Isomerism Test - 20

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 answer is option A
i. Butanone- q) Has three tautomers other than the mentioned compound.
s) Acts as H-bonds acceptor.
ii. 3-pentanone- p) has two tautomer’s other than the mentioned compound itself
s) Acts as H-bonds acceptor.
iii. Methanamide - p) has two tautomer’s other than the mentioned compound itself
r) Form intermolecular H-bonds with its own molecules.
s) Acts as H-bonds acceptor.
Iv. Ethane nitrite - p) has two tautomer’s other than the mentioned compound itself
s) Acts as H-bonds acceptor.