A) \[{{24}^{o}}44\]
B) \[{{29}^{o}}16\]
C) \[{{19}^{o}}22\]
D) \[{{9}^{o}}44\]
Correct Answer: D
Solution :
When carbon is bonded to four other atoms, the angle between any pair of bonds \[={{109}^{o}}\], (tetrahedral angle) but the ring of cyclobutane 28 is square with four angles of \[{{90}^{o}}\]. So, deviation of the bond angle (angle strain) in \[={{109}^{o}}28-{{90}^{o}}/2\]cyclobutane \[={{19}^{o}}28/2={{9}^{o}}44\]
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