Isomerism Test

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Isomerism Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

The number of optically active optical isomers of the compound is:

A
4

B
2

C
0

D
3

Solution

The number of optically active isomers = 2ⁿ⁻¹−2^n−1/2=2³⁻¹−2^3−1/2=2.
Question 2
1 / -0

Find the number of stereoisomers for CH₃ – CHOH – CH = CH – CH₃.

A
1

B
2

C
3

D
4

Solution

The number of stereoisomers for CH₃ – CHOH – CH = CH – CH₃ is four. This is calculated by the formula 2ⁿ⁺¹. where n is number of chiral centres, which is 1 in this case. So, 2¹⁺¹= 2²= 4
Question 3
1 / -0

A hydrocarbon X is optically. X upon hydrogenation gives an optically inactive alkane Y. Which of the following pair of compounds can be X and Y respectively?

A
4-methyl-2-hexene and 3-methyl hexane

B
3-methyl-1-pentene and 3-methyl pentane

C
4-methyl cyclopentene and methyl cyclopentane

D
2-methyl-1-butene and 2-methyl butane

Solution

The correct answer is Option B.
The optically active hydrocarbon X is 3-methyl-1-pentene CH₂=CH−CH(CH₃)CH₂CH₃. On catalytic hydrogenation, it forms 3-methyl pentane CH₃CH₂CH(CH₃)CH₂CH₃, which is optically inactive.
Question 4
1 / -0

What is wrong about enantiomers of 2-chloropropanoic acid?

A
Have same solubility in water

B
Have same pKa value

C
Have same refractive indices

D
Have same rate of reactions with (+)-2-butanol

Solution

The correct answer is option D.
Pair of enantiomers react differently with pure enantiomers of other compounds.
Question 5
1 / -0

How many different stereoisomers exist for the compound below ?

A
2

B
4

C
6

D
8

Solution

If you name the C attached to Me group as C1 and naming the rest clockwise, you observe that at C1 and C2 there are chiral centers and C5 and C2 are symmetrical. So the two chiral centers produce 2×2=4 optically active(and distinguishable) compounds. Hence, B is the answer.
Question 6
1 / -0

How many total cyclic isomers are possible for C₅H₁₀?

A
3

B
5

C
7

D
8

Solution

The correct answer is Option B.

5 isomers are possible. C₅H₁₀(C_nH_2n). Molecules having the CnH2n formulas are most likely to be cyclic alkanes. The five isomers are:

(1) Cyclopentane
(2) 1-Methylcyclobutane
(3) 1-Ethylcyclopropane
(4) 1,1-Dimethylcyclopropane
(5) 1,2-Dimethylcyclopropane
Question 7
1 / -0

How the following pair of isomers cannot be distinguished ?

A
smell

B
optical rotation measurement

C
Circular dichroism

D
Dipole moment measurement
Question 8
1 / -0

Which compound below exhibit stereolsomerism?

A
2-methyl-1-butene

B
3-methyl-1-butene

C
3-methyl butanoic acid

D
2-methyl butanoic acid

Solution
Question 9
1 / -0

Consider all possible isomeric ketones, including stereoisomers of Molecular weight 100. All these isomers are independently reacted with NaBH4
(Note stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are

[JEE Advanced 2014]

A
3

B
4

C
5

D
6

Solution

The general formula of isomeric ketone having molecular mass 100 is C₆H₁₂O [6×12+12×1+16].
The possible structure will be :

The number of ketones that gives racemic mixture with NaBH₄ is 5 as the ketone with chiral center will give diastereoisomer with NaBH₄
Question 10
1 / -0

The number of optical isomers possible for 2, 3-pentanediol is:

A
2

B
3

C
4

D
5

Solution

2 optical centers.
Total optically active isomers =2²
=4
Question 11
1 / -0

The minimum number of C atoms required for a hydrocarbon to exhibit optical isomerism:

A
4

B
5

C
6

D
7

Solution

The minimum number of C atoms required for a hydrocarbon to exhibit optical isomerism = 7
Question 12
1 / -0

Which of the following are not functional isomers of each other?

A
CH₃CH₂NO₂and CH₃CH₂ON=O

B
C₂H₅CHO and CH₃COCH₃

C
CH₃CH₂NH₂ and CH₃NHCH₃

D
C₃H₇NH₂ and (CH₃)₂CHNH₂

Solution

Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar or different functional groups.
10 amine to different 10 amines are not functional isomers.
Hence, option ′d′ is the answer.
Question 13
1 / -0

A pair of enantiomers is possible for _______ isomer of 2,2-dibromobicyclo [2.2.1] heptane.

A
cis

B
trans

C
both cis and trans

D
none of the above

Solution

Only one pair of enantiomers is possible for cis-2,2-dibromobicyclo [2.2.1] heptane. The trans arrangement of one carbon bridge is structurally impossible. Such a molecule would have too much strain.