Question

The reaction of ammonia with a large excess of ethyl chloride will give mainly :

• A. Ethylamine
• B. Tetraethylammonium chloride
• C. Triethylamine
• D. Diethylamine

Hint:

Excess Ammonia leads to a Primary Amine as the main product. Excess Alkyl Halide leads to a Quaternary Ammonium Salt.

Answer 1

The Correct Option is B

Step 1: Describe ammonolysis.
Ammonia ($NH_3$) acts as a nucleophile and attacks the electrophilic carbon of ethyl chloride ($C_2H_5Cl$) via $S_N2$ displacement, forming a primary ammonium salt $C_2H_5NH_3^+Cl^-$, which loses a proton to give the primary amine ethylamine ($C_2H_5NH_2$).
Step 2: Sequential alkylation continues.
The primary amine ($C_2H_5NH_2$) still has two N-H bonds and a lone pair, making it a nucleophile. It reacts with another molecule of $C_2H_5Cl$ to give diethylamine ($(C_2H_5)_2NH$), then triethylamine ($(C_2H_5)_3N$) in successive steps.
Step 3: Final step with excess alkyl halide.
With a large excess of ethyl chloride present, the tertiary amine reacts once more: the lone pair on nitrogen attacks the fourth $C_2H_5Cl$ molecule, forming the quaternary ammonium salt $(C_2H_5)_4N^+Cl^-$. At this stage nitrogen has no lone pair left for further reaction.
Step 4: Conclusion.
Exhaustive alkylation driven by the excess reagent gives tetraethylammonium chloride as the predominant product.
\[ \boxed{\text{Tetraethylammonium chloride}} \]