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Chemistry
The angle strain in cyclobutane is
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Q. The angle strain in cyclobutane is
KCET
KCET 2008
Hydrocarbons
A
$24^\circ 44'$
11%
B
$29^\circ 16'$
16%
C
$19^\circ 22'$
15%
D
$9^\circ 44'$
57%
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 $
$=19^{\circ} 28 \% 2=9^{\circ} 44'$