| (i) \[C{{H}_{3}}-Mg\,Br+C{{H}_{3}}-C\equiv CH\] |
(ii)  |
| (iii) \[C{{H}_{3}}-MgBr+C{{H}_{3}}OH\] |
(iv)  |
A) (i),(ii) & (iii)
B) (i), (ii),(iii) & (iv)
C) (iii) & (iv)
D) (iii) & (i),(iv)
Correct Answer: B
Solution :
All of these reaction gives \[C{{H}_{4}}\]
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