Write the formulas of the products expected to form in the following situations. If no reaction occurs, write N.R. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(\mathrm{aq})+\mathrm{HBr}(\mathrm{aq}) \longrightarrow\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \longrightarrow\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow\)

Short Answer

Expert verified
\(a) CH3CH2NH3^+ + Cl^- \) \(b) (CH3)3NH^+ + Br^- \) \(c) No reaction \) \(d) CH3CH2NH2 + H2O \)

Step by step solution

01

- Reaction of Amine with Aqueous Acid

First we look at the reaction of the amine group with the aqueous solution of hydrochloric acid. Here, the nitrogen atom in the ethylamine (CH3CH2NH2) donates an electron pair to form a bond with a proton (H+) from the hydrochloric acid (HCl). This leads to formation of an ethylammonium ion (CH3CH2NH3+) and chloride ion (Cl-) as the products.
02

- Reaction of tri-methyl-amine with Aqueous Acid

For the reaction between trimethylamine ((CH3)3N) and hydrobromic acid (HBr), a similar process happens. The nitrogen atom in the trimethylamine accepts a proton (H+) from the hydrobromic acid to form trimethylammonium ion ((CH3)3NH+) and bromide ion (Br−).
03

- Interaction with Hydronium Ions

On basis of our understanding of acid–base reactions, the CH3CH2NH3+ and H3O+ don’t react because they are both ions formed from acids. Therefore, we conclude there is 'No Reaction' happening in this case.
04

- Reaction with Hydroxide Ions

With the reaction between CH3CH2NH3+ and OH-, the hydroxide ions (OH−) from water strip off the H+ ion originally added by the acid from the ammonium ion changing it back to the original amine CH3CH2NH2. Thus we get ethylamine and water (H2O) as products.

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