# 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. –Free Chegg Question Answer

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.

Transcribed text From Image:

Answer:

## Step-by-step

### Step 1 of 2

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

The correct arrow representation is shown below:

Explanation | Hint for next step

In {{\rm{S}}_{\rm{N}}}2SN​2 reaction nucleophile act as attacking species and halogen present in haloalkane act as a leaving group. Thus, identify the haloalkane and nucleophile then show the movement of electron by using arrow as shown below:

### Step 2 of 2

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}FormalchargeonBr=7−8−20​=7−8=−1​

\begin{array}{c}\\{\rm{Formal}}\,{\rm{charge on O}} = 6 – 4 – \frac{4}{2}\\\\ = 6 – 4 – 2\\\\ = 0\\\end{array}FormalchargeonO=6−4−24​=6−4−2=0​

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

The final product of the reaction with the correct formal charge on the species is shown below:

Explanation

The substituted product of the given reaction is formed by the attack of lone pair of oxygen on the carbon atom and removal of bromide ion as shown below:

The formal charge is calculated by using the following formula:

{\rm{Formal}}\,{\rm{charge}} = V – N – \frac{B}{2}Formalcharge=VN−2B

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.

As oxygen and bromine are having 6 and 7 valence electrons respectively. In the product formed O is having 4 bonding electrons and 4 nonbonding electrons. similarly, Br is having 8 nonbonding electrons and zero bonding electrons.

These values are substituted in the above formula and formal charge is calculated as follows:

\begin{array}{c}\\{\rm{Formal}}\,{\rm{charge on O}} = 6 – 4 – \frac{4}{2}\\\\ = 6 – 4 – 2\\\\ = 0\\\end{array}FormalchargeonO=6−4−24​=6−4−2=0​

\begin{array}{c}\\{\rm{Formal}}\,{\rm{charge on Br}} = V – N – \frac{B}{2}\\\\ = 7 – 8 – \frac{0}{2}\\\\ = 7 – 8\\\\ = – 1\\\end{array}FormalchargeonBr=VN−2B​=7−8−20​=7−8=−1​

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