Draw the product(s) of the following reaction
CH3-CH2-C(triple bond)C-CH2 ---------> ?
(Na/NH3)
To determine the product(s) of the given reaction, we need to understand the reaction conditions. Na/NH3 refers to sodium amide dissolved in liquid ammonia.
In this reaction, the sodium amide (Na/NH3) acts as a strong base and performs a process known as deprotonation. Since the reactant molecule in this case is an alkyne, the base will remove a proton from the terminal carbon, leading to the formation of a new carbon-carbon double bond.
Let's break down the reaction step by step:
1. Start with the alkyne molecule: CH3-CH2-C(triple bond)C-CH2
2. The sodium amide (Na/NH3) acts as a strong base, abstracting a proton from the terminal carbon of the alkyne. This removal of a proton forms an anion (C≡C-) that has a negative charge on the terminal carbon.
3. With the formation of the anion, the carbon-carbon triple bond is converted into a carbon-carbon double bond:
CH3-CH2-C=C-CH2(-)
Therefore, the product of the reaction is CH3-CH2-C=C-CH2(-).