$1)$ Complete the following acid-base reaction and name the product:

$A)$ $C H_{3} C H_{2} C H_{2} N H_{2} + H C l \to$

$B)$ $(C_{2} H_{5})_{3} N + H C l \to$

Open in App

Solution

$A)$ Reaction of propyl amine with $H C l$

This reaction is an acid-base reaction, so salt is formed.

In this reaction $C H_{3} C H_{2} C H_{2} N H_{2}$ act as base due to availability of lone pair on nitrogen and $H C l$ as acid, so the $N$ will accept $H^{+}$ from $H C l$ and form a salt. Acid-base reaction of n-propyl amine with $H C l$ is shown below.

$C H_{3} C H_{2} C H_{2} N H_{2} H C l - - \to C H_{3} C H_{2} C H_{3} N H_{3}^{+} C l^{-}$
$n-propylamine$ $n-propylammonium chloride$

Hence, the product obtained in the given reaction is $n-propylammonium chloride$

$B)$ Reaction of triethyl amine with $H C l$

This reaction is an acid-base reaction, so salt is formed. In this reaction $(C_{2} H_{5})_{3} N$ act as base due to availability of lone pair of electron on nitrogen and $H C l$ acts asan acid, so the $N$ will accept $H^{+}$ from $H C l$ and form a salt. Acid-base reaction of triethyl amine with $H C l$ is shown below.

$(C_{2} H_{5})_{3} N H C l - - \to (C_{2} H_{5})_{3} N H^{+} C l^{-}$
$Triethylamine$ $Triethylammonium chloride$

Hence, the product obtained in the given reaction is $Triethylammonium chloride$

Suggest Corrections

Similar questions

H I, H B r

H C l

H C l, H B r

H I i s 430.5 k J m o l^{- 1}

363.7 k J m o l^{- 1}

296.8 k J m o l^{- 1}

C H_{3} C H_{2} C H_{2} N H_{2} + H C l \to

(C_{2} H_{5}) N + H C l \to

Name the Acid and Base used to form the following salts:

(a)NaHCO₃

(b)ZnSO₄

(c)AlCl₃

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program