ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<
ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<
10th Edition
ISBN: 9781259917196
Author: Carey
Publisher: MCG CUSTOM
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Textbook Question
Chapter 1, Problem 41P

Write a Lewis formula for each of the following organic molecules:

C 2 H 3 Cl

(vinyl chloride: starting material for the preparation of PVC plastics)

C 2 HBrClF 3

(halothane: a nonflammable inhalation anesthetic; all three fluorines are

bonded to the same carbon)

C 2 Cl 2 F 4

(Freon 114: formerly used as a refrigerant and as an aerosol propellant; each

carbon bears one chlorine)

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

The Lewis formula for each of the given organic molecules is to be written.

Concept introduction:

When writing the Lewis formula for a molecule, the total number of valence electrons in the molecule is first determined based on its molecular formula.

The bonded atoms are connected by a shared pair of electrons which is shown by a dash.

The number of electrons in the bonds is subtracted from the total number of valence electrons. This gives the number of electrons that remain to be added in the formula.

These remaining electrons are added as unshared electron pairs in such a way that the most electronegative atoms have eight electrons around them.

Second-row elements cannot have more than eight valence electrons.

Answer to Problem 41P

Solution:

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  1

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  2

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  3

Explanation of Solution

a) C2H3Cl (vinyl chloride: starting material for the preparation of PVC plastics)

The total number of valence electrons in C2H3Cl (vinyl chloride) is 18. Each carbon atom contributes 4 valence electrons, and each hydrogen atom contributes 1 valence electron. Each chlorine atom contributes 7 valence electrons.

The basic framework of the given molecule can be drawn as follows.

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  4

The above figure shows the presence of 5 bonds in the molecule. This implies that 10 electrons are to be shown as bonded pair in the Lewis formula of the given molecule. 8 electrons will then remain to be shown.

Three pairs of unshared electrons are assigned to the chlorine atom and one pair of unshared electrons is assigned on one of the carbon atoms as follows.

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  5

The octet of each atom in the molecule, except one carbon atom, is complete. Hence, to complete the octet of this one carbon atom, the unshared pair of electrons on the other carbon atom shifts towards the bond between the two carbon atoms resulting in the formation of a double bond.

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  6

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  7

Now the octet of each atom is complete. In this way, the Lewis structure of vinyl chloride was written.

b) C2HBrClF3 (halothane: a nonflammable inhalation anesthetic; all three fluorine atoms are

bonded to the same carbon)

C2HBrClF3 has a total number of valence electrons equal to 44. Each carbon atom contributes 4 valence electrons and each hydrogen atom contributes 1 valence electron to the molecule. Similarly, each halogen atom contributes 7 valence electrons to the molecule. It is given that all three fluorine atoms are bonded to the same carbon atom.

The basic framework can therefore, be drawn as follows.

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  8

It is evident from the figure drawn for the given molecule that there are a total of 7 bonds in the molecule. This implies that 14 electrons are to be shown as bonding electrons in the required Lewis formula. 30 electrons will then remain to be shown as unpaired electrons.

Of these 30 unpaired electrons, three pairs are assigned on each halogen atom to complete their octets.

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  9

In all, 15 pairs of unshared electrons are assigned on the halogen atoms. Thus, all the valence electrons are shown in the structure.

In this way, the Lewis structure of C2HBrClF3 was written.

c) C2Cl2F4 (Freon 114: formerly used as a refrigerant and as an aerosol propellant; each

carbon bears one chlorine)

C2Cl2F4 has a total of 50 number of valence electrons. Each carbon atom contributes 4 valence electrons and each hydrogen atom contributes 1 valence electron to the molecule. Similarly, each halogen atom contributes 7 valence electrons to the molecule. It is given that each carbon atom is bonded to one chlorine atom in the given molecule.

The basic framework of the molecule can be drawn as follows.

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  10

The above figure shows that there are 7 bonds in the given molecule, which means 14 electrons are to be shown as bonding pair of electrons in the Lewis formula. Therefore, 36 electrons will remain to be represented as unpaired electrons.

Three pairs of unshared electrons are assigned on each halogen atom in order to complete their octets.

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<, Chapter 1, Problem 41P , additional homework tip  11

In all, 18 pairs of unshared electrons are assigned on the halogen atoms. Thus, all the valence electrons are shown in the structure.

In this way, the Lewis structure of C2Cl2F4 was written.

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Chapter 1 Solutions

ORGANIC CHEMISTRY(LL)-W/ACCESS >CUSTOM<

Ch. 1.5 - The following inorganic species will be...Ch. 1.5 - Prob. 12PCh. 1.6 - Prob. 13PCh. 1.6 - Problem 1.14 Nitrosomethane and formaldoxime both...Ch. 1.6 - Prob. 15PCh. 1.7 - All of the bonds in the carbonate ion (CO32-) are...Ch. 1.7 - Prob. 17PCh. 1.8 - Prob. 18PCh. 1.8 - Prob. 19PCh. 1.9 - Sodium borohydride, NaBH4, has an ionic bond...Ch. 1.9 - Prob. 21PCh. 1.10 - Which of the following compounds would you expect...Ch. 1.11 - Using the curved arrow to guide your reasoning,...Ch. 1.11 - Prob. 24PCh. 1.11 - Prob. 25PCh. 1.12 - Prob. 26PCh. 1.12 - Prob. 27PCh. 1.12 - Prob. 28PCh. 1.12 - Prob. 29PCh. 1.12 - Prob. 30PCh. 1.13 - Which is the stronger acid, H2O or H2S? Which is...Ch. 1.13 - Prob. 32PCh. 1.13 - Prob. 33PCh. 1.13 - Hypochlorous and hypobromous acid (HOClandHOBr)...Ch. 1.13 - Prob. 35PCh. 1.13 - Prob. 36PCh. 1.14 - What is the equilibrium constant for the following...Ch. 1.14 - Prob. 38PCh. 1.14 - Prob. 39PCh. 1.15 - Write an equation for the Lewis acid/Lewis base...Ch. 1 - Write a Lewis formula for each of the following...Ch. 1 - Prob. 42PCh. 1 - Write structural formulas for all the...Ch. 1 - Prob. 44PCh. 1 - Expand the following structural representations so...Ch. 1 - Each of the following species will be encountered...Ch. 1 - Consider Lewis formulas A, B, and C: H2 C -NN:...Ch. 1 - Prob. 48PCh. 1 - Prob. 49PCh. 1 - Prob. 50PCh. 1 - Prob. 51PCh. 1 - Prob. 52PCh. 1 - Prob. 53PCh. 1 - Prob. 54PCh. 1 - Which compound in each of the following pairs...Ch. 1 - With a pKa of 11.6, hydrogen peroxide is a...Ch. 1 - The structure of montelukast, an antiasthma drug,...Ch. 1 - One acid has a pKa of 2, the other has a pKa of 8....Ch. 1 - Calculate Ka for each of the following acids,...Ch. 1 - Rank the following in order of decreasing acidity....Ch. 1 - Rank the following in order of decreasing...Ch. 1 - Consider 1.0 M aqueous solutions of each of the...Ch. 1 - Prob. 63PCh. 1 - Prob. 64PCh. 1 - Prob. 65PCh. 1 - Prob. 66PCh. 1 - Prob. 67PCh. 1 - Prob. 68PCh. 1 - Amide Lewis Structural Formulas Lewis formulas are...Ch. 1 - Amide Lewis Structural Formulas Lewis formulas are...Ch. 1 - Amide Lewis Structural Formulas Lewis formulas are...Ch. 1 - Prob. 72DSPCh. 1 - Amide Lewis Structural Formulas Lewis formulas are...Ch. 1 - Amide Lewis Structural Formulas Lewis formulas are...

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