Conformational isomerism Questions in English

(B) For $ethane$ $(CH_3-CH_3)$,the $staggered$ conformation is more stable than the $eclipsed$ conformation due to minimum torsional strain.
For $cyclohexane$ $(C_6H_{12})$,the $chair$ conformation is the most stable conformation because it is free from angle strain and torsional strain.
Therefore,the pair of more stable conformations is $staggered$ and $chair$.

(D) The chair conformation of cyclohexane is the most stable conformer due to the absence of angle strain and torsional strain.
The boat conformation is less stable due to flagpole interactions and torsional strain.
The energy difference between the chair and boat conformations of cyclohexane is approximately $44 \ kJ/mol$.
Therefore,the correct statement is that chair and boat conformations differ in energy by $44 \ kJ/mol$.

(B) Conformers - Conformation arises because of free rotation around $C-C$ bond axis.

(A) Conformations arise due to rotation around a single bond.
In $NH_2-NH_2$ (hydrazine),there is a $N-N$ single bond,allowing for rotation and the existence of different conformers (e.g.,gauche and anti).
$CH_4$ has no $C-C$ bond to rotate,and $B_2H_6$ has a bridged structure that does not exhibit simple conformational isomerism like alkanes or hydrazine.
Therefore,$NH_2-NH_2$ is the correct answer.

(A) Isomerism that arises due to the rotation around a $C-C$ single bond is called conformational isomerism or conformers.

(C) The given structure is a Newman projection of butane.
In the initial conformation,the two $CH_3$ groups are anti to each other ($180^{\circ}$ apart).
Rotating the front carbon $(C_2)$ anticlockwise by $120^{\circ}$ moves the $CH_3$ group from the bottom position to the position $120^{\circ}$ away from the back $CH_3$ group.
In this new position,the $CH_3$ group on $C_2$ is at a $60^{\circ}$ dihedral angle with respect to the $CH_3$ group on $C_3$.
This specific arrangement,where the bulky groups are at a $60^{\circ}$ dihedral angle,is known as the Gauche conformation.

(B) In the eclipsed conformation of ethane,the $C-H$ bonds on the front carbon atom are directly aligned with the $C-H$ bonds on the back carbon atom when viewed along the $C-C$ bond axis.
In the provided images,option $B$ (image $282-b23$) represents the eclipsed conformation because the hydrogen atoms on the two carbon atoms are aligned in a way that they eclipse each other.

(C) In ethane $(C_2H_6)$,conformational isomers are represented using Newman projections and sawhorse projections.
Staggered conformation is the most stable form where the dihedral angle between the $C-H$ bonds of adjacent carbon atoms is $60^{\circ}$.
In the provided image:
- Structure $1$ is an eclipsed Newman projection.
- Structure $2$ is a staggered sawhorse projection.
- Structure $3$ is an eclipsed sawhorse projection.
- Structure $4$ is a staggered Newman projection.
Therefore,$2$ and $4$ represent the staggered forms of ethane.

(B) The compound ${C_6H_6Cl_6}$ is known as benzene hexachloride $(BHC)$ or gammaxene.
Among its various stereoisomers,the $\gamma$-isomer (gamma-isomer) has the configuration $aaaeee$ (where $a$ stands for axial and $e$ stands for equatorial).
This specific isomer is the most powerful insecticide among all the conformations of $BHC$.

(B) In $2$-fluoroethanol,the gauche conformer is more stable than the anti conformer due to the formation of an intramolecular hydrogen bond between the $F$ atom and the $OH$ group.
Eclipsed conformers are the least stable due to torsional strain.
Therefore,the increasing order of stability is $Eclipsed < Anti < Gauche$.

(A) Isomers that are formed due to rotation around a $C-C$ single bond are known as conformational isomers or conformers.
These are a type of stereoisomers where the spatial arrangement of atoms changes without breaking any bonds.

(A) The $Twist-boat$ conformation of cyclohexane lacks a plane of symmetry and a center of inversion,making it chiral.
In contrast,the $Chair$ and $Boat$ conformations possess planes of symmetry,rendering them achiral.

(B) The given Newman projection represents $n$-butane in a gauche conformation where the two methyl groups are at a dihedral angle of $60^{\circ}$.
Rotating the front carbon $(C_2)$ by $120^{\circ}$ in the anticlockwise direction will move the front methyl group to a position where it is directly behind the back methyl group (eclipsed).
When the two methyl groups are eclipsed,the conformation is known as the fully eclipsed conformation.

(A) Conformational isomerism arises due to rotation around a single bond. For a molecule to exhibit conformational isomerism,it must have at least three consecutive single bonds (sigma bonds) to allow for rotation around the central bond.
$(1)$ $CH_4$: No central bond for rotation.
$(2)$ $CH_3OH$: Rotation around $C-O$ bond is possible,but it is generally not considered to exhibit distinct conformational isomers due to the symmetry of the methyl group.
$(3)$ $H_2O$: No central bond for rotation.
$(4)$ $H_2O_2$ $(H-O-O-H)$: Has three consecutive bonds ($H-O$,$O-O$,$O-H$),allowing rotation around the $O-O$ bond.
$(5)$ $C_2H_6$ $(CH_3-CH_3)$: Has three consecutive bonds ($H-C$,$C-C$,$C-H$),allowing rotation around the $C-C$ bond.
$(6)$ $C_3H_8$ $(CH_3-CH_2-CH_3)$: Has multiple consecutive bonds,allowing rotation around $C-C$ bonds.
Thus,molecules $(2), (4), (5),$ and $(6)$ exhibit conformational isomerism.

(D) Anti or completely staggered.
The anti-staggered conformation is the most stable because it minimizes the steric repulsion (van der Waals strain) between the two bulky methyl groups,which are placed at a dihedral angle of $180^{\circ}$.

(C) Conformers are different spatial arrangements of atoms in a molecule that can be interconverted by rotation about a single bond.
During this rotation,the bond lengths and bond angles of the molecule remain unchanged because the covalent bonds are not broken or distorted.

(B) The magnitude of torsional strain depends upon the dihedral angle of rotation about the $C-C$ bond.
In the eclipsed conformation,the hydrogen atoms on adjacent carbons are as close as possible,resulting in maximum torsional strain and minimum stability.
In the staggered conformation,the hydrogen atoms are as far apart as possible,resulting in minimum torsional strain and maximum stability.
Therefore,the staggered conformation of ethane is more stable than the eclipsed conformation.

(B) The stability of a conformation depends on the minimization of steric repulsion between the groups attached to adjacent carbon atoms.
In the $anti$ conformation of $n-$butane,the two bulky $-CH_{3}$ groups are at a dihedral angle of $180^{\circ}$,which is the maximum possible distance between them. This minimizes the steric repulsion and torsional strain,making it the most stable conformation.
Therefore,the $anti$ conformation is the most stable.

(B) By rotating the front carbon atom of the first structure by $180^{\circ}$,we obtain the second structure.
Thus,both structures are conformers of each other.
Specifically,one is in the staggered conformation and the other is in the eclipsed conformation.

(A) In $2-$fluoroethanol,the Gauche conformation is unexpectedly more stable than the Anti conformation due to the formation of an intramolecular hydrogen bond between the fluorine atom $(F^{\delta -})$ and the hydroxyl hydrogen atom $(H^{\delta +})$.
The Eclipse conformation is the least stable due to high torsional strain and steric repulsion.
Therefore,the increasing order of stability is: $Eclipse < Anti < Gauche$.

(B) molecule is chiral if it lacks a plane of symmetry or a center of inversion.
The $chair$ and $boat$ conformations of cyclohexane possess planes of symmetry,making them achiral.
The $twist-boat$ conformation of cyclohexane lacks a plane of symmetry,making it chiral.

(B) The $anti$ and $gauche$ conformers of $1, 2-$dichloroethane are stereoisomers that are neither mirror images of each other nor superimposable.
Since they are stereoisomers but not enantiomers,they are classified as diastereomers.

(A) The given compound is $3$-methylpentane $(CH_3-CH_2-CH(CH_3)-CH_2-CH_3)$.
Looking along the $C_2-C_3$ bond,the front carbon $(C_2)$ is attached to two $H$ atoms and one $CH_3$ $(Me)$ group.
The back carbon $(C_3)$ is attached to one $H$ atom,one $CH_3$ $(Me)$ group,and one $Et$ (ethyl) group.
In the provided Newman projection,the front carbon has $H$,$H$,and $Me$ attached.
The back carbon has $H$,$X$,and $Y$ attached.
Comparing the substituents on the back carbon $(C_3)$,we have $H$,$Me$,and $Et$.
Since $H$ is already shown on the back carbon,$X$ and $Y$ must be $Me$ and $Et$.
Based on the standard representation of this specific conformer,$X = Me$ and $Y = Et$.

(D) The first structure is a sawhorse projection and the second is a Newman projection of the same molecule.
To compare them,convert the sawhorse projection into a Newman projection.
Looking down the $C-C$ bond from the front carbon (which has $-CHO$,$-H$,$-OH$ groups) to the back carbon (which has $-CH_3$,$-H$,$-OH$ groups),we see that the groups are identical in their spatial arrangement.
Since the two structures represent the same molecule viewed from different perspectives,they are identical.

(C) In ethylene glycol $(HO-CH_2-CH_2-OH)$,the most stable conformation is the $Gauche$ (skew) form.
This is due to the formation of an intramolecular hydrogen bond between the hydroxyl groups,which stabilizes the $Gauche$ conformation.
Although the $Staggered$ form minimizes steric repulsion,the energy gain from the intramolecular hydrogen bond in the $Gauche$ form makes it more stable.

(A) The initial structure shown is the fully eclipsed conformation of $n$-butane,where the two $CH_3$ groups are directly behind each other.
Rotating the front carbon $(C_2)$ anticlockwise by $120^\circ$ around the $C_2-C_3$ bond moves the front $CH_3$ group to a position $60^\circ$ away from the back $CH_3$ group.
This resulting conformation,where the bulky groups are at a dihedral angle of $60^\circ$,is known as the gauche conformation.

(D) The stability of cyclohexane derivatives in chair conformation depends on the equatorial $(e)$ vs axial $(a)$ orientation of substituents.
Substituents in the equatorial position are more stable due to reduced $1,3-$diaxial interactions.
Analyzing the given structures:
$A$: $1,2$-diaxial $(aa)$
$B$: $1,4$-diaxial $(aa)$
$C$: $1,3$-axial-equatorial $(ae)$
$D$: $1,3$-diequatorial $(ee)$
Structure $D$ has both methyl groups in the equatorial position,which minimizes steric repulsion and results in the lowest energy (highest stability).

(B) The potential energy of a conformer is determined by the steric strain and torsional strain present in the molecule.
In the given Newman projections,we analyze the interactions between the substituents ($Me$ and $Et$ groups) on the two carbon atoms.
$III$: The $Me$ and $Et$ groups are gauche to each other,creating significant steric repulsion.
$I$ and $II$: In both cases,the $Me$ and $Et$ groups are anti to each other,which is more stable than the gauche interaction. Since the structures are equivalent,their potential energies are equal $(I = II)$.
$IV$: The $Me$ and $Et$ groups are anti to each other,but the overall arrangement is the most stable among the given options due to minimal steric hindrance.
Thus,the decreasing order of potential energy is $III > I = II > IV$.

(B) Butane$-1,4-$dioic acid (succinic acid) can form intramolecular hydrogen bonding in its gauche conformation.
In the gauche conformation,the two carboxylic acid groups $(-COOH)$ are positioned such that the hydrogen atom of one group can form a hydrogen bond with the oxygen atom of the other group.
This intramolecular hydrogen bonding stabilizes the gauche conformation,making it the most stable conformation for this specific molecule.

(B) The lowest energy conformation is the one in which the two methyl groups are as far apart as possible,i.e.,$180^{\circ}$ away from each other.
This conformation is maximally staggered and is called the anti conformation.
The order of stability of these conformations is: $\text{Anti} > \text{Gauche} > \text{Partially eclipsed} > \text{Fully eclipsed}$.

(C) To determine the most stable conformer,we analyze the steric hindrance in the Newman projection.
Stability is generally governed by minimizing steric repulsion between bulky groups.
In the given molecule,the bulky groups are the ethyl group $(-C_2H_5)$ and the methyl groups $(-CH_3)$.
The most stable conformation is the one where the largest groups are as far apart as possible (anti-periplanar).
By converting the given structure to a Newman projection,we find that placing the largest group (ethyl) anti to the other large group (methyl) or minimizing the gauche interactions between bulky groups leads to the most stable conformer.
Based on the standard analysis of conformational stability for this molecule,the conformer where the largest groups are anti to each other is the most stable.

(C) In the molecule $HO-CH_2-CH_2-F$,the hydroxyl group $(-OH)$ and the fluorine atom $(-F)$ are electronegative substituents.
In the gauche conformation,the $-OH$ group and the $-F$ atom are at a dihedral angle of $60^{\circ}$.
This allows for the formation of an intramolecular hydrogen bond between the hydrogen of the hydroxyl group and the fluorine atom $(F \cdots H-O)$.
This intramolecular hydrogen bonding stabilizes the gauche conformer,making it more stable than the anti-staggered conformer,which is typically the most stable for other substituted ethanes.

(B) During a ring flip of a cyclohexane chair,all axial substituents become equatorial and all equatorial substituents become axial.
Specifically,the orientation of the substituents relative to the ring (up or down) remains unchanged.
If a substituent is in an axial position pointing down,after the ring flip,it will be in an equatorial position pointing down.
Therefore,the correct structure is the one where the substituent has moved from axial to equatorial while maintaining its relative orientation.

(B) Conformational isomers are typically interconvertible by rotation around a single bond.
However,if we assume that there is no rotation around the $C-C$ single bond,the gauche and anti forms of butane become fixed in their respective spatial arrangements.
Since they are stereoisomers that are not mirror images of each other,they are classified as diastereoisomers.

(C) The stability of butane conformations follows the order: Anti > Gauche > Eclipsed > Fully eclipsed.
The anti conformation is the most stable because the bulky methyl groups are at a dihedral angle of $180^{\circ}$,minimizing steric repulsion.
Therefore,the anti conformation has the lowest energy.

(B) The equilibrium involves the interconversion of two chair conformations of a substituted cyclohexane.
In the reactant,the larger isopropyl group $(-CH(CH_3)_2)$ is in the equatorial position,while the smaller methyl group $(-CH_3)$ is in the axial position.
In the product,the larger isopropyl group is in the axial position,while the smaller methyl group is in the equatorial position.
Axial substituents experience $1,3$-diaxial interactions,which are destabilizing.
The $1,3$-diaxial interaction is significantly stronger for the larger isopropyl group than for the smaller methyl group.
Therefore,the conformation with the isopropyl group in the equatorial position (the reactant) is more stable than the conformation with the isopropyl group in the axial position (the product).
Since the equilibrium favors the more stable reactant,the concentration of the reactant is greater than the concentration of the product,leading to an equilibrium constant $K < 1$.

(B) In the molecule $HO-CH_2-CH_2-CHO$,the $C_2-C_3$ bond rotation allows for different conformations.
The gauche conformation is the most stable because it allows for the formation of an intramolecular hydrogen bond between the hydroxyl $(-OH)$ group and the carbonyl oxygen $(C=O)$.
This stabilization energy outweighs the steric repulsion typically associated with the gauche form,making it more stable than the anti-staggered conformation.

(A) The lowest energy conformer of $n$-butane is the anti-conformer. In this conformation,the two bulky methyl groups are at a dihedral angle of $180^{\circ}$ to each other,which minimizes steric repulsion (van der Waals strain) between them. Other conformations like gauche,eclipsed,and fully eclipsed have higher potential energy due to increased steric or torsional strain.

(C) The given structure is $cis-1,2-dimethylcyclohexane$.
In the chair conformation,the two methyl groups are at adjacent carbons (positions $1$ and $2$).
When viewed along the $C1-C2$ bond using a Newman projection,the dihedral angle between the two methyl groups is $60^{\circ}$.
This specific spatial arrangement,where substituents on adjacent carbons have a dihedral angle of $60^{\circ}$,is referred to as a $Gauche$ interaction.

(A) In $cis-1$-ethyl-$3$-methylcyclohexane $(A)$,the substituents at positions $1$ and $3$ are on the same side of the ring. In the chair conformation,this allows both the ethyl group and the methyl group to occupy equatorial positions simultaneously,minimizing steric repulsion.
In $trans-1$-ethyl-$3$-methylcyclohexane $(B)$,the substituents at positions $1$ and $3$ are on opposite sides of the ring. In the chair conformation,one substituent must be equatorial while the other must be axial.
Since having both substituents in equatorial positions is more stable than having one axial and one equatorial,$compound \, A$ is more stable than $compound \, B$.

(C) In the staggered conformation of ethane,the hydrogen atoms on adjacent carbon atoms are as far apart as possible.
This minimizes the torsional strain (repulsion between electron clouds of $C-H$ bonds).
As a result,the staggered conformation has the lowest potential energy and is the most stable.

(A) The stability of dimethylcyclohexane isomers depends on the number of equatorial methyl groups and steric interactions.
$1$. Structure $(I)$ is $trans-1,3-$dimethylcyclohexane in the diequatorial conformation,which is the most stable.
$2$. Structures $(II)$ and $(III)$ are $cis-1,3-$dimethylcyclohexane (one equatorial,one axial) and $trans-1,4-$dimethylcyclohexane (one equatorial,one axial) respectively. These have similar stability due to one axial methyl group.
$3$. Structure $(IV)$ is $cis-1,4-$dimethylcyclohexane in the diaxial conformation,which is the least stable due to severe $1,3-$diaxial interactions.
Therefore,the decreasing order of stability is $I > II \approx III > IV$.

(B) In acyclic compounds,the bond angles are generally free to rotate,and the carbon atoms can adopt geometries that minimize angle strain. Therefore,angle strain is not a factor that governs the stability of different conformations in acyclic compounds. Conversely,torsional strain,steric interactions,and electrostatic forces are significant factors that determine the relative stability of conformations in acyclic systems.

(C) The dihedral angle is the angle between the $C-H$ bonds on adjacent carbon atoms.
In the given Newman projection,the angle between the front $C-H'$ bond and the back $C-H''$ bond is calculated by considering the standard staggered angle of $120^\circ$ between adjacent bonds and the deviation shown.
From the geometry,the dihedral angle between $H'$ and $H''$ is the sum of the angle between the bonds in the staggered conformation $(120^\circ)$ and the given deviation angle $(29^\circ)$.
Dihedral angle $= 120^\circ + 29^\circ = 149^\circ$.

(C) The stability of conformers of $n$-butane follows the order: Anti > Gauche > Eclipsed.
In the anti-conformer,the two bulky methyl $(-CH_3)$ groups are at a dihedral angle of $180^{\circ}$,which minimizes steric repulsion and torsional strain.
Image $817-c1436$ represents the anti-conformer,where the two $-CH_3$ groups are opposite to each other.

(D) The stability of the Gauche form in molecules like $Z-CH_2-CH_2-OH$ is often enhanced by intramolecular hydrogen bonding between the $-OH$ group and the group $Z$.
For intramolecular hydrogen bonding to occur,$Z$ must be an electronegative atom or group capable of acting as a hydrogen bond acceptor (e.g.,$-F$,$-OH$,$-OCH_3$,$-NH_2$,etc.).
Since $-F$,$-OH$,and $-OCH_3$ can all participate in intramolecular hydrogen bonding with the $-OH$ group in the Gauche conformation,the Gauche form becomes more stable than the anti form in these cases.
Therefore,all of these groups make the Gauche form stable.

(B) In the Newman projection of propane,the front carbon is bonded to one methyl group $(-CH_3)$ and two hydrogen atoms $(-H)$. The back carbon is bonded to three hydrogen atoms $(-H)$.
For the eclipsed conformation to repeat itself,the front carbon must be rotated such that the methyl group aligns with another hydrogen atom on the back carbon.
Since the back carbon has three identical hydrogen atoms arranged at $120^o$ to each other,rotating the front carbon by $120^o$ will bring the methyl group to the position of the next hydrogen atom,resulting in an identical eclipsed conformation.
Thus,the eclipsed form is repeated after a rotation of $120^o$.

(D) Conformational isomers are formed by rotation around a $C-C$ single bond. During this rotation,the hybridization of carbon atoms remains $sp^3$,and the bond lengths and bond angles do not change. Therefore,option $D$ is correct. Conformational isomers cannot be separated at room temperature due to low energy barriers.

(B) In $3$-hydroxypropanal $(HO-CH_2-CH_2-CHO)$,the gauche conformation allows the hydroxyl group $(-OH)$ to come close to the carbonyl oxygen $(C=O)$.
This proximity facilitates the formation of an intramolecular hydrogen bond between the hydrogen of the hydroxyl group and the oxygen of the carbonyl group.
This intramolecular hydrogen bonding stabilizes the gauche conformation,making it the most stable conformation despite the presence of some torsional strain.