Final answer:
Sodium amide (NaNH2) is the strongest base among the provided options and is capable of completely deprotonating 1-octyne due to its high affinity for protons. Option d.
Step-by-step explanation:
To deprotonate 1-octyne, a strong base is required that can remove an H+ ion from the acetylenic hydrogen, which has a pKa of around 25. This requires a base with a greater affinity for protons. Among the given options, NaNH2 (sodium amide) is the strongest base capable of deprotonating 1-octyne.
Sodium amide exists as an amide anion (NH2-) in solution, which is strong enough to deprotonate alkyne molecules.
Comparing the other options:
NaF - It is a weak base and cannot deprotonate 1-octyne effectively.
NaOH - While it is a strong base, it is not strong enough to fully deprotonate 1-octyne.
NH3 - Ammonia is a weak base and is unable to deprotonate 1-octyne.
CH3NH2 - Methylamine, like ammonia, is a weak base and also cannot deprotonate 1-octyne.
Therefore, option d. NaNH2 is correct as it is strong enough to completely remove an H+ from 1-octyne, resulting in an anion with a pKa of 38. The reaction between a strong base such as NaNH2 and 1-octyne is represented by:
R-C≡CH + NH2− → R-C≡C− + NH3
Option d.