Test: Stereochemistry Level - 2 - Chemistry MCQ

In option B and C one of the terminal carbons has the same group, so it is neglected and in A H has least priority. Rotation gives S configuration and since it is at wedge position so we reverse it and S -> R we get the configuration to be R and hence A is correct.

(R)-lactic acid is an optically active form of lactic acid having (R)-configuration. It has a role as an Escherichia coli metabolite and a human metabolite. It is a conjugate acid of a (R)-lactate. It is an enantiomer of a (S)-lactic acid.

The chiral carbon atom has all four different group attached to it.

∴ It has one asymmetric or chiral carbon atom.

There option b) is correct.

The correct answer is B. 1S, 2S, 6R.

To assign the configuration at each chiral center, we need to determine which group is coming out of the page and which group is going into the page. We can do this by drawing a Fischer projection of the molecule.

In a Fischer projection, the horizontal bonds are coming out of the page and the vertical bonds are going into the page. The substituents are then placed on the appropriate bonds according to their priority. The priority of the substituents is determined by the following rules:

1. Atoms with higher atomic numbers have higher priority.
2. If two atoms have the same atomic number, the atom with the greater number of bonds to hydrogen has higher priority.
3. If two atoms have the same atomic number and the same number of bonds to hydrogen, the atom with the lower atomic number has higher priority.

Once the priority of the substituents has been determined, we can assign the configuration of the chiral center by following the Cahn-Ingold-Prelog (CIP) system. The CIP system assigns the configuration of a chiral center as R or S, depending on which group is coming out of the page (R) or going into the page (S).

In the case of bicyclodecanol, the configuration at the first chiral center is S. This is because the group with the highest priority (the hydroxyl group) is coming out of the page. The configuration at the second chiral center is also S. This is because the group with the highest priority (the methyl group) is coming out of the page. The configuration at the third chiral center is R. This is because the group with the highest priority (the hydroxyl group) is going into the page.

Therefore, the configurations at the three chiral centers in bicyclodecanol are 1S, 2S, 6R.

Giving the numbers according to their priority we get:

Since 1 -> 2 -> 3 -> 4 is in the anti clockwise direction, It has S configuration.

Hence D is correct.

Here, basic conditions are given so

1. Most acidic H+ ion is removed which is attached to the carbon atom at benzylic position.

2. Then to stabilize the carbanion, elimination reaction mechanism takes place.

3. Leading to the removal of leaving group -OTs, forming a double bond between the two carbon atoms.

Hence D is correct.

Diastereomers are stereoisomers that are not mirror images of each other. They have different physical properties and different arrangements of their atoms in space, excluding the mirror image scenario. This is the correct answer.

Separation of racemates into their component enantiomers is a process called resolution.

All the options A, B and C are same

1. When A is rotated by 90 degrees two times i.e. 180 degrees in plane of your screen then compound C will be obtained

2. Now by switching A two times or ‘even’ switch compound B will be obtained and configuration remained same as the switch was even

Hence all are the same, so, D is correct.

To determine the most stable conformation of the given compound, we need to consider the concept of steric hindrance and torsional strain in organic chemistry.

The correct answer is C. Here's why:
- In option C, the bulky tert-butyl groups are in the equatorial position, reducing steric hindrance.
- This conformation minimizes torsional strain, making it more stable compared to the other options where the bulky groups are in the axial position.
- Therefore, option C represents the most stable conformation for the given compound.

If minimum priority group on horizontal line is

Clockwise direction ⇒S

Anticlockwise rotation ⇒R

5S, 6R