Question

Which of the following will not increase the acidic nature of $B(OH{)}_3$?

• A. Glycerol
• B. Mannitol
• C. Ethanol
• D. Both (a) and (b)

Answer 1

The correct option is C Ethanol

When certain organic polyhydroxy compounds such as glycerol, mannitol or sugars are added, $B(OH{)}_3$ behaves as a strong monobasic acid.
Here, cis-diol forms a very stable complex with $\left[B\right(OH{)}_4{]}^{–}$, removing it from the solution.


Thus, the removal of one of the products shifts the equilibrium in the forward direction.
$B(OH{)}_3+2H_{2}O\rightleftharpoons B(OH{)}_4^{-}+H_3{O}^{+}$

Hence, it makes the $B(OH{)}_3$ strong monobasic acid.
Ethanol is not a diol so it does not form complex with $\left[B\right(OH{)}_4{]}^{–}$.

THEORY
Properties of orthoboric acid

1. White crystalline solid with soapy touch.
2. Sparingly soluble in water but highly soluble in hot water.
3. It has a layer structure in which planar $B{O}_3$ units are joined by hydrogen bonds.
4. Boric acid is a weak monobasic acid.
5. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion.

$B(OH{)}_3+2H_{2}O\to B(OH{)}_4^{-}+H_3{O}^{+}$
If certain organic polyhydroxy compounds such as glycerol, mannitol, or sugars are added, $B(OH{)}_3$ behaves as a strong monobasic acid.
The cis-diol forms a very stable complex with $\left[B\right(OH{)}_4{]}^{–}$, removing it from the solution. The reaction is reversible. Thus, the removal of one of the products shifts the equilibrium in the forward direction.
$B(OH{)}_3+2H_{2}O\rightleftharpoons B(OH{)}_4+H_3{O}^{+}$

Ethanol does not form similar complexes, but catechol, salicylic acids, and mannitol form similar complexes.



Polymeric metaborate species are formed at higher concentrations.
$3B(OH{)}_3\rightleftharpoons H_3{O}^{+}+[B_3{O}_3(OH{)}_4{]}^{–}+H_{2}O$

On heating, orthoboric acid above 370K forms metaboric acid, $HB{O}_2$ which on further heating yields boric oxide, $B_2{O}_3$.
$H_{3}B{O}_3\stackrel{△}{\to }HB{O}_2\stackrel{△}{\to }{B}_2{O}_3$

Test for Borate radical:
When $H_{3}B{O}_3$ is heated with $C_2{H}_{5}OH$, the evolved gas is burned to form a green-edged flame.
$H_{3}B{O}_3+3C_2{H}_{5}OH\to \underset{\text{ethyl borate}}{B(O{C}_2{H}_5{)}_3}+3H_{2}O$