1. Tardigrade
  2. Question
  3. Chemistry
  4. An unknown compound A has a molecular formula C4H6, when A is treated with an excess of Br2, a new substance B with formula C4H6Br4 is formed. A forms a white precipitate with ammoniacal silver nitrate solution. A may be

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Q. An unknown compound A has a molecular formula $ {{C}_{4}}{{H}_{6}}, $ when A is treated with an excess of $ B{{r}_{2}}, $ a new substance B with formula $ {{C}_{4}}{{H}_{6}}B{{r}_{4}} $ is formed. A forms a white precipitate with ammoniacal silver nitrate solution. A may be

AMUAMU 1998

Solution:

: Since A $ ({{C}_{4}}{{H}_{6}}) $ adds two moles of $ B{{r}_{2}} $ to give B $ ({{C}_{4}}{{H}_{6}}B{{r}_{4}}) $ therefore, it is a butyne. Further since A forms a white ppt. with ammoniacal $ AgN{{O}_{3}} $ solution therefore, A is 1-butyne. $ C{{H}_{3}}C{{H}_{2}}C\equiv CH+AgN{{O}_{3}}+N{{H}_{4}}OH\xrightarrow[{}]{{}} $ $ \underset{(white\text{ }ppt.)}{\mathop{C{{H}_{3}}C{{H}_{2}}C}}\,\equiv CAg+N{{H}_{4}}N{{O}_{3}}+{{H}_{2}}O $