61 Notes

Chapter 1-1 Chem 61

What is it? Why is it a two semester course? Why is it important to careers in health care?

Organic chemistry is essential to the understanding of the intricate details of life The interactions and reactions of organic molecules are what define living systems.

One needs to understand bonding, structure, properties, reactions, and synthesis to understand natural systems.

CHAPTER 1

INTRODUCTION

Organic Chemistry: the study of carbon compounds...

organic compounds contain the elements C, H, N, O, S, Cl, Br, etc.

Organic compounds were originally thought to come only from living organisms...thus the term organic (until about 1830)

Inorganic compounds were all those which came from non-living sources

Scientists thought a vital force was necessary to produce organic compounds and that they could not be synthesized in the lab.

In 1828 Fredreich Woeller synthesized urea, an organic compound excreted as waste, from ammonium cyanate, and vitalism slowly died out.

Organic compounds range from methane , CH4 (natural gas) to DNA, the genetic coding material, and taxol, a plant derived substance which is a potential anticancer agent.

Organic chemistry is fundamental to many scientific disciplines...biochemistry, polymer chemistry, microbiology, botany, pharmacy, medicine...since living systems are composed primarily of organic compounds and water.

What is Organic Chemistry?

Organic Chemistry

Chem 61Chapter 1-2 Electronic Structure of Carbon

STRUCTURAL THEORY AND BONDING

Carbon is intermediate in electronegativity, therefore it neither completelydonates or completely accepts electrons. As a result it forms covalent bonds to itself and other atoms. It can bond to itself to form chains andrings...catenation. This allows the formation of a staggering number of organiccompounds. 95% of all known compounds are organic.

ATOMIC STRUCTURE

It is important to understand molecular structure to understand reactivity oforganic compounds. Molecular structure depends on atomic structure.

Atomic structure of carbon C: 1s22s22p2

that is, carbon has 2 electrons in the lowest energy level, the 1s orbital 2 electrons in the next energy level, the 2s orbital 2 electrons in the third energy level, the 2p orbital

1s, 2s, 2p orbitals

1s is spherical with the same phase throughout

2s is spherical with a node node is where ψ2 = 0

2p... three orbitals of equal energy (degenerate)

2px 2py 2pz

node

1s

2s

Chapter 1-3 Chem 61Rules for Electronic Configuration

Pauli Exclusion Principle: Maximum of two electrons per orbital.

Electrons have a spin of +1/2 or -1/2 which gives rise to a small magnetic field since electrons are charged.

Repulsion between the electrons is reduced if they have opposite spins and thus opposite magnetic moments.

If two electrons occupy the same orbital they must have opposite spins or be paired.

Aufbau Principle: orbitals are filled from lower to higher energy

4s __3p __ __ __3s __2p __ __ __ order of filling of atomic orbitals2s __1s __

THUS: Li 1s22s1

Hund's Rule: orbitals of equal energy (degenerate) receive one electron until there is one electron in each orbital: then pairing of electrons begins.

THUS: C 1s22s22px12py1

Chapter 1-4 Chem 61Atomic Radius and Electronegativity

ATOMIC RADIUS

Atomic Radius: distance between the nucleus and the outermost electrons (valence electrons) or one half the bondlength of a diatomic molecule

H—H bond length is 0.74Å, atomic radius is 0.37Å Atomic radius increases with increasing number of electron shells within an atom and decreases with the increase in the number of protons within an atom

Thus atomic radius decreases from left to right within the same row of the periodic table (increasing number of protons in the nucleus) and increases from top to bottom within a group of the periodic table (increasing number of electronic shells)

ELECTRONEGATIVITY Electronegativity: a measure of an atom's attraction for outer bonding electrons

Electronegativity increases with increasing charge on the nucleus and with decreasing distance between the nucleus and the electrons

Thus electronegativity increases from left to right within a row of the periodic table and from bottom to top within a group in the periodic table

C HH

HH

Chapter 1- 5

C::OH

H

Chem 61Chemical Bonding

4H + :C:

CHEMICAL BONDING

G.N. Lewis put forth the first explanation of the nature of the chemical bond.

Ionic Bond..one or more electrons is completely transferred from one atom to another creating ions: a cation (positively charged) and an anion (negatively charged).

the ions are held together by electrostatic attractions

ionic bonds generally occur between atoms of highly different electronegativies. e.g.

C O N H

Covalent Bond...one or more electrons is shared by atoms...atomic orbitals merge into shared or molecular orbitals covalent bonds are usually formed between atoms of similar electronegativities

since carbon is intermediate in electronegativity it usually forms covalent bonds

Generally...atoms not having the noble gas configuration tend to form bonds such that each atom may obtain the stable noble gas electronic configuration [OCTET RULE]

Nao – e– Na+ + e–

Clo + e– Cl –

Nao + Clo Na+Cl –

Atoms of the elements of organic compounds can form a fixed number of bonds.

single bond double bond triple bond

H:C:::C:H

..

..

..

Chapter 1- 6 Chem 61Chemical Bonding

.

COVALENT BONDS

Bond lengths: the distance between two covalently bonded nuclei Bond angle: the angle formed between two covalent bonds

Bond dissociation energy: the energy required for homolytic cleavage of a bond;

Cl Cl

H3C H H3C + H

Cl + Cl

Heterolytic cleavage: cleavage of a bond to give a cation and an anion; one atom which is part of the covalent bond retains both electrons from the bond

∆H (change in enthalpy) for homolytic cleavage of many covalent bonds has been determined

H3C—H H—H Br—Br H—OH

two radicals

104 kcal/mole 104 kcal/mole 46 kcal/mole 110 kcal/mole

88 kcal/mole163 ckal/mole 230 kcal/mole

CH3—CH3 CH2=CH2 HC≡CH

Cl Cl

H3C H H3C+ + H –

Cl+ + Cl –

cation anion

...

C

Cl

Cl Cl

ClCCl

Cl

H

Cl

O

H H

δ+ δ− δ+ δ−

Chapter 1-7 Chem 61Polar Covalent Bonds

POLAR COVALENT BONDS

covalent bonds between atoms of similar electronegativities are said to be nonpolar bonds since each atom shares the electrons of the bond approximately equally

H—H CH3—CH3

if atoms of different electronegativities form a covalent bond, the more electronegative atom will have a stronger attraction for the electrons and polarize the bond giving a polar covalent bond in which one atom is partially negatively (δ–)charged and one atom is partially positively charged (δ+)

H3C—Cl H—Cl H2C=O H3C—OH CH3—NH2

If any entire molecule has an overall dipole, then the molecule is said to be polar.

CCl4...nonpolar, polar covalent bonds cancel each other because of tetrahedral symmetry

HCCl3...polar

H2O, NH3...very polar

δ+ δ− δ+ δ− δ+ δ−

δ_

δ+ δ+

δ_

δ+

δ+

δ_

δ_

δ_

dipoles cancel

dipoles add dipoles addδ+δ_

Chapter 1-8 Chem 61Polar Covalent Bonds

O H :N

N H :O

ATTRACTIONS BETWEEN MOLECULES

ion-ion forces: the attraction between oppositely charged ions and repulsion between like charges very strong...not often encountered in organic compounds

van der Waals forces: all dipole-dipole forces are both repulsive and attractive... the distance at which the repulsive forces are minimized and the attractive forces are maximized is called the van der Waals radius

Induced dipole-dipole interactions or London forces: small temporary dipoles occur and induce dipoles in another molecule due to small uneven distribution of electron density.

Dipole-dipole interactions: permanent dipoles in molecules attract or repel

Hydrogen bonding: a specific type of dipole-dipole interaction; very strong

occur only between a hydrogen atom bonded to an electronegative atom O, N, S and a lone pair on O, N, S

N H :N

Intermolecular hydrogen bonds increase the boiling points of compounds and increase their solubility in water.

Hydrogen bonds also can influence the shapes of biomolecules by internal hydrogen bonding as well as hydrogen bonding between molecules.

O H :O

7 kcal /mole

2 kcal/mole

3 kcal/mole

5 kcal/mole

Chapter 1-9 Chem 61Formulas and Formal Charge

CHEMICAL FORMULAS

empirical formula: gives the types and ratios of atoms in a molecule

molecular formula: gives the type and actual number of atoms

structural formula: gives the type, number and attachment of atoms...the actual structure

for example: hexane: C6H14 C3H7 CH3CH2CH2CH2CH2CH3

molecular formula empirical formula structural formula

LEWIS STRUCTURES

1. draw the molecular skeleton2. count the number of available valence electrons (be sure to account for any overall charge on the species)3. draw covalent bonds between all the atoms giving as many as possible an octet (duet for hydrogen)4. assign charges in the molecule

FORMAL CHARGE

formal charge on an atom = # valence electrons - # shared pairs of electrons - # unshared electrons

HNO3 Lewis structure: N...5-4-0 = +1O1...6-2-4 = 0O2...6-1-6 = -1O3...6-2-4 = 0H...1-1 = O

....3

2

1

..

..

..

:O:

..:O:

N:O:H

....

..

..SO4

-2

:O:....

..

..:O:

:O:S:O:H2SO4 Lewis Structure

O...6-1-6 = -1S...6-4-0 = +2 overall charge (+2) + 4(-1) = -2

C OH

H

H C C H

C CH

H

H

H

Chapter 1-10 Chem 61Formulas and Formal Charge

HC≡CH

Structural Formulas

Lewis Structures: as described above using dots for electrons

Line-bond formulas: a line is used to represent two electrons forming a bond (a shared pair)

Condensed formulas: bonds are not always shown and atoms of the same type bonded to another atom are grouped together

Lewis condensed

CH2=CH2H

H......

..H

HC::C

line-bond Lewis condensed

CC

CCC

CH

H

H H H

H

HHH H

H

H

H:C:::C:H

condensedLewisline-bond

H2C=O..

H

H....C::O

Polygon formulas: polygon formulas are often used to represent cyclic compounds for simplicity;

each carbon is assumed to have enough additional hydrogens to give each carbon four bonds

line-bond

decalincyclopentanecyclobutanecyclopropane

benzenecyclohexane

..

......

..

..

Chem 61Chapter 1-11 Acids and Bases

ACIDS AND BASES

Bronsted-Lowry: acid: a proton donor base: a proton acceptor

Strong acid: completely ionized or dissociated in water e.g., HCl, H2SO4, HNO3, HBr

Weak acid: only partially dissociated in water:

carboxylic acids are weak acid

conjugate acidconjugate basebaseacid

CH3CO2-

+ H3O+CH3CO2H + H2O

amines are weak bases

conjugate acid conjugate baseacidbase

CH3NH3+ + HO – CH3NH2 + H2O

Generally: strong acids have weak conjugate bases and weak acids have strong conjugate bases

that is, as acid strength increases, the basicity of the conjugate base decreases Thus the ability of the conjugate base to stabilize a negative charge determines the strength of an acid

Conjugate Acids

increasing acid strength

H2O HCN CH3CO2H H3PO4 HCl

15.75 6.37 4.75 2.12 -7pKa

Conjugate BasesHO – NC – CH3CO2

– H2PO4 – C l –

decreasing base strength


Factors affecting Acidity:

the electronegativity and the size of the atom which carries the negative charge influence its ability to stabilize the negative charge

electronegativity

size of the atom

increasing electronegativity of atom, increasing acid strength

(CH3)3C—H (CH3)2N—H CH3O—H F—H

increasing size of halogen, increasing acid strength

H—F H—Cl H—Br H—I

pKa 3.45 -7 -9 -9.5

pKa 50 35 15.5 3.45

ACIDITY CONSTANTS, Ka's

for acetic acid

CH3CO2H + H2O CH3CO2– + H3O+

Ka =[CH3CO2

–] [H+][CH3CO2H]

since stronger acids are more ionized, the larger Ka, the stronger the acid

pKa = -logKa, the lower the pKa, the stronger the acid

also, the higher the pKa, the weaker the acid or the stronger the base

the stronger the acid, the more stable the anion produced by ionization of the acid.

Chem 61Chapter 1-13 Lewis Acids and Bases

LEWIS ACIDS AND BASES

Lewis acid: electron pair acceptor: any species with an electron deficient atom

Lewis base: electron pair donor; any species with an unshared pair of electrons

BBr3, AlCl3, H3C+

..

.. ..H3N:, CH3CH2OH, H2C=O:

Chem 61Chapter 1-14 Quantum Mechanics

QUANTUM MECHANICS...Molecular Orbitals

In 1923 Louis De Broglie postulated that electrons have properties of three dimensional waves

Later a wave equation was developed. Solutions (Ψ) to this wave equation give the various electronic states known as atomic orbitals.

Ψ = probability of finding an electron in a certain space = electron probability densityplots of Ψ give the familiar s,p,d...orbitals

WAVE PROPERTIES

The amplitude of a wave may be above the resting state (positive) or below (negative)...no charge implied

A node is a point at which the amplitude is zero

Waves reinforce creating a wave of higher amplitude if they they are in phase.

+

–

+

–

+

–

Waves interfere if they are out of phase and create a wave which is of lower amplitude. Complete interference results in the cancelling of one wave by another.

+

–

+

–

+

–

2

2

Chem 61Chapter 1-15 Quantum Mechanics

1s, 2s, 2p orbitals

1s is spherical with the same phase throughout

2s is spherical with a node node is where ψ2 = 0

2p... three orbitals of equal energy (degenerate)

4s __3p __ __ __3s __2p __ __ __ order of filling of atomic orbitals2s __1s __

2pz2py2px

+

+-

node

node

Chem 61Chapter 1-16 Molecular Orbitals

MOLECULAR ORBITALS

Molecular orbitals = Linear combination of atomic orbitals

2 Atomic Orbitals must produce 2 Molecular Orbitals (the number of molecular orbitals equals the number of atomic orbitals which were combined to form them)

the hydrogen molecule

the energy of the hydrogen molecule with two electrons in the sigma orbital is 104 kcal/mole more stable than the separate hydrogen atoms; ∆E = 52 kcal/mole

if one electron is in the sigma and one in the sigma*, the molecule is of higher energy than the two separate atoms because the s* is slightly >∆E higher than the s orbitals while the s is ∆E lower

Bonding orbital...high electron density between nuclei

Antibonding orbital..node between nuclei (zero electron density)

σ bond is cylindrically symmetrical

Aufbau principle...fill lowest energy orbitals first

Hund's Rule...place one electron in each degenerate orbital first, then pair up

Pauli Exclusion Principle...two electrons in the same orbital must have opposite spins

H1s H1s

σ

σ*

bonding

antibonding

.

. .H2

∆E

>∆E

.Ψ1

Ψ2

Ψ1 − Ψ2

Ψ1 + Ψ2

4-sp3's

Chapter 1-17 Chem 61Molecular Orbitals of Carbon

MOLECULAR ORBITALS ON CARBON

Overlap between atomic orbitals in complex molecules often results in electron repulsions which destabilize the molecule. Hybrid orbitals allow for better overlap and a more accurate prediction of molecular structure.

Methane CH4 has a central carbon with four equivalent bonds to hydrogen

bond angles = 109.5°bond lengths = 1.09Å

sp3 Hybridization

carbon has electronic configuration 1s22s22p2

2p

2s

1s1s

2s

2p

sp3

1s

96 kcal hybridize

- +

2s3-3p's

C

H H

H H

+ 109.5° apart+

sp3 orbitals point toward the corners of a tetrahedron, 109.5° apart

any carbon bonded to four other atoms is sp3 hybridized, e.g. CH4, H3CCH3,

C—C bond length = 1.54Å

C-H σ bonds require 104 kcal/mol to be broken

-

methane

bond angles = 109.5°; tetrahedral

Chapter 1-18 Chem 61sp2 Molecular Orbitals of Carbon

sp2 Hybridization

sp2 hybridized carbons are trigonal planar with atoms 120° apart

sigma σpi π

C CHH

H H

H

HH

H

C

H

HC

H

H

C

H

HC

H

H

C

H

HC

H

H

C

H

HC

H

H

sigma σ*pi π*

ethylene C=C double bond: one sigma, one piE

hybridize96 kcal

1s

2p2p

2s

1s 1s

2s

2psp2

ethylene: trigonal planar

sigma* orbitalpi* orbital

pi orbital

sigma orbital

sp22p

120°

bond angles = 120°; C=C bond length = 1.34Å

pi (π) bonds are formed by the side by side overlap of two p-orbitals (approx. 68

kcal/mol)

pi bonds are above and below the plane where the sigma bond is located

pi bonds make the molecule rigid between the two atoms preventing rotation


2-2p's on each carboncombine to form pi-orbitals

2p

2s

1s1s

2s

2psp

1s

96 kcal hybridize

H H

H H

C C HH

C C HH

Chapter 1-19 Chem 61sp2 Molecular Orbitals of Carbon

sp Hybridization

sp hybridized carbons are linear with atoms 180° apart

1-2p 2-sp's

180° apart; the remaining two 2porbitals are 90° to the sp andeach other

+

-

+

2s

2p

C—C σ*C—C π*C—C π

E

C—C σ

Acetylene: linear; bond angles 180° C=C bond length = 1.20Åacetylene has two perpendicular pi bonds and one sigma bond

acetylene π-orbitals

acetylene σ-orbital2-sp's

CH3 OH

CH3 CO

OCH3

CH3 CO

OHCH3 C

O

HCH3 C

O

CH3

CH3 OCH3 CH3 NH2 CH3 Br

CH3CH3 CH2 CH2 HC CH

N

HHH

N

HCH3H

Chapter 1-20 Chem 61Functional Groups

FUNCTIONAL GROUPs

alkenesalkanes

alkyl halidesamines

O

HH

O

CH2CH3

HO

CH2CH3

CH2CH3

hybridize

one sp3 orbital already filledwith an unshared pair of electronsbonding can occur to three other atoms

N: 1s22s22p3

methyl amineammonia

....

1s

Oxygen: sp3

2s

2p sp3

1s

ethanol

esterscarboxylic acidsketonesaldehydes

ethersalcohols

Hybrid orbitals of oxygen and nitrogen

Nitrogen: sp3

alkynes

2s

2p sp3

1s

hybridize

sp3

acetone

..

sp2

..

diethyl ether

..O

CCH3

CH3....

water

1s

..

two sp3 orbitals already filled with an unshared pair of electronsbonding can occur to two other atoms

O: 1s22s22p4

....

– O C O –O

– O C O –O –

– O C OO –

2/3– O C O –2/3O –2/3

Chapter 1-21 Chem 61Resonance Structures

RESONANCE STRUCTURES

Chapter 6; Bruice, Pages 260 - 282:

Some molecules cannot be accurately represented by one simple line-bond formula:

they are "hybrids" of two or more structures

1. Resonance structures exist only on paper...the actual structures are hybrids of all the resonance structures

2. Resonance structures differ only in the position of electron pairs....not atoms

3. All structures should be proper Lewis structures (exceptions)

4. All resonance structures should have the same number of unpaired electrons

Nonequivalent Resonance Structures

1. Structures with a maximum number of octets is preferred.

2. Charges should be located on atoms with compatible electronegativity.

3. Minimize charge separation.

4. Charge separation may be enforced by the octet rule (atoms may be charged if they have an octet.)

CH2 CH2 CH3 CH3

C CCH3 CH3H3C–H2C CH2–CH3

CH3 CCH3

CH3

CH3 CH3 C C CH3

H

H

CH3

HCH3 C C C

H

H

H

HCH3

H

H

CHAPTER 2

Hydrocarbons: compounds containing only hydrogen and carbon: alkanes, alkynes, alkenes

alkanes contain only C—H and C—C single bonds CnH2n+2; alkenes CnH2n contain C— C double bonds and alkynes CnH2n-2 contain C—C triple bonds

Alkanes do not react with hydrogen, but alkenes and alkynes can react with hydrogen under certain conditions

2H2, catalyst

H2, catalyst

alkene

Chapter 2-1 Chem 61Hydrocarbons

alkane

alkyne

ISOMERISM

Structural Isomers: compounds with the same molecular formula that differ in the order in which the atoms are bonded to one another

C5H12n-pentane

C5H122-methylbutane

C5H122,2-dimethylpropane

CH3 CCH3

CH3 C C CH3

H

H

CH3

HCH3 C C CH3

H

H

H

CH3 CH

CH3

NOMENCLATURE

Alkanes are named by the following parent names

CH4 CH3CH3 CH3CH2CH3 CH3(CH2)2CH3 CH3(CH2)3CH3CH3(CH2)4CH3 CH3(CH2)5CH3 CH3(CH2)6CH3 CH3(CH2)7CH3 CH3(CH2)8CH3

Cyclic alkanes are named the same but with cyclo as a prefix

methaneethanepropanebutanepentanehexaneheptaneoctanenonanedecane

C1C2C3C4C5C6C7C8C9C10

cyclobutanecyclohexane

Branched hydrocarbons are named from the parent with a substituent as a prefix hydrocarbon

branches are named by dropping ane from the parent and adding -yl

thus methane: methyl ethane: ethyl propane: propyl , etc.

Chapter 2-2 Chem 61Nomenclature

tert-butyl isobutyl sec-butylisopropyl

special trivial names for branches

– 23 –

CH3 C C CH3

H

H

CH3

H

CH3 C

H

H3C

C

H

H

C C

CH3

CH3

H

H

CH3


methyl

butane is the parent

the branch is CH3; therefore methane

4

BASIC RULES OF NOMENCLATURE

1. Find the longest continuous chain (not necessarily drawn as a straight line) and name the parent

2. Number the parent chain starting at the end nearest the branch

3. Identify the branch and its position

4. Attach the number and the name of the branch to the parent name.

3

C H

OC

O

2 1

2-methylbutane

3,3,5-trimethylhexane

Other Functional Substitutents

increasingpriority -ol

-amine-ene-yne

prefixsubstituents

-one

-al

-oic acid

—OH—NR2—C=C——C≡C—R—,C6H5—, Cl—, Br—, —NO2

—CO2H

– 24 –

CH3 C C C

CH2CH3

CHO

H H

H CH3H

CH3 C C C

CH2CH2CH3

CH3

H O

H H

H C C C

H

COOH

H H

Br H H

CH3 C C C

CH3

H

H H

HO H H

CH3 C C C C CH3

H H

CH3H H CH3

4-methyl-3-heptanone2-ethyl-3-methylpentanal

2-pentanol


2,5-dimethyl-2-hexene

4-bromobutanoic acid

alkenes

alcohols

carboxylic acids

aldehydes and ketones

– 25 –

Chapter 2-5 Chem 61Alkanes

ALKANES

Physical Properties

nonpolar compounds

C1 to C4 are gases; C5 to C17 are liquids; >C17 are solids

Boiling points increase about 30°C for each additional CH2 unit

branching lowers the boiling point due to disruption of van der Waals attractions

insoluble in water; soluble in organic solvents like diethyl ether, benzene

Chemical Properties

very unreactive compounds

Halogenation

CH3CH2Cl + HCl + other productslight

CH3CH3 + Cl2

Oxidation of Alkanes

Combustion

5CO2 + 6H2Ospark

CH3CH2CH2CH2CH3 + 8O2

oxidation: a reaction that either removes a hydrogen atom from a carbon or adds an electronegative element to the molecule (O, N, S, halogen).

CO2

CCl4

CH3CO2HCH3C≡NCH3CCl3

CH3CH3

HC≡CHCH3CH=OCH3CH=NHCH3CHCl2

CH2=CH2CH3CH2OH CH3CH2NH2CH3CH2Cl

high oxidation levellow oxidation level

– 26 –

Chapter 2-6 Chem 61Oxidation and Reduction

"Complete"/"incomplete" oxidation (combustion) of propane

3 O=C=O + 4 H2O "complete" combustion

6 CO + 8 H2O "incomplete" combustion

3 C + 4 H2O "incomplete" combustion2O2

7O2

5O2CH3CH2CH3

2 CH3CH2CH3

CH3CH2CH3

Heat of Combustion: energy released when a compound is completely oxidized to CO2 and water; depends mostly on number of CH2 units: approximately 157 kcal/ methylene unit

Reduction of Alkenes, Alkynes

reduction: a reaction that either adds H atoms or removes an electronegative atom from the molecule

heat

CH3CH3Pd, Ni, PtH2

no reaction (NR)

CH3CH3

CH2=CH2

Pd, Ni,or Pt; H2CH3CH3

CH2=CH2

HC≡CH

Pd, Ni,or Pt; H2

Pd, Ni,or Pt; H2

Pd, Ni,or Pt; H2

0 60 120 180 240 300 360

0 kcal

Eeclipsed

staggered

Ethane

0 60 120 180 240 300 360

E

gauche anti

eclipsed methyls

eclipsed

CC

H

H

HH

HH H

H

H H

H

H

HH

H

H

H

H

HCH3

H H

H

CH3

H CH3H

H

H

CH3

HH

H CH3

H

CH3

H HH

H3C

H

CH3

H HCH3

H

H

CH3

HH

CH3 H

H

CH3

CONFORMATIONS OF OPEN CHAIN COMPOUNDS

Molecular Mechanics

Steric Energy: (isolated molecule in gas phase at 0° K): relative energy of a conformation or stereoisomer calculated using classical mechanics (atoms and bonds treated as balls and springs)

Stretch (bond length): energy associated with stretching or compressing bonds from their optimal length

Bend (bond angle): energy associated with deforming bond angles from their optimal angle

Stretch-Bend: energy required to stretch two bonds involved in a severely compressed bond angle

dipole-dipole: energy associated with interaction of bond dipoles

out of plane: energy required to distort a trigonal center out of planarity

torsional strain: destabilization from eclipsing of bonds on adjacent atoms

van der waals strain: destabilization from two atoms being too close together

Ethane

Chapter 2-7 Chem 61Conformations of Acyclic Hydrocarbons

Newman projection

staggered eclipsed

dimensional

∆H = 3 kcal/mole

Butane

3.8 kcal

4.5 kcal

0.9 kcal

anti eclipsed1 gauche eclipsed2 gauche eclipsed3

staggered

Chapter 2-8 Chem 61Conformations of Cyclic Hydrocarbons

H

H

H

H

H

H

H

H

H

HH

H

H

H

H

H

H

H

-∆H (kcal/mole)

499.8655.9 793.5 944.5

-∆H per CH2

166.6 164.0158.7 157.4

strain energy per CH2

9.2 6.6 1.3 0

total strain energy

27.626.46.50

cyclopropanecyclobutane cyclopentanecyclohexane

cyclopropane: bond angles 60°; tetrahedral 109.5°

cyclobutane and cyclopentane

sp3 orbitals are 109.5° apartcyclopropane bond angles 60°maximum overlap cannot be achieved

envelope cyclopentanepuckered cyclobutane

puckering allows bond angles to be at or close to the tetrahedral angle and minimizes torsional strain (electron—electron repulsions in eclipsed bonds) between adjacent C—H bonds

CYCLIC COMPOUNDS

Strain energy

H

H

H

HHH

H

H

H

HH

H

H

H

H

HHH H

H

H

HH

H H

HH

HH

HH

H

H

HH

H

H

H

H

HH

H

H

H

HH

H

H

axial bonds

half chairchair

chair

boat

Chapter 2-9 Chem 61Conformations of Cyclohexane

twist boat

cyclohexane

equatorial bonds

chair cyclohexane

E

chair

half chairboat

twist boat

10.8 kcal5.57.1

0 60 120 180 240 300 360

H

H

H

H

CH

H

HH

H

HCH3

HH

H

HH

H C

CCH3

H

H

H

HH

H C

CH

HCH3

H

HH

H

H

CH3

H

CH

H

CH3CH3

H

HC(CH3)3

HH

H

Substituted Cyclohexanes

substitutents on cyclohexanes preferntially occupy equatorial positions due to 1,3 diaxial interactions in axially substituted cyclohexanes

Chapter 2-10 Chem 61Conformations of Cyclohexane

more stable by 1.8 kcal/mole

anti

gauche

axial substitution similar to gauche butane equatorial substitution similar to anti butane

5.6 kcal/molemore stable

1,3-diaxial interactions

∆G = -RTlnK

∆G = -(1.98 kcal/mol°)(298)(2.94) = - (1.98 kcal/mol°)(298)(ln19) = -1.74 kcal/mole for methylcyclohexane

E

OH

OH

CH3

OH

OH

Br

CH3

CH3

H

OH

OH

H

H

Br

H

Br

C CCH3

H

H

CH3

C CH

CH3

H

CH3

C CCH3

H

H

CH3CH2

C CH

Cl

H

Cl

Chapter 3STEREOCHEMISTRY

Geometric isomerism in cyclic compounds

cis

cis

trans-1,3-dimethylcyclobutane

Geometric Isomerism in alkenes

68 kcal/mole to cleave a carbon-carbon pi bond thus no "free" rotation

trans-1,2-cyclohexanediol

cis-2-butene

cis-1,2-dichloroethenetrans-2-pentene

trans-2-butene

Chapter 3-1 Chem 61Stereochemistry: Geometric Isomers

cis

Z zusammen (together)E entgegen (across)

Cahn-Ingold-Prelog Sequence rules

1. if the atoms are different, highest atomic number gets highest priority

2. if two isotopes of the same element, the one with the higher mass gets higher priority

3. if the atoms are the same, the atomic numbers of the next atoms are used to assign priority

4. atoms attached by double or triple bonds are given single bond equivalencies

C CCl

H

CH3

CH3CH2C C

F

I

Cl

Br

R C

O

R' R C

(O)

R'

O (C)

R C

O

OH R C

(O)

OH

O (C)

R C CR' R C (C)

(C)

C (C)R'

(C)

R C C—R' R C H

(C)

C R'H

(C)

H H

Chapter 3-2 Chem 61Absolute Configuration

Stereoisomers: compounds with the same structures differing only in their arrangement of atoms in space.

not cis or trans

=

=

=

000

000

000

000

I

000

000

000

000 Br

0000

0000

0000

0000

Br

000

000

000

000

I

H

0000

0000

0000

Cl

H

000

000

000

000 Cl

0000

0000

0000

0000

I

0000

0000

0000

0000

Br 0000

0000

0000

Br

0000

0000

0000

0000

I

H

000

000

000

000

Cl

H

0000

0000

0000

Cl

Chapter 3-3 Chem 61Stereochemistry: Chirality

H

C*

CH2CH3

CH3Cl

H

C*

CH2CH3

ClCH3

OH

C*

CH2CH2CH3

CH2CH3CH3

Br

C*

CH3

H

HCCCH3

OH

CH3

C

CH2CH3

OHH

CH3

CH2CH3

OHHHH OH

H

CH3

OHHH OH

CHIRALITY

An object or molecule which cannot be superimposed on its mirror image is said to be chiral

If an object or molecule can be superimposed on its mirror image, it is achiral.

Enantiomers: isomers which are nonsuperimposable mirror images.

Stereogenic carbon atom: a carbon with four different groups bonded to it (designated *).

Fischer Projections

by convention: horizontal bonds come out of the paper vertical bonds go back into the paper

enantiomers

mirror plane

Chapter 3-4 Chem 61Stereochemistry: Chirality

OPTICAL ROTATION

enantiomers have almost all the same physical and chemical properties

properties which differ are:

1. interaction with other chiral substances

2. interactions with polarized light

Polarimeter

lamp

ordinarylight

polarizer

polarized light

solution ofsample

rotated light

if plane polarized light is passed through a solution of a single enantiomer, the light is rotated either to the right or the left; the opposite enantiomer will rotate the light in the opposite direction

optically active: a compound which rotates plane polarized light optical isomers: enantiomers

dextrorotatory: rotates plane polarized light to the right, also (+) or dlevorotatory: rotates plane polarized light to the left, also (–) or l

racemic mixture: a 1:1 (50:50) mixture of two enantiomers

does not rotate plane polarized light; therefore optically inactive

Chapter 3-5 Chem 61Stereochemistry: Absolute Configuration

Absolute configuration: the order of arrangement of the four groups around a stereogenic center

enantiomers have opposite configurations

C

H

Br

CH3

Cl C

H

CH3CH2

CH3

OH

C

H

Br

Cl

CH3 C

H

CH3

Cl

Br

(R) and (S) Cahn-Ingold-Prelog System

R: rectus or right S: sinister or left

To assign R and S to an asymmetric atom

1. Rank the four attached groups from 1 (highest) to 4 (lowest) priority based on the Cahn-Ingold -Prelog Sequence rules

2. Project the molecule with the lowest priority group to the rear

3. Draw a semicircle from 1 to 2 to 3

4. If the direction of the semicircle is clockwise; configuration is R; if counterclockwise; configuration is S

R

If lowest priority group is out; assign the configuration as usual and then reverse it

R

lowest priority group out S SR

1 2

3

4

12

3

4

1

2

3

4

1

2

3

4

CH2OH

C

C

CH3

OHH

OHH

CH2OH

C

C

CH3

HHO

HHO

CH2OH

C

C

CH3

HHO

OHH

CH2OH

C

C

CH3

OHH

HHO

CH2OH

C

C

CH2OH

OHH

OHH

CH2OH

C

C

CH2OH

HHO

HHO

CH2OH

C

C

CH2OH

HHO

OHH

CH2OH

C

C

CH2OH

OHH

HHO

CH3CH3

HH

CH3CH3

HHH H

HO OH

H OH

HO H

Molecules with Two or more Asymmetric centers

If a molecule has n asymmetric carbon atoms, it contains a maximum of 2n isomers; may not have that many

diastereomers

enantiomersenantiomers

two asymmetric carbons; 4 isomers

diastereomers: stereoisomers which are not enantiomers; may have different physical and chemical properties

Meso compounds

meso compounds contain an internal plane of symmetry and are achiral: that is the mirror image is identical to the original

diastereomers

Chapter 3-6 Chem 61Stereochemistry: Diastereomers

meso: identical

enantiomerssuperimposable

diastereomers

chiralmeso

meso chiral

mirror plane mirror plane

Chapter 3-7 Chem 61Stereochemistry: Asymmetric Synthesis, Resolution

OEt

O O

OEt

HO OH

OH OH

OH

H

Preparation of Enantiomerically Enriched Compounds

Generating chiral compounds from achiral compounds

Enzymes

yeast

Asymmetric Reagents

>95% one enantiomer

>98% R

t-BuOOH

(+)-diethyl tartrateTi(i-OPr)4

Resolution of a Racemic Mixture

racemic

(R) R*COOH + (S) R*NH2

(S) R*COOH (S) R*COO–(S) R*NH3

+

(R) R*COO–(S) R*NH3+

+ +

diastereomeric salts (separable)

separated salt converted back to acid

(R) R*COOH

(S) R*COOH


ACIDS AND BASES

Bronsted-Lowry: acid: a proton donor base: a proton acceptor

Strong acid: completely ionized or dissociated in water e.g., HCl, H2SO4, HNO3, HBr

Weak acid: only partially dissociated in water:

carboxylic acids are weak acid

conjugate acidconjugate basebaseacid

CH3CO2-

+ H3O+CH3CO2H + H2O

amines are weak bases

conjugate acid conjugate baseacidbase

CH3NH3+ + HO – CH3NH2 + H2O

Generally: strong acids have weak conjugate bases and weak acids have strong conjugate bases

that is, as acid strength increases, the basicity of the conjugate base decreases Thus the ability of the conjugate base to stabilize a negative charge determines the strength of an acid

Conjugate Acids


H2O HCN CH3CO2H H3PO4 HCl

15.75 6.37 4.75 2.12 -7pKa

Conjugate BasesHO – NC – CH3CO2

– H2PO4 – C l –

decreasing base strength


Factors affecting Acidity:

the electronegativity and the size of the atom which carries the negative charge influence its ability to stabilize the negative charge

electronegativity

size of the atom

increasing electronegativity of atom, increasing acid strength

(CH3)3C—H (CH3)2N—H CH3O—H F—H

increasing size of halogen, increasing acid strength

H—F H—Cl H—Br H—I

pKa 3.45 -7 -9 -9.5

pKa 50 35 15.5 3.45

ACIDITY CONSTANTS, Ka's

for acetic acid

CH3CO2H + H2O CH3CO2– + H3O+

Ka =[CH3CO2

–] [H+][CH3CO2H]

since stronger acids are more ionized, the larger Ka, the stronger the acid

pKa = -logKa, the lower the pKa, the stronger the acid

also, the higher the pKa, the weaker the acid or the stronger the base

the stronger the acid, the more stable the anion produced by ionization of the acid.

Chem 61Chapter 4-3 Lewis Acids and Bases

LEWIS ACIDS AND BASES

Lewis acid: electron pair acceptor: any species with an electron deficient atom

Lewis base: electron pair donor; any species with an unshared pair of electrons

BBr3, AlCl3, H3C+

..

.. ..H3N:, CH3CH2OH, H2C=O:

Chapter 5-1 Chem 61Alkenes: Structure and Isomerism

Alkene Structure

sp2 hybridized carbons are trigonal planar with atoms 120° apart

sigma σpi π

C CHH

H H

H

HH

H

C

H

HC

H

H

C

H

HC

H

H

C

H

HC

H

H

C

H

HC

H

H

ALKENES

sigma σ*pi π*

ethylene C=C double bond: one sigma, one piE

hybridize96 kcal

1s

2p2p

2s

1s 1s

2s

2psp2


sigma* orbitalpi* orbital

pi orbital

sigma orbital

sp22p

120°

bond angles = 120°; C=C bond length = 1.34Å

pi (π) bonds are formed by the side by side overlap of two p-orbitals (approx. 68 kcal/mol)

pi bonds are above and below the plane where the sigma bond is located

pi bonds make the molecule rigid between the two atoms preventing rotation

Cahn-Ingold-Prelog Sequence rules

1. if the atoms are different, highest atomic number gets highest priority

2. if two isotopes of the same element, the one with the higher mass gets higher priority

3. if the atoms are the same, the atomic numbers of the next atoms are used to assign priority

4. atoms attached by double or triple bonds are given single bond equivalencies

not cis or trans

Chapter 5-2 Chem 61Alkenes: Structure and Isomerism

Geometric Isomerism in alkenes

68 kcal/mole to cleave a carbon-carbon pi bond thus no "free" rotation

cis-1,2-dichloroethene

cis-2-butene

trans-2-pentene

Stereoisomers: compounds with the same structures differing only in their arrangement of atoms in space.

trans-2-butene

Z zusammen (together)E entgegen (across)

C CCH3

H

H

CH3

C CH

CH3

H

CH3

C CCH3

H

H

CH3CH2

C CH

Cl

H

Cl

C CCl

H

CH3

CH3CH2C C

F

I

Cl

Br

C CHH

H H

H

HH

H

C

H

HC

H

H

C

H

HC

H

H

C

H

HC

H

H

C

H

HC

H

H

C

H

HC

H

H

Alkenes: Reactivity

The reactivity of alkenes is due to their ability to donate a pair of electrons: their Lewis basicity.

pi* orbitalpi orbital

sigma orbital

sp22p

120°

Consider Ethylene:

The site of reactivity is the pi-bond due to the exposed nature of the pi electrons

sigma* orbital

H2C CH2

Br H

The pi electrons act as a Lewis base (electron pair donor).These electrons react with electron deficient species (Lewis acids)

Alkenes undergo electrophilic additions reactions.

Electrophile: an electron deficient ion or molecule

Nucleophile: an electron rich ion or molecule.

Reaction Mechanisms of Electrophilic Addition Reactions

reaction mechanism: a detailed description of how a chemical reaction occurs.

A roadmap of a reaction....curved arrows show which bonds are formed or broken. A mechanism includes the transition states involved in making and breaking bonds and reactive intermediates that are formed along the pathway from reactants to products.

H+ Br+

Chapter 5-3 Chem 61

HH

H

H

Alkenes: Electrophilic Addition Reactions

BH3 +CH3

I– CH3NH2 H2O

H

H–BrCH2—CH2–H

HH

CH2=CH2

Br–

HH

H

Br

H

+

Br

2p

Br–

Chapter 6-1 Chem 61Thermodynamics

Thermodynamics

describes the properties of a system at equilibrium

∆G = – RT ln Keq where Keq = [reactants]

reactants products

transition stateEnergy Diagram

E

progress of reaction

reactants

reactants

products

products

– ∆G

+ ∆G

∆G ‡

exergonic

endergonic

[products]

R = 1.986 X 10-3 kcal K-1 mol-1 (gas costant)T = temperature in degrees Kelvin

∆Go = ∆Ho – T∆So

free energy = enthalpy – T x entropy

enthalpy = heat of reactionentropy = state of disorder

Chapter 6-2 Chem 61Kinetics

Kinetics

describes the rate of progression of a reaction

depends on the energy of activation (the stability of the transition state):

the lower the transition state energy, the faster the reaction will be.

reactants products

transition state

E


∆G ‡

first order reaction

rate = k[A] rate is proportional to the concentration of one reactant

second order reaction

rate = k[A] [B] rate is proportional to the concentration of two reactants

reactants products

transition state

E


∆G ‡slower

faster

reactants products

transition statetransition state

intermediateE

A two step reaction

Rate limiting step:

is the slowest step (step with the highest energy of activation)

∆G‡ ∆G‡

C C

C C C C

E

C C

EA

CH3 C CH2

CH3

CH3 C CH3

CH3

CH3 C CH3

CH3

X

Electrophilic Addition: Addition of H–X

Electrophilic Addition Reactions

Nucleophilic addition does not occur with alkenes unless an electron-attracting group is attached to one of the carbon atoms to cause a polarity difference.

+ Nu:

A–

However, in the electrophilic addition reaction, the reagent is E+A– (E+ = electrophile, A– = some anion)

Chapter 6-3 Chem 61

+E+ A–+

Note the alkene acts as a Lewis base (or nucleophile) toward the Lewis acid (or electrophile)

The intermediate, a carbocation, reacts with A– to yield a product in which E and A have added to the C=C double bond.

Addition of HX (E+ = H– A– = F–, Cl–, Br–, I–)

No Reaction

++ H—X

Reactivity: H—I > H—Br > H—Cl > H—F

X–

The reaction is said to be regioselective since an unsymmetrical alkene gives a predominance of one of two possible electrophilic addition products. The term regiospecific is used if one product is formed exclusively.

In these reactions, the halogen (A–) is found attached to the most substituted carbon atom of the alkene (Markovnikoff's Rule):

CH3 C CH2

CH3

CH3 C CH3

CH3

CH3 C CH3

CH3

X

R C CHR

RR C CHR

R

R C CHR

R

X

R CH C—R

RR C CHR

R

HR C CHR

R

X

CH CH2 CH CH3 CH CH3

Cl OCOCH3

Chapter 6-4

++ H—X

Chem 61Electrophilic Addition: Addition of H–X

(CH3)2CH CH CH2 (CH3)2CH CH CH3

(CH3)2CH CH CH3 (CH3)2C CH2 CH3

Cl Cl

X –

3° carbocationmore favored

X –

+ H—X+

X –++ H—X

Thus regioselectivity is explained by the lower Eact leading to the 3° carbocation intermediate.

If a nucleophilic solvent is employed in the electrophilic addition reaction solvent may compete with A – for the intermediate carbocation.

2° carbocationless favored

HCl

Note the C=C's of the aromatic ring do not undergo this type of reaction.

Since a carbocation intermediate is involved, rearrangement may sometimes occur.

+CH3COOH(solvent)

HCl

40%+

+

60%

Chapter 6-5 Chem 61Carbocation Stability

CH2+

CHCH3

C(CH3)2

CH

C

Name

methyl

primary 1°

secondary 2°

tertiary 3°

vinyl

allyl

secondary allyl

tertiary allyl

phenyl

benzyl

secondary benzyl

tertiary benzyl

diphenylmethyl

triphenylmethyl

Structure

CH3+

CH3CH2+

(CH3)2CH+

(CH3)3C+

CH2=CH+

CH2=CH—CH2+

CH2=CH—CH+(CH3)

CH2=CH—C+(CH3)2

+

+

+

+

+

Hydride Affinity

(kcal/mol)

314

274

247

230

287

256

237

225

298

233

226

220

215

210

Carbocation stabilityCarbocations are highly reactive intermediates ; they cannot be observed directly in the reaction mixture since they react as soon as they are formed.

Any structural feature that will disperse a positive charge will stabilize the carbocation.

This stabilization has been quantitated by so-called hydride H:– affinity measurements (gas phase).

The lower the value of the hydride affinity, the more stable the carbocation.

Chapter 6-6 Chem 61Carbocation Stability

CCH3

CH3 C

H

H

CH

HC

CH

H

H

CH2 CH2CH CH2 CHCH2

Thus the order of carbocation stability is triphenyl methyl > diphenylmethyl > 3° ≈ benzyl ≈ allyl > 2° > 1° >> methyl

Carbocation centers can be stabilized by overlap with adjacent pi orbitals (resonance) or adjacent sigma orbitals (hyperconjugation).

σ

p

H sigma donation

adjacent sigma bond donates some electron density to the empty p orbital on the carbocation center

more adjacent sigma orbitals result in a more stable carbocation thus 3° > 2° > 1° > methyl

..

equivalent resonance structures

+ +

empty p orbital ovelaps with adjacent pi orbital to disperse the positive charge

pip

Since the pi orbital is closer in energy to the empty p than the sigma orbital, the pi overlap stabilizes the cation more efficiently.

+

+

Chapter 6-7 Chem 61Electrophilic Addition: Addition of H2O

Addition of Water (E+ = H+; A– = HOH)

R2C CHR R2C CH2 R

R2CR2C CH2 R

CH2 R

:OH

H

R2C CH2 R

O:+HH

R2C CH2 R

:O H

..

use H2SO4, H2O

Step 1

Step 2

Step 3:OH2

R2C CHR

R2C CH2 R

R2C CH2 R

OH

H

Eact

+

H+

R2C CH2 R

O H

+

+..

+

TS1

TS2

TS3

Addition of water to alkenes is a reversible reaction and whether the alkene orthe alcohol predominates at equilibriuim depends on the reaction conditions.

Low temperatures and high concentrations of water favor the alcohol.

Higher temperatures and removal of water favor the alkene.

:OH2

..

+ H3O+

+

bridged carbocation

CH3 C

O

O O C

O

CH3Hg –O C

O

CH3O C

O

CH3+Hg

C CH2

CH3

CH3

O C

O

CH3+Hg C CH2CH3

CH3

O C

O

CH3Hg

C CH2CH3

CH3

O C

O

CH3+Hg

C CH2

CH3CH3 O C

O

CH3Hg

+OH2

C CH2

CH3CH3C CH2HgOCOCH3

CH3

HgOCOCH3

HO

CH3

+OH2

C CH3 + Hg°

CH3CH3

HO

+

mercury acetate E+

carbocation

++

Chapter 6-8

C CH2

CH3

CH3

C CH3

CH3CH3

RO

Chem 61Electrophilic Addition: Oxymercuration

Since no rearrangement is observed in reactions with +HgOCOCH3 thebridged carbocation is believed to be the true intermediate. Attack by H2O on the more electron deficient carbon opens the bridged cation (backsideattack).

NaBH4 reduction of the C—Hg bond yields the alcohol in a separate step.

or

Addition of Mercury acetate and Water ( E+ = +HgOCOCH3; A– = OH2)

To avoid carbocation rearrangements under the conditions used in addition of H2O (H2SO4/H2O) a better system for the addition of water to a C=C has been devised.

H2O

NaBH4

If alcohol is used as the solvent in this process ethers are obtained as the products.

1. Hg(OCOCH3)2, ROH

2. NaBH4

H2O:

C CH H

R H

H B HH

C CH H

R H

H B HH

C C

H HR H

H B HH

4 BH3 (borane)

2 B2H6 (diborane)+ 3 NaBF4

Addition of BH3 to C=C, C≡C is termed hydroboration.

R—CH=CH2 + BH3

Note B is the electrophile and H acts as A-. The addition of H– and B occur

from the same face of the alkene (syn addition). Further reaction of

R—CH2–CH2—BH2 with more alkene yields first the dialkyl borane and then

the trialkyl borane as the stable product.

R—CH=CH2

( R—CH2—CH2)2BH

‡

Chapter 6-9 Chem 61Electrophilic Addition: Hydroboration

Brown discovered that trialkylboranes were easily oxidized by alkaline hydrogen peroxide (H2O2 in HO–). The oxidation reaction proceeds asshown.

3 NaBH4 + 4 BF3

R—CH=CH2 + R—CH2—CH2—BH2

Hydroboration (E+ = B, A– = H–)

Discovered by H.C. Brown in the 1950's. The reagent BH3 (borane) is used in electrophilic addition reactions.

Borane is generated by

( R—CH2—CH2)3B


RCH2 B – O

RCH2

OH

CH2R

RCH2 B O RCH2CH2CH2R

RCH2

B – O CH2R

CH2R

O OH

RCH2 B O CH2R

OCH2R

RCH2O B O CH2R

OCH2R

1. BH3

2. H2O2, – OH

– OH

The overall reaction is

– OOH

R—CH2—CH2—OH

– OOH

– OOH

+ —OH

( R—CH2)3B

3 R—CH2—OH + B(OH)3

R—CH=CH2

Note the reaction is regioselective and yields an alcohol that is isomeric to that obtained by Hg(OCOCH3)2—NaBH4 or H2SO4—H2O.

R—CH—CH3H3O+

or Hg(OCOCH3)2then NaBH4

R—CH=CH2

Treatment of a trialkylborane with CH3CO2D yields a monodeuterated alkane

CH3CH2DCH3CO2D

(CH3CH2)3B

OH


Treatment of a trialkylborane with CH3CO2D yields a monodeuterated alkane

CH3CH2DCH3CO2D

(CH3CH2)3B

CH3

H

B

CH3

HH

OH

CH3

HH

OH

CH3

HH

Bromination of a trialkylborane yields the alkyl bromide.

C6H5CH2CH2BrBr2

HO–

(C6H5CH2CH2)3B

Stereochemical Results of Hydroboration

Hydroboration is regioselective and stereospecific

cis-addition

BH3 H2O2

HO– +

enantiomers

Thus H and OH add cis to the C=C.

C C C C

Br

C C

Br

Br

C CH H

CH3CH3

CC C

H

HCH3

CH3

X

XC

H

H

CH3

CH3

X

X

C CH CH3

HCH3C C

H

CH3H

CH3

X

X

X

X

HH

X

X

HH

Stereochemistry of Halogenation

Alkenes containing alkyl substituents react with Br2 via a bridged bromonium ion intermediate. Nucleophilic opening of the bridged ion by backside attackof Br – at carbon gives overall anti (trans) addition of E+(Br+) and A– (Br–) tothe double bond. Chlorination reactions proceed by a similar pathway.

Some examples:

Br—

+

Chapter 6-12 Chem 61Electrophilic Addition: Stereochemistry of Halogen Addition

+

mesoBr2

or Cl2

(enantiomers)

+

Br2

or Cl2

CH3CO2H(solvent)

Br2 or Cl2

CCl4

C C

C

C C

Br+ CBr

Br

CCCl

Br

C C CC C

X+

COH

X

C C

C C

C C

S+

C C C C

I+

CC

Cl

RS

CC

NO

Cl

CC

Cl

I

R

CC

NO

Chapter 6-13 Chem 61Electrophilic Addition: Mixed Addition

Mixed Addition

Cl—

Br—

Br+

Br2 or Cl2 with H2O, HO– present:

HO–X+

Mixed Reagents

I—Cl +

O=N—Cl +

RS—Cl +

+

Chapter 6-14 Chem 61Hydrogenation of Alkenes

CH3

CH3 CH3

CH3

H

H

C C CH3CH3 C CCH3

H

CH3

H

C C

CH3

H

C

CH3

H

C

CH3

CH3

H

H

C C

CH3CH2

H

H

H

Hydrogenation of Alkenes

addition of H2 is cis (syn)

H2/Pd

H2/Pd

ii) Heat of Hydrogenation Studies The heat of hydrogenation of an alkene is the energy difference between the starting alkene and the product alkane. It is calculated by measuring theamount of heat released in a hydrogenation reaction.

CH3CH2CH2CH3 + heat

E

progressprogressprogress

∆Hh∆Hh

∆Hh

E-CH3CH=CHCH3Z-CH3CH=CHCH3CH3CH2CH=CH2

Alkene Heat of Hydrogenation, ∆Hh (kcal/mol)

CH3CH2CH=CH2 -30.3CH3CH=CHCH3, Z -28.6CH3CH=CHCH3, E -27.6

Thus the stability of the three alkenes is

CH3CH=CHCH3, E > CH3CH=CHCH3, Z > CH3CH2CH=CH2

These comparisons indicate

i) increasing alkyl substitution stabilizes an alkene ii) conjugated dienes are more stable than non-conjugated dienes,iii) trans alkenes are more stable than cis alkenes (steric repulsions in cis)

CH3 C CH2

CH3

CH3 C CH3

CH3

CH3 C CH3

CH3

X

CH3 C CH2

CH3

CH3 C CH2

CH3

CH3 C CH2

CH3

H

R C CHR

RR C CHR

R

R C CHR

R

H

termination

.Br2

3° carbocationmore favored

.

.

ROOR+ H—Br

+

RO–ORperoxide

heat or light

2 RO .

Br Br

propagation

+ Br

RO . RO–H + Br .

Br

Br

Chem 61Radical Addition Reactions

Br

.

Addition of HBr in the presence of a radical initiator such as peroxides effectsso called anti Markovnikov addition of HBr

++ H—XX –

Reversal of regioselectivity (position of bromine and H addition) results from the inital addition of Br radical rather than H+.

initiation

CH3 C CH2

CH3

Br

propagation

Chapter 6-15

+ Br

+ H—Br

. . H–Br

+ Br

+ Br

CH3 C CH2

CH3

Br

.

Br.

RO–OR2 RO .

combination of any two radicals

Free Radical: any atom or group of atoms that contains one or more unpaired

electrons.

A free radical is symbolized as a single dot representing the unpaired electron: Cl.,

Br ., H3C., etc.

Free radicals are usually encountered as high energy short-lived, non-isolable

intermediates in certain reactions.

Stability of Carbon Free Radicals (Which H atom is abstracted?)

Stability of carbon free radicals follows the same pattern as carbocations:

CH2=CH—CH2. , C6H5CH2

. > (CH3)3C

. > (CH3)2CH

. > CH3CH2

. >

CH3.

allyl benzyl tertiary secondary primary methyl

b)

+ Br –

Note that the carbon in a) has lost an electron pair to the bromine atom, thus thecarbon has an sp2 configuration and a single "+" charge resulting from the empty p orbital.

In b) the carbon atom has lost one electron; the carbon is also in the sp2

configuration with the unpaired electron in the p orbital.

CH3 C Br

CH3

CH3

CH3 C Br

CH3

CH3

CH3 C

CH3

CH3

CH3

C.CH3

CH3

Free Radical Substitution Reactions

Free Radical: any atom or group of atoms that contains one or more unpaired

electrons

A free radical is symbolized as a single dot representing the unpaired electron:

Cl., Br., H3C., etc.

Free radicals are usually encountered as high energy short-lived, non-isolable

intermediates in certain reactions. (Recall that carbocations were seen to be

intermediates in certain reactions in Chapter 5).

a)

CH3C

CH3

CH3

CH3

C CH3CH3

single barbed arrows indicate the movement of one electron

free radical intermediate

carbocation intermediate

Br.

Chapter 7-1 Chem 61Radical Reactions of Alkanes

equal probability of electron being in either lobe

(CH3)3C.(CH3)3C+

.

double barbed arrows indicate the movement of two electrons

Chapter 7-2 Chem 61Radical Reactions of Alkanes

Cl Cl

Free Radical Substitution Reactions

Chlorination of methane in the presence of light is the classic example

a) Initiation of the radical chain reaction

CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCllightCH4 + Cl2

Reaction mechanism involves three general steps: Initiation, propagation, and termination

H C H

H

light

heat or2Cl

. requires 58 kcal/mole

b) Propagation (self-perpetuating the chain reaction)

two chlorine radicals

i)

a methyl radical

H—Cl + CH3. requires 1 kcal/mole

Cl.

chlorine radical can now react with more CH4and propogate the chain

CH3—Cl + Cl.CH3

. + Cl—Clii)

Note: Cl has substituted for a H and therefore free radical substitution

iii) As the concentration of CH3—Cl increases in the reaction mixture Cl. starts to

react with both CH4 and CH3Cl.

Cl—CH2—Cl + Cl. .CH2—Cl + Cl—Cl

H—Cl + .CH2—Cl (chloromethyl radical)Cl. + CH3—Cl

Hence CH2Cl2, CHCl3, and CCl4 are formed

c) Termination of the Chain Reaction

Any reaction of Cl. or of the carbon free radical intermediates (CH3.,

.CH2Cl, .CHCl2, .CCl3) that disrupts the propagation of the chain will

terminate the reaction.

Cl—CH2—CH2—Cl.CH2Cl +

.CH2Cl

CH3—CH3CH3. + CH3

.Cl

. + .CH2Cl CH2Cl2

CH3—ClCl. + CH3

.

Chapter 7-3 Chem 61Reactivity of the Halogens

Reactivity of the Halogens

Bond Dissociation Energy

F2

37

Cl2

58

Br2

46

I2

36 kcal/mole

Thus the energy of activation for formation of F. and I. is lower than Cl. and Br.

however, the order of reactivity of X. with alkanes is: F2 >> Cl2 > Br2 >> I2 (I2does not normally react; F2 reacts explosively) Thus the rate determining step is

not:

2X.X—X

H—X + .CH3X. + CH4

H—Cl + .CH3Cl. + CH4

Br . + CH4 H—Br +

.CH3

Rather, hydrogen atom abstraction is the rate determining step.

Since chlorination is faster than bromination, Eact for

must be lower than the Eact for

[CH3. + HCl +Cl2]

CH3Cl + Cl.

Cl. + CH4

Eact

Progress

E

Eact

CH3Br + Br .

[CH3. + HBr + Br2]

Br. + CH4

CH3 C

CH3

CH3

CH CH3 CH3 C

CH3

CH3

CH CH3 CH3 C

CH3

CH3

CH CH3

OH

CH3 C

CH3

CH3

CH CH3 CH3 C

CH3

CH3

CH CH3

Cl

H

CCH2Cl

CH3

CH3CH2

Br

CCH2Cl

CH3

CH3CH2

Cl

CCH2Cl

CH3

CH3CH2

Cl

CCH2Cl

CH3

CH3CH2

OCH3

CCH2Cl

CH3

CH3CH2OCH3

CCH2Cl

CH3

CH3CH2

Stability of Carbon Free Radicals (Which H atom is abstracted?)

Stability of carbon free radicals follows the same pattern as carbocations:

CH2=CH—CH2. , C6H5CH2

. > (CH3)3C

. > (CH3)2CH. > CH3CH2

. > CH3.

allyl benzyl tertiary secondary primary methyl

Unlike carbocations, free radicals do not rearrange to a more stable free radical.

Chapter 7-4 Chem 61Reactivity of the Halogens

2°

3°2°

Cl2.

H2O

++

Since both carbocations and free radicals involve planar sp2 carbon atoms, racemization will be observed in the products resulting from both intermediates.

(50:50)

(50:50)

RSS

R RS

+

+

CH3OH

Cl2

c) allylb) 3° d) allyl e) 2°

CH3 CH2 CH3 CH3 CH3 CH3

CH3 CH3

Br

Br

Selectivity of Hydrogen Atom Abstraction

The more stable free radical should determine the nature of the product

Br.

CH3 CH CH3

Cl

CH3 CH CH3

Br

.. .

..

a) 1°

Thus intermediates c) and d) should be more stable and should lead to

andas products.

Chlorination is much less specific than bromination

Chapter 7-5 Chem 61Selectivity of Halogenation

100%

45%55%

CH3CH2CH3

CH2CH3CHCH3

CHCH3

CH2CH2Cl

Cl

Br

Br2

Cl2 + CH3CH2CH2ClCH3CH2CH3

The difference in selectivity of H atom abstraction by Cl. and Br

. is explained by

the Hammond postulate. The transition state in the chlorination reaction is less

influenced by the stability of the intermediate free radical; thus both (CH3)2CH.

and CH3CH2CH2. result.

In bromination the more stable free radical intermediate is highly favored; thus

(CH3)2CH. is formed exclusively.

Other examples of this are:

Br2, light

Cl2, light

100%

(56:44)

+

NBr

O

O

NH

O

O Br

C6H5 C O O C C6H5

O O

C6H5 C O.

H O O

O

H H O.

CH3

CCH3

C≡N.

N N C CH3

CH3

C≡N.

CH3

C.

CH3

C≡N.

Bromination with N-Bromosuccinimide (NBS)

CCl4

light or peroxide

NBS acts as a Br2 source; the reaction is initiated by either light or a peroxide (ROOR). NBS is used to introduce a Br atom at allylic or benzylic positions.

+ +

Chapter 7-6 Chem 61Halogenations

a) dibenzoyl peroxide

b) hydrogen peroxide

c) alkyl hypochlorites

d) azobisisobutyrylnitrile

2

2

R—O—Cl RO. + Cl

.∆

+ N≡N2

Other Sources of Free Radicals

Although light (hν) and heat (∆) are used for halogenation reactions other reagents are often useful for initiating free radical reactions. Some of these reagents include:

Clues to Whether a Reaction Involves Free Radicals

a) Reaction requires high temperatures (>200°)b) Reaction requires light energy (hν)c) Reaction requires an initiator (a-d above) or oxygen.

Chapter 8-1 Chem 61Nucleophilic Substitution Reactions

X

Cl

Br

CH3

SH

OCH3

CH3

Nucleophilic Substitution Reactions (SN1, SN2)

alkyl halides: CH3—X R3C—XR2CH—X RCH2—X

tertiary, 3°secondary, 2°primary, 1° methyl

CH2=CH—Xaryl halide vinyl halide

do not easily undergo nucleophilic substitution reactions

Examples

+ HBr

+ NaCl

CH3CH2CH2—OH + Na+ – Br

CH3—OCH3 + Na+ – I

+ Na+ – SH

CH3 CH CH3 CH3

I

CH CH3

SH

+ CH3OH

CH3OH

CH3OH

CH3OH

CH3CH2CH2—Br + Na+ – OH

CH3—I + Na+ – OCH3

Anatomy of a nucleophilic substitution reaction

+ Na+ I –+ Na+ – SHCH3OH


CH3 CH CH3 CH3

I

CH CH3

SH

(L) Leaving Group: any group that can be displaced from a carbon atom (in thiscase I—)

(Nu) Nucleophile: the species that attacks the carbon atom bearing L anddonates the electron pair to form the nucleophile—C bond (in this case —SH)

(R—I) Substrate: the molecule, containing the leaving group that is acted on bythe nucleophile (here: 2-iodopropane)

Solvent: the medium used to dissolve the substrate and the nucleophile (in this case methanol, CH3OH); solvent can sometimes be the nucleophile (solvolysis)

Curved arrows are used to indicate the movement of electrons during anucleophilic substitution reaction. By convention electron movement is writtenfrom negative to positive.

electron pair that accompanies L

electron pair fornew bond at C

R—CH2—Nu + L:– Nu:– + R—CH2—L

δ+ δ −

Anatomy of a nucleophilic substitution reaction

+ Na+ – I+ Na+ – SHCH3OH

energy of transition state

potential energy E

Eact

products

reactants



CH

CH3

H

Br CH

CH3

H

CH3OC

HCH3

H

CH3O Br

C L

HH

H

Nu Nu C L

HH

H

Nu C

HH

H

L

Reaction Mechanisms of Nucleophilic Substitution Reactions

reaction mechanism: a detailed description of how a chemical reaction occurs. A roadmap of a reaction....curved arrows show which bonds are formed or broken. Amechanism includes the transition states involved in making and breaking bondsand reactive intermediates that are formed along the pathway from reactants toproducts.

CASE 1:SN2 Substitution Nucleophilic 2nd order (bimolecular):

CH3O–

partial bonds in transition state

sp2sp3

sp3

+ Br–

Orbital Picture

L nonbonding orbitalNu—C sigma

δ−δ−

+

C—L sigma*Nu nonbonding orbital

Energetics of SN2

∆H of reaction


Orientation of Nu: and L in SN2:

Nucleophile approaches substrate from the backside of L

Point of highest energy along the reaction coordinate is the transition state: theO—C bond is partially formed and the C—Br bond is partially broken

The Nu is bonded to the carbon on the opposite side of that occupied by theleaving group L: Net inversion of the carbon atom is observed in all SN2reactions

Reaction Rate: the time required for all substrate molecules to be converted toproduct

The rate of an SN2 reaction is proportional to the concentrations of the substrate and the nucleophile, thus the reaction is second order

Rate = k[substrate][nucleophile]

k is the proportionality constant called the rate constant

Rate and Eact

Under the same reaction conditions, the reaction with the lower Eact has a fasterrate

Reaction 2 below is faster

Rxn 2Rxn 1

EactEact

E

Increasing alkyl group substitution at the carbon atom bonded to the leaving grouphinders approach of the nucleophile. The Eact increases due to steric hindrance of nucleophile approach and thus the SN2 rate decreases


Increasing alkyl group substitution at the carbon atom bonded to the leaving grouphinders approach of the nucleophile. The Eact increases due to steric hindrance of nucleophile approach and thus the SN2 rate decreases

Relationship between Substrate Structure and SN2 Rate

Alkyl halide Relative rate of SN2

CH3—X 30

CH3CH2—X 1

CH3CH2CH2—X 0.4

(CH3)2CH—X 0.025

(CH3)3C—X ~0

Relative SN2 rate

93

1

0.0076

CH3—I + Cl–

CH3CH2—I + Cl–

(CH3)2CH—I + Cl–

CH3—Cl + I–

CH3CH2—Cl + I–

(CH3)2CH—Cl + I–

SUMMARY

SN2 reactions occur by the attack of a nucleophile on substrates containing the leaving group attached to a methyl, primary carbon, or secondary carbon.

SN2 reactions do not occur with tertiary alkyl halides

SN2 exhibit a single transition state (concerted reaction, i.e. not stepwise, no intermediates)

Inversion of configuration at the carbon results from backside attack of the nucleophile

The rate of SN2 reaction depends on the concentration of the nucleophile and the substrate

The rate of SN2 follows the order:

CH3—X > CH3CH2—X > CH3CH2CH2—X > (CH3)2CH—X >> (CH3)3C—X

Step 2: attack by the nucleophile

Step 3: deprotonation (loss of proton)

Chapter 8-6

(CH3)3C O+ CH3

H

Chem 61Nucleophilic Substitution Reactions

(CH3)3C—Br + CH3OH

fast

1

slow(CH3)3C+ + Br –(CH3)3C—Br + CH3OH

Step 1 involves ionization of the alkyl halide to the halide ion and the tertiary carbocation.

The step with the highest Eact is the slowest step (rate determining step) in the pathway.

Ionization is aided by polar solvents (H2O, ROH) which solvate and thus stabilize the carbocation and the leaving group anion.

The transition state is pictured as (CH3)3Cδ+------δ+OCH3

(CH3)3C O+ CH3

H

(CH3)3C O CH3

Step 1: ionization (loss of halide ion)

CASE 2: SN1: Substitution Nucleophilic First Order (Unimolecular)

Substrates containing the leaving group attached to a tertiary carbon atom reactwith weakly basic nucleophiles by an alternative nucleophilic substitution reaction mechanism

The process involves three steps:

(CH3)3C—OCH3 + HBr

3

H

Br–

+ HBrO+ H O

H

CH3

CH3

(CH3)3C

The transition state is pictured as (CH3)3Cδ+ ------Br δ–.

fast

..

..CH3O—H

2

(CH3)3C + Br –

The transition state is pictured as

Step 3 involves acid-base reaction between either Br– or more likely solvent CH3OH in a very rapid reaction

CCH3

CH3CH3

C

CH3CH2

CH3 CH2CH2CH3

Br

C

CH3CH2

CH3 CH2CH2CH3

OH

C

CH3CH2

CH3CH2CH2CH3

OH

Stereochemistry of SN1 reaction

the intermediate carbocation contains an sp2 hybridized carbon atom which is planar

+

Thus attack of CH3OH on the carbocation can occur equally from the top and bottom. If the carbon atom containing the leaving group were asymmetric,racemization of the product would be observed


R

+H2O

R S(50:50)

Reaction Rate of SN1

The rate of the SN1 reaction does not depend on the concentration of thenucleophile, but depends only on the concentration of the substrate

SN1 Rate = k[substrate]

The reaction is first order because the rate is proportional to the concentration of only one reactant

Relationship Between Substrate Structure and SN1 Rate

Alkyl halide Relative SN1 rate

CH3—Br

CH3CH2—Br

(CH3)2CH—Br

(CH3)3C—Br

*These observed reaction rates probably proceed through SN2 not SN1

SN1 rates reflect the relative Eact leading to the different carbocations (transition state 1)

Since the relative stability of carbocations is 3° > 2° > 1° > methyl; tertiary alkyl halides proceed at the faster rate

1.0*

1.0*

11.6

1.2 X 106

R+ + Nu:— or Nu:

+

3. Rearrangement to a more stable carbocation: occurs whenever a more stable carbocation can be formed

CH3 CH CH2

H

CH3 CH C

CH3

CH3 CH3 CH2 C

CH3

CH3

H

Allyl and benzyl primary halides

Allyl and benzyl primary halides are reactive substrates in both SN1 and SN2 type reactions.

Substrate Realtive SN1 Rate Relative SN2 Rate

CH3—X

CH3CH2—X

CH2=CH—CH2—X

C6H5—CH2—X

1.0

1.0

33

380

30

1

40

120

Reactions of Carbocations

1. Reaction with a nucleophile: SN1 reactions

2. Elimination of H+ on an adjacent carbon: E1 reactions

SUMMARY

SN1 reactions involve an intermediate carbocation which is attacked by the nucleophile to yield the product of nucleophilic substitution.

SN1 reactions occur with substrates that yield stabilized carbocations(allyl, benzyl, 3°, 2°, not primary or methyl)

SN1 reactions are first order in rate.

SN1 reactions lead to racemization.


+

CH3—CH=CH2 + H++

Chapter 8-9 Chem 61Elimination Reactions

(CH3)3C Nu +CH3 C

CH3

CH

H

H

C CH2

CH3

CH3

Elimination Reactions (E1, E2)

Case 1: E1: Elimination First order

Formation of a carbocation intermediate in a reaction is a first order process.

Once formed the carbocation may react with a nucleophile (SN1) or

lose a proton from an adjacent carbon to form an alkene (E1)

Thus SN1 and E1 reactions are competitive

Br – (SN1)

+ HBr (E1)

Br –(CH3)3C—Br +

E1 reactions predominate when the reaction contains only poor nucleophiles; otherwise SN1 reactions are the more likely pathway

Since a carbocation is involved, rearangement may occur

TS2‡

TS1‡

E

R–X

R+

alkene

Eact

H C Br

H

H

C

H

H HC C

H

H

H

H C C

Brδ−

HH

H

H

ROδ−

H C C

Brδ−

HH

H

H

ROδ−

CH3CH2CH2 CHCH3

Br

CH3CH2CH CHCH3

CH3CH2CH2CH CH2

CASE 2: E2: Elimination Second Order

Bimolecular elimination results when an alkyl halide is treated with a strong base (HO –, RO –, etc) at elevated temperatures (80°-120°).

The E2 reaction does not involve an intermediate as in the E1 reactionE2 reactions are concerted as in the SN2 reaction

ROH + + Br–

transition state for E2 Elimination

Chapter 8-10

CH3CH2 CHCH3

Br

CH3CH CHCH3 CH3CH2CH CH2

CH3CH CHCH3 CH3CH2CH CH2

Chem 61Elimination Reactions

becoming more sp2

Reaction rate

E2 rate = k[base][substrate] tertiary halides > secondary halides > primary halides

Direction of Elimination

The more substituted (and more stable) product is normally the predominant E2 product

CH2=CH2 RCH=CH2 RCH=CHR R2C=CH2 R2CH=CHR R2C=CR2

unsubstituted mono- di- di- tri- tetra- increasing alkene stability

RO–

Thus

69%

+

RO–

80°, ROH

31%

If RO– is bulky (CH3)3CO– vs. CH3O–; a higher percentage of the less substituted alkene results

+

+

(CH3)3CO –

CH3O –

50%50%

20%80%

H

c d

b a

L

C Cc

d

b

a

Ph C C

Br

H

Ph

H

CH3Ph C C

Br

HPh

H

CH3

C C

Ph

H

Ph

CH3

Stereochemistry of E2

Generally, in E2 reactions the proton and the leaving group must be in an anti orientation.This allows for the best overlap of the developing pi orbitals.

RO–

In order to predict the stereochemistry of the alkene resulting from an E2 reaction:

a) rotate the C atoms so that the H to be removed by RO– and the leaving group are in the anti conformationb) the stereochemistry of the alkene resulting from the E2 reaction is as shown

Chapter 8-11

+ ROH + L–

H

Cl

D

HD

H

Chem 61Elimination Reactions

RO–

In cyclohexane systems the H— and the leaving group must be trans and diaxial

Z

rotate

RO–

SUMMARY

E1 reactions result from carbocation intermediates and are competivive with SN1 reactions.

E1 reactions may be accompanied by carbocation rearrangements.

E1 reactions generally yield the most substituted (most stable) alkene.

E1 reactions are not stereoespecific and yield a mixture of E,Z alkenes.

Relative rates: 3° > 2°.

E2 reactions result from a concerted reaction and are competitive with SN2 reactions.

E2 reactions generally produce the more stable alkene (except with bulky bases or leaving groups).

E2 reactions are stereospecific and give anti elimination of H and L.

Relative rates: 3° > 2° > 1°.

Chapter 8-12 Chem 61Substitution vs. Elimination Reactions

Factors Governing Substitution and Elimination Reactions

Structure of Alkyl Halide

Leaving group attached to methyl or primary carbon: SN2 reaction observed unless Nu: is a strong base and elevated temperature employed; then observe E2.

Leaving group attached to tertiary carbon: E2 reaction observed unless Nu: is weak base; then SN1 observed.

Solvent nucleophilicity increases with increasing electron releasing capacity ofmolecule. If the solvent has low nucleophilic properties; i.e. H2SO4, H3PO4 then E1reaction results CH3CH2OH > H2O > CH3CO2H > CF3CH2OH > CF3CO2H increasing solvent nucleophilicity

Leaving group attached to secondary carbon: if the nucleophile is less basic than HO – (e.g. – CN, – N3, – SR) then the reaction will follow SN2 path

If the Nu: is HO – or more basic (e.g. RO–, R2N–, RC≡C–), then the reaction will follow E2 path

If the Nu: is a weak nucleophile (e.g. H2O, ROH, RCO2H) the reaction will follow SN1path with some E1

The leaving group is attached to an allylic or benzylic carbon: SN1

Solvent

Highly polar solvents (high dielectric constant) favor SN1, E1 reactions

Concentration of Nucleophile or Base

Increasing the concentration of the nucleophile has no effect on SN1, E1 but increases SN2,E2 rates.

Temperature

Increase in temperature increases the rates of all reaction types but has a larger effect on E2 reactions.

Chapter 8-13 Chem 61Nucleophilicity

NUCLEOPHILICITY

Nucleophilic Constants of Various Nucleophiles

pKa of conjugate acid

-1.7 (CH3OH2)

-1.3

3.45

4.8

5.7

9.25

4.74

9.9

-7.7

15.7

15.7

5.8

7.9

10.7

9.3

-10.7

8.69

8.5

n(CH3I)a

0.0

1.5

2.7

4.3

4.4

5.3

5.5

5.8

5.8

5.8

6.3

6.5

6.6

6.6

6.7

6.7

7.4

7.8

8.7

9.9

10.7

11.5

Nucleophile

CH3OH

NO3–

F–

CH3CO2–

Cl–≠

(CH2)2S

NH3

N3–

PhO –

Br –

CH3O–

HO–

NH2OH

NH2NH2

(CH3CH2)3N NC–

I –

HOO–

(CH3CH2)3P

PhS–

PhSe–

Ph3Sn–

an(CH3I) = log(knucleophile/kCH3OH) in CH3OH 25°Cn = nucleophilic constant

General comments

1. Note that nucleophilicity toward CH3I does not correlate directly with basicity. (N3

–, PhO–, Br- are equivalent nucleophilicity but differ greatly in basicity. Also N3

– and CH3CO2– are nearly identical in basicity but N3

– is 30 times [1.5 log units] more nucleophilic).

2. Among neutral nucleophiles while (CH3CH2)3N is more basic than (CH3CH2)3P (pKa 10.7 vs. 8.69) the phosphine is 100 times (n = 8.7 vs 6.7) more nucleophilic.

3. Correlation of nucleophilicity with basicity is better if the attacking atom is the same. Thus CH3O– > PhO– > CH3CO2

–> NO3–

4. Nucleophilicity usually decreases going across a row in the periodic table. (HO– > F– ; PhS–> Cl–). This order is determined by electronegativity.

5. Nucleophilicity usually increases going down the periodic table (I– > Br– > Cl– > F–; and PhSe– > PhS– > PhO–) Related to weaker solvation and greater polarizability of the heavier atoms.

R CH CH2 L

H

R CH CH2 Nu

H

R CH CH2

C C

H L

C C C

H

C C C

H Nu+ H

C

C C C C C

H Nu

CC C C

C C

H

H

L

C

C C

H

H

C

C C

H

C

C C

H

H

Nu

C

Examples

Chapter 8-14 Chem 61Nucleophilicity

hindered

strong baseheat

Nu:–

E2 (anti elimination of HL)

SN2 (inversion

1°

E2 E1SN1

NuH

Nu:– actsas base – H+– H+

+

3°

SN1 or E1

poor Nu or base

+

E2

SN2

Nu:– or base:–2°

E1: 3° > 2° > 1°E2: 3° > 2° > 1°SN1: 3° > 2° > 1°SN2: 1° > 2° > 3°

Chapter 9-1 Chem 61Alcohols

HO

H CH3CH2

OH

Alcohols are similar to water: contain sp3 hybridized oxygen, —OH functional group bonded to an sp3 carbon, with two lone pairs on oxygen

CH3CH2O

H CH3CH2O+

H

H

water

.. . .... .

Alcohols can hydrogen bond thus their boiling points are higher thansimilar polar compounds which cannot hydrogen bond.

Low molecular weight alcohols are soluble in water

Alcohols contain a hydrophobic alkyl group and a hydrophilic hydroxyl group

As the alkyl group increases in length and size, the solubility in waterdecreases

Acidity and Basicity

ethanol

alcohols act as bases in the presence of strong acids

acidbase H+

..

(CH3)3CO

H (CH3)3C O – K+

....

and as acids in the presence of strong bases

CH3CH2O– Na+ + H2CH3CH2O—H + Na0

....

. + H2+ K+H–

acid base

.....

Chapter 9-2 Chem 61Alcohols

CH2OH

IUPAC Names

For alcohols drop —e from the alkane name and add —ol

CH3CH2OH: ethan + ol = ethanol

CH3CH2CHOHCH3: butan + ol; —OH on carbon 2: 2-butanol

Classification

Alcohols are classified as:

methyl, primary, secondary, tertiary, allylic, benzylic

benzylicallylicCH2=CHCH2OH

tertiary 3°

R OH R O+H

H

secondary 2°primary 1°methyl

CH3OH CH3CH2CH2CH2CH2CH2OH CH3CHOHCH3 (CH3)3COH

REACTIONS OF ALCOHOLS

Substitution Reactions: Reaction with Hydrogen Halides (HX)

Alcohols do not undergo nucleophilic substitution by X– since HO– is apoor leaving group.

Alcohols do undergo substitution by X – in acidic solution.

Here the alcohol is protonated and H2O, a neutral species, is the leaving group.

+ H2OX – R—XH+

..

....

increasing reactivity of ROH toward HX

All alcohols react readily with HBr and HI; 3°, allylic and benzylic react rapidly with HCl2° and 1° alcohols require the addition of ZnCl2 for rapid reaction with HCl

Mechanism of Alcohol Substitutions

Methyl and primary alcohols follow the SN2 mechanism

H+

SN2+ H2O

CH3CH2CH2 OH CH3CH2CH2+O

H

CH3CH2CH2 BrCH3CH2CH2 O+H

H

Reactivity of Hydrogen Halides

The reactivity of hydrogen halides in alcohol substitution reactions is as follows:

HF < HCl < HBr < HIpKa 3.45 -7 -9 -9.5

The reactivity is explained since the acidity of HX increases and the nucleophilicity of X– increases going from HF to HCl to HBr to HI.

Reactivity of Alcohols Toward HX

methyl primary secondary tertiary allylic and benzylic

H

Br –

Step 1 protonation of the alcohol

CH3CH2CH2 OH

Step 2 SN2 dsiplacement of Water

CH3CH2CH2

+

O

H

Chapter 9-3

H

CH3CH2CH2 Br

Chem 61Alcohols

E


E

Chapter 9-4 Chem 61

CH3 C CH2CH3

OH

CH3

CH3 C CH2CH3

+OH2

CH3

CH3 C CH2CH3

CH3

CH3 C CH2CH3

I

CH3

Step 2 loss of water to for the carbocation

– H2O

H+

CH3 C CH2CH3

+OH2

CH3

CH3 C CH2CH3

CH3

+

SN1

X –

Secondary and tertiary alcohols follow the SN1 mechanism

Step 1 protonation of the alcohol

+

Step 3 attack of halide ion on the carbocation

Rearrangement can occur

R–OH

R–O+H

R+

R–XH

Alcohols

Other Reagents for the Conversion of Alcohols to Alkyl Halides

R—Br + HOPBr2R—OH + PBr3

CCH3

CH3CH2

H

OH

CCH3

CH3CH2

H

OSOClC

CH3

CH3CH2

H

Cl

O

SClCl

CCH3

CH3CH2

H

O+S

O –

ClCl

H

H

CCH3

CH3CH2

H

OS

O

Cl

R—Cl + HCl + SO2R—OH + SOCl2

+

CH3CH2CH2 C

O

OH

CH3CH2CH2 C

O

OCH2CH2CH3

Chapter 9-5 Chem 61Conversion of Alcohols to Alkyl Halides

R3N:

Cl –

+ SO2

Esters of Alcohols

Reaction of alcohols with carboxylic acids in the presence of acid produces esters of carboxylic acids

ester

alcohol carboxylic acid

H+

heat

+ CH3CH2CH2OH

C

C

C

H

ONO2

ONO2

ONO2

H

H

H

H

C

C

C

H

O

H

H

H

H

H

H

P

O

O

OH

P

O

OH

OHCH3O S OCH3

O

O

CH3O S CH3

O

O

Inorganic esters

mineral acids such as H2SO4, HNO3 and H3PO4 form esters with alcohols to produce important compounds

dimethyl sulfate,a sulfate ester

CH3 S

O

O

Cl

OH O CH3S

O

O

a diphosphate esternitroglycerin, a nitrate ester

P-toluenesulfonates (tosylates) and methanesulfonates (mesylates) are excellent leaving groups in nucleophilic substitution reactions. They are readily prepared form an alcohol and the corresponding sulfonyl chloride.

Chapter 9-6 Chem 61Reactions of Alcohols

(CH3CH2)3N+HCl –+

a methanesulfonate (mesylate)

OTs OCH2CH2CH3

Reaction of sulfonates with nucleophiles

CH3CH2CH2O– Na+

cyclohexyl tosylate

(CH3CH2)3NCH3OH + ClSO2CH3

p-toluenesulfonylchloride(tosyl chloride)

+

CH3 C

H

CH3

C

CH3

CH3

OH CH3 C

H

CH3

C

CH3

CH3

O+H2

CH3 C

H

CH3

C +

CH3

CH3

CH3

C

CH3

C

CH3

CH3

CH2 CHCH2CH3

OH

CH CHCH2CH3

CH3

CH

OH

CH3

CH3

CH3

CH3 C

CH3

CH3

C

CH3

H

OH CH3 C

CH3

CH3

C

CH3

H

O+H2

CH3 C

CH3

CH3

C CH3

H

CH3

CH3

C

C

CH3

C

CH3

CH3

CH3

C CH3

H

CH3

Elimination (Dehydration) of Alcohols

Alcohols undergo elimination much like alkyl halides

Tertiary alcohols readily undergo dehydration by an E1 pathwaySecondary alcohols also follow an E1 path, but primary alcohols probably eliminate by an E2 mechanism

ease ofdehydration

conc. H2SO4

180° C

conc. H2SO4

100° C

conc. H2SO4

60° C(CH3)2C=CH2 +H2O

CH3CH=CH2 +H2O

CH3CH=CH2 +H2O

(CH3)3COH

(CH3)2CHOH

CH3CH2CH2OH

Mechanism:

– H+

– H2OH+

ROH2+

E

ROH

R+


The most stable alkene predominates in dehydration reactions.Rearrangements can occur.

Examples:

alkene

Chapter 9-7 Chem 61Dehydration of Alcohols

H+

H2SO4

heat

H+– H2O

– H++1,2 shift+

Chapter 9-8 Chem 61Dehydration of Alcohols

O

O

CH2 CH2

ETHERS, EPOXIDES AND SULFIDES�

Ethers are derivatives of water where both hydrogens have been replaced by an alkyl group.

Ethers are less polar than alcohols and water and are not capable of hydrogen bonding to themselves because of the lack of an —OH group.

The boiling points of ethers are also much lower than alcohols of comparable molecular weight.

ethylene oxidetetrahydrofuran

CH3CH2OCH2CH3

diethyl ether

Preparation of Ethers

Williamson Ether Synthesis (SN2 Reaction of an Alkoxide with an Alkyl Halide)

SN2R1—O—R2 + X–R1O– + R2—X

Best results are obtained if the alkyl halide is methyl or primary (2° and 3° give mostly elimination)

There are few limitations on the alkoxide

O – OCH2CH3

CH3CH2CH2O—CH2CH(CH3)2CH3CH2CH2O – + (CH3)2CHCH2—Br

+ CH3CH2—I

(CH3)3CO—CH3(CH3)3CO – + CH3—I

Chapter 9-9 Chem 61Reactions of Ethers and Epoxides

OCH2CH(CH3)2 OH

H

O

C CCH3

CH3

H

H

– O

C CCH3

H3C

H

H

OCH3

HO

C CCH3

H3C

H

H

OCH3

Substitution Reactions of Ethers

Ethers are relatively unreactive compounds, but they do undergo substitution reaction when heated with hydrogen halides, particularly HI and HBr.

This is a very similar reaction to the reaction of alcohols with HX

HBrCH3CH2CH2—Br CH3CH2CH2—Br + CH3CH2CH2—OH

Br –

+CH3CH2CH2—O—CH2CH2CH3

H—BrCH3CH2CH2—O—CH2CH2CH3

+ I—CH2CH(CH3)2

HI

Reactions of Epoxides

Epoxides are strained much like cyclopropanes because their bond angles (60°) are far removed from the normal tetrahedral angle (109.5°).

The orbitals have poor overlap and the bonds are weakened. The C—O bond is also polarized and consequently, epoxides are highly reactive compounds.

Base Catalyzed Cleavage of Epoxides

Alkoxides:

unreactive with HI

nucleophile attacks at the least hindered carbon in an SN2 reaction

HOCH3

CH3O –

CH3O –

CH3OH

....

O

C CH

H

H

H

BrMgO

C CH

H

H

H

CH3

HO

C CH

H

H

HCH3

O

C CCH3

CH3

H

H

H

H

O

C CCH3

CH3

H

H

OH

CC H

H

CH3

CH3

CH3O

CH3

C CCH3

CH3O

OH

H

H

Grignard Reagents

Chapter 9-10 Chem 61Reactions of Epoxides

+O

H

H

H

OH

HH

OH

CH3MgBr

CH3–MgBr+

H+

Acid Catalyzed Cleavage

H+

CH3OH

+

CH3O—H....

+

– H+

δ+

Nucleophile attacks at the most hindered carbon since the carbon is partially positive and the C—O bond is partially broken

trans diaxial openingin cyclohexanesH2O

Chapter 9-11 Chem 61Crown Ethers and Thiols

O

K+

O

O

O

O

Na+

O

O

O

O

OO

SHBr

Crown Ethers

Crown ethers are macrocyclic ethers with repeating –OCH2CH2— units. Depending on the ring size, they effectively chelate alkali metal ions such as K+, Na+ or Li+

12-crown-518-crown-6

I SCH2CH2CH3

Thiols and Sulfides

The sulfur analog of an alcohol is a thiol, or mercaptan. Thiols are stronger acids than alcohols (pKa = 8, ROH = 16).

Thiols form weaker hydrogen bonds than alcohols due to the lower electronegativity of sulfur.

Thiols and thioethers (sulfides) are prepared by substitution reaction in the same way that alcohols and ethers are prepared.

HS–

CH3CH2CH2S–

Disulfides are formed by the oxidation of thiols and are an important structural feature in some proteins.

I2 or

K3Fe(CN)6

a disulfideRS—SRRSH

R1

Cδ+

Oδ−

R1 R1

C

O –

R1NuR1

C+

O –

R1

R2 Mg-X

R1

C

O– MgX+

R1R2R1

C+

O –

R1 R1

Cδ+

Oδ−

R1

HC

O

H

CH3 Mg-X

H C

OH

H

CH3

Grignard Reactions

Grignard reagents are organometallic reagents derived from an alkyl halide and magnesium

Since the carbon carries a partial negative charge, the carbon is a strong base and a good nucleophile.

Because carbonyl pi bonds are polarized, they can undergo a reaction called nucleophilic addition: the addition of a nucleophile to an electron deficient pi bond.

diethyl etherRδ−—Mgδ+X Grignard reagent R—X + Mg

CH3CH2—HHOHdiethyl etherCH3CH2

δ−—Mgδ+Br CH3CH2—Br + Mg

Nu:

NucleophilicAddition

Chapter 9-12 Chem 61Organometallic Compounds

NucleophilicAddition

A Grignard reaction with

1. formaldehyde produces a primary alcohol

2. H2O, H+

1.

Chapter 9-13

2. an aldehyde produces a secondary alcohol

3. a ketone produces a tertiary alcohol

4. an ester produces a tertiary alcohol (addition of two molecules of Grignard reagent)

5. ethylene oxide produces a primary alcohol

CH3CH2

C

O

H

(CH3)CHMgBrH C

OH

CH2CH3

CH(CH3)22. H2O, H+

1.

CH3CH2

C

O

CH3

(CH3)CHMgBrCH3 C

OH

CH2CH3

CH(CH3)22. H2O, H+

1.

C

O

OCH3

CH3MgBrCH3 C

OH

CH32. H2O, H+

1. 2

O

OH

Chem 61Organometallic Compounds

1. C6H5MgBr

2. H2O, H+

Chapter 10-1 Chem 61Alkynes: Structure and Bonding


2-2p's on each carboncombine to form pi-orbitals

2p

2s

1s1s

2s

2psp

1s

96 kcal hybridize

H H

H H

C C HH

C C HH

sp Hybridization

sp hybridized carbons are linear with atoms 180° apart

1-2p 2-sp's

180° apart; the remaining two 2porbitals are 90° to the sp andeach other

+–+

2s

2p

C—C σ*C—C π*C—C π

E

C—C σ

Acetylene: linear; bond angles 180° C=C bond length = 1.20Åacetylene has two perpendicular pi bonds and one sigma bond

acetylene π-orbitals

acetylene σ-orbital2-sp's

C C RR C CR

C CR

E+

R

ER

C C RR C CR

C CR

E+

R

ER

Reaction of Alkynes with E+A– Reagents

Reaction of alkynes with E+A– reagents proceed in the same manner as alkenes except different intermediates are possible

vinyl carbocation+E+

For H—X the vinyl carbocation is more stable than the bridged intermediate. Thus:

bridged intermediate

Chapter 10-2 Chem 61Alkynes: Electrophilic Addition

vinyl carbocation+E+

secondary vinyl carbocation is about the same stability as a primary carbocation

bridged intermediate

Thus the order of carbocation stability is

triphenyl methyl > diphenylmethyl > 3°≈ benzyl ≈ allyl > 2° > 1° ≈ 2° vinyl >> methyl

Chapter 10-3 Chem 61Alkynes: Electrophilic Addition

C C HCH3

C C CH3CH3

C C

C C

CH3

Cl

H

H

CH3

Cl

CH3

HC C

CH3

Cl

H

CH3

C C

CH2CH3

HgOCOCH3CH3CH2

CH3CO2

C C

R

C C

C6H5

XR

XX

C6H5

X

C C

X

C6H5X

C6H5

C C

R

X+

C

R

C

C6H5

X

C6H5

HCl+

HCl

C2H5C≡CC2H5

For Hg(OCOCH3)2:

+

Hg(OCOCH3)2

Thus the choice of intermediate depends on structure; alkyl groups tend to favor the bridged ion; groups such as phenyl which stabilize the free carbocation tend to proceed via the vinyl carbocation.

E

+but,

C6H5—C≡C—C6H5

R—C≡C—R

For Cl2, Br2:

For H–X

X2

X2

R—CR—C C—R R—C CH—R CH2—R

OOH

R—C C—R

Hg2+

R—C C—R

Hg2+

+OHH

R—C C—R

Hg2+

OH

R—C C—R

Hg2+

OH

R—C C—R

H

OHR—C C—R

H

OH

Addition of Water

R—C C—H R—C CH

H

H2SO4

Hg(II)H2O

H2O:

keto

H2O:

BH2

Chapter 10-4

R—C CH

H

enol

With alkynes this electrophilic addition reaction generates a vinyl alcohol (also called an enol). Hg(II) ion is often used.

OH

Chem 61

R—C C–H

H

Alkynes: Electrophilic Addition

Hydroboration

H+

O

enol

H

BH3

keto (aldehyde)

The relatively high acidity of the alkyne —C≡C—H bond is associated with the large degree of s character in the sp C—H bond (50% compared with33% in sp2 bonds). The carbon atom is more electronegative in the spstate; thus the C—H bond is more acidic.

The acetylide ion may be formed by such strong bases as —:NH2 (pKa33), RMgX or RLi (pKa 45-50).

No reactionNaNH2

NH3

C CH

H

C C H

C CH

C C

R C C H

CH

R C C:– Na+

CH2R

pKa = 45

+ base—H

+ base—H.

–

–

..

.

+ base

+ base

+ NH3

acetylide ion

NaNH2

NH3

pKa = 25

Chapter 10-5 Chem 61Alkynes: Acidity

sp > sp2 > sp3

Electronegativities

N < O < F

NH3 < H2O < HF

Electronegativities

acid strength

pKa 36 15.7 3.2

HC≡CH H2C=CH2 H3C–CH3

pKa 25 44 50

HC≡CH H2C=CH2 H3C–CH3HF H2O NH3

15.73.2 36

HC≡C– H2C=CH– H3C–CH2–HO– NH2

–F–


increasing base strength

R' C C:– MgBr+ R C

O

H R' C C C

O – MgBr+

H

R

R' C C C

OH

H

RR' C C—H

R' C C:– MgBr+

O

C C R'O – MgBr+

C C R'OH

H+

Nucleophilic addition reaction with acetylide ion.

Chapter 10-6 Chem 61Alkynes: Acetylides

H+CH3CH2MgBr

+

SN2 reaction with acetylide ion

R' C C:– MgBr+O

R' C C—CH2—CH2—O – MgBr+

R' C C—CH2—CH2—OH

NH3R—CH2—C≡C—R'

+

H+

R'—C≡C:– Na+ + R—CH2—L

Chapter 12-1 Chem 61Spectroscopy

Infrared and Nuclear Magnetic Resonance Spectroscopy

electromagnetic radiation: energy that is transmitted through space in the form of waves

wavelength: (λ): the distance from the crest of one wave to the crest of the next wave

frequency: (ν): the number of complete cycles per second

where c = speed of lightλcν =

Electromagnetic radiation is transmitted in particle-like packets called photons or quanta. The energy is inversely proportional to thewavelength and directly proportional to frequency.

where c = speed of light; h = Planck's constantλhcΕ =

h = Planck's constanthνΕ =

ultraviolet visible infrared radio

decreasing energy

Absorbtion of ultraviolet light results in the promotion of an electron to a higher energy orbital.

Absorbtion of infrared results in increased amplitudes of vibration of bonded atoms.

Intensity of radiation is proportional to the number of photons.

frequency

%T

100

0

Chapter 12-2 Chem 61Infrared Spectroscopy

Infrared is recorded as %T versus wavelength or frequency

When a sample absorbs at a particular wavelength or frequency, %T is reduced and a peak or band is displayed in the spectrum.

O CH3

CH3

O CH3

CH3

Infrared Spectroscopy

Infrared is recorded as %T versus wavelength or frequency

When a sample absorbs at a particular wavelength or frequency, %T is reduced and a peak or band is displayed in the spectrum.

Nuclei of bonded atoms undergo vibrations similar to two ballsconnected by a spring. Depending on the particular atoms bonded toeach other (and their masses) the frequency of this vibration will vary.

Infrared energy is absorbed by molecules resulting in an excitedvibrational state. Vibrations occur in quantized energy levels and thus a particular type of bond will absorb only at certain frequencies.

Both stretching and bending vibrations can be observed by infrared.

bendingstretching

800100015002000250030003500

C—C str

C—N str

C—O str

OH bend

CH bendNH bend

C=C str

C=N str

C=O str

C≡N strCH str

OH and NH str

Chapter 12-3 Chem 61Infrared Spectroscopy

Interpretation of Infrared Spectra

Correlation tables

Infrared spectra of thousands of compounds have been tabulated and general trends are known. Some common functional groups areshown below.

sp3 C—Csp2 C=Csp2 C—C (aryl)sp C≡Csp3 C—Hsp2 C—H

sp C—HC(CH3)2

weak, not useful

1600-1700 cm–1

1450-1600 cm–1

2100-2250 cm–1

2800-3000 cm–1

3000-3300 cm–1

3300 cm–1

1360-1385 cm–1 (two peaks)

Alcohols and Amines

O—H or N—H C—O or C—N

C—C and C—H Bonds

Ethers

C—O

3000-3700 cm–1

900-1300 cm–1

1050-1260 strong

Carbonyls One of the most useful absorbtions in infrared 1640-1820 cm-1

Ketones (saturated) C=O

Aldehydes C=O; C—H(O)

Carboxylic acids C=O;

C(O)—OH

Esters C=O ;

C(O)—OR

1640-1820 cm–1

1640-1820 cm–1

2820-2900 and 2700-2780 cm–1 (weak but characteristic)

1640-1820 cm–1

3330-2900 cm–1

1640-1820 cm–1

1100-1300 cm–1

Chapter 13-1 Chem 61Nuclear Magnetic Resonance Spectroscopy

Nuclear Magnetic Resonance (NMR) Spectroscopy

Some atomic nuclei (1H, 13C, others) behave as if they are spinning...theyhave a nuclear spin.

Spinning of a charged particle creates a magnetic moment.

If an external magnetic field is applied, these small magnetic moments (of the nuclei) either align with the field (α) or against the field (β), about 50% withand 50% against the field at any one time.

HoHo

α

β

∆Ehν

∆E

β

α

Ho = the external magnetic field

Resonance: the flip of the magnetic moment from parallel to antiparallel tothe external magnetic field.

Irradiation at the frequency equal to the energy difference, ∆E, causesresonance.

∆E depends on the external magnetic field.

Protons (or other nuclei) in different magnetic environments resonate atdifferent field strengths.

A proton which resonates at a higher field is in a stronger magneticenvironment or shielded.

A proton which resonates at a lower magnetic field is said to be deshielded.

Different magnetic environments are created by different electron densities in the vicinity of a proton.

2.1 2.7 3.0

Pi electron effects

Magnetic fields created by pi electrons are directional and said to have an anisotropic effect.

Chapter 13-2 Chem 61

ppm

In methyl halides, the more electronegative the halogen, the more deshielded the prIn methyl halides, the more electronegative the halogen, the more deshielded the pr

In methyl halides, the more electronegative the halogen, the more deshielded the protons on the methyl. This is because F is inductively more electronwithdrawing, causing the carbon to be more positive and thus pulling moreelectrons away from the hydrogen and causing it to be less shielded.

H3C—F H3C—Cl H3C—Br H3C—I

C OH

R

H

δ 4.3

Nuclear Magnetic Resonance Spectroscopy

distance from TMS in HzMHz of spectrumδ =

Adjacent electron withdrawing groups, highly electronegative atoms, or the hybridization of the carbon to which the proton is bonded can alter the magnetic environment.

The local electrons create a small electric and magnetic field around a proton and shield it.

The more electron density present around the proton, the greater the field and the greater the shielding.

Resonances are reported in chemical shifts (δ) downfield from tetramethylsilane (TMS) (CH3)4Si.

H deshielded

H deshielded

Ho

The pi system of benzene creates a magnetic field or ring current which deshields the protons attached to the ring.

Similarly, pi electrons in a C=O bond create a field which deshields theproton bonded to the C=O of an aldehyde. This is also affected by theinductive effect of the C=O.

H C

H

H

C

H

H

OHH C

H

H

C

H

Cl

C

H

H

H

Equivalent and Nonequivalent Protons

Protons that are in the same magnetic environment are equivalent and havethe same chemical shifts.

Protons in different magnetic fields are nonequivalent and have differentchemical shifts.

Magnetic equivalence is usually the same as chemical equivalence.

Equivalence can be established by symmetry operations such as rotation,mirror planes and centers of symmetry

Chemically equivalent protons have the same chemical shifts.

To determine if protons are chemically equivalent, replace one by a different group, e.g. D or Br.

Then replace a different one by the same group and compare the twocompounds. If they are identical, the protons are equivalent.

equivalent, but not to CH3 protons


equivalent Equivalent protons can be on different carbons.

all six are equivalent

Protons which are homotopic or enantiotopic resonate at the same chemical shift in the NMR.

If protons are interconverted by rotation about a single bond, they will average out on the NMR time scale and a single resonance will be observed.

ClH2CCH2Cl anti and gauche forms rapidly interconvert and a single resonance is observed.

Axial and equatorial hydrogens in cyclohexane average to a single peak because of rapid ring inversion.

Diastereotopic hydrogens are chemically nonequivalent and thus give different chemical shifts in the NMR


Intergration

The spectrometer can integrate and determine the relative number of hydrogens associated with each resonance in the NMR spectrum by determining the area under the peaks.

Spin-Spin Coupling

for example...

CH3CH2OCH3

TMS

33

2

If a proton (Ha) is bonded to a carbon which is bonded to a carbon that hasone proton (Hb), Ha will appear as a doublet

Since in half the molecules, Hb will be in the α state and in half will be in the β state, Ha will experience two different magnetic fields and two peaks (adoublet) will appear for Ha.

Ha without an adjacent hydrogen

Ha with Hb adjacent in the α state

Ha with Hb adjacent in theβ state

For one adjacent hydrogenα or β


For two adjacent hydrogens: Hb, Hc

At any one time Hb or Hc could be in the α or β state (50:50) thus 4 combinations for Hb, Hc exist:

αbαc αbβc βbβc gives 1:2:1 triplet βbαc

When both Hb and Hc are α, a different field is observed than if both are β or one is α and one is β.

When one is α and one is β, the field is the same. That is, βbαc and αbβc produce the same field and a single signal for Ha is observed with twice the intensity.

Thus three signals are observed in a 1:2:1 ratio: a so-called triplet

For three adjacent protons:

ααα ααβ αββ βββ 1:3:3:1 quartet αβα βαβ βαα ββα

Thus the splitting pattern of a particular proton or equivalent protons will be a pattern with n+1 lines where n is the number of adjacent equivalent protons.

singlet 0 neighboring protonsdoublet 1 neighboring protonstriplet 2 neighboring protonsquartet 3 neighboring protonsquintet 4 neighboring protonssextet 5 neighboring protonsseptet 6 neighboring protons

The separation of the peaks in a splitting pattern is called the coupling constant, J.


Splitting Diagrams

Splitting patterns for protons can be constructed in diagram form by startingwith one line to represent the unsplit proton resonance.

If an adjacent proton Hb affects Ha it is split into a doublet; if anotherequivalent proton to Hb is present, each line of the double will be split into a doublet, since the coupling constant J is the same, the two center linesoverlap and a only three lines are observed with the center line twice theheight.

This can be repeated for additional adjacent protons.

Ha without an adjacent hydrogen

Ha split by one adjacent hydrogen

Ha split by a second adjacent hydrogen

Ha split by a third adjacent hydrogen1 3 3 1

splitting diagram

1 1

1 2 1

Chemical Exchange and Hydrogen Bonding

CH3OH, methanol would be expected to give an NMR spectrum of adoublet for the CH3 and a quartet for the OH. For a dilute sample at -40° inCCl4 this is the case.

If the NMR spectrum is run at 25° as a more concentrated sample only twosinglets are observed. This is because the intermolecular hydrogenbonding in methanol allows the rapid exchange of the OH proton from oneCH3OH molecule to another, effectively averaging the spin states of the OH proton and resulting in no change in the magnetic field due to the OH.

Amines and other compounds which can undergo hydrogen bonding canalso show this effect. Thus the NMR spectra of alcohols, amines andcarboxylic acids are temperature, concentration and solvent dependent.


CHEMICAL SHIFTS

Functional Group Shift,δ

Primary alkyl, RCH3 Secondary alkyl, RCH2R Tertiary alkyl, R3CH

Allylic, R2C=C—CH2R

Benzylic, ArCH2R Iodoalkane, RCH2I Bromoalkane, RCH2Br Chloroalkane, RCH2Cl Ether, RCH2OR Alcohol, RCH2OH Ketone, RCH2C(=O)R

Aldehyde, RCH(O)

Terminal alkene, R2C=CH2 Internal alkene, R2C=CHR

Aromatic, Ar—H

Alkyne, RC≡C—H

Alcoholic hydroxy, ROH Amine, RNH2

0.8-1.01.2-1.41.4-1.7

1.6-1.9

2.2-2.53.1-3.33.4-3.63.6-3.83.3-3.93.3-4.0

2.1-2.6

9.5-9.6

4.6-5.05.2-5.7 6.0-9.5

1.7-3.1

0.5-5.0 (variable)0.5-5.0 (variable)

1

CHEMISTRY 61 Name___________________________________________Exam 1Dr. M.T. Crimmins Pledge: I have neither given nor received aid on this exam.September 15, 1998

Signature________________________________________

I. Nomeclature (12 points) Give the IPUAC name for the following compounds: Indicate R, S, cis, trans,E, or Z where appropriate.

1.

CH2CH3 C H C H CH2 CH2 C H CH3

CH3

CH2CH2CH2CH3

CH3 ______________________________

2.

H3CC

CCH2CH3

H

CH3 ______________________________

3.

CH3

H______________________________

4.

CH(CH3)2______________________________

II. A. Write valid Lewis structures for the following species. Show all nonbonding (unshared) electronsand indicate any formal charges . (6 points).

5. [H2COH]+ 6. IO4 –

7. Give the hybridization of the indicated atoms in the species below (6 points)

H3C C N:H3C N.. CH2

H3C CO

O CH3

2

8. Draw three structural isomers for C3H6O. Indicate what type of functional group is represented byeach compound (e.g. carboxylic acid) (6 points).

9. Draw all the possible stereoisomers of 3-bromo-2-butanol. Indicate if they are chiral, meso or achiral andindicate their relationship to each other. (i.e. enantiomers, diastereomers) (8 points)

10. Draw both chair conformations of trans-1,2-dimethylcyclohexane. If one is more stable than theother, circle it. (6 points)

11. Draw an energy diagram for one 360° rotation about the C3-C4 C–C bond of hexane. Also draw aNewman Projection of the most stable conformation. (8 points).

12. Circle the molecule(s) which have a permanent dipole. In those which have a permanent dipole, showthe direction of the overall dipole. (6 points)

C

O

H H ClB

Cl

Cl

H3C O H

3

13. In the space to the right, indicate if each of the pairs of molecules below are identical compounds,enantiomers, diastereomers, structural isomers, or conformational isomers. (9 points).

a.

H3C

CH3

H3CCH3 CH3

CH3 ________________________

b.

BrH

O HH

Br

H

H

O H

________________________

c.

C CH3C

H CH3

HC C

H

H3C CH3

H________________________

14. What two effects cause cyclobutane and cyclopropane to be higher in energy than cyclohexane? (3pts)

15. Label the species below as Lewis Acids or Lewis Bases (4 points)

Br + H3CO–

_____________ _____________

16. Circle the following which has the highest heat of combustion per CH2 unit. (4 points)

a. cyclopentane

b. cyclopropane

c. cyclohexane

d. cyclobutane

17. Indicate the geometry of carbon in the molecules below (e.g. trigonal bipyramidal). (6 points)

a. CH3CH3 b. H2C=O c. HC≡CH

____________ ______________ ______________

4

18. Circle the statement(s) which are true of enantiomers. (4 points)

a. They have a non-superimposable mirror image.

b. They have no asymmetric carbon atoms.

c. They are chiral.

d. They do not rotate the plane of polarized light.

19. Draw an energy diagram of the molecular orbitals of the C=C bond of ethylene (H2C=CH2) and labelthem (e.g. σ) and indicate their relative energies. Indicate the ground state electronic configuration ofthe C=C electrons. (6 points)

20. What kind of molecular orbital results (σ, σ*, π, π*) results when the pairs of orbitals show below arecombined in the indicated manner? (6 pts)

+ + +

a. b . c.

1

CHEMISTRY 61 Name___________________________________________Exam 1Dr. M.T. Crimmins Pledge: I have neither given nor received aid on this exam.September 21, 1999

Signature________________________________________

I. Nomeclature (12 points) Give the IPUAC name for the following compounds: Indicate R, S, cis, trans,E, or Z where appropriate.

1.

CH2CH3CH2 C H C H C H CH2 C H CH3

CH3

CH2CH2CH2CH3

CH3CH2CH3 ____________________________

2.C(CH3)3

CH3

_____________________________

3.

CH2CH2CH3CCH2CH3

H3C

H _____________________________

4.

O H

_____________________________

5. A. Write valid Lewis structures for the following species. Show all nonbonding (unshared) electronsand indicate any formal charges . (6 points).

CH2N2 CH3–

6. Give the hybridization of the indicated atoms in the species below (6 points)

H2C C C(CH3)2H3C N.. CH3

CH3

H3C CO

O CH3

2

7. Write structures for the each of the following having a molecular formula of C4H8O (6 points).

a. a n aldehyde b. a cyclic alcohol c. an ether

8. What intermolecular forces exist between molecules of each of the following. (6 points).

a. CH3CH2CH2CH2CH3 _________________________

b.

CH3 S CH3

O

..

_________________________

c. CH3CH2OH _________________________

9. Draw all the possible stereoisomers of 3,4-dibromohexane. Indicate if they are chiral, meso or achiraland indicate their relationship to each other. (i.e. enantiomers, diastereomers) (8 points)

10. Label the following molecules as chiral or achiral.

CH3

OH H C

OH

CH2CH3

CH3CH2OH

H

HO

H

_____________ ______________ ______________

11. Draw both chair conformations of trans-1,2-dimethylcyclohexane. If one is more stable than theother, circle it. (6 points)

3

12. Draw an energy diagram for one 360° rotation about the C2-C3 C–C bond of butane. Also draw aNewman Projection of the most stable conformation. (8 points).

13. Circle the molecule(s) which have a permanent dipole. In those which have a permanent dipole, showthe direction of the overall dipole. (6 points)

14. In the space to the right, indicate if each of the pairs of molecules below are identical compounds,enantiomers, diastereomers, structural isomers, or conformational isomers. (9 points).

a.

H3C

CH3

H3CCH3 CH3

CH3

H3C

CH3

H3CCH3 CH3

CH3

________________________

b.

BrH

O HH

Br

H

H

O H

________________________

c.

C CH3C

H CH3

HC C

H

H3C CH3

H________________________

15. What effect(s) cause cyclobutane and cyclopentane to be non-planar? (3 pts)

4

16. Label the species below as Lewis Acids or Lewis Bases (4 points)

H3C+ H2C=CH2

_____________ _____________

17. Indicate the geometry of carbon in the molecules below (e.g. trigonal bipyramidal). (6 points)

H2C C CH2C

CH3

H3CCH3

CH4

____________ ______________ ______________

18. Draw and label the atomic orbitals which combine to form the molecular orbitals of formaldehyde,H2C=O. . (6 points)

19. What kind of molecular orbital results (σ, σ*, π, π*) results when the pairs of orbitals show below arecombined (mathematically) in the indicated manner? (6 pts)

– + +

a. b . c.

1

CHEMISTRY 61 Name___________________________________________Exam 2Dr. M.T. Crimmins Pledge: I have neither given nor received aid on this exam.October 22, 1998

Signature________________________________________

I. REACTIONS: Predict the major organic products of the following reactions.INDICATE STEREOCHEMISTRY AS NEEDED. (4 points each)

1.CH3 HI

2.HBr

peroxideCH2

3.

C C

H

H H

CH2CH3

1. BH3

2. H2O2, NaOH

4.CH3

CH3

H2, Pd/C

5.

C

CCH2

CH2H3C

H3C

C

CCO2CH3H

+heat

H3C H

6.

C C CH3H3CCl2

1 equivalent

7.

C

CCH2

HH2C

H

HBr, -80°C

8.CH3 1. Hg(OAc)2, H2O

2. NaBH4

2

II. Multiple Choice: Place the letter in the blank and Circle the best answer (only one). (4 points each)

___9. The rate limiting step for hydration of an alkene with water and acid is

a. protonation of the alkene by a strong acidb. addition of water to a carbocation to form the protonated alcoholc. loss of a proton from the protonated alcohol to form the alcohol.d. simultaneous addition of H+ and HO – to the alkene.

___10. Which of the following free radicals is the most stable?

a. c. c. d.

H3C C

CH3

H. H3C C

CH3

CH3. H C

CH3

H.

H3C C

CH3

.___11. Which of the following indicated hydrogens is the most acidic?

a. CH3CH=CH2 b. CH3C≡C–Hc. CH3CH2CH3 d. H–CH2CH2OCH3

___12. In the addition of HBr to 1,3-butadiene the 1,2 product predominates at -80°C while the 1,4 product predominates at 40°C. The 1,4 product is said to result from

a. kinetic control.b. thermodynamic control.c. a Diels Alder reaction.d. the s-cis diene.

___13. Which of the following alkenes would have the lowest heat of hydrogenation?

a. b. c. d.

CC

H3C CH3

CH2CH3H3C

CHC

H2C CH3

CH2CH3H3C

CHCH

H3C CH3

HCH3C CH2

CCH

H3C CH2

CH3C CH3H

H

___14. What is the stereochemical relationship of the products of the following reaction?

C

CCH2CH3

HH

H3C

Br2

a. diastereomersb. enantiomersc. identical (only one stereosiomer of the product is formed).d. cis-trans isomerse. conformational isomers

___15. Which of the carbocations below is the most stable?

a. b. c. d.

CCH3

HH + C

CH3

H

+ CCH2+

CCH3

CH3

+

3

___16. What is the hybridization of the positively charged carbon in the carbocation below?

CCH2+

a. sp3 b. sp2 c. sp d. s

V. Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the following reactions. Be sure to account for stereochemistry as needed. (6 points each)

17.

C

CCH3

HH3C

H

Br2C C

Br

CH3

Br

H3C

H H

18.

C CH2

CH3

CH3 X + Y –

C CH2CH3

Y

CH3

X

Syntheses: Give reagents to carry out the transformations below. (5 points each)

19.

C C

H

H H

H

ClCH2–CH2Cl

20.

C C HH3C H3C C

O

CH2CH3

4

21. Consider the energy diagram below and answer the questions using the letters on the diagram. (10 points).

A

B

C

D

EF

G

HJ

K

L

ENERGY

PROGRESS OF REACTION

MN

a. What point(s) in the diagram represent transition states? _______________

b. What point(s) in the diagram represent intermediates? _______________

c. What is the energy of activation for the reaction? _______________

d. What is the rate-limiting step in the reaction? _______________

e. Is the reaction endergonic or exergonic? _______________

f. What is the free energy change of the reaction? _______________

g. Does G or C form faster from E? _______________

22. Draw the HOMO (highest occupied molecular obrital) and the LUMO (lowest unoccupied molecularobrital) of 1,3-butadiene and label them. Circle the one which would interact favorably with theethylene orbital below in a Diels-Alder reaction. (4 points)

1

CHEMISTRY 61 Name___________________________________________Exam 2: October 21, 1999Dr. M.T. Crimmins Pledge: I have neither given nor received aid on this exam.

Signature________________________________________

I.REACTIONS: Predict the major organic products of the following reactions.INDICATE STEREOCHEMISTRY AS NEEDED. (4 points each)

1.1. Hg(OAc)2, H2O

2. NaBH4

CH3

2.

1. BH3

2. H2O2, HO –C C

H

CH3CH2CH3

CH3

3.HCl (1 equiv)

C C HCH3CH2

4.CH3 Br2

H

5.H2

poisoned catalyst(Pd/BaSO4)

C CCH3CH2 CH2CH3

6.

CH2CH3C

H3C

Br2, H2O

7.

HBr, peroxideC

CCH3

H

H

8.

CH2C

H3C

H3C

H2O, H+

9.NaNH2

CH3CH2BrC CCH3CH2 H

2

II. Multiple Choice: Circle the best answer (only one). (3 points each)

10. Which of the following alkenes is the most stable?

CH3

CH3

CH3

CH3

CH2

CH3

CH3

CH3

a. b. c. d.

11. Which of the following is the least stable carbocation?

a. H2C=CHCH2+ c. C6H5(CH3)2C+b. (CH3)3C+ d. CH3CH2+

12. In the following reaction what is the relationship of the products formed?

CH3

H Br2

a. enantiomers c. structural isomersb. meso compound d. diastereomers

13. The carbon -carbon triple bond of an alkyne is composed of_________

a. two σ bonds and one π bondb. three σ bondsc. one σ bond and two π bondsd. three π bonds

14. The free energy of reaction is

a. the difference in energy between the reactants and an intermediate in the reactionb. the difference in energy between the reactants and the transition statec. the difference in energy between the reactants and the productsd. the difference in energy between the transition state and the productse. the difference in energy between the intermediate and the products

15. What is the hybridization of the positively charged carbon in H3C+

a. p c. sp e. d2sp3

b. sp2 d. sp3 f. s

16. A secondary cation is more stable than a primary carbocation because of

a. overlap of a filled p orbital with an adjecent σ* antibonding orbitalb. overlap of an empty p orbital with adjacent σ bonding orbitalsc. resonanced. deduction

17. Which of the following free radicals is the most stable?

H3C CH

CH3

.CH3

.H3C C CH3

.H C CH3

.

CH3 H

a) b) c) d)

3

18. Which of the following is not true of H2C=CH2?

a) It contains 5 σ bonds.b) It has bond angles of 120°.c) All the atoms are in the same plane.d) It has free rotation about the C=C bond.


19.

C CH2

CH3

CH3

C CH3CH3

OCH3

CH3

20.H3CC C–H CH3CH2CH2CH2CH3

21.

HC CCH2CH3 CH3CH2 C

O

CH3

22. Draw an energy diagram for the hypothetical exergonic reaction below where B is an unstableintermediate.Label the positions for A, B, and C on the diagram and indicate the energy of activation on thediagram. (5 points).

A B Cslow fast

23. Draw resonance structures (4 pts each) for

a) benzyl cation

b) acetate ion (CH3CO2–)

4

V. Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the following reactions. (4 points each)

24.

CH

CH2

H2O, H+

C CHCH3

CH3

OH

C

CH3

H3C

CH3 CH3

CH3

25.

HC

CH3C Br2

H

CH3 H3C

H3C Br

Br

H

H

26.

HBr

C CH2

H3C

H3C

C CH3H3C

CH3

Br

1

CHEMISTRY 61 Name___________________________________________Exam 3Dr. M.T. Crimmins Pledge: I have neither given nor received aid on this exam.November 24, 1998

Signature________________________________________

I. Multiple Choice: Place the letter in the blank and Circle the best answer (only one). (4 points each)

___1. The rate limiting step for free radical halogenation is

a. initiationb. hydrogen atom abstraction from carbon by the halogen radicalc. attack of carbon radical on molecular halogend. termination

___2. Which of the following are "concerted" reactions?

a. SN1 d. E2b. SN2 e. SN1 and E1c. E1 f. SN2 and E2

___3. Reaction of a strong base with a tertiary alkyl halide is most likely to result in:

a. no reaction c. SN1 substitutionb. E2 elimination d. E1 elimination

___4. Which of the following statements is correct?

a. Free radical bromination is more selective than chlorination because the transition state is more reactant-like.

a. Free radical bromination is more selective than chlorination because the transition state is more product-like.

c. Free radical chlorination is more selective than bromination because the transition state is more product-like.

d. Free radical chlorination is more selective than bromination because the transition state is more reactant-like.

___5. Which of the folowing species is the most nucleophilic?

a. NH3 b. H3P c. H2S d. H2O

___6. SN1 reactions lead to

a. formation of free radicalsb. retention of stereochemistryc. racemizationd. inversion of stereochemistry

___7. Which of the following would undergo the fastest dehydration reaction in the presence of acid?

a. b. c. d.

CH3

O H

CH2OHO H

CH3

CH3

O H

2

___8. Which of the following reactions will proceed the fastest?

a. b. CH2Cl

NaOHCH2Br NaOH

c. d.

CH2BrNaOH

CH2BrNaOH

CH3

.

II. REACTIONS: Predict the major organic products of the following reactions.INDICATE STEREOCHEMISTRY AS NEEDED. (4 points each)

9.CH2CH3 Br2, light

10.

C HCH3CH2CH2OH, H+CH2

O

11.CH3

O H

1. NaH

2. CH3I

12.

CH(CH3)2H

(CH3)3CO–K+

BrH

13.

C(CH3)3

OH

H

H

NaOH, H2O

3

14.

CH3

Br

H

H CH3Se– Na+

15.

H3CC

CC

CH3

CH3 CH3Br

HCH3OH

16.

CH3CH2CH2Br1. Mg

2. H2C=O3. H2SO4

V. Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the following reactions. Be sure to account for stereochemistry as needed. (6 points each)

17.CH3

CH3

O H

H3PO4, heat CH3

CH3

18.

CH4 + Br2 CH3Br + HBrheat or light

4

19.

C CH3

CH3

CH3

OCH3CH2 ICH3CH2 C CH3

CH3

CH3

IHI

+excess


20.I

21.

OCH2CH2CH3

OCH3

22. Only one monochlorination product is obtained from an alkane with the molecular formula C5H12. Whatis the structure of the alkane? (4 points)

23. Draw the transition state for the reaction of CH3Br with HO–. (4 points)

– 1 –

CHEMISTRY 61 Name___________________________________________Exam 3Dr. M.T. Crimmins Pledge: I have neither given nor received aid on this exam.November 23, 1999

Signature________________________________________

II. Reactions: Predict the major organic product of the following reactions. If more than one product isformed give both and indicate the major product. Indicate stereochemistry where necessary. (4points each)

1.

+H CO2CH3

H CO2CH3

heat

2.

HBr

-80 °C

3.

H3C C

CH3

H

CH2CH2CH3

Br2, light

4.Br

CH3

H

H

NaI

5.

CH3

H3CH

Br

H2O

low temperature

6.

C C

H

BrPh

HPh

CH3 (CH3)3CO – K +

7.

C CH

CH3

Ph

Br

CH2

CH3OH

– 2 –

8. Indicate if the following compounds are aromatic, non-aromatic or anti-aromatic. (2 points each)

H

++

a. b. c. d.

H

e.

9. List three criteria for aromaticity. (6 points)

1.______________________________________

2.______________________________________

3.______________________________________

II. Multiple Choice: Place the letter in the blank and Circle the best answer (only one). (3 points each)

___10. In the following solvolysis reaction what is the relationship of the products formed?

H3C C

CH2CH3

Br

CH2CH2CH3CH3OH

a. enantiomers c. structural isomersb. meso compound d. diastereomers

___11. Which of the following reactions would proceed the fastest?

a. CH3CH2CH3 + Br2 + light → CH3CHBrCH3

b. CH3CH2CH3 + F2 + light → CH3CHFCH3

c. CH3CH2CH3 + I2 + light → CH3CHICH3

d. CH3CH2CH3 + Cl2 + light → CH3CHClCH3

___12. Which of the following is the strongest nucleophile?

a. CH3NH– c. CH3O-

b. Cl- d. CH3Se-

___13. Which of the following alkyl halides would undergo the fastest SN2 reaction?

CH2I CH2Cla. b.

b. CH3CH2CH2CH2—I d. (CH3CH2)2CH—I

___14. Which of the following are "concerted" reactions?

a. SN1 d. electrophilic additionb. Diels-Alder reaction e. SN1 and E1c. E1

– 3 –

___15. Reaction of a hydroxide ion (HO–) with a primary alkyl halide is most likely to result in:

a. SN2 substitution c. SN1 substitutionb. E2 elimination d. E1 elimination

___16. Which of the following statements is correct?

a. Free radical bromination is more selective than chlorination because the transition state for bromination is more reactant-like.

a. Free radical bromination is more selective than chlorination because the rate limiting step for bromination is more endothermic than for chlorination.

c. Free radical bromination is more selective than chlorination because the rate limiting step for bromination is more exothermic than for chlorination.

d. Free radical chlorination is more selective than bromination because the transition state for chlorination is more reactant-like.

___17. Which of the following would be the rate limiting step in a free radical halogenation?

a. CH3CH2. + Br2 → CH3CH2Br + Br.

b. Br2 → 2 Br.

c. 2 CH3CH2. → CH3CH2CH2CH3d. CH3CH3 + Br. → CH3CH2. + HBr

___18. Ionization to give a carbocation and a leaving group is the rate determining step for

a. SN1 c. E1 and E2 e. SN1 and SN2b. SN2 d. E1 f. SN1 and E1

Syntheses. Give reagents to show how to synthesize the compounds on the right from the compoundson the left. They may require more than one step. (4 pts each)

19.CH3 CH2OH

20.

H3C C

CH3

CH3

CH

CH3

CH3 H3C C

CH3

CH3

C

CH3

CH2

– 4 –

V. Mechanisms. Give a stepwise, detailed mechanism with arrows and intermediates for the following reactions. (5 points each)

21.Br2, light

CH3CH2CH3 CH3CHBrCH3

22.

CH3Br

CH3

OCH3CH3

+CH3OH

23.HBr

40 °CH2C=CH CH=CH2 H3C–CH CH=CH2 H3C–CH CH–CH2Br+

Br

– 5 –

24. The energy diagram for the hypothetical reaction A + B → D → G + H is shown below. (6pts).

E

reaction coordinate

A + B

C

D

E

G

F

H

a. What is the rate determining step?_____________

b. What happens to the rate if the concentration of A is doubled? _____________

c. What is the rate expression? rate = _______________

d. If D is a charged species and A and B are neutral, what effect will a polar protic solvent have onthe rate of reaction?___________________

e. What is the kinetically favored product?__________

f. What is the thermodynamically favored product?_________

61 Notes

Documents

ch3ch3 ch2ch2

pauli exclusion

single dot

2p 2s 1s

r1 r1

rc cr hg2

negative charge

free radical