Transcript Practice Problem

Chapter 10
Introduction
• Alkyl halides – are compounds containing
halogen bonded to a saturated,
sp3-hybridized carbon atom
– are haloalkanes
Properties and some uses
It is a pain reliever
It is more potent than morphine
A.
Naming Alkyl Halides
• Alkyl halides are named according to the system of
nomenclature devised by the International Union of
Pure and Applied Chemistry (IUPAC):
Steps to naming alkyl halides1
1. Find the parent hydrocarbon
•
Include the double or triple bond if present
2. Number the atoms in the main chain
a. The correct sequence is when the substituents have the
lowest possible number
b. Use multiplicity prefixes di-, tri-, tetra-, …
c. Put the substituents in alphabetical order (Do not
consider multiplier prefixes)
Steps to naming alkyl halides2
3.
If the parent chain can be properly numbered from
either end by step 2, begin at the end nearer the
substituent that has alphabetical precedence
2,4,5
3,4,6
2,4,5
3,4,6
2,3,4
1,3,4
3,4,5
2,3,5
Practice Problem: Give IUPAC names for the following alkyl
halides:
Practice Problem: Draw structures corresponding to the
following IUPAC names:
a) 2-Chloro-3,3-dimethylhexane
b) 3,3-Dichloro-2-methylhexane
c) 3-Bromo-3-ethylpentane
d) 1,1-Dibromo-4-isopropylcyclohexane
e) 4-sec-Butyl-2-chlorononane
f) 1,1-Dibromo-4-tert-butylcyclohexane
B.
Structure of Alkyl Halides
• As you go down the periodic table,
– C-X bond is longer
– C-X bond is weaker
• Due to DEN, C-X bond is polarized with:
– slight positive on carbon and
– slight negative on halogen
• The C-X carbon atom can behave as an electrophile
C.
Preparing Alkyl Halides
• Alkyl halide - is synthesized from addition of HCl,
HBr, HI to alkenes to give Markovnikov product
• Alkyl dihalide - is synthesized from anti addition
of Br2 or Cl2
Reaction of Alkanes with Halogens
• Alkane + Cl2 or Br2, heat or light, replaces C-H
with C-X but gives mixtures
– hard to control
– via free radical mechanism
• It is usually not a good idea to plan a synthesis
that uses this method
Mechanism of the radical chlorination of methane
D.
Radical Halogenation of Alkanes
• Radical halogenation of alkanes is not a good method
of alkyl halide synthesis because mixtures of products
usually result.
• If there is more than one type of hydrogen in an
alkane, reactions favor replacing the hydrogen at
the most highly substituted carbons
30% / 6 = 5%
70% / 4 = 17.5%
Primary H’s
Secondary H’s
35% / 1 = 35%
Tertiary H’s
65% / 9 = 7.2%
Primary H’s
Relative Reactivity
• Based on quantitative analysis of reaction
products, relative reactivity is estimated
• Order parallels stability of radicals
• Increasing alkyl substitution stabilizes the transition
state and radical intermediate
 The more stable radical forms faster
• Reaction distinction is more selective with
bromine than chlorine
• Reaction distinction is more selective with
bromine than chlorine
• Reaction with Br. is much less exergonic
• T.Sbromination resembles the alkyl radical more closely
than does T.Schlorination
Practice Problem: Draw and name all monochloro products you
would expect to obtain from radical chlorination
of 2-methylpentane. Which, if any, are chiral?
Practice Problem: Taking the relative reactivities of 1o, 2o, and 3o
hydrogen atoms into account, what product(s)
would you expect to obtain from
monochlorination of 2-methylbutane? What
would the approximate percentage of each
product be? (Don’t forget to take into account
the number of each type of hydrogen)
E.
Allylic Bromination of Alkenes
• N-bromosuccinimide (NBS) selectively brominates
allylic positions
– It requires light for activation
– It is a source of dilute bromine atoms
NBS allylic bromination
Br. radical abstracts
an allylic hydrogen
Allylic radical forms as an intermediate and reacts with Br2
• Bromination with NBS occurs exclusively at an
allylic position because:
– an allylic radical is more stable than a typical alkyl
radical by about 40 kJ/mol
Stability Order
• Allylic radical is more stable than tertiary alkyl radical
F.
Stability of the Allyl Radical
• Allyl radical is resonance-stabilized
– Allyl radical has two resonance forms
– Alkyl radical has only one structure
The greater the number of resonance forms, the greater
the stability
• Allyl radical is delocalized
– The unpaired electron is spread out over an
extended p orbital network
– The unpaired electron is equally shared between
the two terminal carbons
• Allylic bromination of an unsymmetrical alkene
often leads to a mixture of products
less hindered
• Allylic bromination can be used to convert alkenes
into dienes by dehydrohalogenation with base.
Practice Problem: Draw three resonance forms for the
cyclohexadienyl radical
Practice Problem: The major product of the reaction of
methylenecyclohexane with N-bromo
succinimide is 1-(bromomethyl)cyclohexene.
Explain.
Practice Problem: What products would you expect from the
reaction of the following alkenes with NBS? If
more than one product is formed, show the
structures of all.
G.
Preparing Alkyl halides from
Alcohols
• Alcohols react with HX to form alkyl halides, but the
reaction works well for tertiary alcohols, R3COH
• Reaction of tertiary C-OH with HX is fast and effective
– Add HCl or HBr gas into ether solution of tertiary alcohol
– Primary and secondary alcohols react very slowly and
often rearrange, so alternative methods are used
Preparation of Alkyl Halides from Primary and Secondary Alcohols
• Primary and secondary alkyl halides are normally
prepared from alcohols using either
– thionyl chloride (SOCl2)
SOCl2 : ROH  RCl
– phosphorus tribromide (PBr3)
PBr3 : ROH  RBr
– These reactions use mild conditions – less acidic and
less likely to cause rearrangements
Practice Problem: How would you prepare the following alkyl
halides from the corresponding alcohols?
H.
Reaction of Organohalides:
Grignard Reagents
• Alkyl halides react with magnesium in ether solution
to form organomagnesium halides, Grignard reagents
(RMgX)
• Reaction of RX with Mg in ether or THF
• Product is RMgX – an organometallic compound
(alkyl-metal bond)
– R is alkyl 1°, 2°, 3°, aryl, alkenyl
– X = Cl, Br, I
• The carbon-magnesium bond is polarized
 The carbon atom is thus both nucleophilic and
basic
Reactions of Grignard Reagents
• Many useful reactions
– RMgX behaves as R– R- (carbon anions) are very strong bases
– RMgX react with such weak acids as H2O, ROH,
RCO2H and RNH2 to abstract H+
– RMgX + H3O+  R-H
Practice Problem: Just how strong a base would you expect a
Grignard reagent be? Look at Table 8.1, and
then predict whether the following reactions
will occur as written. (The pKa of NH3 is 35)
a) CH3MgBr
+
H-CΞC-H
b) CH3MgBr
+
NH3


CH4 +
CH4 +
H-CΞC-MgBr
H2N-MgBr
Practice Problem: How might you replace a halogen substituent
by a deuterium atom if you wanted to prepare
a deuterated compound?
I.
Organometallic Coupling Reactions
1. Alkyllithium (RLi) forms from RBr and Li metal
2. RLi reacts with copper iodide to give lithium
dialkylcopper (Gilman reagents)
3. Lithium dialkylcopper reagents react with alkyl
halides to give alkanes
1. Alkyllithium (RLi) forms from RBr and Li metal
Alkyllithiums are both nucleophiles and bases
2. RLi reacts with copper iodide to give lithium
dialkylcopper (Gilman reagents)
3. Lithium dialkylcopper reagents react with alkyl
halides to give alkanes
Gilman reagents undergo organometallic coupling reactions
with chlorides, bromides and iodides (but not fluorides)
Utility of Organometallic Coupling in Synthesis
• Coupling of two organometallic molecules produces
larger molecules of defined structure
• Aryl and vinyl organometallics are also effective
• Coupling of lithium dialkylcopper molecules
proceeds through trialkylcopper intermediate
Practice Problem: How would you carry out the following
transformations using an organocopper
coupling reaction? More than one step is
required in each case.
J.
Oxidation and Reduction in Organic
Chemistry
• Oxidation is a reaction that results in loss of
electron density at carbon either by
• bond formation between carbon and a more electronegative
atom (oxygen, nitrogen, or halogen)
• bond breaking between carbon and a less electronegative
atom (usually hydrogen)
– Not necessarily defined as loss of electrons by an atom
as in inorganic chemistry
Oxidation: break C-H (or C-C) and
: form C-O, C-N, C-X
• Organic Reduction is the opposite of oxidation. It
results in increase of electron density at carbon
either by
• bond breaking between carbon and a more electronegative
atom (oxygen, nitrogen, or halogen)
• bond formation between carbon and a less electronegative
atom (usually hydrogen)
Reduction: form C-H (or C-C) and
break C-O, C-N, C-X
Halogenation
Conversion of alkyl halide to alkane
• Functional groups are associated with specific levels
Maximum number of C-H
Maximum number of C-X
Practice Problem: Rank each of the following series of
compounds in order of increasing oxidation
level:
Practice Problem: Tell whether each of the following reactions is
an oxidation, a reduction, or neither. Explain
your answers.
Chapter 10