Does a non-terminal alkyne react with sodamide?


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> 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.
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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?




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.


[![Mechanism](https://i.stack.imgur.com/dKoSO.jpg)](https://i.stack.imgur.com/dKoSO.jpg)


Secondly,Ozonolysis will yield an aldehyde which can get oxidised to acid in oxidative medium.


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