The total number of aromatic species generate | vedclass

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

Question

(C) To determine the aromaticity, we use the Huckel rule: a cyclic, planar, fully conjugated system with $(4n+2)$ $\pi$-electrons is aromatic, where $

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) To determine the aromaticity, we use the Huckel rule: a cyclic, planar, fully conjugated system with $(4n+2)$ $\pi$-electrons is aromatic, where $n$ is an integer $(0, 1, 2, \dots)$.$(i)$ Cyclopropenyl chloride reacts with $SbCl_5$ to form the cyclopropenyl cation, which has $2$ $\pi$-electrons $(n=0)$. This is aromatic.$(ii)$ Cyclopentadiene reacts with sodium metal to form the cyclopentadienyl anion, which has $6$ $\pi$-electrons $(n=1)$. This is aromatic.$(iii)$ $7$-Bromocyclohepta-$1,3,5$-triene reacts with water to form the tropylium cation, which has $6$ $\pi$-electrons $(n=1)$. This is aromatic.$(iv)$ Cyclopentadienylamine reacts with $HNO_2$ to form a $4$ $\pi$-electron system (cyclopentadienyl cation), which is anti-aromatic (not aromatic).Therefore, there are $3$ aromatic species generated.

The total number of aromatic species generated in the following reactions is:

Similar Questions

What is meant by deactivating groups? Explain with suitable examples.

The number of $sp^2-s$ sigma bonds in benzene are

$CH_3-CH_2-CH_2-Cl$ $\xrightarrow{Na, \text{Dry ether}} P$ $\xrightarrow[V_2O_5, Al_2O_3]{\Delta} Q$. $Q$ is

What type of difference is observed in the hybridization of carbon in benzene during nitration and the formation of the $\sigma$-complex?

Vedclass Test Series

Exam Paper Generator

Online Exam Module