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Question

Chemistry Question on Some basic concepts of chemistry

Angle strain in cyclopropane is

A

24∘44′24^{\circ} 44'

B

9∘44′9^{\circ}44'

C

44'

D

−5∘6′-5^{\circ}6'

Answer

24∘44′24^{\circ} 44'

Explanation

Solution

Angle strain in cyclopropane is 24∘44′24^{\circ} 44^{'}
θ=[180−360n]\theta =\left[180-\frac{360}{n}\right]
=[180−3603]=180−120=60∘=\left[180-\frac{360}{3}\right]=180-120=60^{\circ}

Angle strain α=12[109∘28∘−θ]\alpha =\frac{1}{2}\left[109^{\circ} 28^{\circ}-\theta\right]
=12[109∘28′−60∘]=\frac{1}{2}\left[109^{\circ} 28^{\prime}-60^{\circ}\right]
=24∘44′=24^{\circ} 44'