Alcohols and Esters: Question 9

Worked solution

The reaction of an alcohol with phosphorus(V) chloride

Phosphorus(V) chloride, $\text{PCl}_5$, reacts readily with the $-\text{OH}$ group of an alcohol at room temperature by substitution, replacing it with a chlorine atom:

$\text{CH}_3\text{CH}_2\text{OH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{POCl}_3 + \text{HCl}$

The hydrogen chloride gas produced is immediately visible as steamy white fumes, which is the standard qualitative test for the presence of an $-\text{OH}$ group in an unknown compound. The organic product is the halogenoalkane chloroethane, $\text{CH}_3\text{CH}_2\text{Cl}$.

Why the other options are wrong

• B is wrong because ethanol reacts vigorously with phosphorus(V) chloride at room temperature; it does not require heating.
• C correctly identifies the steamy white fumes but wrongly gives an alkene as the product. $\text{PCl}_5$ carries out substitution, not elimination, so no carbon–carbon double bond is formed.
• D is wrong on both counts: the gas evolved is acidic (hydrogen chloride, seen as steamy white fumes), not colourless and odourless, and $\text{PCl}_5$ does not oxidise the alcohol to a carboxylic acid.

Final answer

A. Steamy white fumes of hydrogen chloride gas are seen, and chloroethane, $\text{CH}_3\text{CH}_2\text{Cl}$, is formed.