CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
84
You visited us 84 times! Enjoying our articles? Unlock Full Access!
Question

Which is the correction expression for the magnitude of the equilibrium constant K for the following equilibrium?

A
K = 1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
K < 1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
K > 1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
No estimate of K can be made
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B K < 1
The given reaction is a ring fliping of a conformer.
It is an equilibrium process, where all axial bonds are converts to equatorial and all equatorial bonds are converted to axial bond.
We know,
When a substitutent is present at axial position, the conformer will be less stable because it has 1, 3 diaxial interaction which is a steric interaction of axial group. This will increase the energy of the conformer and make it less stable.
When a substituent is presnt at equatorial position, the conformer is stable because the substituent has more room and fewer steric interactions when it is at equatorial position.

In reactant conformer, the methyl group is at axial position and bulkier isopropyl group is at equatorial position.
In product conformer, the isopropyl group is at axial position and methyl group is at equatorial position.
Since, isopropyl group is bulkier than methyl group it has more 1, 3- diaxial interaction compared to methyl. Thus product conformer is less stable compared to reactant conformer. So the equilibrium will be towards the reactant conformer. Therefore, the K value will be less than 1.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon