A) \[{{C}_{2}}{{H}_{4}}\]
B) \[{{C}_{3}}{{H}_{6}}\]
C) \[{{C}_{3}}{{H}_{3}}\]
D) \[{{C}_{3}}{{H}_{4}}\]
Correct Answer: D
Solution :
The successive members of a homologous series differ by \[aC{{H}_{2}}\]group or by \[12+2\times 1=14\]mass unit. Hence, the homologue of\[HC\equiv CH\](ethyne, \[{{C}_{2}}{{H}_{2}})\]is\[C{{H}_{3}}-C\equiv CH\](propyne,\[{{C}_{3}}{{H}_{4}}\]).
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