119 mL of 4.0 M H2SO4 with 229 mL of H2O
1 answer:
You might be interested in
Answer:
I
Explanation:
The complete question can be seen in the image attached.
We need to understand what is actually going on here. In the first step that yields product A, the sodamide in liquid ammonia attacks the alkyne and abstracts the acidic hydrogen of the alkyne. The second step is a nucleophilic attack of the C6H5C≡C^- on the alkyl halide to yield product B (C6H5C≡C-CH3CH2).
Partial reduction of B using the Lindlar catalyst leads to syn addition of hydrogen to yield structure I as the product C.
