KCET · Chemistry · General Organic Chemistry
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 & Solution
Correct Answer
(D) \(9^{\circ} 44^{\prime}\)
Step-by-step Solution
Detailed explanation
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 \(=109^{\circ} 28^{\prime}-90^{\circ} / 2\)
\[
=19^{\circ} 28^{\prime} / 2=9^{\circ} 44^{\prime}
\]
So, deviation of the bond angle (angle strain) in cyclobutane \(=109^{\circ} 28^{\prime}-90^{\circ} / 2\)
\[
=19^{\circ} 28^{\prime} / 2=9^{\circ} 44^{\prime}
\]
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