Test: Geometrical Isomerism - 1 - NEET MCQ

The correct option is D Molecule in meso form have a plane of symmetry
Meso forms of isomers are single compound and their molecules are achiral and hence they cannot be separated into optically active enantiometric pairs.
Molecule in meso form have a plane of symmetry due to which the optical rotations of upper and lower parts are equal and in the opposite direction which balanced internally and compound become optically inactive. this property is called internal compensation.

To determine which compounds have the Z-configuration, we need to evaluate the priority of the groups attached to the double-bonded carbon atoms according to the Cahn-Ingold-Prelog (CIP) rules. The Z-configuration occurs when the higher priority groups on each carbon are on the same side of the double bond.

Here’s how we assess each compound:

a) CH₃CH₂CH₃ > C=C < HCH₃

• For the left carbon:

  • CH₃CH₂CH₃ (propyl) vs H
  • Propyl group has higher priority than hydrogen.

• For the right carbon:

  • CH₃ (methyl) vs H
  • Methyl group has higher priority than hydrogen.

Conclusion: The higher priority groups (propyl and methyl) are on opposite sides, so this compound has an E-configuration, not Z.

b) H₂NCl > C=C < COOHBr

• For the left carbon:

  • H₂N (amino group) vs Cl
  • Cl has higher atomic number, so it has higher priority.

• For the right carbon:

  • COOH (carboxylic acid) vs Br
  • Br has a higher atomic number than C in COOH, so Br has higher priority.

Conclusion: The higher priority groups (Cl and Br) are on opposite sides, so this compound has an E-configuration, not Z.

c) HPh > C=C < COOHPh

• For the left carbon:

  • H vs Ph (phenyl group)
  • Ph (phenyl group) has higher priority than hydrogen.

• For the right carbon:

  • COOH vs Ph (phenyl group)
  • COOH (carboxylic acid) has higher priority because the C in COOH is double-bonded to oxygen atoms, giving it higher priority over Ph.

Conclusion: The higher priority groups (Ph and COOH) are on the same side, so this compound has a Z-configuration.

d) CH₃Ph > C=C < CH₂CH₃Br

• For the left carbon:

  • CH₃ (methyl) vs Ph (phenyl group)
  • Ph has higher priority than methyl.

• For the right carbon:

  • CH₂CH₃ (ethyl) vs Br
  • Br has higher priority than ethyl.

Conclusion: The higher priority groups (Ph and Br) are on opposite sides, so this compound has an E-configuration, not Z.

Final Answer:

The compound c) HPh > C=C < COOHPh has the Z-configuration.

A sawhorse projection can reveal staggered and eclipsed conformations.

Only option b exists with 5 constitutional + geometrical isomers without altering carbon skeleton.

The given picture starts with an enol form of 3-hexanone.

So, there are 4 isomers of given compound.

The correct answer is Option B.
 
 
CH₃−CH=CH−CH=CH−CH₃
          2,4-hexadiene
 
Three different geometrical isomers of cis-cis, trans-trans and sic-trans are possible.
 

Alkenes like 1-hexene when flipped from top to bottom they have identical structures and also they have C=CH2 unit which does not exist as cis- trans isomers.

The correct answer is option C

In chemistry, protonation (or hydronation) is the addition of a proton (or hydron, or hydrogen cation), (H+) to an atom, molecule, or ion, forming the conjugate acid. Some examples include. the protonation of water by sulfuric acid: 
H₂SO₄ + H₂O ⇌ H₃O⁺ + HSO ⁻
 

The correct answer is option c

The cis isomer is more likely to exhibit intramolecular hydrogen bonding between the two hydroxyl groups. This can make the molecule less soluble in water because hydrogen bonding with water molecules becomes less favorable when intramolecular hydrogen bonding is present.Considering the factors influencing solubility in water, the correct answer is option c. The cis stereoisomer has a higher dipole moment and is more likely to exhibit intramolecular hydrogen bonding, making it less soluble in water compared to the trans isomer.

These 3 are the possible structures.
Hence, B is correct.
 

(a) 
We have the same arrangement if the -CH₃ and -H are interchanged. So, no GI.
(b) 
Also in this compound we will get same arrangement across double bond if the positions of both -CH₃ are interchanged around  a single bond
(c) 
Same case here as in the above two.
(d) 
Here, if we interchange -CH₃ and -H around a double bond, then we will get a different arrangement with respect to other double bond. So it shows GI.

However, there is no such arrangement in structure i, it won’t have any non polar structure.

According to the definition of intramolecular H bonding, a H atom attached with F, O and N is attracted by highly electronegative atoms (N, O, and F) within the molecule. So, by following the definition, only c and d are expected to form intramolecular H bonding.

Statement a) true, GI have the same constitutions. Only the position of groups is changed.
Statement b) False, Even without pi bond we have GI. eg
 
Statement c) True, Smaller cycloalkenes have only cis configuration at double bonds to maintain the planarity of bond
Statement d) True, cyclodecene exists with both cis and trans configuration at double bond.

The correct answer is Option A and C.

CH₃ - CH = H - CH₃ shows the property of geometrical isomerism.

 

CH₃ - CH = CH₂ does not show the property of geometrical isomerism. 

CH₃ - CH = CH - C₆H₅ shows the property of geometrical isomerism.

(CH₃)₂ - C = CH - CH₃ does not show the property of geometrical isomerism.

The correct answers are option B,C,D
The correct statements concerning the  structures E, F and G are:
(B) E, F and  G are tautomers.
E is in keto form (C=O) and F and G are in enol form (C=C−OH)
(C) F and G are geometrical isomers.
F is Z isomer and E is E isomer.
(D) F and G are diastereomers. They are not mirror images of each other and non-superimposable.
 

The correct answer is option a

Statement I: Butanone, upon protonation at the oxygen atom, results in a pair of stereoisomers. This statement is correct. When butanone (C4H8O) is protonated at the oxygen atom of the carbonyl group, it forms two stereoisomers.

Statement II: Presence of a lone pair of electrons at oxygen is responsible for protonation.This statement is also correct. The lone pair of electrons on the oxygen atom of the carbonyl group is responsible for accepting a proton (H⁺) during the protonation process. Statement II is a correct explanation for Statement I. The presence of a lone pair of electrons on oxygen enables protonation, leading to the formation of stereoisomers.Therefore, the correct answer is option (a): Both Statement I and Statement II are correct, and Statement II is the correct explanation of Statement I.

In order to show stereoisomers, a compound should not contain plain of symmetry,centre of symmetry,or any
In cis- form of this compound plain of symmetry will be present making the molecule incapable to show sterioisomerism
So statement 2 is correct, while statement 1  is incorrect.

The correct answer is option c

If Statement I is referring to geometric isomers (cis-trans isomers), then it is correct. If Statement II is claiming that two out of the four stereoisomers are polar, it might be inaccurate unless it is considering other aspects of the molecule (such as the presence of polar functional groups). Geometric isomers, on their own, are usually not distinctly polar or non-polar.

So, Statement I seems correct if it is referring to geometric isomers, but Statement II may be inaccurate or incomplete without further details on the criteria for considering a stereoisomer as polar.