A) \[{{24}^{o}}44'\]
B) \[{{9}^{o}}44'\]
C) \[44'\]
D) \[-{{5}^{o}}16'\]
Correct Answer: A
Solution :
Angle strain in cyclopropane is \[{{24}^{o}}44'\] \[\theta =\left[ 180-\frac{360}{n} \right]\] \[=\left[ 180-\frac{360}{3} \right]=180-120={{60}^{o}}\] Angle strain \[\alpha =\frac{1}{2}[{{109}^{o}}{{28}^{o}}-\theta ]\] \[=\frac{1}{2}[{{109}^{o}}28'-{{60}^{o}}]\] \[={{24}^{o}}44'\]
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