EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
2nd Edition
ISBN: 9780393630817
Author: KARTY
Publisher: W.W.NORTON+CO. (CC)
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Chapter 1, Problem 1.46P
Interpretation Introduction

(a)

Interpretation:

Lewis structure for the given molecule is to be completed.

Concept introduction:

Lewis structures involve only valence electrons. When drawing a Lewis structure, the first step is to calculate the total number of valence electrons. For a complete Lewis structure of a molecule, the atoms must complete their normal valency by bond formation and lone pairs of electrons. Maximum number of covalent bonds formed by any neutral atom with maximum number of lone pairs is

Atom Number of bond Number of lone pairs
C 4 0
H 1 0
O 2 2
N 1 1
F 1 3

Expert Solution
Check Mark

Answer to Problem 1.46P

The complete Lewis structure for the given molecule is

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 1

Explanation of Solution

The given structure is

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 2

Total valence electron count for the given molecule is 54 valence electrons. In the given structure, there are, in all, 14 single bonds, which means 28 valence electrons are used. Two lone pairs are placed on the oxygen atom attached to the C and H as H-O-C and three lone pairs are placed on the fluorine atom. This means 4 + 6 = 10 more electrons are used. This leaves 16 electrons.

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 3

The other oxygen atom has formed only one bond with carbon. This is converted to a double bond and two lone pairs are placed on the oxygen atom so that its octet is complete. A double bond is placed between C and N atom to complete the octet of carbon and a lone pair is placed in nitrogen to complete its octet.

A triple bond is placed between the other C and N to complete the octet of carbon and a lone pair is placed in the nitrogen to complete its octet.

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 4

This structure now accounts for all 54 electrons and the octet of each atom, except hydrogen, is complete. The duet for all hydrogens is complete.

Conclusion

The Lewis structure for the given molecule is completed from total valence electron count.

Interpretation Introduction

(b)

Interpretation:

Lewis structure for the given molecule is to be completed.

Concept introduction:

Lewis structures involve only valence electrons. When drawing a Lewis structure, the first step is to calculate the total number of valence electrons. For a complete Lewis structure of a molecule, every carbon atom must form four covalent bonds whereas the hydrogen atom forms one bond.

Expert Solution
Check Mark

Answer to Problem 1.46P

The complete Lewis structure for the given molecule is

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 5

Explanation of Solution

The given structure is

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 6

Total valence electron count for the given molecule must be 38. In the given structure, there are, in all, 14 single bonds which means 28 valence electrons are used. A triple bond is placed between the carbon atoms outside the ring, so that the octet of both carbon atoms is complete. Alternate double bonds are placed in the ring so that each carbon atom in the ring has a complete octet.

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 7

This structure now accounts for all 38 electrons and the octet of each atom, except hydrogen, is complete. The duet for all hydrogen atoms is complete.

Conclusion

The Lewis structure for the given molecule is completed from total valence electron count.

Interpretation Introduction

(c)

Interpretation:

Lewis structure for the given molecule is to be completed.

Concept introduction:

Lewis structures involve only valence electrons. When drawing a Lewis structure, the first step is to calculate the total number of valence electrons. For a complete Lewis structure of a molecule, the atoms must complete their normal valency by bond formation and lone pairs of electrons. Maximum numbers of covalent bonds formed by any neutral atom with maximum number of lone pair are

Atom Number of bond Number of lone pairs
C 4 0
H 1 0
O 2 2
N 1 1

Expert Solution
Check Mark

Answer to Problem 1.46P

The complete Lewis structure for the given molecule is

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 8

Explanation of Solution

The given structure is

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 9

Total valence electron count for the given molecule is 28 valence electrons. In the given structure, there are in all 9 single bonds which means 18 valence electrons are used.

The oxygen atom has formed only one bond with nitrogen. This is converted to a double bond and two lone pairs are placed on the oxygen atom so that its octet is complete. Another lone pair is placed on the nitrogen atom so that its octet is complete.

A double bond is placed between the C atoms attached to one hydrogen each. This completes the octet of both carbon atoms

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.46P , additional homework tip 10

This structure now accounts for all the 28 electrons, and the octet of each atom, except hydrogen, is complete. The duet for all hydrogen atoms is complete.

Conclusion

The Lewis structure for the given molecule is completed from total valence electron count.

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Chapter 1 Solutions

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M

Ch. 1 - Prob. 1.1PCh. 1 - Prob. 1.2PCh. 1 - Prob. 1.3PCh. 1 - Prob. 1.4PCh. 1 - Prob. 1.5PCh. 1 - Prob. 1.6PCh. 1 - Prob. 1.7PCh. 1 - Prob. 1.8PCh. 1 - Prob. 1.9PCh. 1 - Prob. 1.10P
Ch. 1 - Prob. 1.11PCh. 1 - Prob. 1.12PCh. 1 - Prob. 1.13PCh. 1 - Prob. 1.14PCh. 1 - Prob. 1.15PCh. 1 - Prob. 1.16PCh. 1 - Prob. 1.17PCh. 1 - Prob. 1.18PCh. 1 - Prob. 1.19PCh. 1 - Prob. 1.20PCh. 1 - Prob. 1.21PCh. 1 - Prob. 1.22PCh. 1 - Prob. 1.23PCh. 1 - Prob. 1.24PCh. 1 - Prob. 1.25PCh. 1 - Prob. 1.26PCh. 1 - Prob. 1.27PCh. 1 - Prob. 1.28PCh. 1 - Prob. 1.29PCh. 1 - Prob. 1.30PCh. 1 - Prob. 1.31PCh. 1 - Prob. 1.32PCh. 1 - Prob. 1.33PCh. 1 - Prob. 1.34PCh. 1 - Prob. 1.35PCh. 1 - Prob. 1.36PCh. 1 - Prob. 1.37PCh. 1 - Prob. 1.38PCh. 1 - Prob. 1.39PCh. 1 - Prob. 1.40PCh. 1 - Prob. 1.41PCh. 1 - Prob. 1.42PCh. 1 - Prob. 1.43PCh. 1 - Prob. 1.44PCh. 1 - Prob. 1.45PCh. 1 - Prob. 1.46PCh. 1 - Prob. 1.47PCh. 1 - Prob. 1.48PCh. 1 - Prob. 1.49PCh. 1 - Prob. 1.50PCh. 1 - Prob. 1.51PCh. 1 - Prob. 1.52PCh. 1 - Prob. 1.53PCh. 1 - Prob. 1.54PCh. 1 - Prob. 1.55PCh. 1 - Prob. 1.56PCh. 1 - Prob. 1.57PCh. 1 - Prob. 1.58PCh. 1 - Prob. 1.59PCh. 1 - Prob. 1.60PCh. 1 - Prob. 1.61PCh. 1 - Prob. 1.62PCh. 1 - Prob. 1.63PCh. 1 - Prob. 1.64PCh. 1 - Prob. 1.65PCh. 1 - Prob. 1.66PCh. 1 - Prob. 1.67PCh. 1 - Prob. 1.68PCh. 1 - Prob. 1.69PCh. 1 - Prob. 1.70PCh. 1 - Prob. 1.71PCh. 1 - Prob. 1.72PCh. 1 - Prob. 1.73PCh. 1 - Prob. 1.74PCh. 1 - Prob. 1.75PCh. 1 - Prob. 1.76PCh. 1 - Prob. 1.77PCh. 1 - Prob. 1.78PCh. 1 - Prob. 1.79PCh. 1 - Prob. 1.80PCh. 1 - Prob. 1.81PCh. 1 - Prob. 1.82PCh. 1 - Prob. 1.1YTCh. 1 - Prob. 1.2YTCh. 1 - Prob. 1.3YTCh. 1 - Prob. 1.4YTCh. 1 - Prob. 1.5YTCh. 1 - Prob. 1.6YTCh. 1 - Prob. 1.7YTCh. 1 - Prob. 1.8YTCh. 1 - Prob. 1.9YTCh. 1 - Prob. 1.10YTCh. 1 - Prob. 1.11YTCh. 1 - Prob. 1.12YTCh. 1 - Prob. 1.13YTCh. 1 - Prob. 1.14YTCh. 1 - Prob. 1.15YTCh. 1 - Prob. 1.16YTCh. 1 - Prob. 1.17YT
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