Question

2-chloropropane can be prepared by different methods:
First method: $C{H}_3-\underset{OH}{\underset{|}{C}}H-C{H}_3\stackrel{KCl}{\to }$
Second method: $C{H}_3-CH=C{H}_2\stackrel{C{l}_2+H_{2}O}{\to }$
Third method: $C{H}_3-\underset{OH}{\underset{|}{C}}H-C{H}_3\underset{pyridine}{\overset{SOC{l}_2}{\to }}$

• A. All three methods are correct
• B. Only the third method is correct
• C. Only the second method is correct
• D. Both first and second methods are correct

Answer 1

The correct option is B Only the third method is correct

Third method:
$C{H}_3-\underset{OH}{\underset{|}{C}}H-C{H}_3\underset{pyridine}{\overset{SOC{l}_2}{\to }}C{H}_3-\underset{Cl}{\underset{|}{C}}H-C{H}_3+S{O}_2\left(g\right)\uparrow +HCl\left(g\right)\uparrow$
The by-products $S{O}_2\left(g\right)$ and $HCl\left(g\right)$ are gaseous and hence 2-chloropropane is easily separated.