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Hydrogen is the most abundant element in the universe, making up ~75% of all matter.
In the reaction H − C ≡ H → 2. C H 3 C H 2 B r 1. N a N H 2 / l i q . N H 3 X → 2. C H 3 C H 2 B r 1. N a N H 2 / l i q . N H 3 Y . H-C \equiv H \xrightarrow[2. CH_3 CH_2Br]{1. NaNH_2/liq. NH_3} X \xrightarrow[2. CH_3CH_2Br]{1. NaNH_2/liq. NH_3} Y. H − C ≡ H 1. N a N H 2 / l i q . N H 3 2. C H 3 C H 2 B r X 1. N a N H 2 / l i q . N H 3 2. C H 3 C H 2 B r Y . X and Y are
medium
Hydrocarbons
2016
chemistry
X = 1-Butyne; Y = 3-Hexyne
X = 2-Butyne; Y = 3-Hexyne
X = 2-Butyne; Y = 2-Hexyne
X = 1-Butyne; Y = 2-Hexyne
Explanation To solve this problem, we need to understand the reaction sequence given: \\
1. Start with acetylene ( H − C ≡ C − H ) . (H-C \equiv C-H). \\
( H − C ≡ C − H ) . 2. The first step involves the use of N a N H 2 \mathrm{NaNH_2} NaN H 2
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in liquid ammonia, which is a strong base.
This base will deprotonate the terminal hydrogen of acetylene, forming an acetylide ion:
H − C ≡ C − H-C \equiv C^- H − C ≡ C − (acetylide ion).
3. The acetylide ion then reacts with
C H 3 C H 2 B r \mathrm{CH_3CH_2Br} C H 3 C H 2 Br (ethyl bromide) in an
reaction.
This results in the formation of a new carbon-carbon bond:
H − C ≡ C − C H 2 C H 3 H-C \equiv C-CH_2CH_3 H − C ≡ C − C H 2 C H 3 (1-butyne).
4. The compound formed, 1-butyne, is denoted as
5. The second part of the reaction sequence repeats the same steps:
a. Deprotonation of the terminal hydrogen of 1-butyne using
N a N H 2 \mathrm{NaNH_2} NaN H 2 in liquid ammonia,
forming a new acetylide ion:
H − C ≡ C − C H 2 C H 2 − H-C \equiv C-CH_2CH_2^- H − C ≡ C − C H 2 C H 2 − .
b. Reaction of this acetylide ion with another equivalent of
C H 3 C H 2 B r \mathrm{CH_3CH_2Br} C H 3 C H 2 Br (ethyl bromide).
This results in the formation of a new carbon-carbon bond:
H − C ≡ C − C H 2 C H 2 C H 2 C H 3 H-C \equiv C-CH_2CH_2CH_2CH_3 H − C ≡ C − C H 2 C H 2 C H 2 C H 3 (3-hexyne).
6. The compound formed, 3-hexyne, is denoted as
Therefore, the correct answer is:
Option 1:
1-Butyne;
3-Hexyne.