Question

The products for the following reaction will be:

$3C_2{H}_{5}OH+PC{l}_3\stackrel{\text{Pyridine}}{\to }\text{Products}$

• A. $3C{H}_{3}OH+PCl$
• B. $3C_2{H}_{5}Cl+H_{3}P{O}_3$
• C. $C_6{H}_{14}+H_{3}P{O}_3$
• D. None of the above

Answer 1

The correct option is B $3C_2{H}_{5}Cl+H_{3}P{O}_3$

Reaction of Alcohol with $PC{l}_3$:
$3R-OH+P{X}_3\stackrel{\text{Pyridine}}{\to }3R-X+H_{3}P{O}_3$
Here,
$X=Cl,Br$
Steps Involved in the Mechanism
Protonation of the alcohol
Attack of the nucleophile
$C{l}^{-}+C{H}_3-C{H}_2-OPC{l}_2\to C{H}_{3}C{H}_{2}Cl{+}^{-}OPC{l}_2$
The same mechanism will be followed by the rest 2 molecules of alcohol.
So,
$3C_2{H}_{5}OH+PC{l}_3\stackrel{\text{Pyridine}}{\to }3C_2{H}_{5}Cl+H_{3}P{O}_3$