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(B) The rate of electrophilic aromatic substitution depends on the electron density of the aromatic ring. Higher electron density makes the ring more

Vedclass · Accepted Answer

$iii > ii > i$

Vedclass · Answer

(B) The rate of electrophilic aromatic substitution depends on the electron density of the aromatic ring. Higher electron density makes the ring more reactive towards electrophiles.$(i)$ Benzene: It is the reference compound with no substituents.$(ii)$ Biphenyl: One phenyl ring acts as a $+R$ (resonance donating) group to the other. However,due to steric hindrance between the ortho-hydrogens of the two rings,they are not perfectly coplanar,which reduces the effectiveness of the resonance effect.$(iii)$ Fluorene: The two benzene rings are locked in a coplanar conformation by the methylene bridge. This allows for maximum resonance stabilization and electron donation between the rings,making it the most reactive among the three.Therefore,the decreasing order of reactivity is: $iii > ii > i$.

$I. -OH$	$II. -CN$
$III. -CO_2H$	$IV. -OCH_3$
$V. -NHCOCH_3$	$VI. -CHO$

Arrange the following in decreasing order of the rate of electrophilic aromatic substitution:

Similar Questions

Give the name and structure of the aromatic compound with the molecular formula $C_{10}H_8$. Also,determine the number of $\sigma$ and $\pi$ bonds present in the structure.

Identify the ortho and para directing groups towards aromatic electrophilic substitution reactions from the following list:

$I. -OH$ $II. -CN$

$III. -CO_2H$ $IV. -OCH_3$

$V. -NHCOCH_3$ $VI. -CHO$

Choose the best method to prepare the given compound:

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

Are the given structures representations of $Benzene$?

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