Question

In the following reaction,ether linkage is cleaved by ${PCI}_5$
$R-O-R^{\prime }+{PCI}_5⟶RCI+R^{\prime }CI+{POCI}_3$
If $C_5{H}_{12}O$ (ethre) forms 2-chloropropane as one of the products,then ether

• A. 1-ethoxypropane
• B. 2-ethoxypropane
• C. 1-methoxybutane
• D. 2-methoxypropane

Answer 1

The correct option is B 2-ethoxypropane

Option B is the correct answer.
The molecular formula of the given ether is $C_5{H}_{12}O$
If 2-chloropropane $[C{H}_3-CH(Cl)-C{H}_3]$ is one of the products, then we are left with $C_2{H}_5$.
Then, the ether molecule must be 2-ethoxypropane.