromobutane),the second carbon atom is bonded to four different groups: $-H$,$-CH_3$,$-Br$,and $-CH_2CH_3$.
Since it has a chiral center,it is optically active and exhibits enantiomerism.
The other options do not contain any chiral carbon atoms.

(B) Enantiomers are stereoisomers that are non-superimposable mirror images of each other.
They possess identical physical properties such as boiling point,density,and bond lengths in an achiral environment.
However,they differ in their interaction with plane-polarized light,which is measured as specific rotation.
Therefore,the two enantiomers of secondary butyl chloride differ in specific rotation.

(A) Enantiomers are non-superimposable mirror images of each other. For a molecule with multiple chiral centers,the enantiomer is formed by inverting the configuration at every chiral center.
For $2,3-$butanediol,the chiral centers are at positions $2$ and $3$.
The enantiomer of the $(2R, 3R)$ isomer is the $(2S, 3S)$ isomer.
Therefore,the pair $(2R, 3R)$ and $(2S, 3S)$ represents enantiomers.

(C) The formula for specific rotation is $[\alpha] = \frac{\alpha}{l \times c}$,where $\alpha$ is the observed rotation in degrees,$l$ is the path length in decimeters $(dm)$,and $c$ is the concentration in $g/mL$.
Given: $\alpha = -1.2^{\circ}$,$l = 5 \ cm = 0.5 \ dm$,and $c = 6.15 \ g / 100 \ mL = 0.0615 \ g/mL$.
Substituting these values: $[\alpha] = \frac{-1.2}{0.5 \times 0.0615} = \frac{-1.2}{0.03075} = -39^{\circ}$.

(B, C) compound is optically active if it possesses at least one chiral center and lacks a plane of symmetry or center of symmetry.
$(A)$ Glycine $(H_2N-CH_2-COOH)$ has no chiral center.
$(B)$ Lactic acid $(CH_3-CH(OH)-COOH)$ has one chiral center at the central carbon atom,making it optically active.
$(C)$ Glyceraldehyde $(HOCH_2-CH(OH)-CHO)$ has one chiral center at the central carbon atom,making it optically active.
$(D)$ Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ has two chiral centers,but the meso form has a plane of symmetry and is optically inactive. However,the chiral isomers of tartaric acid are optically active.
Given the options provided,$(B)$ and $(C)$ are clearly optically active.

(B) To find the enantiomer of $Y$,we must look for a structure that is a non-superimposable mirror image of $Y$.
$1$. The molecule $Y$ contains a chiral center. The priority order of groups attached to the chiral center is: $-Br (1) > -CH=CHCH_3 (2) > -CH_3 (3) > -H (4)$.
$2$. In $Y$,the configuration of the chiral center is '$R$'.
$3$. An enantiomer must have the opposite configuration ('$S$') at the chiral center while maintaining the same geometry at the double bond $(trans)$.
$4$. Analyzing the options:
- Option $A$ has the same configuration as $Y$ (identical).
- Option $B$ has the '$S$' configuration at the chiral center and maintains the $trans$ geometry,making it the enantiomer of $Y$.
- Option $C$ has the '$S$' configuration but the structure is different.
- Option $D$ has a different geometry at the double bond $(cis)$,so it is a diastereomer,not an enantiomer.

(B) The given pair of compounds are non-superimposable mirror images of each other.
By assigning $R/S$ configurations using the Cahn-Ingold-Prelog $(CIP)$ priority rules:
$1$. The priority order is $-OH > -CO_2H > -CH=CH-CH_3 > -H$.
$2$. In the first structure,the configuration is $R$.
$3$. In the second structure,the configuration is $S$.
Since they are mirror images and have opposite configurations at the chiral center,they are enantiomers.

(C) The enantiomeric excess $(ee)$ is defined as the difference between the percentages of the two enantiomers in a mixture.
$ee = |\% \text{ of enantiomer } 1 - \% \text{ of enantiomer } 2|$
Given the percentages are $85 \%$ and $15 \%$,the calculation is:
$ee = 85 \% - 15 \% = 70 \%$

(D) The structures represent butane-$2,3$-diol isomers.
$I$ is $(2R, 3R)$-butane-$2,3$-diol,which is chiral.
$II$ is $(2R, 3S)$-butane-$2,3$-diol,which is a meso compound (achiral) due to an internal plane of symmetry.
$III$ is $(2S, 3S)$-butane-$2,3$-diol,which is chiral.
Therefore,only $I$ and $III$ are chiral.

(C) An optically active molecule must lack a plane of symmetry or a center of inversion.
In option $(A)$,the molecule has a plane of symmetry,making it a meso compound.
In option $(B)$,the molecule has a plane of symmetry,making it a meso compound.
In option $(C)$,the molecule has different functional groups at the ends ($COOMe$ and $COOH$),which prevents the existence of a plane of symmetry,thus making it optically active.
In option $(D)$,the molecule has a plane of symmetry,making it a meso compound.
Therefore,only molecule $(C)$ is optically active.

(D) $(+)$-lactic acid and $(-)$-lactic acid are non-superimposable mirror images of each other.
They possess a chiral carbon atom and rotate the plane of plane-polarized light in opposite directions.
Therefore,they are enantiomers,which is a type of optical isomerism.

(B) molecule exhibits optical isomerism if it contains a chiral carbon atom,which is a carbon atom bonded to four different groups.
$A$. Glycolic acid $(HO-CH_2-COOH)$: The central carbon is bonded to two identical hydrogen atoms. It is achiral.
$B$. Lactic acid $(CH_3-CH(OH)-COOH)$: The central carbon is bonded to four different groups: $-H$,$-OH$,$-CH_3$,and $-COOH$. It is chiral and shows optical isomerism.
$C$. Isobutyric acid $((CH_3)_2CH-COOH)$: The central carbon is bonded to two identical methyl groups. It is achiral.
$D$. $2-$chloro$-2-$methylpropanoic acid $(CH_3-C(Cl)(CH_3)-COOH)$: The central carbon is bonded to two identical methyl groups. It is achiral.
Therefore,lactic acid is the correct answer.

(A) The structures shown are mirror images of each other and are non-superimposable.
They represent the $(R)$ and $(S)$ configurations of the chiral center.
Since they are non-superimposable mirror images,they are enantiomers.