Consider the tetrahedral intermediate and the two leaving groups, methoxide and amide anion. does this reaction favor reactants
1 answer:
The reaction of tetrahedral intermediate and the two leaving groups, methoxide and amide anion favour product.
<h3>What is leaving group? </h3>A leaving group is a group of atoms or an atom which is able to break away from a molecule as a stable species or with a lone pair, breaking the bond between the molecule and itself.
<h3>What is intermediate? </h3>A stage is come before forming product. This stage is termed as transition stage and the reactant at this stage is termed as intermediate.
According to Le Châtelier's Principle, a reactions always tend towards the equilibrium, the reaction produces more products from the excess reactant, that's why it causing the system to shift to the left which allows the system to reach equilibrium.
Due to this reason, this reaction shift towards product.
Thus, we concluded that the reaction of tetrahedral intermediate and the two leaving groups, methoxide and amide anion favour product.
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Answer : The molar concentration of sucrose in the tea is, 0.0549 M
Explanation : Given,
Mass of sucrose = 3.765 g
Volume of solution = 0.200 L
Molar mass of sucrose = 342.3 g/mole
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
Now put all the given values in this formula, we get:
Therefore, the molar concentration of sucrose in the tea is, 0.0549 M