Organic Chemistry - Standalone book
Organic Chemistry - Standalone book
10th Edition
ISBN: 9780073511214
Author: Francis A Carey Dr., Robert M. Giuliano
Publisher: McGraw-Hill Education
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Chapter 6, Problem 23P
Interpretation Introduction

Interpretation:

Two most stable Lewis structures for thiocyanate ion are to be written and the atom in each of them that bears a formal charge 1 is to be determined. Structures for the two compounds isolated in the given reaction are to be suggested.

Concept introduction:

The formal charge of an atom is calculated by the following formula:

Formal charge = number of valence electrons - electron count

Electron count =12(Number of shared electrons) + (Number of unshared electrons)

Thiocyanate ion is a good nucleophile and reacts with a primary alkyl halide in the presence of polar aprotic solvent via SN2 mechanism.

Constitutional isomers have the same molecular formula but differ in the connectivity of atoms in their structure.

Expert Solution & Answer
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Answer to Problem 23P

Solution:

a) The stable Lewis structures for thiocyanate ion are:

Organic Chemistry - Standalone book, Chapter 6, Problem 23P , additional homework tip  1

In structure I, the sulfur atom has a formal charge of 1 and in structure II, the nitrogen atom has a formal charge of 1.

b) The structures of the two constitutional isomers of C5H9NS are:

Organic Chemistry - Standalone book, Chapter 6, Problem 23P , additional homework tip  2

Explanation of Solution

a) Thiocyanate ion, SCN-.

The total number of valence electrons in SCN- are 16. Each carbon atom contributes four valence electrons; each nitrogen atom contributes five valence electrons, and each sulfur atom contributes six valence electrons. The negative charge on the ion also contributes one electron.

The two stable Lewis structures of the thiocyanate ion are shown below:

Organic Chemistry - Standalone book, Chapter 6, Problem 23P , additional homework tip  3

Organic Chemistry - Standalone book, Chapter 6, Problem 23P , additional homework tip  4

Carbon belongs to Group 4A; nitrogen belong to group 5A, and sulfur belongs to group 6A, which means their group numbers are 4,5and6, respectively.

In structure I, the electron counts and formal charge for each atom are as follows:

ElectroncountforS=12(2)+(6)=1+6=7

FormalchargeonS=6 - 7 = -1

ElectroncountforC =12(8)+(0)=4 + 4

FormalchargeonC=4 - 4 = 0

ElectroncountforN=12(6)+(2)=3+2=5

FormalchargeonN=5 - 5 = 0

Hence, in structure I, the sulfur atom has a formal charge of 1.

In structure II, the electron counts and formal charge for each atom are as follows:

ElectroncountforS=12(4)+(4)=2+4=6

FormalchargeonS=6 - 6 = 0

ElectroncountforC=12(8)+(0)=4 + 4

FormalchargeonC=4 - 4 = 0

ElectroncountforN=12(4)+(4)=2+4= 6

FormalchargeonN=5 - 6 = -1

Hence, in structure II, the nitrogen atom has a formal charge of 1.

b)

In the reaction given, the alkyl halide is a primary alkyl bromide. The nucleophile is thiocyanate ion, formed by the dissociation of KSCN and the solvent is DMF, which is a polar aprotic solvent. All these conditions suggest SN2 reaction mechanism.

Thiocyanate ion has two nucleophilic centers, which means it can attack through either sulfur or nitrogen atom. Because of this, when thiocyanate is used as a nucleophile, two possible products are obtained. In one product, the nucleophile attacks through the sulfur atom and gets attached to the carbon atom in the alkyl halide. In the other product, the nucleophile attacks through the nitrogen atom and gets attached to the carbon atom in the alkyl halide. The molecular formula of both the products remains the same; only the connectivity of atoms is different. Hence, they form constitutional isomers as follows:

Organic Chemistry - Standalone book, Chapter 6, Problem 23P , additional homework tip  5

Organic Chemistry - Standalone book, Chapter 6, Problem 23P , additional homework tip  6

Thus, the structures of the two constitutionally isomeric products of the molecular formula C5H9NS are as follows:

Organic Chemistry - Standalone book, Chapter 6, Problem 23P , additional homework tip  7

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