Question

Complete each acid-base reaction and predict whether the position of equilibrium lies toward the left or toward the right. (b) $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Na}^{+} \mathrm{NH}_{2}{ }^{-} \rightleftharpoons$ (c) $\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+}+\mathrm{CH}_{3} \mathrm{COH} \rightleftharpoons$

Short answer

(a) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{Na}^{+} \mathrm{NH}_{2}{ }^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+} + \mathrm{CH}_{3} \mathrm{COH}\) Answer: (a) The position of equilibrium lies towards the right side. (b) The position of equilibrium lies towards the right side.

Step-by-step solution

In this reaction, the acid is \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) and the base is \(\mathrm{Na}^{+} \mathrm{NH}_{2}{ }^{-}\). #Step 2: Carry out the acid-base reaction#

The base \(\mathrm{NH}_{2}^{-}\) will abstract a proton (H+) from the acid \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\), forming \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+}\) and \(\mathrm{NH}_{3}\). #Step 3: Write the complete balanced reaction#

\(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{Na}^{+} \mathrm{NH}_{2}{ }^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-} \mathrm{Na}^{+} + \mathrm{NH}_{3}\). #Step 4: Predict the position of the equilibrium#

As \(\mathrm{NH}_{2}{ }^{-}\) is a stronger base than the resulting \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{O}^{-}\), the equilibrium will likely be on the right side. (c) #Step 1: Identify the acid and base in the reaction#

In this reaction, the acid is \(\mathrm{CH}_{3} \mathrm{COH}\) and the base is \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+}\). #Step 2: Carry out the acid-base reaction#

The base \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-}\) will abstract a proton (H+) from the acid \(\mathrm{CH}_{3} \mathrm{COH}\), forming \(\mathrm{CH}_{3} \mathrm{CO}^{-} \mathrm{Na}^{+}\) and \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\). #Step 3: Write the complete balanced reaction#

\(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-} \mathrm{Na}^{+} + \mathrm{CH}_{3} \mathrm{COH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}^{-} \mathrm{Na}^{+} + \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\). #Step 4: Predict the position of the equilibrium#

As \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}^{-}\) is a stronger base than the resulting \(\mathrm{CH}_{3} \mathrm{CO}^{-}\), the equilibrium will likely be on the right side.