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Answer c is correct because c have 3 alpha hydrogen while d have only 2.


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The thermodynamics are always the same. If you go from carbon monoxide to methane, there is only one final enthalpy, and if you have the choice of going to methane or methanol the thermodynamically preferred product is always the same. The difference indeed less within the different pathways and the activation barriers needed to get from one intermediate to ...


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The terminal alkyne has acidic $\ce{H}$ and therefore will give acid base reaction with $\ce{NaNH3}$ and go $\ce{NH4+}$ and $\ce{RC#C- Na+}.$


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THF is a polar aprotic solvent, that is, it can stabilize positive charges but not negative ones since it won't form hydrogen bonds with the help of any of its own hydrogen atoms. Thus the carbocationic intermediate, that is the mercurinium ion is still stabilized, allowing the reaction to proceed. And in Hydroboration, THF acts as the solvent as well but ...


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