A) \[C{{H}_{4}}\]
B) \[{{C}_{2}}{{H}_{2}}\]
C) \[{{C}_{2}}{{H}_{6}}\]
D) \[C{{O}_{2}}\]
Correct Answer: A
Solution :
\[A{{l}_{4}}{{C}_{3}}\]is aluminium carbide, which upon hydrolysis gives \[C{{H}_{4}}\] \[A{{l}_{4}}C{{l}_{3}}+2{{H}_{2}}O\xrightarrow{{}}4Al{{(OH)}_{3}}+3C{{H}_{4}}\] \[{{C}_{2}}{{H}_{2}}\]is formed on hydrolysis of \[Ca{{C}_{2}}\] (calcium carbide) \[Ca{{C}_{2}}+2{{H}_{2}}O\xrightarrow{{}}Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}\] \[{{C}_{2}}{{H}_{6}}\]can be prepared as follows (a) \[C{{H}_{3}}X+Na\xrightarrow[Wurtz\,reaction]{dry\,ether}{{C}_{2}}{{H}_{6}}+Na\,X\] (b) \[C{{H}_{3}}COOK+2HOH\xrightarrow{Electrolysis}{{C}_{2}}{{H}_{6}}\] \[+2C{{O}_{2}}+{{H}_{2}}+2KOH\] (c) \[{{C}_{2}}{{H}_{5}}Br+Mg\xrightarrow{dry\,ether}{{C}_{2}}{{H}_{5}}MgBr\] \[\xrightarrow{HOH}{{C}_{2}}{{H}_{6}}+MgBr(OH)\]
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