Principles of Stereochemistry MCQ Quiz in मल्याळम - Objective Question with Answer for Principles of Stereochemistry - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:-

Elimination reactions:

• The elimination reaction is an organic reaction in which two substituents are removed from a molecule to form a new product.
• The process takes place in the presence of acid, base, metal, and sometimes through heating.

Mechansim of elimination reaction:

1. E1 Mechansim:

• Depending on the leaving group, followed by ionization of the molecule results in the formation of a carbocation.
• Removal of the hydrogen atom through deprotonation in the presence of base to form the C=C bond.

2. E2 Mechansim:

• Simultaneous removal of the leaving group and hydrogen atom in presence of a base to form C=C bond.

Explanation:-

For the reaction of sodium iodide with dibromo cyclohexanes to give cyclohexane, the two bromine atoms must be in anti-orientation.

• The reaction pathway is shown below:

• In the reaction we can see that, the vinicial anti-compound contains the two Br atom that are anti to the each other and can undergo E2 elimination reaction with sodium iodide.

Conclusion:-

• Hence, option 3 is the correct option.

CONCEPT:

Conformation Stability in Pyranose Rings

• The stability of a conformation in six-membered rings like pyranose is governed by steric strain, torsional strain, and 1,3-diaxial interactions.
• Chair conformations are generally the most stable because they minimize steric hindrance and torsional strain.
• In the most stable conformation, bulky substituents like hydroxyl groups prefer to be in equatorial positions to reduce 1,3-diaxial interactions.

Explanation:-

Hydrogen bonding and no 1, 3-diaxial interaction

• For compound A (shown in the question), the most stable conformation corresponds to the chair form where the hydroxyl (OH) group occupies an equatorial position, reducing steric hindrance.
• In the given options, we identify the chair conformation where the hydroxyl group is equatorial and the methoxy group is in the axial position.
• This is represented in the second structure (from the top), which corresponds to the most stable conformation.

The correct option is the option 1 structure.

Concept: 

In a Newmann projection, the three lines in the shape of a Y represent the three bonds of the first carbon that are sighting down; where the three lines connect is where the front carbon is. A circle represents the back carbon; the three lines coming out of the circle represents the three bonds that come off of that carbon. Note that the fourth bond for each of these carbons is the carbon-carbon bond that is looking down. A Newmann projection can help to analyze the rotation around a particular carbon-carbon bond.

Explanation:

→Find the longest continuous chain of carbon atoms.

→There is one C atom at the centre of the circle and a second one hidden at the back of the circle.

→ A methyl group off the back adds a third C atom.

→Each atom of cyclohexane is bonded to two hydrogens. Hydrogen lying in the plane of the ring are called equatorial hydrogen because they are situated essentially along the ring and the hydrogen lying above and below the ring are called the axial hydrogens because they are present to the axis along an axis perpendicular to the plane of the ring.

Conclusion: 
Option A is correct.

Explanation:- 

Specific rotation is a property found in chiral chemical compounds, which is remarked by the change in the angle of plane-polarized monochromatic light.

Calculation:-

Given,

Specific rotation of optically pure (R)-2-bromobutane = -112.00.

The specific rotation of the sample mixture = -82.88.

Using, ee = observed rotation / actual rotation x 100 ⇒

ee = -82.88 / -112.00 x 100 ⇒ ee = 74% (R).

Percentage of Racemic Mixture = 100 - R% = 100 - 74 % = 26%.⇒

26% = 13%(S)+13%(R).

Concept:

Stereoisomerism -

• This isomerism arises in compounds having the same chemical formula but different orientations of the atoms belonging to the molecule in three-dimensional space. 
• The compounds that exhibit stereoisomerism are often referred to as Stereoisomers. 
• This phenomenon can be further categorized into 2 subtypes. Both these subtypes are -
  • Optical Isomerism
  • Geometric Isomerism

Geometrical isomerism:

• Geometrical isomerism is shown in molecules that have a carbon-carbon double bond C = C.
• Geometrical isomerism is shown only when each carbon atom of the double bond is attached to two different atoms or groups.
• Compounds of the type abC = Cad, abC = Cab, and abC = Cde will show geometrical isomerism.
• The cause of geometrical isomerism is restricted rotation about a C = C bond.

Criteria to show optical isomerism-

• Absence of centre of Symmetry - when a line is drawn from the centre of a molecule towards the corner of each atom, it should not encounter similar atoms.
• Absence of plane of symmetry - A real or imaginary plane, vertical or horizontal when passed through a molecule, bisects it so that the one half of the molecule should not be the mirror image of the other half.
• The presence of a Chiral centre - An object or molecule which has no plane of symmetry and is not superimposable on its mirror image is said to be chiral or dissymmetric.

​Explanation:

• The analysis of all the given compounds is shown below:

​

Hence, the total number of compounds showing stereoisomerism is six.

Concept:

Diastereomers 

• Diastereomers are stereoisomers that are not mirror images of each other.  They have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of those atoms.  

Explanation:

Therefore, the correct option is 4.

The correct answer is option 4.

Explanation:

Based on the nomenclature of the A, B, C and D. The correct relationships between the four structures are:

• A and D are Diastereomers.
• A and C are Enantiomers.
• B and D are Enantiomers.
• B and C are Diastereomers

Conclusion:

The correct statements regarding the four structures are Statements A, B, C and D.

Concept:

• The number of stereoisomers in an asymmetrical molecule is \(2^{n}\).
  • Where n is the number of asymmetric carbon atoms in the molecule. This is known as the Le Bel-van't Hoff rule.

Explanation:

This molecule has 5 asymmetric centers.

\(\text{The number of stereoisomer} = 2^n\)

\(\text{The number of stereoisomer} = 2^5\)

\(\text{The number of stereoisomer} = 32\)

The number of possible stereoisomers is 32.

CONCEPT:

Specific Rotation

• Specific rotation is a characteristic property of chiral compounds that quantifies their ability to rotate plane-polarized light.
• The specific rotation ( \(\alpha\) ) is calculated using the formula:

  \([\alpha] = \frac{\alpha_{\text{obs}}}{l \cdot c}\)

  where:
  • \(\alpha_{\text{obs}}\) = observed rotation in degrees
  • l = length of the polarimeter tube in decimeters (dm)
  • c = concentration of the solution in grams per milliliter (g/mL)

EXPLANATION:

Given:

• \(\alpha_{\text{obs}}\) = \(+12^\circ\)
• l = 2 dm
• Mass of the compound = 10 g
• Volume of solution = 100 mL

Calculate the concentration (c):

\(c = \frac{\text{Mass of compound}}{\text{Volume of solution}}\)

\(c = \frac{10}{100}\)

\(c = 0.1 \, \text{g/mL}\)

Calculate the specific rotation ( \(\alpha\) ):

\([\alpha] = \frac{\alpha_{\text{obs}}}{l \cdot c}\)

\([\alpha] = \frac{+12}{2 \times 0.1}\)

\([\alpha] = \frac{+12}{0.2}\)

\([\alpha] = +60.0 \, \text{°}\)

Therefore, the specific rotation of the compound is \(+60.0^\circ\).

Concept:

The concept of enol content in ketones revolves around keto-enol tautomerism, where a ketone can exist in equilibrium with its enol form. Enol content is primarily governed by factors that stabilize the enol form compared to the keto form. These factors include hydrogen bonding (both intra- and intermolecular), conjugation, solvent effects, steric considerations, electronic effects from substituents, the acidity of α-hydrogens, temperature, and the presence of catalysts. Understanding these parameters allows chemists to predict and manipulate the equilibrium between keto and enol forms in different chemical environments.

Explanation:

Least stabilized enol form, as there is no resonance or intramolecular hydrogen bonding to stabilize the enol form significantly.

The enol form is stabilized by conjugation with the furan ring, providing a higher enol content compared to non-conjugated systems.

The enol form can be stabilized moderately by the lactone, but without the additional conjugation seen in B. Therefore, it is expected to have an intermediate enol content.

Therefore, the order of enol content among these compounds is: B>C>A.

Conclusion:

So, the correct option is 2.