Draw a mechanism for the reaction of methylamine with propanoic acid. In the box to the left, draw any necessary curved arrows.
Show the products of the reaction in the box to the right. Include any nonzero formal charges and all lone pairs of electrons. Finally, check the box to indicate which side of the reaction is favored at equilibrium.
1 answer:
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Answer: Option (d) is the correct answer.
Explanation:
According to Bronsted-Lowry, species which donate a proton are known as acid. The species which accept a proton are known as a base.
In the given reaction, acids and bases are as follows.
HI +
+
Acid Base Conjugate acid Conjugate base
Therefore, the acid HI loses a proton to form a conjugate base that is .
Thus, we can conclude that HI and is an acid conjugate base pair.