
On heating acetamide in presence of \[{{P}_{2}}{{O}_{5}}\] which of the following is formed?
(A) Ammonium acetate
(B) Acetonitrile
(C) \[N{{H}_{3}}\]
(D) Methylamines
Answer
164.4k+ views
Hint: Acetamide is an organic compound having the molecular formula \[C{{H}_{2}}CON{{H}_{2}}\]. The compound \[{{P}_{2}}{{O}_{5}}\] is a dehydrating agent. Acetamide reacts with \[{{P}_{2}}{{O}_{5}}\] to give an intermediate with a positive charge on the nitrogen of NH2 of acetamide and a negative charge on any one oxygen of \[{{P}_{2}}{{O}_{5}}\]which undergoes deprotonation to give the resultant compound.
Complete Step by Step Solution:
When Acetamide reacts with \[{{P}_{2}}{{O}_{5}}\], nitrogen in acetamide give sits lone pair in the bond (carbon and nitrogen bond) due to which oxygen of acetamide (double bonded with carbon) form bond with the phosphorus of \[{{P}_{2}}{{O}_{5}}\]. This result in a negative charge on one oxygen of \[{{P}_{2}}{{O}_{5}}\]and a positive charge on the nitrogen of\[N{{H}_{2}}\]in acetamide.
Now negatively charged oxygen of \[{{P}_{2}}{{O}_{5}}\]is protonated by one hydrogen of \[N{{H}_{2}}\]so that the nitrogen charge gets balanced such as
Now this intermediate undergo an elimination reaction followed by deprotonation (proton remove from\[{{N}^{+}}H\]to neutralise the charge on nitrogen and this hydrogen get attached to negatively charged oxygen atom in \[{{P}_{2}}{{O}_{5}}\]) to give acetonitrile compound such as
Thus, the correct option is B.
Note: Acetamide reacts with \[{{P}_{2}}{{O}_{5}}\]in three steps. In the first step, protonation of oxygen in \[{{P}_{2}}{{O}_{5}}\]takes place when hydrogen electron is attracted by nitrogen (\[N{{H}_{2}}\]) in acetamide and an intermediate forms. This intermediate undergoes an elimination reaction to separate anion and cation. In the last step, deprotonation takes place (release of the proton) to give nitrile compound.
Complete Step by Step Solution:
When Acetamide reacts with \[{{P}_{2}}{{O}_{5}}\], nitrogen in acetamide give sits lone pair in the bond (carbon and nitrogen bond) due to which oxygen of acetamide (double bonded with carbon) form bond with the phosphorus of \[{{P}_{2}}{{O}_{5}}\]. This result in a negative charge on one oxygen of \[{{P}_{2}}{{O}_{5}}\]and a positive charge on the nitrogen of\[N{{H}_{2}}\]in acetamide.
Now negatively charged oxygen of \[{{P}_{2}}{{O}_{5}}\]is protonated by one hydrogen of \[N{{H}_{2}}\]so that the nitrogen charge gets balanced such as
Now this intermediate undergo an elimination reaction followed by deprotonation (proton remove from\[{{N}^{+}}H\]to neutralise the charge on nitrogen and this hydrogen get attached to negatively charged oxygen atom in \[{{P}_{2}}{{O}_{5}}\]) to give acetonitrile compound such as
Thus, the correct option is B.
Note: Acetamide reacts with \[{{P}_{2}}{{O}_{5}}\]in three steps. In the first step, protonation of oxygen in \[{{P}_{2}}{{O}_{5}}\]takes place when hydrogen electron is attracted by nitrogen (\[N{{H}_{2}}\]) in acetamide and an intermediate forms. This intermediate undergoes an elimination reaction to separate anion and cation. In the last step, deprotonation takes place (release of the proton) to give nitrile compound.
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