Question

The number of non-ionisable protons present in the product $B$ obtained from the following reactions is .
$C_2{H}_{5}OH+PC{l}_3\to C_2{H}_{5}Cl+A$
$A+PC{l}_3\to B$

• A. 2.0
• B. 2
• C. 2.00
• D. (02.00)

Answer 1

Answer: (A), (B), (C), (D)

$PC{l}_3+C_2{H}_{5}OH\to C_2{H}_{5}Cl+H_{3}P{O}_3$
$\left(A\right)$

$H_{3}P{O}_3+PC{l}_3\to H_4{P}_2{O}_5$
$\left(A\right)$ $\left(B\right)$
Structure of $H_4{P}_2{O}_5$


Total $2$ non-ionizable protons are present