If NaH is taken in excess then the product of the following reaction is?
HC≡C-CH2-CH2-Br
HC≡C-CH2-CH2-OH
HC≡C-CH2-CH2-C≡C-CH2-CH2-OH
All of these are formed
HC≡CH→ExcessNaH -C≡C- → HO-CH2-CH2-BrHC≡C-CH2-CH2-OH