The angle strain in cyclobutane is:

24^{\circ} 44^{'}

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29^{\circ} 16^{'}

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19^{\circ} 22^{'}

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9^{\circ} 44^{'}

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Solution

The correct option is D $9^{\circ} 44^{'}$
The angle strain in cyclobutane is $d = \frac{109^{o} 28^{'} - 90^{0}}{2} = + 9^{0} 44^{'}$

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The angle strain in cyclobutane is:

The correct option is D 9∘44′The angle strain in cyclobutane is d=109o28′−9002=+9044′

The correct option is D $9^{\circ} 44^{'}$
The angle strain in cyclobutane is $d = \frac{109^{o} 28^{'} - 90^{0}}{2} = + 9^{0} 44^{'}$