Isomerism Test - 10

According to option A chiral compound at least have one chiral carbon but compounds other than carbon and also chiral like NCl Br I is chiral. According to option b a compound containing only one ka kalakaron is always is always correct that is statement is true it is the minimum requirement for a carbon atom to become chiral. Efficiency is not necessary to be true all the time compound will not be superimposable on its Mirror image. And for option d a plane of symmetry is sufficient but not a mandatory to give achiral compound.

Presence of either plane of symmetrical or centre of symmetry. Optical activity of one half of the molecule is exactely cancelled by other half and is internally compensated.

The given compound is superimposable on its mirror image, so it is achiral. Achiral molecules which have chirality centers are called meso compounds. Meso compounds are possible if two (or more) chirality centers in a molecule have the same set of four substituents.

The configuration is done as follow:-
i) We first select the left carbon and hen prioritize its valencies. Since the least prioritised group( Here Hydrogen) is on dashed line, we don’t need any modification.

So the configuration comes R.
For the other carbon atom, we again prioritise its valencies. Here, the least prioritised group( Here Hydrogen) is not  on the dashed line, we need to put it on the dashed line. So by interchanging H and OH and then applying the configuration, itcomes S.
But we have done odd times interchange, so the above configuration will be reverse.
The configuration comes out R.
SO the configuration of both carbon atoms are R,R.

The correct answer is Option A.

Out of the given four compounds only the first one has a chiral carbon. Hence compound A is optically active and it can rotate the plane polarised light.
 

Here all compounds except in option c are achiral. So automatically option c is correct.

1 and 2 are non-superimposable mirror images of each other. So, according to the options given, 1 and 2 are enantiomers of each other.

The configuration of the first compound is (R,S) and that of the second compound is (S,S). And we know that the isomers which are having different configurations at one or more equivalent chiral centers and the same configuration at the remaining chiral centers are called diastereomers.

If we rotate one off the molecule in a plane perpendicular to molecule, then we will get the mirror image of others which are non superimposable. So, they are enantiomers.

I have drawn only skeleton. It can be known that for C₅H₁₂ and C₆H₁₄, we have no isomerism. For C₇H₁₆ we have one isomer. So, the minimum carbon required for an alkane is C₇H₁₆.

Compound I has a plane of symmetry along the plane of cyclopropane. Compound II also has a plane of symmetry along the plane which is diagonal to carbon atoms having substituent perpendicular to cyclobutane. However, compounds 3 and 4 have not any plane of symmetry. Compound 5 has a plane of symmetry perpendicular to the plane of cyclohexane dividing outer –CH₃. So, the correct option is d 

The correct answer is option B
  
2 are optically active

The correct Option is B.

Two diastereomeric forms are possible for  1,3-dimethylcyclobutane. These are as shown.
 
Note:
Diastereomers have different configurations at one or more stereocenters and are not mirror images of each other.

We can see that in option iii, we have a chiral centre. So we have an optical somer only in option iii.

The positions are numbered. Note that 4 and α are the same for this purpose.

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.