He structure of the product, C, of the following sequence of reactions would be:
1 answer:
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.
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Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in
<u>= 4.75 × 10²³ molecules of Propane</u>.
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No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =
<u>= 34.72 g</u>