To convert cyclononatetraene into an aromatic | vedclass

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(C) Cyclononatetraene has $9$ carbon atoms and $4$ $\pi$ bonds. To be aromatic,it must follow $H$ückel's rule,which states that a planar,cyclic,conjug

Vedclass · Accepted Answer

Loss of $H^{+}$ from $sp^3$ hybridised carbon

Vedclass · Answer

(C) Cyclononatetraene has $9$ carbon atoms and $4$ $\pi$ bonds. To be aromatic,it must follow $H$ückel's rule,which states that a planar,cyclic,conjugated system must have $(4n + 2)$ $\pi$ electrons. For $n = 2$,the system requires $10$ $\pi$ electrons. Cyclononatetraene has $8$ $\pi$ electrons. If we remove a proton $(H^+)$ from the $sp^3$ hybridized carbon (the methylene bridge),the carbon becomes $sp^2$ hybridized and carries a negative charge (a lone pair). This lone pair contributes $2$ electrons to the $\pi$ system,resulting in a total of $8 + 2 = 10$ $\pi$ electrons. Thus,the cyclononatetraenyl anion is aromatic.

To convert cyclononatetraene into an aromatic ion,the process needed is

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