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Q.
When $ PC{{l}_{5}} $ reacts with sulphuric acid, sulphuryl chloride $ (S{{O}_{2}}C{{l}_{2}}) $ is formed as the final product. This shows that sulphuric acid:
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Solution:
$ \overset{O}{\mathop{\overset{||}{\mathop{\underset{O}{\mathop{\underset{||}{\mathop{PC{{l}_{5}}+HO-S-OH\xrightarrow{{}}}}\,}}\,}}\,}}\, $ $ Cl-\underset{O}{\mathop{\underset{||}{\mathop{\overset{O}{\mathop{\overset{||}{\mathop{S}}\,}}\,}}\,}}\,-Cl+POC{{l}_{3}}+{{H}_{2}}O $ $ PC{{l}_{5}} $ attacks $ -OH $ group and replace it by $ -Cl $ group. Hence, reaction of $ PC{{l}_{5}} $ with $ {{H}_{2}}S{{O}_{4}} $ shows the presence of 2 $ -OH $ groups in $ {{H}_{2}}S{{O}_{4}} $ .