Conformational isomerism Questions in English

(A) In $3-$Hydroxypropanal $(HO-CH_2-CH_2-CHO)$,the most stable conformation is the $Gauche$ form.
This is due to the formation of an intramolecular hydrogen bond between the hydroxyl group $(-OH)$ and the carbonyl oxygen $(C=O)$.
Although the $Anti$ conformation is generally more stable due to reduced steric hindrance,the stabilization provided by the intramolecular hydrogen bond in the $Gauche$ conformation makes it the most stable conformer for this specific molecule.

(A) The given structures are Newman projections of the same molecule,$n$-butane.
Structure $I$ represents the staggered conformation,while structure $II$ represents the eclipsed conformation.
Since these structures are interconvertible by rotation around the $C-C$ single bond,they are conformational isomers.

(D) The stability of $n-$butane conformations is determined by torsional strain and steric hindrance.
$1$. The $Anti$ conformation is the most stable because the bulky methyl groups are $180^{\circ}$ apart,minimizing steric repulsion.
$2$. The $Gauche$ conformation is less stable than $Anti$ due to steric repulsion between methyl groups at a $60^{\circ}$ dihedral angle.
$3$. The $Eclipsed$ conformation (where methyl and hydrogen are eclipsed) has higher energy due to torsional strain.
$4$. The $Fully \text{ } eclipsed$ conformation is the least stable because the two bulky methyl groups are directly aligned ($0^{\circ}$ dihedral angle),resulting in maximum steric repulsion and torsional strain.

(B) The stability of $n$-butane conformers follows the order: $Anti-staggered > Gauche > Eclipsed > Fully eclipsed$.
In the $anti-staggered$ conformation,the two bulky methyl groups are at a dihedral angle of $180^{\circ}$,which minimizes steric repulsion between them.
In the $gauche$ conformation,the methyl groups are at a dihedral angle of $60^{\circ}$,leading to some steric hindrance.
In the $eclipsed$ and $fully eclipsed$ conformations,the groups are closer,leading to significant torsional and steric strain.
Therefore,the $anti-staggered$ conformer is the most stable.

(A) In $n$-butane,the conformations are generated by rotation around the $C_2-C_3$ bond.
The stability order of these conformations is: $Anti$ (Staggered) $>$ $Gauche$ $>$ $Eclipsed$.
The $Anti$ conformation is the most stable because the bulky methyl groups are at a distance of $180^{\circ}$ from each other,minimizing steric repulsion and torsional strain.

(C) The given structures are Newman projections of $n$-butane.
These structures represent different spatial arrangements of atoms that can be interconverted by rotation around the $C-C$ single bond.
Such structures are known as conformational isomers or conformers.
Therefore,the correct option is $C$.

(B) Isomers that arise due to rotation around a $C-C$ single bond are known as conformational isomers or conformers.
These are different spatial arrangements of atoms that can be converted into one another by rotation about a single bond.

(B) The stability of butane conformations is determined by torsional strain and steric hindrance (van der Waals repulsion).
$I$ represents the gauche conformation (staggered,relatively stable).
$II$ represents the anti conformation (staggered,most stable).
$III$ represents the fully eclipsed conformation (highest energy due to $CH_3-CH_3$ repulsion,least stable).
$IV$ represents the eclipsed conformation (less stable than staggered,but more stable than fully eclipsed).
Therefore,the least stable conformation is $III$.

(B) In the staggered conformation of ethane $(C_2H_6)$,the hydrogen atoms on one carbon atom are positioned as far as possible from the hydrogen atoms on the adjacent carbon atom.
This arrangement results in a dihedral angle of $60^o$ between the $C-H$ bonds of the two carbon atoms.
Therefore,the correct option is $B$.

(B) The stability of cyclohexane conformations is determined by the minimization of torsional strain and steric strain.
In the $chair$ conformation,all $C-C-C-C$ dihedral angles are staggered,and there is minimal steric hindrance between hydrogen atoms.
This makes the $chair$ conformation the most stable form of cyclohexane,with the lowest potential energy.
In contrast,the $boat$,$twist-boat$,and $half-chair$ conformations possess higher energy due to eclipsing interactions and transannular strain.

(B) In ethylene glycol $(HO-CH_2-CH_2-OH)$,the gauche conformation is the most stable.
This is due to the formation of an intramolecular hydrogen bond between the two hydroxyl $(-OH)$ groups,which stabilizes the gauche form.
Although the anti conformation minimizes steric repulsion,the stabilizing effect of the intramolecular hydrogen bond in the gauche conformation makes it more stable.

(N/A) Alkanes contain carbon-carbon sigma $(\sigma)$ bonds. The electron distribution of the sigma molecular orbital is symmetrical around the internuclear axis of the $C-C$ bond,which is not disturbed due to rotation about its axis.
This permits free rotation about the $C-C$ single bond. This rotation results in different spatial arrangements of atoms in space which can change into one another.
Such spatial arrangements of atoms that can be converted into one another by rotation around a $C-C$ single bond are called conformations,conformers,or rotamers.

(N/A) The differences between eclipsed and staggered ethane are as follows:
| Feature | Eclipsed Ethane | Staggered Ethane |
| :--- | :--- | :--- |
| $(i)$ Sawhorse projection | $C-H$ bonds are aligned. | $C-H$ bonds are staggered. |
| $(ii)$ Newman projection | $H$ atoms are directly behind each other. | $H$ atoms are at maximum distance. |
| $(iii)$ Dihedral angle | $0^{\circ}$ | $60^{\circ}$ |
| $(iv)$ Stability | Less stable due to high torsional strain. | More stable due to minimum repulsion. |
| $(v)$ Energy | Higher energy. | Lower energy (by $12.5 \ kJ \ mol^{-1}$). |

(N/A) In alkanes,the $C-C$ bond is represented by the sawhorse projection as follows:
$(i)$ In this projection,the molecule is viewed along the molecular axis. It is then projected on paper by drawing the central $C-C$ bond as a somewhat longer straight line.
$(ii)$ The upper end of the line is slightly tilted towards the right or left-hand side.
$(iii)$ The front carbon is shown at the lower end of the line,whereas the rear carbon is shown at the upper end.
$(iv)$ Each carbon has three lines attached to it,corresponding to three hydrogen atoms.
$(v)$ The lines are inclined at an angle of $120^{\circ}$ to each other.
$(vi)$ Sawhorse projections of eclipsed and staggered conformations of ethane are shown below:
$923$-s104

(N/A) Newman projection :
$(i)$ In this projection,the molecule is viewed along the $C-C$ bond head-on.
$(ii)$ The carbon atom nearer to the eye is represented by a point. Three hydrogen atoms attached to the front carbon atom are shown by three lines drawn at an angle of $120^{\circ}$ to each other.
$(iii)$ The rear carbon atom (the carbon atom away from the eye) is represented by a circle and the three hydrogen atoms are shown attached to it by lines drawn at an angle of $120^{\circ}$ to each other. At the end of all lines,$H$ is written.
$(b)$ Example: Eclipsed ethane and staggered ethane are shown below:

(3 ECLIPSED AND 3 STAGGERED) During a full $360^{\circ}$ rotation about the $C-C$ bond in ethane,the molecule passes through various conformational states.
$1$. Eclipsed conformations: These occur at $0^{\circ}$,$120^{\circ}$,and $240^{\circ}$. Thus,there are $3$ eclipsed conformations.
$2$. Staggered conformations: These occur at $60^{\circ}$,$180^{\circ}$,and $300^{\circ}$. Thus,there are $3$ staggered conformations.
Therefore,a total of $3$ eclipsed and $3$ staggered conformations are obtained.

(N/A) The internal rotation of ethane about the $C-C$ bond leads to different conformational isomers.
$1$. The staggered conformation is the most stable form with minimum energy due to minimum torsional strain.
$2$. The eclipsed conformation is the least stable form with maximum energy due to maximum torsional strain.
$3$. $A$ rotation of $60^{\circ}$ from staggered to eclipsed increases the energy by $12.5 \ kJ \ mol^{-1}$.
$4$. $A$ further rotation of $60^{\circ}$ (total $120^{\circ}$) brings the molecule back to a staggered conformation,which is equivalent to the initial state in terms of energy.
Thus,after a $120^{\circ}$ rotation,the molecule returns to a staggered conformation,which is a minimum energy state.

(A) In ethane $(CH_3-CH_3)$,the bond angles (e.g.,$H-C-H$ and $C-C-H$) remain constant because the hybridization of carbon atoms $(sp^3)$ does not change during rotation.
However,the dihedral angle (the angle between the planes of $H-C-C$ bonds on adjacent carbons) changes continuously from $0^{\circ}$ to $360^{\circ}$ as the molecule undergoes rotation around the $C-C$ sigma bond.

(N/A) The magnitude of torsional strain depends upon the angle of rotation about a $C-C$ bond. This angle is known as the dihedral angle or torsional angle.

(N/A) The repulsive interaction between the electron clouds of adjacent bonds in a molecule,which affects the stability of a conformation,is called torsional strain.
Due to these increased repulsive forces,the molecule possesses higher energy and consequently exhibits lower stability.

(N/A) There are an infinite number of conformations of ethane. However,there are two extreme cases. One such conformation in which hydrogen atoms attached to two carbons are as close together as possible is called eclipsed conformation,and the other in which hydrogens are as far apart as possible is known as the staggered conformation. Any other intermediate conformation is called a skew conformation.

(C) In general,the anti-staggered conformer is the most stable for most molecules due to minimal steric hindrance. However,for ethane-$1,2$-diol,the gauche conformer is more stable than the anti-staggered form. This is due to the formation of an intramolecular hydrogen bond between the two hydroxyl $(-OH)$ groups,which stabilizes the gauche conformation. The structure showing this intramolecular hydrogen bonding is given in the reference image.

(A) The stability of conformational isomers is inversely proportional to their potential energy. More stable conformers have lower potential energy.
The stability order for the conformers of $n$-butane is:
$I$ (Anti) > $III$ (Gauche) > $IV$ (Eclipsed) > $II$ (Fully Eclipsed).
Therefore,the order of increasing potential energy is the reverse of the stability order:
$I < III < IV < II$.

(A) The least stable conformer of ethane is the eclipsed form.
In the eclipsed conformation,the hydrogen atoms on the front carbon are directly aligned with the hydrogen atoms on the back carbon,resulting in maximum torsional strain.
This corresponds to a dihedral angle of $0^{\circ}$.

(B) Staggered and eclipsed conformers of ethane are generated by rotation around the $C-C$ single bond.
These different spatial arrangements are known as conformational isomers or rotamers.
Since they are interconvertible by rotation about a single bond,they are classified as rotamers.

(D) In the Newman projection of $1,1,1-$trichloroethane $(CCl_3-CH_3)$,the staggered conformation is formed when the substituents on the front carbon are placed at an angle of $60^{\circ}$ relative to the substituents on the back carbon.
As shown in the provided figure,the dihedral angle $(\phi)$ between the $C-Cl$ bond and the $C-H$ bond in the staggered conformation is $60^{\circ}$.

(C) In an ethane molecule $(CH_3-CH_3)$,the carbon-carbon bond is a single sigma $(\sigma)$ bond.
Because the sigma $(\sigma)$ bond is formed by the head-on overlap of orbitals,it possesses cylindrical symmetry along the internuclear axis.
This cylindrical symmetry allows for free rotation around the carbon-carbon single bond,leading to different conformational isomers.

(C) Dihedral angle $:$ It is the angle between $2$ specified groups ($-CH_3$ here) in a Newman projection.
Staggered conformation refers to the arrangement where the groups on the front and back carbons are as far apart as possible.
In option $(A)$,the dihedral angle between the two $-CH_3$ groups is $60^{\circ}$ (gauche).
In option $(B)$,the structure is eclipsed.
In option $(C)$,the two $-CH_3$ groups are in the anti-position,meaning the dihedral angle is $180^{\circ}$. This is the staggered conformation with the maximum dihedral angle.
In option $(D)$,the structure is eclipsed.

(D) $X$ and $Y$ are conformers of each other.
Conformers are different spatial arrangements of atoms in a molecule that can be interconverted by rotation around a $C-C$ single bond.
In the given Newman projections,$X$ represents an eclipsed conformation,while $Y$ represents a staggered conformation.
Since they represent the same molecule in different rotational states,they are classified as conformers.

(C) The most stable conformation of a molecule is the one with the minimum steric hindrance and electrostatic repulsion. In the case of $2,3-$dibromobutane,the anti-staggered conformation is the most stable because the two bulky bromine atoms are placed at a dihedral angle of $180^{\circ}$ to each other,minimizing steric repulsion. Among the given options,the structure where the bromine atoms are furthest apart is the anti-staggered form.

(A) The most stable conformation for $n$-butane is the anti-staggered form,where the two bulky methyl groups are at a dihedral angle of $180^{\circ}$ to each other.
In this conformation,the steric repulsion between the methyl groups is minimized,making it the most stable conformer.
This structure is represented in option $(A)$.

(D) The correct answer is $(D)$.
In the given Newman projection of $n$-butane,the two methyl $(-CH_3)$ groups are at a dihedral angle of $60^{\circ}$ to each other.
This specific staggered conformation,where the bulky groups are adjacent to each other,is known as the gauche conformer.

(A) The stability of conformations of butane is determined by the magnitude of van der Waals strain and torsional strain.
$1$. The anti-conformation (where the two $Me$ groups are $180^{\circ}$ apart) has the lowest energy because it minimizes steric repulsion (van der Waals strain) between the bulky methyl groups and minimizes torsional strain.
$2$. The gauche conformation has higher energy than the anti-conformation due to steric repulsion between the two $Me$ groups.
$3$. Eclipsed conformations have the highest energy due to significant torsional strain and steric repulsion.
$4$. Therefore,the anti-conformation,as shown in option $A$,is the most stable.

(C) The eclipsed conformation of ethane has maximum torsional strain due to the proximity of $C-H$ bonds on adjacent carbon atoms,making it the least stable conformation. The staggered conformation is the most stable.

(B) The structure of $2,2$-dimethylbutane is $CH_3-CH_2-C(CH_3)_2-CH_3$.
For the given Newman projection,the front carbon has two $CH_3$ groups and one $H$ atom,while the back carbon has two $H$ atoms and one group $Y$. The group $X$ is attached to the front carbon.
Considering the $C_2-C_3$ bond axis:
Front carbon $(C_2)$ has two $CH_3$ groups and one $H$ atom. Back carbon $(C_3)$ has two $H$ atoms and one $C_2H_5$ group.
If $X = CH_3$ and $Y = C_2H_5$,this matches the structure.
Considering the $C_1-C_2$ bond axis:
Front carbon $(C_1)$ has three $H$ atoms. Back carbon $(C_2)$ has two $CH_3$ groups and one $C_2H_5$ group.
If $X = H$ and $Y = C_2H_5$,this matches the structure.
Comparing with the options,$X=H$ and $Y=C_2H_5$ (Option $B$) is a valid possibility,and $X=C_2H_5$ and $Y=H$ (Option $C$) is also a valid possibility depending on the rotation and projection chosen. Thus,$B$ and $C$ are correct.

(B) The most stable conformation of meso-butane-$2,3$-diol is determined by the presence of intramolecular hydrogen bonding between the two hydroxyl $(-OH)$ groups.
In the Newman projection of the meso form,when the two $-OH$ groups are in a gauche position relative to each other,they can form an intramolecular hydrogen bond,which stabilizes the conformation.
Comparing the given options,the structure that shows the $-OH$ groups in a gauche orientation allowing for this stabilization is represented in option $B$.

(C) The given compound is $2,3$-dibromo-$2,3$-dichlorobutane.
To find the stable conformers,we draw the Newman projections by rotating one carbon atom relative to the other.
There are three staggered conformers for this molecule.
In a staggered conformation,the dipole moment $\mu$ is non-zero if the molecule lacks a center of inversion or a plane of symmetry that cancels out the individual bond dipoles.
For this specific molecule,all three staggered conformers possess a non-zero dipole moment $(\mu \neq 0)$ because the substituents on the two carbon atoms are not arranged in a way that allows for complete cancellation of the dipole moments in any of the staggered forms.
Therefore,the total number of stable conformers with a non-zero dipole moment is $3$.

(B) The given compound is $1$-chloro-$2$-methylpropane. Looking down the $C_1-C_2$ bond:
$C_1$ has two hydrogen atoms and one chlorine atom attached.
$C_2$ has one hydrogen atom and two methyl $(Me)$ groups attached.
In a Newman projection,the front carbon $(C_1)$ is represented by the center point,and the back carbon $(C_2)$ is represented by the circle.
$C_1$ has $-Cl$ and two $-H$ atoms.
$C_2$ has two $-Me$ groups and one $-H$ atom.
Option $A$ shows $C_1$ with $-H, -Me, -Me$ and $C_2$ with $-Cl, -H, -H$,which is incorrect.
Option $B$ shows $C_1$ with $-Cl, -H, -H$ and $C_2$ with $-Me, -Me, -H$,which matches the structure of $1$-chloro-$2$-methylpropane.
Therefore,the correct Newman projection is $B$.

(B) In ethylene glycol $(HO-CH_2-CH_2-OH)$,the gauche conformation is more stable than the anti conformation.
This is due to the formation of intramolecular hydrogen bonding between the two hydroxyl $(-OH)$ groups in the gauche form.
In the anti conformation,the distance between the two $-OH$ groups is too large for hydrogen bonding to occur.
Therefore,the stability order is $anti < Gauche$.

(A) Cyclohexane exists in various conformations,primarily the chair and boat forms.
In the chair conformation,all $C-H$ bonds on adjacent carbon atoms are in a staggered (skew) position,which minimizes torsional strain.
In the boat conformation,some $C-H$ bonds are eclipsed,leading to torsional strain. Additionally,there is steric repulsion between the hydrogen atoms at the $1$ and $4$ positions (flagpole interactions).
Because the boat conformation has higher total strain than the chair conformation,it is less stable.
Therefore,the chair conformation is the most stable and least energetic conformation of cyclohexane.

(D) The stability of cyclohexane conformations is determined by their potential energy. The order of stability is: $Chair > Twisted \ boat > Boat > Half \ chair$.
Conversely,the order of energy is: $Half \ chair > Boat > Twisted \ boat > Chair$.
The $Half \ chair$ conformation is the most energetic (least stable) due to significant torsional strain and steric interactions.

(A) The stability order of cyclohexane conformations is $Chair > Twist-boat > Boat > Half-chair$.
The $Half-chair$ conformation is the least stable due to high torsional strain and angle strain.

(B) In a sawhorse projection,the staggered conformation of ethane is characterized by the hydrogen atoms on the front carbon being positioned between the hydrogen atoms on the back carbon,minimizing steric repulsion.
Looking at the provided images:
Image $A$ shows an eclipsed conformation.
Image $B$ shows a staggered conformation where the $C-H$ bonds on the front carbon are staggered with respect to the $C-H$ bonds on the back carbon.
Image $C$ shows a staggered conformation,but Image $B$ is the standard representation.
Image $D$ is a Newman projection,not a sawhorse projection.
Therefore,the correct representation for the staggered conformation of ethane in sawhorse projection is $B$.

(D) Conformational isomers are stereoisomers that can be interconverted simply by rotation about a single bond. These are also known as $rotamers$ or $conformers$,and the phenomenon is known as $rotamerism$.

(D) The eclipsed conformation of ethane is the least stable due to torsional strain caused by the repulsion between the hydrogen atoms on adjacent carbon atoms.
Conversely,the staggered conformation is the most stable as the hydrogen atoms are as far apart as possible,minimizing repulsion.
The energy barrier for rotation about the $C-C$ single bond in ethane is approximately $12.5 \ kJ / mol$ $(3 \ kcal / mol)$.
Therefore,the energy difference between the staggered and eclipsed conformations is $12.5 \ kJ / mol$.

(A) The given structures represent the Newman projections of $n$-butane.
Structure $II$ is the anti-conformer,where the two bulky $-CH_3$ groups are at a dihedral angle of $180^{\circ}$,minimizing steric repulsion.
Structures $I$ and $III$ are gauche-conformers,where the two $-CH_3$ groups are at a dihedral angle of $60^{\circ}$,leading to higher steric repulsion compared to the anti-form.
Since both $I$ and $III$ are equivalent gauche-conformers,they have equal stability.
Thus,the order of stability is $II > I = III$.

(D) The conformations of $n$-butane (eclipsed,gauche,and anti) are generated by the rotation around the $C2-C3$ sigma bond. This rotation changes the dihedral angle between the two methyl groups attached to the $C2$ and $C3$ carbons,leading to different energy states.
Hence,the correct option is $(d)$.

(D) Statement $I$ is true: In $n$-butane,the fully eclipsed conformer (dihedral angle $0^{\circ}$) has maximum steric and torsional strain,making it the least stable. The anti-staggered conformer (dihedral angle $180^{\circ}$) has minimum steric and torsional strain,making it the most stable.
Statement $II$ is true: Torsional strain is caused by the repulsion between bonding electrons of adjacent carbons. As the dihedral angle increases from $0^{\circ}$ (fully eclipsed) to $60^{\circ}$ (gauche),the eclipsed interactions decrease,thereby reducing the torsional strain.
Therefore,both statements are correct.