Question:

The angle strain in cyclobutane is

Updated On: Apr 2, 2024
  • $24^\circ 44'$
  • $29^\circ 16'$
  • $19^\circ 22'$
  • $9^\circ 44'$
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The Correct Option is D

Solution and Explanation

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