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Chapter 1 1
Chapter 1
© 2010, Prentice Hall
Organic Chemistry, 7th
Edition
L. G. Wade, Jr.
Introduction and Review
Chapter 1 2
Organic Chemistry
• Organic chemistry is the chemistry of carbon
compounds.
H5C6 N
H
O
O C6H5
OH
O
O
O
O OH
O
O
O
CH3
O
O
H5C6
HO
Taxol
O
CH3O
H
NCH3
HO
Codeine
2-Pentenylpenicillin
N
S
CHCH2CONH
H H
O
CH3
CH3
COONa
CH3CH2CH
Chapter 1 3
Electronic Structure of the Atom
• An atom has a dense,
positively charged
nucleus surrounded by a
cloud of electrons.
• The electron density is
highest at the nucleus
and drops off
exponentially with
increasing distance from
the nucleus in any
direction.
Chapter 1 4
The 2p Orbitals
• There are three 2p
orbitals, oriented at
right angles to each
other.
• Each p orbital
consists of two lobes.
• Each is labeled
according to its
orientation along the
x, y, or z axis.
Chapter 1 5
Isotopes
• Isotopes are atoms with the same number of
protons but different number of neutrons.
• Mass number is the sum of the protons and
neutrons in an atom.
C
12
6 6C
14
Chapter 1 6
Electronic Configurations of Atoms
• Valence electrons are electrons on the
outermost shell of the atom.
Chapter 1 7
Electronic Configurations
• The aufbau principle
states to fill the lowest
energy orbitals first.
• Hund’s rule states that
when there are two or more
orbitals of the same energy
(degenerate), electrons will
go into different orbitals
rather than pairing up in the
same orbital.
Electronic configuration of carbon
Chapter 1 8
Ionic Bonding
• To obtain a noble gas
configuration (a full valence
shell), atoms may transfer
electrons from one atom to
another.
• The atoms, now bearing
opposite charges, stay
together by electrostatic
attraction.
Chapter 1 9
Covalent Bonding
• Electrons are shared between the atoms to complete the
octet.
• When the electrons are shared evenly the bond is said to
be nonpolar or pure covalent.
• When electrons are not shared evenly between the
atoms, the resulting bond will be polar.
Chapter 1 10
CHCH44 NHNH33
HH22O ClO Cl22
Lewis Structures
C
H
H
H
H
N H
H
H
O HH ClCl
Carbon: 4 e
4 H@1 e ea: 4 e
8 e
Nitrogen: 5 e
3 H@1 e ea: 3 e
8 e
Oxygen: 6 e
2 H@1 e ea: 2 e
8 e
2 Cl @7 e ea: 14 e
Chapter 1 11
Double and Triple Bonds
Chapter 1 12
Bonding Patterns
ValenceValence
electronselectrons
# Bonds# Bonds # Lone Pair# Lone Pair
ElectronsElectrons
CC
NN
OO
HalidesHalides
(F, Cl, Br, I)(F, Cl, Br, I)
44 44 00
55 33 11
66 22 22
77 11 33
Chapter 1 13
Lone Pairs
Chapter 1 14
Dipole Moment
• Amount of electrical charge x bond length.
• Charge separation shown by electrostatic
potential map (EPM).
• Red indicates a partially negative region and
blue indicates a partially positive region.
Chapter 1 15
Electronegativity and Bond
Polarity
• Electronegativities can be used to predict
whether a bond will be polar.
• Since the electronegativity of carbon and
hydrogen are similar, C—H bonds are
considered to be nonpolar.
Chapter 1 16
Charged Species
H3O+
NO+
HCO3
-
OH H
H
Formal charge = number of valence electrons – (Formal charge = number of valence electrons – (ee in lone pairs + # bonds)in lone pairs + # bonds)
C
O
OHO
N O
6 – (2 + 3) = +1 6 – (2 + 3) = +1
5 – (2 + 3) = 0
6 – (6 + 1) = -1
++
++
Chapter 1 17
Compute the formal charge (FC) on each atom in H3N – BH3.
Solved Problem 1
Solution
Chapter 1 18
Common Bonding Patterns
Chapter 1 19
Resonance Forms
• In a resonance form, only the electrons are
moved. Connectivity between atoms stay the
same.
• The real structure is a hybrid of the different
resonance forms.
• Arrows connecting resonance forms are double
headed.
• Spreading the charges over two or more atoms
stabilize the ion.
Chapter 1 20
Resonance Forms
Resonance Forms can be compared using
the following criteria, beginning with the
most important:
– Has as many octets as possible.
– Has as many bonds as possible.
– Has the negative charge on the most
electronegative atom.
– Has as little charge separation as possible.
Chapter 1 21
Major and Minor Contributors
• The major contributor is the one in which
all the atoms have a complete octet of
electrons.
C
O
H H
C
O
H H
The carbon atom does
not have a complete
octet of electrons.
MAJOR MINOR
Chapter 1 22
Major and Minor Contributors
(Continued)
• When both resonance forms obey the octet
rule, the major contributor is the one with
the negative charge on the most
electronegative atom.
N C O N C O
MAJOR MINOR
The oxygen is more electronegative,
so it should have the negative
charge.
Chapter 1 23
• Opposite charges should be on adjacent
atoms.
Non-Equivalent Resonance
CH3 O N O CH3 O N O
The most stable one is the one with the smallest
separation of oppositely charged atoms.
MAJOR MINOR
Solved Problem 2
Chapter 1 24
Draw the important resonance forms for [CH3OCH2]+
. Indicate which structure is major and minor
contributor or whether they would have the same energy.
The first (minor) structure has a carbon atom with only six electrons around it. The second (major)
structure has octets on all atoms and an additional bond.
Solution
Solved Problem 3
Chapter 1 25
Draw the resonance structure the compound below. Indicate which structure is major and minor
contributor or whether they would have the same energy.
Both of these structures have octets on oxygen and both carbon atoms, and they have the same number
of bonds. The first structure has the negative charge on carbon; the second has it on oxygen. Oxygen is
the more electronegative element, so the second structure is the major contributor.
Solution
Chapter 1 26
Resonance Forms
• The structure of some compounds are not adequately represented
by a single Lewis structure.
• Resonance forms are Lewis structures that can be interconverted by
moving electrons only.
• The true structure will be a hybrid between the contributing
resonance forms.
Chapter 1 27
Resonance Forms for the
Acetate Ion
• When acetic acid loses a proton, the resulting acetate
ion has a negative charge delocalized over both of the
oxygen atoms.
• Each oxygen atom bears half of the negative charge,
and this delocalization stabilizes the ion.
• Each of the carbon–oxygen bonds is halfway between a
single bond and a double bond, and they are said to
have a bond order of 1½.
Chapter 1 28
Structures and Formulas
Extended Condensed
CH3CH3H C C H
H
H
H
H
H C C C
H
H
H
H
C
H
H
H
H
H
1 2 3 41 2 3 4
1 21 2
CH3CH2CH2CH3
Chapter 1 29
Structures and Formulas
Extended Condensed
H C C C
H
H
H
H
C
Br
H
H
H
H
CC C
H
H
HC
H
H OH
OH
H
1 2 3 41 2 3 4
1 2 3 41 2 3 4
CH3CH2CHCH3
Br
HOCH2CH2CCH3
O
Line-Angle Formulas
Chapter 1 30
H C C C
H
H
H
H
C
H
H
H
H
H
C
C
C
C
H C C C
H
H
H
H
C
H
C
H
H
H H
H
C
O
H
O
H
1 2 3 4
1 2 3 4 5 6
1 2 3 4 5 6
Chapter 1 31
Line-Angle Formulas (Continued)
Line-Angle Formulas (Continued)
Chapter 1 32
Line-Angle Formulas (Continued)
Chapter 1 33
Chapter 1 34
Calculating Empirical Formulas
The following are items that need to be
considered when calculating empirical
formulas:
– Given % composition for each element,
assume 100 grams.
– Convert the grams of each element to
moles.
– Divide by the smallest moles to get ratio.
– Molecular formula may be a multiple of the
empirical formula.
Chapter 1 35
Arrhenius Acids
• Arrhenius acids are substances that
dissociate in water to give H3O+
ions.
Chapter 1 36
Arrhenius Bases
• Arrhenius bases are substances that
dissociate in water to give hydroxide ions.
Chapter 1 37
Brønsted-Lowry Acids and Bases
Brønsted-Lowry acids are any species that donate a proton.
Brønsted-Lowry bases are any species that can accept a proton.
Chapter 1 38
Conjugate Acids and Bases
• Conjugate acid: when a base accepts a proton, it
becomes an acid capable of returning that proton.
• Conjugate base: when an acid donates its proton, it
becomes capable of accepting that proton back.
Chapter 1 39
Effect of Electronegativity on pKa
• As the bond to H becomes more polarized, H becomes
more positive and the bond is easier to break.
Chapter 1 40
Effect of Size on pKa
• As size increases, the H is more loosely
held and the bond is easier to break.
• A larger size also stabilizes the anion.
Chapter 1 41
Effect of Resonance on pKa
• If the negative charge on an atom can be delocalized over two or
more atoms, the acidity of that compound will be greater than when
the negative charge cannot be delocalized.
• The ethoxide anion is less acidic than the acetate ion simply
because the acetate ion can delocalize the negative charge.
• Methanesulfonic acid can delocalize the charge in three different
resonance forms, making it more acidic than the acetate ion.
Chapter 1 42
Nucleophiles and Electrophiles
• Nucleophile: Donates electrons to a
nucleus with an empty orbital.
• Electrophile: Accepts a pair of electrons.
• When forming a bond, the nucleophile
attacks the electrophile, so the arrow goes
from negative to positive.
• When breaking a bond, the more
electronegative atom receives the
electrons.
Chapter 1 43
Nucleophiles and Electrophiles
(Continued)

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Chapter01wade7thcgd 140409011227-phpapp01

  • 1. Chapter 1 1 Chapter 1 © 2010, Prentice Hall Organic Chemistry, 7th Edition L. G. Wade, Jr. Introduction and Review
  • 2. Chapter 1 2 Organic Chemistry • Organic chemistry is the chemistry of carbon compounds. H5C6 N H O O C6H5 OH O O O O OH O O O CH3 O O H5C6 HO Taxol O CH3O H NCH3 HO Codeine 2-Pentenylpenicillin N S CHCH2CONH H H O CH3 CH3 COONa CH3CH2CH
  • 3. Chapter 1 3 Electronic Structure of the Atom • An atom has a dense, positively charged nucleus surrounded by a cloud of electrons. • The electron density is highest at the nucleus and drops off exponentially with increasing distance from the nucleus in any direction.
  • 4. Chapter 1 4 The 2p Orbitals • There are three 2p orbitals, oriented at right angles to each other. • Each p orbital consists of two lobes. • Each is labeled according to its orientation along the x, y, or z axis.
  • 5. Chapter 1 5 Isotopes • Isotopes are atoms with the same number of protons but different number of neutrons. • Mass number is the sum of the protons and neutrons in an atom. C 12 6 6C 14
  • 6. Chapter 1 6 Electronic Configurations of Atoms • Valence electrons are electrons on the outermost shell of the atom.
  • 7. Chapter 1 7 Electronic Configurations • The aufbau principle states to fill the lowest energy orbitals first. • Hund’s rule states that when there are two or more orbitals of the same energy (degenerate), electrons will go into different orbitals rather than pairing up in the same orbital. Electronic configuration of carbon
  • 8. Chapter 1 8 Ionic Bonding • To obtain a noble gas configuration (a full valence shell), atoms may transfer electrons from one atom to another. • The atoms, now bearing opposite charges, stay together by electrostatic attraction.
  • 9. Chapter 1 9 Covalent Bonding • Electrons are shared between the atoms to complete the octet. • When the electrons are shared evenly the bond is said to be nonpolar or pure covalent. • When electrons are not shared evenly between the atoms, the resulting bond will be polar.
  • 10. Chapter 1 10 CHCH44 NHNH33 HH22O ClO Cl22 Lewis Structures C H H H H N H H H O HH ClCl Carbon: 4 e 4 H@1 e ea: 4 e 8 e Nitrogen: 5 e 3 H@1 e ea: 3 e 8 e Oxygen: 6 e 2 H@1 e ea: 2 e 8 e 2 Cl @7 e ea: 14 e
  • 11. Chapter 1 11 Double and Triple Bonds
  • 12. Chapter 1 12 Bonding Patterns ValenceValence electronselectrons # Bonds# Bonds # Lone Pair# Lone Pair ElectronsElectrons CC NN OO HalidesHalides (F, Cl, Br, I)(F, Cl, Br, I) 44 44 00 55 33 11 66 22 22 77 11 33
  • 14. Chapter 1 14 Dipole Moment • Amount of electrical charge x bond length. • Charge separation shown by electrostatic potential map (EPM). • Red indicates a partially negative region and blue indicates a partially positive region.
  • 15. Chapter 1 15 Electronegativity and Bond Polarity • Electronegativities can be used to predict whether a bond will be polar. • Since the electronegativity of carbon and hydrogen are similar, C—H bonds are considered to be nonpolar.
  • 16. Chapter 1 16 Charged Species H3O+ NO+ HCO3 - OH H H Formal charge = number of valence electrons – (Formal charge = number of valence electrons – (ee in lone pairs + # bonds)in lone pairs + # bonds) C O OHO N O 6 – (2 + 3) = +1 6 – (2 + 3) = +1 5 – (2 + 3) = 0 6 – (6 + 1) = -1 ++ ++
  • 17. Chapter 1 17 Compute the formal charge (FC) on each atom in H3N – BH3. Solved Problem 1 Solution
  • 18. Chapter 1 18 Common Bonding Patterns
  • 19. Chapter 1 19 Resonance Forms • In a resonance form, only the electrons are moved. Connectivity between atoms stay the same. • The real structure is a hybrid of the different resonance forms. • Arrows connecting resonance forms are double headed. • Spreading the charges over two or more atoms stabilize the ion.
  • 20. Chapter 1 20 Resonance Forms Resonance Forms can be compared using the following criteria, beginning with the most important: – Has as many octets as possible. – Has as many bonds as possible. – Has the negative charge on the most electronegative atom. – Has as little charge separation as possible.
  • 21. Chapter 1 21 Major and Minor Contributors • The major contributor is the one in which all the atoms have a complete octet of electrons. C O H H C O H H The carbon atom does not have a complete octet of electrons. MAJOR MINOR
  • 22. Chapter 1 22 Major and Minor Contributors (Continued) • When both resonance forms obey the octet rule, the major contributor is the one with the negative charge on the most electronegative atom. N C O N C O MAJOR MINOR The oxygen is more electronegative, so it should have the negative charge.
  • 23. Chapter 1 23 • Opposite charges should be on adjacent atoms. Non-Equivalent Resonance CH3 O N O CH3 O N O The most stable one is the one with the smallest separation of oppositely charged atoms. MAJOR MINOR
  • 24. Solved Problem 2 Chapter 1 24 Draw the important resonance forms for [CH3OCH2]+ . Indicate which structure is major and minor contributor or whether they would have the same energy. The first (minor) structure has a carbon atom with only six electrons around it. The second (major) structure has octets on all atoms and an additional bond. Solution
  • 25. Solved Problem 3 Chapter 1 25 Draw the resonance structure the compound below. Indicate which structure is major and minor contributor or whether they would have the same energy. Both of these structures have octets on oxygen and both carbon atoms, and they have the same number of bonds. The first structure has the negative charge on carbon; the second has it on oxygen. Oxygen is the more electronegative element, so the second structure is the major contributor. Solution
  • 26. Chapter 1 26 Resonance Forms • The structure of some compounds are not adequately represented by a single Lewis structure. • Resonance forms are Lewis structures that can be interconverted by moving electrons only. • The true structure will be a hybrid between the contributing resonance forms.
  • 27. Chapter 1 27 Resonance Forms for the Acetate Ion • When acetic acid loses a proton, the resulting acetate ion has a negative charge delocalized over both of the oxygen atoms. • Each oxygen atom bears half of the negative charge, and this delocalization stabilizes the ion. • Each of the carbon–oxygen bonds is halfway between a single bond and a double bond, and they are said to have a bond order of 1½.
  • 28. Chapter 1 28 Structures and Formulas Extended Condensed CH3CH3H C C H H H H H H C C C H H H H C H H H H H 1 2 3 41 2 3 4 1 21 2 CH3CH2CH2CH3
  • 29. Chapter 1 29 Structures and Formulas Extended Condensed H C C C H H H H C Br H H H H CC C H H HC H H OH OH H 1 2 3 41 2 3 4 1 2 3 41 2 3 4 CH3CH2CHCH3 Br HOCH2CH2CCH3 O
  • 30. Line-Angle Formulas Chapter 1 30 H C C C H H H H C H H H H H C C C C H C C C H H H H C H C H H H H H C O H O H 1 2 3 4 1 2 3 4 5 6 1 2 3 4 5 6
  • 31. Chapter 1 31 Line-Angle Formulas (Continued)
  • 34. Chapter 1 34 Calculating Empirical Formulas The following are items that need to be considered when calculating empirical formulas: – Given % composition for each element, assume 100 grams. – Convert the grams of each element to moles. – Divide by the smallest moles to get ratio. – Molecular formula may be a multiple of the empirical formula.
  • 35. Chapter 1 35 Arrhenius Acids • Arrhenius acids are substances that dissociate in water to give H3O+ ions.
  • 36. Chapter 1 36 Arrhenius Bases • Arrhenius bases are substances that dissociate in water to give hydroxide ions.
  • 37. Chapter 1 37 Brønsted-Lowry Acids and Bases Brønsted-Lowry acids are any species that donate a proton. Brønsted-Lowry bases are any species that can accept a proton.
  • 38. Chapter 1 38 Conjugate Acids and Bases • Conjugate acid: when a base accepts a proton, it becomes an acid capable of returning that proton. • Conjugate base: when an acid donates its proton, it becomes capable of accepting that proton back.
  • 39. Chapter 1 39 Effect of Electronegativity on pKa • As the bond to H becomes more polarized, H becomes more positive and the bond is easier to break.
  • 40. Chapter 1 40 Effect of Size on pKa • As size increases, the H is more loosely held and the bond is easier to break. • A larger size also stabilizes the anion.
  • 41. Chapter 1 41 Effect of Resonance on pKa • If the negative charge on an atom can be delocalized over two or more atoms, the acidity of that compound will be greater than when the negative charge cannot be delocalized. • The ethoxide anion is less acidic than the acetate ion simply because the acetate ion can delocalize the negative charge. • Methanesulfonic acid can delocalize the charge in three different resonance forms, making it more acidic than the acetate ion.
  • 42. Chapter 1 42 Nucleophiles and Electrophiles • Nucleophile: Donates electrons to a nucleus with an empty orbital. • Electrophile: Accepts a pair of electrons. • When forming a bond, the nucleophile attacks the electrophile, so the arrow goes from negative to positive. • When breaking a bond, the more electronegative atom receives the electrons.
  • 43. Chapter 1 43 Nucleophiles and Electrophiles (Continued)

Editor's Notes

  1. Copyright © 2006 Pearson Prentice Hall, Inc.
  2. Copyright © 2006 Pearson Prentice Hall, Inc.
  3. Copyright © 2006 Pearson Prentice Hall, Inc.
  4. Figure: 01_07-020T1-3.jpg Title: Condensed Formulas for Double and Triple Bonded Compounds Caption: Condensed Structural Formulas for Double and Triple Bonds Notes: In condensed formulas double and triple bonds are drawn as they would be in a Lewis structure showing two dashes for a double bond and three dashes for a triple bond.
  5. Figure: 01_07-033T1-4.jpg Title: Line-angle Drawings Caption: Examples of Line-angle Drawings Notes: Line-angle formula (a.k.a. skeletal structure or stick figure) is a shorthand way of drawing organic compounds. Bonds between carbons are represented by lines. Carbons are present where the lines begin and end, and where the lines meet. Nitrogen, oxygen, and halogen atoms are shown but hydrogens are not drawn unless they are bonded to a drawn atom.