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Q. The angle strain in cyclobutane is

KCETKCET 2008Hydrocarbons

Solution:

When carbon is bonded to four other atoms, the angle between any pair of bonds $=109^{\circ}, 28'$ (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'-90^{\circ} / 2 $

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