A) SiC
B) \[A{{l}_{4}}{{C}_{3}}\]
C) \[M{{g}_{2}}{{C}_{3}}\]
D) \[Ca{{C}_{2}}\]
Correct Answer: C
Solution :
\[M{{g}_{2}}{{C}_{3}}\] has \[{{(}^{-}}C\equiv C-{{C}^{3-}})\] ion and gives propyne on reaction with \[{{H}_{2}}O\]. \[M{{g}_{2}}{{C}_{3}}+4{{H}_{2}}O\xrightarrow{{}}2Mg{{(OH)}_{2}}+\underset{propyne}{\mathop{C{{H}_{3}}C\equiv CH}}\,\]
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