Test: Optical Isomerism - NEET MCQ

The number of optically active isomers = 2ⁿ⁻¹−2^{n−1/2 ​}=2³⁻¹−2^3−1/2 ​=2.

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

The correct answer is Option B.
The optically active hydrocarbon X is 3-methyl-1-pentene CH₂=CH−CH(CH₃)CH₂CH_3​. On catalytic hydrogenation, it forms 3-methyl pentane CH₃CH₂CH(CH₃)CH₂CH₃, which is optically inactive.

The correct answer is option D.
Pair of enantiomers react differently with pure enantiomers of other compounds.
 

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.

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
 

The general formula of isomeric ketone having molecular mass 100 is C_6​H₁₂​O [6×12+12×1+16].

The possible structure will be :

The number of ketones that gives racemic mixture with NaBH_4​ is 5 as the ketone with chiral center will give diastereoisomer with NaBH_4​

•  The given compound has a total of six stereoisomers 

• It's meso form upon ozonolysis followed by Zn-hydrolysis gives racemic mixture 

• It's optically active isomers, each upon ozonolysis followed by Zn-hydrolysis gives a single enantiomer. 

There are 4 final products, they are as follow:-

2 products of the structure above due to presence of a chiral carbon.

1 product of structure above due to absence of Chiral Carbon.

1 product of structure above due to absence of Chiral Carbon.

Hence, Options A, B are correct.

• Trichlorocyclobutane has total seven isomers
• It has two meso isomers
• One isomer is devoid of any chiral carbon 

The correct options are A & D
Since a carbon-carbon double bond is present, the molecule can have either cis isomer or trans isomer.
In cis isomer (which is symmetrical), the number of enantiomers is
2⁽ⁿ⁻¹⁾=2⁽²⁻¹⁾=2
The number of meso compounds possible is 2^(n/2)−1=2^(2/2)−1=1
Hence, the total number of optical isomers possible is 2+1=3.
Trans isomer is also symmetrical.
Hence, the total number of optical isomers possible for trans isomers is also 3.
Hence, the total number of stereoisomers possible for X is 3+3=6.
Hence, option A is correct and option B is incorrect.
If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4.
Thus, the option C is incorrect.
If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2.
Thus the option D is correct.
 

The correct answer is option A & D
The given image is Self-Explanatory.
We know that following are the cases when a compound is Optically Inactive:
1. When it does not have a chiral Carbon Atom eg.: Methyl bromide
2. When it has a Plane or Axis of Symmetry eg.: Tartaric Acid (these compounds are called as Meso Compounds)
3. When it forms a Racemic Mixture eg.: Equimolar mixture of d-Lactic Acid and l-Lactic Acid.
The given compound in the question possesses an Axis of Symmetry (shown in the attached image). Hence, it is Optically Inactive.
 

The Fischer Projections of the given compounds are below.

• M and N are non-mirror image stereoisomers
• M and O are identical
• M and P are enantiomers

2 optical centers.
Total optically active isomers =2²
=4

The minimum number of C atoms required for a hydrocarbon to exhibit optical isomerism = 7

Correct option is
A: It is optically inactive due to the plane of symmetry &

B: It is optically inactive due to the center of symmetry

The presence of any kind of symmetry makes the compound optically inactive.

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.

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.