Question

The angle strain in cyclobutane is

• A. $24^{\circ }44^{\prime }$
• B. $29^{\circ }16^{\prime }$
• C. $19^{\circ }22^{\prime }$
• D. $9^{\circ }44^{\prime }$

Answer 1

Answer: (D)

When carbon is bonded to four other atoms, the angle between any pair of bonds $=109^{\circ },28^{\prime }$ (tetrahedral angle) but the ring of cyclobutane is square with four angles of $90^{\circ }$.

So, deviation of the bond angle (angle strain) in

cyclobutane $=109^{\circ }28^{\prime }-90^{\circ }/2$

$=19^{\circ }28\%2=9^{\circ }44^{\prime }$