# Does a non-terminal alkyne react with sodamide?

Hydrocarbon (A), $\ce{C6H10}$, on treatment with $\ce{H2/Ni}$, $\ce{H2}$/Lindlar catalyst and $\ce{Na}$/liquid $\ce{NH3}$ forms three different reduction products (B),(C) and (D) respectively. (A) does not form any salt with ammoniacal $\ce{AgNO3}$ solution but forms a salt (E) on heating with $\ce{NaNH2}$ in an inert solvent. Compund (E) reacts with $\ce{CH3I}$ to give (F). Compound (D) on oxidative ozonolysis gives n-butanoic acid along with other other products. Give structures of (A) to (F) with proper reasoning.

From the formula of (A) and the details given, it is clear that (A) is an alkyne. I could make out that (B) would be an alkane while (C) and (D) would be cis and trans alkene respectively. Now, as (A) does not react with ammoniacal $\ce{AgNO3}$ solution, it is not a terminal alkyne. But, then I got confused as the question told, (A) reacts with $\ce{NaNH2}$ to give a salt. Till now, I knew that only terminal alkynes react with $\ce{NaNH2}$. What is it that am I missing?

• It is a part of the assignments that I had received, so i don't know as to which book it was but a quick google search, and I found this link : link – Rajdeep Dutta Feb 19 '18 at 12:48
• Use of a strong base at elevated temperature allows for alkyne-1,2diene equilibrium. Among other things, it allows migration of triple bonds into terminal position. – permeakra Feb 19 '18 at 12:49
• @permeakra That means one can use $\ce{NaNH2}$ to convert an internal alkyne into a terminal alkyne ? – Rajdeep Dutta Feb 19 '18 at 12:51
• @RajdeepDutta and vice versa, it's an equilibrium. The product is guarded by thermodynamics. – permeakra Feb 19 '18 at 13:04
• The ozonolysis question is a duplicate: chemistry.stackexchange.com/questions/90256/ozonolysis-reaction/… – Waylander Feb 19 '18 at 15:36

By the given question conditions you should easily guess that $(A)$ is $\ce {Hex-2-yne}$. When it is reacted with $\ce{NaNH_2}$ in presence of heat it can isomerise to terminal alkyne by following this mechanism and then $\ce{CH_3I}$ addition will simply follow an $\ce{S_N2}$ mechanism.