Test Bank for Organic Chemistry 3rd Edition by Janice Smith Study Guide And Solutions Manual

Test Bank for Organic Chemistry 3rd Edition by Janice Smith Study Guide And Solutions Manual

Test Bank for Organic Chemistry 3rd Edition by Janice Smith Study Guide And Solutions Manual
Test Bank for Organic Chemistry 3rd Edition by Janice Smith Study Guide And Solutions Manual
Test Bank for Organic Chemistry 3rd Edition by Janice Smith Study Guide And Solutions Manual
Count up the number of valence electrons.
=
=
=
=
3 electrons
8 electrons
5 electrons
+ 6 electrons
22 electrons total
1 electron per H
4 electrons per C
5 electrons per N
6 electrons per O
3H's
2C's
1N
1O
x
x
x
x
[3] Use the shortcut to figure out how many bonds are needed.
• Number of electrons needed if there were no bonds:
3 H's
4 second-row elements
x
x
2 electrons per H
8 electrons per element
=
= 6 electrons
+ 32 electrons
38 electrons needed if
there were no bonds
• Number of electrons that must be shared:
38 electrons
– 22 electrons
16 electrons must be shared
• Since every bond takes two electrons, 16/2 = 8 bonds are needed.
[4] Draw all possible Lewis structures.
• Draw the bonds to the H’s first (three bonds). Then add five more bonds. Arrange them between
the C’s, N, and O, making sure that no atom gets more than eight electrons. There are three
possible arrangements of bonds; i.e., there are three resonance structures.
• Add additional electron pairs to give each atom an octet and check that all 22 electrons are used.
HCN
H
H
C O HCN
H
H
C O
HCN
H
H
C O
HCN
H
H
C O
HCN
H
H
C O
HCN
H
H
C O
HCN
H
H
C O
All bonds drawn in.
Three arrangements possible.
Electron pairs drawn in.
Every atom has an octet.
Bonds to H's added.
• Calculate the formal charge on each atom.
HCN
H
H
HCN C O
H
H
C O HCN
H
H
C O
+1 –1 –1 +1
• You can evaluate the Lewis structures you have drawn. The middle structure is the best
resonance structure, since it has no charged atoms.
Note: This method works for compounds that contain second-row elements in which every element gets
an octet of electrons. It does NOT necessarily work for compounds with an atom that does not have an
octet (such as BF3), or compounds that have elements located in the third row and later in the periodic
table.
Study Guide/Solutions Manual to accompany Organic Chemistry, Third Edition 5
Smith: Study Guide/
Solutions Manual to
accompany Organic
Chemistry, Third Edition
1. Structure and Bonding Text © The McGraw−Hill
Companies, 2011
Chapter 1–6
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1.1 The mass number is the number of protons and neutrons. The atomic number is the number of
protons and is the same for all isotopes.
Nitrogen-14 Nitrogen-13
a. number of protons = atomic number for N = 7 7 7
b. number of neutrons = mass number – atomic number 7 6
c. number of electrons = number of protons 7 7
d. The group number is the same for all isotopes. 5A 5A
1.2 The atomic number is the number of protons. The total number of electrons in the neutral atom
is equal to the number of protons. The number of valence electrons is equal to the group number
for second-row elements. The group number is located above each column in the periodic table.
[1] 31P
[2] 19F
[3] 2H
a. atomic number
15
9
1
b. total number of e–
15
9
1
c. valence e–
5
7
1
d. group number
5A
7A
1A
15
9
1
1.3 Ionic bonds form when an element on the far left side of the periodic table transfers an electron to
an element on the far right side of the periodic table. Covalent bonds result when two atoms
share electrons.
a. F F b. Li+ Br c. H C C
H
H
N H
H
d. Na+
Both N–H bonds
are covalent.