Test: Geometrical Isomerism - 2 - NEET MCQ

Since in compound a, b and d we have symmetry in the molecule, so both double bonds are identical. THerefore, stereoisomerism is counted from only one bond. This will give us 2 isomers.
In compound c, both double bonds are different, so both will involve GI. N has a lone pair which will also be counted. So, we have 4 isomers in all.

The correct answer is option B
CH₃ – CH = C = C = CH – CH₃ will show geometrical isomerism because it is odd cumulene (3 pie bond) and odd cumulene always shows geometrical isomerism 

The correct answer is Option C.
If C6H3Cl has 4 positional isomers then it’s structure can be;
So, X can be

5 isomers are possible. C₅H₁₀(C_nH_2n). Molecules having the C_nH_2n  formula are most likely to be cyclic alkanes. The five isomers are:

The correct answer is option B
 
Threegeometrical isomers exits of the molecule given cis-cis, tran-trans and cis-trans.

The correct answer is option D
There will be three constitutional isomers i.e. 1,2 dichloro cyclohexane,1,3 dichloro cyclohexane and1,4 dichloro cyclohexane.Each constitutional isomer will have cis and trans configuration thereby making six isomers in all.Also there will be two optically active isomers.
You can understand it better through these infographics
1,2 dichloro cyclohexane

The correct answers are Option B,C and D.
C2H2C12 => 3 isomers possible.

1.  1,2-dichloroethane ---> cis
2.  1,1-dichloroethane ---> trans

(1)
Net dipole moment=0
So, no polar isomer.
(2)

The correct answers are option B & C.
Compound is dibromoethene, it can have 3 isomers (two geometrical isomers (cis & trans) of 1,2 dibromoethene and one is 1,1 dibromoethene).

The correct answers are Options C and D.
 
Enolization involves the α-carbon atom. The ketone 2,4-dimethyl-3-pentanone gives a single enol, since the two α carbons are equivalent. An enol is not a resonance form of a carbonyl compound; the two are constitutional isomers of each other.
 

a) Nitroethanal

• Nitroethanal contains a nitro group (-NO2) and an aldehyde group (-CHO).
• In the enol form, the aldehyde could potentially convert to a hydroxyethene structure.
• However, the distance and orientation between the enol hydroxyl and the nitro oxygen atoms do not allow for the formation of a stable six-membered ring through intramolecular hydrogen bonding.

b) Cyanoethanal

• Cyanoethanal has a cyano group (-CN) and an aldehyde group (-CHO).
• When enolized, the resulting structure has a hydroxyl group adjacent to a carbon-carbon double bond.
• The cyano group does not participate effectively in forming intramolecular hydrogen bonds, and the molecular geometry does not favor the formation of a stable six-membered ring.

c) 2,4-Pentanedione

• 2,4-Pentanedione contains two keto groups separated by a methylene group.
• In its enol form, the molecule can exhibit intramolecular hydrogen bonding.
• However, this bonding does not lead to the formation of a six-membered ring but rather a five-membered ring involving the hydroxyl and one of the keto oxygens.

d) 2-Oxocyclohexane Carbaldehyde

• 2-Oxocyclohexane carbaldehyde has a keto group at the 2-position and an aldehyde at the 1-position of a cyclohexane ring.
• In the enol form, the enol hydroxyl group can hydrogen bond with the carbonyl oxygen, forming a six-membered ring structure.
• This arrangement is energetically favorable and stabilizes the enol form through intramolecular hydrogen bonding.

Therefore, the correct answer is option d) 2-oxocyclohexane carbaldehyde, as it can form a stable six-membered ring through intramolecular hydrogen bonding in its enol form.

a) Geometric isomerism occurs when two structures with the same connectivity are not inconvertible.
b) The boiling points of the isomers might be different enough that you can separate them by fractional distillation. The affinities for the stationary phase may be different enough that they can be separated by gas chromatography.
c) Geometric isomers are diastereomers, i.e. they are stereoisomers that are not enantiomers. The prefixes cis and trans refer to the relative dispositions of the substituents attached to the doubly bonded carbon atoms. In the cis diastereomer the hydrogen atom attached to one carbon is on the same side of the double bond as the hydrogen attached to the other carbon. In the trans stereoisomer they are on opposite sides.
 

The correct answer is option c

Pentene has two positional isomers, two geometrical isomers and more than four cyclic isomers

1-methoxypropene has two geometrical isomers and more than four cyclic isomers

Dichlorocyclopropane has two positional isomers and two geometrical isomers

2,4,6-octatriene has six geometrical isomers and more than four cyclic isomers

1. Pentene has two positional isomers, two geometrical isomers, and more than four
cyclic isomers.
Pentene (C₅H₁₀) is an unsaturated hydrocarbon with the formula C₅H₁₀.It has two positional isomers: 1-pentene and 2-pentene. It has two geometrical isomers due to the presence of a double bond. For cyclic isomers, we can have a cyclopentane ring, so the statement is correct.

2.1-methoxypropene has two geometrical isomers and more than four cyclic isomers.
• 1-methoxypropene (CH₃O-CH= CH-CH₂CH₃) has two geometrical isomers due to the presence of a double bond. For cyclic isomers, it can form a cyclic structure by including the oxygen atom, so the statement is correct.
3. Dichlorocyclopropane has two positional isomers and two geometrical isomers.
• Dichlorocyclopropane (C₃H₄Cl₂) has two positional isomers: 1,2- dichlorocyclopropane and 1,3-dichlorocyclopropane. It has two geometrical isomers due to the cis-trans isomerism in the cyclopropane ring. So the statement is correct.

4. 2,4,6-octatriene has six geometrical isomers and more than four cyclic isomers.
• 2,4,6-octatriene (CH₃-CH = CH- CH₂ - CH=CH- CH₂-CH = CH₂) has six geometrical isomers because of the three double bonds. For cyclic isomers, it can form various cyclic structures, so the statement is correct.

The correct answer is option A

Butanone has three tautomers other than the mentioned compound itself and acts as a H-bond acceptor3-pentanone has two tautomers other than the mentioned compound itself and acts as a H-bond acceptorMethanamide has two tautomers other than the mentioned compound itself, forms intermolecular H-bonds with its own molecules and acts as a H-bond acceptor Ethane nitrile  has two tautomers other than the mentioned compound itself and acts as a H-bond acceptor

1. Butanone has three tautomers other than the mentioned compound itself and acts as an H-bond acceptor.
• Butanone (CH₃- CO - CH₂CH₃) can exist in different tautomeric forms. The keto form is the most stable, but it can also undergo keto-enol tautomerism. However, saying it has three tautomers may be misleading. It primarily exists as the keto form with a minor contribution from the enol form. Also, ketones like butanone can act as H-bond acceptors due to the oxygen atom in the carbonyl group. So the statement is correct.
2. 3-pentanone has two tautomers other than the mentioned compound itself and acts as an H-bond acceptor.
• 3-pentanone (CH₃- CO - CH₂CH₂- CH₃) can undergo keto-enol tautomerism, so it has an enol form as a tautomer. Also, ketones can act as H-bond acceptors. So the statement is correct.
3. Methanamide has two tautomers other than the mentioned compound itself, forms intermolecular H-bonds with its own molecules, and acts as an H-bond acceptor.
• Methanamide (HCONH₂) can exist in different tautomeric forms due to the possibility of hydrogen migration. It can form intermolecular H-bonds with its own molecules, and as an amide, it acts as an H-bond acceptor. So the statement is correct.
4. Ethane nitrile has two tautomers other than the mentioned compound itself and acts as an H-bond acceptor.
• Ethane nitrile (CH₃CH₂- CN) can undergo tautomerism involving the movement of a hydrogen atom. Also, nitriles like ethane nitrile can act as H-bond acceptors due to the electronegative nitrogen atom. So the statement is correct

the number of stereogenic centres which comprises of optically active carbon and the carbon showing geometrical isomerism. In the above picture, there is only one geometric isomer and chiral carbon. So there is 2 stereogenic centre.
stereoisomers = 2ⁿ
=2²
= 4

C₅H₈ has the following three isomers :

CH3CH2CH2C≡CH

CH3CH2C≡CCH3

(CH3)2CH−C≡CH

The correct answer is 9