Add bent arrows come the reactant side of the following SN2 reaction to indicate the flow of electrons. Draw the product species to display the well balanced equation, including nonbonding electrons and also formal charges.
You are watching: Add curved arrows to the reactant side of the sn2 reaction shown.
The correct arrow representation is presented below:
Explanation | Hint for following step
In \\rmS_\\rmN2SN2 reaction nucleophile act as attacking varieties and halogen current in haloalkane act together a leaving group. Thus, identify the haloalkane and also nucleophile then display the activity of electron through using arrowhead as presented below:
Step 2 of 2
Write the product and show the formal charge on the compound formed.
Write the product of the offered reaction through the exactly formal fee on the species as shown below:
The formal fee on Br is calculate as shown below:
\\beginarrayc\\\\\\rmFormal\\,\\rmcharge ~ above Br = 7 – 8 – \\frac02\\\\\\\\ = 7 – 8\\\\\\\\ = – 1\\\\\\endarrayFormalchargeonBr=7−8−20=7−8=−1
\\beginarrayc\\\\\\rmFormal\\,\\rmcharge on O = 6 – 4 – \\frac42\\\\\\\\ = 6 – 4 – 2\\\\\\\\ = 0\\\\\\endarrayFormalchargeonO=6−4−24=6−4−2=0
Hence, the formal fee on O is zero and on Br is – \\rm1−1 .
The last product of the reaction through the correct formal charge on the varieties is displayed below:
The substituted product that the provided reaction is created by the attack of lone pair of oxygen ~ above the carbon atom and also removal that bromide ion as displayed below:
The formal charge is calculation by making use of the adhering to formula:
\\rmFormal\\,\\rmcharge = V – N – \\fracB2Formalcharge=V−N−2B
Here, V is valence electron in neutral atom, N is the number of nonbonding electrons, and B the variety of bonding electrons present in one atom.
As oxygen and bromine are having 6 and also 7 valence electron respectively. In the product developed O is having actually 4 bonding electrons and also 4 nonbonding electrons. Similarly, Br is having 8 nonbonding electrons and zero bonding electrons.
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These values room substituted in the over formula and also formal fee is calculated as follows:
\\beginarrayc\\\\\\rmFormal\\,\\rmcharge top top O = 6 – 4 – \\frac42\\\\\\\\ = 6 – 4 – 2\\\\\\\\ = 0\\\\\\endarrayFormalchargeonO=6−4−24=6−4−2=0
\\beginarrayc\\\\\\rmFormal\\,\\rmcharge on Br = V – N – \\fracB2\\\\\\\\ = 7 – 8 – \\frac02\\\\\\\\ = 7 – 8\\\\\\\\ = – 1\\\\\\endarrayFormalchargeonBr=V−N−2B=7−8−20=7−8=−1
Hence, the formal charge on O is zero and also on Br is – \\rm1−1 .
The correct arrow representation is displayed below:
The last product of the reaction with the exactly formal fee on the species is presented below: