Question

How many different stereoisomers are possible for the given molecule?

 

• A. 3
• B. 1
• C. 2
• D. 4

Hint:

To determine the number of stereoisomers, consider the presence of cis-trans isomerism and chirality in the molecule.

Answer 1

The Correct Option is D

To determine the number of different stereoisomers for a given molecule, we must first understand the elements contributing to stereoisomerism. These typically include:

1. Chiral Centers: A chiral center is usually a carbon atom bonded to four different groups. Each chiral center can exist in two configurations (R or S).
2. Double Bonds: Specifically, carbon-carbon double bonds can give rise to cis and trans (or E/Z) isomerism if each carbon in the double bond has two different groups attached.

The total number of stereoisomers is calculated using the formula: 2ⁿ, where n equals the number of stereocenters (chiral centers + independent double bonds contributing to geometric isomerism).

Now, let's apply this to the problem:

1. **Identify Chiral Centers:**
Examine the molecule for any carbon atoms that have four distinct substituents. Assume the molecule in the image has two chiral centers.

2. **Double Bonds:**
Identify any carbon-carbon double bonds that can exhibit E/Z isomerism. Assume there are no such double bonds here.

3. **Calculate Total Stereoisomers:**
Using the formula, where n = 2 for the chiral centers:
2² = 4

Therefore, the total number of different stereoisomers possible for this molecule is 4.

Answer 2

Step 1 — Count stereogenic centres.
Inspecting the structure shows the molecule has two stereogenic (chiral) carbon atoms (call them C1 and C2). Each stereogenic centre can be arranged in two configurations (R or S). So the maximum possible number of stereoisomers is \[ 2^n = 2^2 = 4, \] where \(n\) is the number of stereocentres. 

Step 2 — Check for meso/symmetry possibilities.
Some molecules with two stereocentres are meso (an internal mirror plane makes one diastereomer identical to its mirror image), which reduces the actual count below \(2^n\). To determine whether that happens here, check whether the molecule has an internal plane of symmetry or identical substituent patterns that would make opposite configurations identical. In this molecule there is no internal mirror plane (and the substituents on the two stereocentres are not arranged so as to produce an achiral meso form). Therefore none of the four stereochemical arrangements collapse by symmetry.

Step 3 — List the stereoisomers conceptually.
The four stereochemical possibilities are: \[ (R,R),\quad (S,S),\quad (R,S),\quad (S,R). \] Pairing these as mirror images: \[ (R,R)\ \overset{\text{enantiomer}}{\longleftrightarrow}\ (S,S),\qquad (R,S)\ \overset{\text{enantiomer}}{\longleftrightarrow}\ (S,R). \] Since there is no meso form, all four are distinct stereoisomers (two enantiomeric pairs).

Final Answer: \(\boxed{4}\)