Question

Friedel-Craft’s reaction using $MeCl$ and anhydrous $AlC{l_3}$ will take place most efficiently with:
(A) benzene
(B) nitrobenzene
(C) acetophenone
(D) toluene

Answer 1

Hint: Friedel-Craft’s is an electrophilic aromatic substitution reaction in which a carbocation attacks on an aromatic ring.
There are two primary friedel-Crafts reactions i.e., alkylation and acylation.
Example:

Complete step by step answer:
Let us write Friedel-Crafts reactions one by one with the given option.
Diagram:
It takes place at a slower rate.
Friedel-Crafts reaction with acetophenone takes place at high pressure and not possible.

With nitrobenzene it is not possible because electrons are a withdrawing group present in the benzene ring.

With toluene it takes place at a faster rate due to hyper conjugation.
Hyper conjugation is interaction of electrons in $\sigma $-bond (usually C-H or C-C) with an adjacent empty or partially filled P-orbital or $\pi $-orbital.
This increases the stability of molecules.
This is also known as no-bond.
This is a permanent effect.

Hyperconjugation stabilizes carbocation as it helps in dispersal of positive charges.
As the number of $\alpha - H - $atoms increases. The relative stability on the basis of hyperconjugation increase

Therefore, form the above explanation the correct option toluene.

So, the correct answer is “Option D”.

Note:
Friedel-Crafts reactions always occur in the presence of anhydrous $AlC{l_3}. $Anhydrous $AlC{l_3}$ converts nucleophile and electrophile.
In Aromatic compounds only electrophilic substitution takes place.