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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 \(\text { . } C _{2} H _{5} OH + PCl _{3} \rightarrow C _{2} H _{5} Cl + A\)  \(A + PCl _{3} \rightarrow B\)

  1. A \(9\)
  2. B \(2\)
  3. C \(4\)
  4. D \(3\)
Verified Solution

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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