Isomers JEE Advanced Previous Year Questions with Solutions are available here for JEE aspirants to help them prepare effectively for the exam. Basically, we have compiled a set of questions that have appeared in the various sessions of the entrance exam. Additionally, aspirants will find detailed solutions that will help them to not only learn the correct answers but they will also be able to have a quick revision of concepts. Alternatively, they will gain a much better idea of the question pattern, types and difficulty level.
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JEE Advanced Previous Year Questions on Isomers
Question 1. Which of the given statement(s) about N, O, P, Q with respect to M is/are correct?
A. M and N are non-mirror image stereoisomers
B. M and O are identical
C. M and P are enantiomers
D. M and Q are identical
Solution: (A, B and C)
Since, M and N have – OH on the same side and opposite side respectively, they cannot be mirror images, they are diastereomers.
M and O are identical
M and P are enantiomers
Hence, the correct options are a,b and c.
Question 2. Among the complex ions given below,
[Co(NH2 – CH2 – CH2 – NH2)2Cl2]+,
[CrCl2(C2O4)2]3-,
[Fe(H2O)4(OH)2]+,
[Fe(NH3)2(CN)4]–,
[Co(NH2 – CH2 – CH2 – NH2)2(NH3)Cl]2+ and
[Co(NH3)4(H2O)Cl]2+],
The number of complex ion(s) that show(s) cis-trans isomerism is(are).
Solution: (6)
All the complexes given show cis-trans isomerism.
Question 3. The number of isomers of C6H14 is,
A. 4
B. 5
C. 6
D. 7
Solution: (B)
The IUPAC name of C6H14 is hexane.
It has 5 structural isomers.
Question 4. The given structure shows:
A. Geometrical isomerism
B. Optical isomerism
C. Geometrical isomerism and Optical isomerism
D. Tautomerism
Solution: (B)
The compound doesn’t show geometrical isomerism because one carbon atom attached to a double bond has the same substituent group (−CH3), however, it can show optical isomerism, because it has one optically active chiral carbon atom.
Question 5. Which of the following compounds will exhibit geometrical isomerism?
A. 1 – phenyl – 2 – butene
B. 3 – phenyl – 1 -butene
C. 2 – phenyl – 1 – butene
D. 1,1 – diphenyl – 1 – propene
Solution: (A)
If we look in 1-phenyl-2-butene (C6H5CH2CH = ChC⋅H3) we will see that both the carbon atoms of the double bond have different substituents which as a result will show geometrical isomerism. Due to Cab = Cad two isomers are possible.
Question 6. Which kind of isomerism is exhibited by octahedral Co(NH3)4Br2Cl?
A. Geometrical and ionisation
B. Geometrical and optical
C. Optical and ionisation
D. Geometrical only
Solution: (A)
Octahedral Co(NH3)4Br2Cl shows ionisation and geometrical isomerism.
In ionisation isomerism, the ligands that are shown in the coordination sphere and the anions present outside the coordination sphere are exchanged with each other. It can be represented in the following manner.
The coordination number of the central atom (cobalt) is six and the shape is octahedral
Thus, both geometrical and ionization isomerism are exhibited by octahedral Co(NH3)4Br2Cl.
Question 7. The pair of coordination complexes/ions exhibiting the same kind of isomerism is:
A. [Cr(NH3)5Cl]Cl2 and [Cr(NH3)4Cl2]Cl
B. [Co(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]+
C. [CoBr2Cl2]2- and [PtBr2Cl2]2-
D. [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]Br
Solution: (B and D)
The pairs of coordination complexes/ions exhibiting the same kind of isomerism are:
Octahedral [Co(NH3)4Cl2]+and square planar [Pt(NH3)2(H2O)Cl]+ show geometrical (cis-trans) isomerism.
Square planar [Pt(NH3)3(NO3)]Cl and square planar [Pt(NH3)3Cl]Br shows ionization isomerism.
Question 8. The compound(s) that exhibit(s) geometrical isomerism is:
A. [Pt(en)Cl2]
B. [Pt(en)2]Cl2
C. [Pt(en)2Cl2]Cl2
D. [Pt(NH3)2][Cl2]
Solution: (C and D)
The possible structures by [Pt(en)2Cl2]Cl2 are as follows.
So, option C shows geometrical isomerism.
For option D which is [Pt(NH3)2][Cl2], the possible structures by this compound are as follows.
This also shows geometrical isomerism.
Question 9. Which of the following compounds exhibits, stereoisomerism?
A. 2-methyl butene-1
B. 3-methyl butyne-1
C. 3-methyl butanoic acid
D. 2-methyl butanoic acid
Solution: (D)
2–methyl butanoic acid has a chiral carbon (4 different groups are attached) hence optically active.
Question 10. The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are):
A. 2
B. 4
C. 3
D. 5
Solution: (C)
Three stable (staggered) conformers exist (with μ ≠ 0)
Isomerism in Organic Compounds
Isomerism: General Organic Chemistry
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