
Which one of the following compounds forms a quaternary salt on reacting with an excess of methyl iodide?
(A) ${{C}_{2}}{{H}_{5}}OC{{H}_{3}}$
(B) ${{(C{{H}_{3}})}_{2}}CHO{{C}_{2}}{{H}_{5}}$
(C) ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$
(D) ${{C}_{6}}{{H}_{5}}N{{O}_{2}}$
Answer
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Hint: An important class of organic compound derived by replacing one or more hydrogen atoms of $N{{H}_{3}}$ a molecule with alkyl or aryl groups are amines. The amines are named as alkamines according to IUPAC nomenclature. $-N{{H}_{2}}$ the functional group attached to alkyl groups are named as alkyl amines and the same attached to the benzene ring named as arylamines.
Complete step by step solution:
Regarding the derivatives of ammonia in which one, two, or three hydrogen atoms are replaced by alkyl or aryl group formed amines.
If one H atom replaces with alkyl group in ammonia forms primary amine; $RN{{H}_{2}}({{1}^{0}})$
If two H atoms replaced with alkyl groups in ammonia forms secondary amine; ${{R}_{2}}NH({{2}^{0}})$
If three H atoms replaced with alkyl groups in ammonia forms tertiary amine; ${{R}_{3}}N({{3}^{0}})$
One lone pair of electrons of nitrogen atom contains primary amine. So, primary amine acts as a strong nucleophile, and hydrogen atoms on nitrogen are easily replaceable. Hence, primary amines are easily reacted with alkyl halides due to its nucleophilic character. If excess alkyl halides react with a primary amine, one amine molecule can rapidly react with many alkyl halides and result in the formation of a quaternary ammonium salt.
If an excess of methyl iodide reacts with a primary amine $RN{{H}_{2}}$ , the reaction follows:
$RN{{H}_{2}}+C{{H}_{3}}I\to {{[RN{{(C{{H}_{3}})}_{3}}]}^{+}}$ (Quaternary ammonium salt)
Hence, primary amines are reacting with excess methyl iodide to form Quaternary ammonium salts.
Given organic compounds, ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$ only belongs to a primary amine.
The correct answer is option C.
Note: The formation of Quaternary ammonium salts from primary amines reacting with an excess of alkyl halides is known as ammonolysis of alkyl halides. The disadvantage of ammonolysis yielding a mixture of a primary, secondary, tertiary, and quaternary ammonium salt. However, primary amine is obtained as a major product by taking large excess of $N{{H}_{3}}$ , and the order of reactivity of halides with amines is RI>RBr>RCl.
Complete step by step solution:
Regarding the derivatives of ammonia in which one, two, or three hydrogen atoms are replaced by alkyl or aryl group formed amines.
If one H atom replaces with alkyl group in ammonia forms primary amine; $RN{{H}_{2}}({{1}^{0}})$
If two H atoms replaced with alkyl groups in ammonia forms secondary amine; ${{R}_{2}}NH({{2}^{0}})$
If three H atoms replaced with alkyl groups in ammonia forms tertiary amine; ${{R}_{3}}N({{3}^{0}})$
One lone pair of electrons of nitrogen atom contains primary amine. So, primary amine acts as a strong nucleophile, and hydrogen atoms on nitrogen are easily replaceable. Hence, primary amines are easily reacted with alkyl halides due to its nucleophilic character. If excess alkyl halides react with a primary amine, one amine molecule can rapidly react with many alkyl halides and result in the formation of a quaternary ammonium salt.
If an excess of methyl iodide reacts with a primary amine $RN{{H}_{2}}$ , the reaction follows:
$RN{{H}_{2}}+C{{H}_{3}}I\to {{[RN{{(C{{H}_{3}})}_{3}}]}^{+}}$ (Quaternary ammonium salt)
Hence, primary amines are reacting with excess methyl iodide to form Quaternary ammonium salts.
Given organic compounds, ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$ only belongs to a primary amine.
The correct answer is option C.
Note: The formation of Quaternary ammonium salts from primary amines reacting with an excess of alkyl halides is known as ammonolysis of alkyl halides. The disadvantage of ammonolysis yielding a mixture of a primary, secondary, tertiary, and quaternary ammonium salt. However, primary amine is obtained as a major product by taking large excess of $N{{H}_{3}}$ , and the order of reactivity of halides with amines is RI>RBr>RCl.
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