JEE Mains · Chemistry · STD 12 - p -Block elements - ll
The number of non-ionisable protons present in the product \(B\) obtained from the following reaction is \(\text { . } C _{2} H _{5} OH + PCl _{3} \rightarrow C _{2} H _{5} Cl + A\) \(A + PCl _{3} \rightarrow B\)
- A \(9\)
- B \(2\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(C _{2} H _{5} OH + PCl _{3} \longrightarrow C _{2} H _{5} Cl + H _{3} PO _{3}\) \(H _{3} PO _{3}+ PCl _{3} \longrightarrow H _{4} P _{2} O _{5}+ HCl\)
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