Question

Which reactions will produce the desired product in good yield? You may assume that aluminum chloride is added as a catalyst in each case. For the reactions that will not give a good yield of the desired product, predict the major products.

Reagents Desired Product

(a) benzene + n-butyl bromide n-butylbenzene

(b) ethylbenzene + tert-butyl chloride p-ethyl-tert-butylbenzene

(c) bromobenzene + ethyl chloride p-bromoethylbenzene

(d) benzamide (PhCONH₃) + CH₃ CH₂ CI p-ethylbenzamide

(e) toluene + HNO₃, H₂SO₄ , heat 2,4,6-trinitrotoluene (TNT)

Short answer

⦁ Benzene reacted with n butyl bromide in the presence of Lewis acid AICI₃ , showing Friedel– Crafts alkylation reaction, thus giving the desired product n-butylbenzene as a major product.

Formation of n-butylbenzene

⦁ tert-butyl chloride in the presence of Lewis acid AICI₃ , making stable tertiary carbocation (electrophile). Ethylbenzene reacting with tert-butyl chloride (tertiary carbocation) shows Friedal-craft alkylation reaction giving desired product p-ethyl-tert-butylbenzene as a major product.

Formation of p-ethyl-tert-butylbenzene

⦁ Benzene reacted with ethyl chloride in the presence of Lewis acid , showing Friedel– Crafts alkylation reaction, thus giving the desired product p-bromoethylbenzene as a major product and, along with it, ortho product also formed.

Formation of p-bromoethylbenzene and o-bromoethylbenzene

⦁ Nitrating reagent nitric acid (HNO₃ ) and concentrated sulphuric acid (H₂SO₄) formed the nitronium ion as the electrophile, which reacts with toluene to form the ortho para nitrotoluene isomer because of the ring activating alkyl group present in the toluene.

If the reaction is heated, it forms the dinitrotoluene and finally gives the desired product 2,4,6-trinitrotoluene (TNT) as the major product.

Formation of 2,4,6-trinitrotoluene

Step-by-step solution

Reaction producing the desired product.

⦁ Benzene reacted with n butyl bromide in the presence of Lewis acid AICI₃ , showing Friedel– Crafts alkylation reaction, thus giving the desired product n-butylbenzene as a major product.

Formation of n-butylbenzene

⦁ tert-butyl chloride in the presence of Lewis acid AICI₃ , making stable tertiary carbocation (electrophile). Ethylbenzene reacting with tert-butyl chloride (tertiary carbocation) shows Friedal-craft alkylation reaction giving desired product p-ethyl-tert-butylbenzene as a major product.

Formation of p-ethyl-tert-butylbenzene

⦁ Benzene reacted with ethyl chloride in the presence of Lewis acid , showing Friedel– Crafts alkylation reaction, thus giving the desired product p-bromoethylbenzene as a major product and, along with it, ortho product also formed.

Formation of p-bromoethylbenzene and o-bromoethylbenzene

⦁ Nitrating reagent nitric acid (HNO₃ ) and concentrated sulphuric acid (H₂SO₄) formed the nitronium ion as the electrophile, which reacts with toluene to form the ortho para nitrotoluene isomer because of the ring activating alkyl group present in the toluene.

If the reaction is heated, it forms the dinitrotoluene and finally gives the desired product 2,4,6-trinitrotoluene (TNT) as the major product.

Formation of 2,4,6-trinitrotoluene

Reaction not producing the desired product.

⦁ Benzamide has the amide functional group (-CONH₂ ), the strong deactivating group, as it deactivates the ring toward electrophile aromatic substitution.

Ethyl chloride reacted with a Lewis acid (AICI₃ )form the primary carbocation (electrophile), but due to deactivating group present in the benzamide, it will not show Friedel– Crafts alkylation reaction when reacted with an electrophile (CH₃C⁺H₂) ; thus, the desired product p-ethyl benzamide will not be formed.

Benzamide not showing Friedel– Crafts alkylation reaction