(a) How many pairs of eclipsed hydrogen are present in planar cyclobutane? (b) In puckered cyclobutane ?

In a planar conformation, all the atoms lie on the same plane. So, in this case, all carbon atoms and their connected hydrogen atoms of cyclobutane will be on the same plane.

In a planar cyclobutane, the molecule looks like a square, with each corner being a carbon atom. Each carbon atom is attached to two hydrogen atoms. If we think about the connectivity, every pair of neighboring carbon atoms is joined by a single bond. You can imagine placing a cyclobutane molecule flat on a table, with each carbon atom at the corner of an imaginary square. The hydrogen pairs will be on either side of these carbon atoms.

With the planar cyclobutane molecule visualized, we can now identify the pairs of eclipsed hydrogen atoms. A pair of eclipsed hydrogen atoms occurs when two hydrogen atoms on neighboring carbon atoms are aligned with each other. In planar cyclobutane, the hydrogen atoms paired to one carbon atom will be eclipsed by all four neighboring hydrogen atoms. So, then, for (a): 1. One hydrogen atom from Carbon 1 will be eclipsed by one hydrogen atom from Carbon 2. 2. The other hydrogen atom from Carbon 1 will be eclipsed by the other hydrogen atom from Carbon 2. We do not need to repeat this step for all Carbon atoms, given that it is symmetric, and each one will have two pairs in an eclipsed position. Thus, we have found the number of pairs of eclipsed hydrogen atoms in the planar cyclobutane: Answer (a): There are 2 pairs of eclipsed hydrogen atoms present in the planar cyclobutane.

In a puckered conformation, the cyclobutane molecule is bent slightly out of the plane, with two carbon atoms moving above the plane and two carbon atoms moving below the plane. This reduces the number of eclipsed hydrogen pairs, as they no longer perfectly align with one another.

To visualize the puckered conformation, imagine the cyclobutane molecule twisted out of a flat plane. Two carbon atoms are now above the table while the other two are below. The eclipsed hydrogen pairs present in the planar conformation are now not perfectly eclipsed, as they are no longer in one plane.

In the puckered conformation, there are no hydrogen pairs left in a perfect alignment (eclipsed). The strain caused due to the eclipsed position is reduced because of the torsional angles between the neighboring hydrogen atoms. Answer (b): There are 0 pairs of eclipsed hydrogen atoms present in the puckered cyclobutane.