Question

Add curved arrows to the reactant side of the following SN2 reaction to indicate the flow of electrons. Draw the product species to show the balanced equation, including nonbonding electrons and formal charges.

Add curved arrows to the reactant side of the following SN2 reaction to indicate the flow of electrons. Draw the product species to show the balanced equation, including nonbonding electrons and formal charges.

Answer 1

Concepts and reason

Two compounds are given which are undergoing ${{\rm{S}}_{\rm{N}}}2$ reaction whose product needs to be written by using arrow mechanism. In ${{\rm{S}}_{\rm{N}}}2$ reaction nucleophile attacks the haloalkane and halogen leave as a leaving group. Now, mark the arrows showing the attack of nucleophile and departure of the leaving group.

Formal charge on the species can be calculated by knowing the valence electrons, bonding electrons, and nonbonding electrons.

Fundamentals

${{\rm{S}}_{\rm{N}}}2$ reaction is bimolecular nucleophilic substitution reaction which takes place between a haloalkane and a nucleophile. The reaction takes place in one step through a transition state. Consider the following scheme of a ${{\rm{S}}_{\rm{N}}}2$ reaction:

The formal charge can be calculated by the following formula:

${\rm{Formal}}\,{\rm{charge}} = V - N - \frac{B}{2}$

Here, V is valence electrons in neutral atom, N is the number of nonbonding electrons, and B the number of bonding electrons present in an atom.

Identify the given compounds as nucleophile and haloalkane then, show the movement of electron by using arrow as shown below:

Write the product and show the formal charge on the compounds formed.

Write the product of the given reaction with the correct formal charge on the species as shown below:

The formal charge on Br is calculated as shown below:

$\begin{array}{c}\\{\rm{Formal}}\,{\rm{charge on Br}} = 7 - 8 - \frac{0}{2}\\\\ = 7 - 8\\\\ = - 1\\\end{array}$

$\begin{array}{c}\\{\rm{Formal}}\,{\rm{charge on O}} = 6 - 4 - \frac{4}{2}\\\\ = 6 - 4 - 2\\\\ = 0\\\end{array}$

Hence, the formal charge on O is zero and on Br is $- {\rm{1}}$ .

Ans:

The correct arrow representation is shown below: