To convert cyclononatetraene into an aromatic | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

Out of benzene,$m$-dinitrobenzene and toluene,which will undergo nitration most easily and why?

The correct arrangement for the decreasing order of electrophilic substitution for the above compounds is:

Product $(B)$ in the above reaction is

Which of the following compounds has the lowest bond dissociation energy for a $C-H$ bond?

Benzene on nitration gives nitrobenzene in the presence of a mixture of $HNO_3$ and $H_2SO_4$,where:

Vedclass Test Series

Exam Paper Generator

Online Exam Module