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Problem:

Which of these molecules have stereoisomers?

Solution:

(The molecules with a red X beside them are the ones with stereoisomers.)

enter image description here

Question:

What is the method used to identify which molecules have stereoisomers? I can see the commonality between the three molecules with stereoisomers being that the two atoms that can exchange side are right in the middle if that makes any sense. But is that the correct way to think?

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    $\begingroup$ ii and iv have optical isomerism and iii has geometrical isomerism, both come under stereoisomerism. $\endgroup$ Commented Mar 3 at 10:42

1 Answer 1

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Stereoisomerism is a type of isomerism in which molecules have same molecular formula and structural formula (sequence of bonded atoms), but differ in orientation in space.

Stereoisomerism is of two types:

  1. Conformational : The type of stereoisomerism in which isomers can be converted into one another by free rotation about $\sigma$ bond.

  2. Configurational : It is of two types-


(Note: When in questions, stereoisomerism is mentioned, it's self understood that configurational stereoisomerism has been asked, unless mentioned specifically.)


Geometrical Isomerism (GI)

Geometrical isomers are compounds having same molecular and structural formula, but different geometry across restricted rotation system.

Conditions for GI:

  • restricted rotation system (like $\pi$ bonds)
  • Distance between terminal atoms or groups must be different in both isomers.

Optical Isomerism (OI) Optical isomers are compounds having same molecular and structural formula, but different optical behavior.

Conditions for OI:

[First check for Asymmetric/Chiral Centre (CC)]:

  • If CC Present -- Compound would surely show OI.
  • If CC Absent then: Check for Plane of symmetry(POS), Centre of symmetry(COS) & Alternating axis of Symmetry(AAOS)/n-fold axis of improper rotation symmetry.

If all three are absent then compound would show OI otherwise not.

(For details on symmetry elements, click here.


Now Answering your question:

Compound(i) has a $\pi$ bond, but both the groups on oxygen are same (i.e. Lone pairs) hence no GI. It has POS as shown, so no OI.

Compound(ii)has two chiral centers as shown, so it shows OI.

Compound(iii) has a $\pi$ bond, and the distance between terminal groups is different as shown, hence it shows GI.

Compound(iv) has a chiral center as shown so it shows OI.

enter image description here


Hope helpful.


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  • $\begingroup$ Thank you so much. And thank you for making the added effort of drawing the molecules. Very helpful :) $\endgroup$
    – Nora
    Commented Mar 4 at 1:08

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