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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Complete question:
A spirit burner used 1.00 g methanol to raise the temperature of 100.0 g water in a metal can from 28.00C to 58.0C. Calculate the heat of combustion of methanol in kJ/mol.
Answer:
the heat of combustion of the methanol is 402.31 kJ/mol
Explanation:
Given;
mass of water, = 100 g
initial temperature of water, t₁ = 28 ⁰C
final temperature of water, t₂ = 58 ⁰C
specific heat capacity of water = 4.184 J/g⁰C
reacting mass of the methanol, m = 1.00 g
molecular mass of methanol = 32.04 g/mol
number of moles = 1 / 32.04
= 0.0312 mol
Apply the principle of conservation of energy;
Therefore, the heat of combustion of the methanol is 402.31 kJ/mol