A) \[C{{H}_{3}}-C\equiv CH\]
B) \[\underset{OH}{\mathop{\underset{|}{\mathop{C{{H}_{3}}-C=C{{H}_{2}}}}\,}}\,\]
C) \[\underset{OH}{\mathop{\underset{|}{\mathop{C{{H}_{3}}-CH=\underset{OH}{\mathop{\underset{|}{\mathop{C{{H}_{2}}}}\,}}\,}}\,}}\,\]
D) \[\underset{N{{H}_{2}}N{{H}_{2}}}{\mathop{\underset{||}{\mathop{C{{H}_{3}}-CH-C{{H}_{2}}}}\,}}\,\]
Correct Answer: A
Solution :
Excess of silica is absorbed by basic lining of the converter and part of cuprous sulphide is oxidised which combines with remaining cuprous sulphide to form free copper metal. \[A{{B}_{2}}\] \[1.0M\text{ }HCl\]
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