Number of geometrical isomers possible for the given structure is/are ___________.
Correct Answer: 4
Solution and Explanation
To determine the number of geometrical isomers for the given structure:
Identifying Stereocenters: The given structure contains three stereocenters, which influence the overall geometric configurations.
Counting Geometrical Isomers: The maximum potential geometric isomers can be calculated using the formula:
Total Geometrical Isomers = 2n
where n is the number of stereocenters. In this case:
23 = 8
However, due to symmetry in the molecule, some configurations are equivalent, reducing the number of unique geometrical isomers.
Final Count: Considering the symmetry and equivalent configurations, the total number of unique geometrical isomers for the given structure is 4.
Top Questions on Isomerism
- What is metamerism?
- What are chiral and achiral compounds?
Draw the possible isomers of:
\[ [ \text{Co}(\text{en})_2\text{Cl}_2 ]^+ \]- Given below are two statements: Statement (I):
Statement (II):
In the light of the above statements, choose the correct answer from the options given below: - Which of the following is not a chiral molecule?
- KEAM - 2025
- Chemistry
- Isomerism
Questions Asked in JEE Main exam
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
- JEE Main - 2025
- Relations and functions
- The area of the region enclosed by the curves \( y = x^2 - 4x + 4 \) and \( y^2 = 16 - 8x \) is:
- JEE Main - 2025
- Area between Two Curves
- Let the foci of a hyperbola be \( (1, 14) \) and \( (1, -12) \). If it passes through the point \( (1, 6) \), then the length of its latus-rectum is:
- JEE Main - 2025
- Hyperbola
- Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:
- JEE Main - 2025
- Speed and velocity
- Let \(E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1\) be an ellipse. Ellipses \(E_i\) are constructed such that their centers and eccentricities are the same as that of \(E_1\), and the length of the minor axis of \(E_{i+1}\) is the length of the major axis of \(E_i\). If \(A_i\) is the area of the ellipse \(E_i\), then \(\frac{5}{\pi} \sum_{i=1}^{\infty} A_i\) is equal to:
- JEE Main - 2025
- Calculus