Explain why the following systems are not aromatic?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) For a compound to be aromatic,it must satisfy the following conditions: $(1)$ It must be cyclic,$(2)$ It must be planar,$(3)$ It must have a complete conjugation of $\pi-$electrons,and $(4)$ It must follow Huckel's rule,i.e.,it must have $(4n+2)$ $\pi-$electrons,where $n$ is an integer $(n = 0, 1, 2, \dots)$.
$(i)$ The compound is heptafulvene. It is not aromatic because the $sp^3$ hybridized carbon atom at the exocyclic double bond breaks the continuous conjugation of the ring.
$(ii)$ The compound is cyclopentadiene. It is not aromatic because it has a $sp^3$ hybridized carbon atom in the ring,which prevents the continuous conjugation of $\pi-$electrons. Additionally,it has $4$ $\pi-$electrons,which does not satisfy Huckel's rule $(4n+2)$.
$(iii)$ The compound is cyclooctatetraene. It is not aromatic because it is non-planar (it adopts a tub-shaped conformation) and it has $8$ $\pi-$electrons,which does not satisfy Huckel's rule $(4n+2)$.

Explore More

Similar Questions

What is the product of the following reaction?
$C_6H_6 + CH_2=CH_2 \xrightarrow{HCl, AlCl_3} \text{Product}$

Which of the following are not aromatic?

Nitrobenzene can be prepared from benzene by using a mixture of conc. $HNO_3$ and conc. $H_2SO_4$. In the nitrating mixture,$HNO_3$ acts as a

Which of the following groups increases the reactivity of electrophilic aromatic substitution?

For a compound with molecular formula $C_7 H_8 O$,how many aromatic structures are possible?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo