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CHM 2211C
th
6 edition Notes
Chapter 10
Alkyl Halides
By
Dr. Andrea Wallace
Coastal Georgia Community College
Edited by
John T. Taylor
Florida Community College at Jacksonville
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Chapter 10: Alkyl Halides
Alkyl halides occur widely in nature and have many uses in industrial processes.
Uses of Alkyl Halides – see p. 316 – solvents, anesthetic, refrigerant, fumigant
Epibatidine (p. 317) – is found on the skin of an Ecuadorian frog and is 200 x more
potent than morphine in blocking pain.
10.1 Naming Alkyl Halides
Rule 1) Find the parent chain – longest continuous chain (that contains a double or triple
bond if one is present).
Rule 2) Assign lowest numbers to the branches – alkyl and halo.
a) Apply prefixes – di, tri, tetra
b) Alphabetize
Rule 3) If the parent chain can be properly numbered from either end by Rule 2, assign
them alphabetically.
Common names: Alkyl group + Halide
Example: Iodomethane (IUPAC) or methyl iodide (common)
Example: 2-Chlorpropane (IUPAC) or isopropyl chloride (common)
Problem 10.1, p. 318
Give the IUPAC names of the following alkyl halides.
a)
d)
f)
Problem 10.2, p. 318
Draw the structures corresponding to the following IUPAC names:
d) 1,1-Dibromo-4-isopropylcyclohexane
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10.2 Structures of Alkyl Halides
Table 10.1, p. 319
CH3F
CH3Cl
CH3Br
CH3I
----C-X Bond Length ________________ Why? ______________________________
---C-X Bond Strength _______________ Why? ______________________________
--Dipole Moment (Bond Polarity) ___________ Why? _________________________
Halogens are more electronegative than Carbon. All have substantial dipole moments.
Figure:
10.3 Preparation of Alkyl Halides
Electrophilic Addition
1) Alkene
+ X2

2) Alkene + HX 
X2 = Cl2 or Br2,
HX = HCl, HBr, or HI
Free Radical Substitution of Alkanes
See Figure 10.1, p. 320 for Mechanism
Alkane
+ X2

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10.4 Radical Halogenation of Alkanes
Why is this not necessarily the best choice?
What are the possible products of CH4 + Cl2?
There are even more products possible when more than one type of Hydrogen is present.
Example:
+
Cl2
--h
Butane
Ratios:
(only showing monochlorinated products – di, tri, tetra, etc, are possible)
Another Example:
+
Cl2
--h
2-Methylpropane
Ratios:
Compare equivalent Hydrogen’s and Ratios:
Butane
2-Methylpropane
____eq. Primary
____eq. Secondary
___ eq. Primary
___ eq. Tertiary
Primary accounts for ___% of pdt.
Secondary accounts for ___% of pdt.
Primary accounts for ___% of pdt.
Tertiary accounts for ___% of pdt.
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Secondary is _____ times more likely.
See Reactivity Figure on p. 321.
Tertiary is _____ times more likely.
Primary
Tertiary
<
Secondary
<
This shows relative reactivity towards Chlorination.
Why? Table 5.3, p. 154
Energy needed to break bond
Primary
Secondary
Tertiary
Radical Stability
-------_____________________ Stability
The more stable radical forms faster. See Figure 10.2, p. 322.
Bromination is even more selective.
+
2-methylpropane
Br2 --h
2-bromo-2methylpropane
> 99%
+
1-bromo-2methylpropane
< 1%
Why?
Hammond Postulate – the transition state most closely resembles the species (reactant
or product ) to which it is closest in energy.
H = -50 kJ for X = Cl
H = +13 kJ for X = Br
The bromine reaction is more ____________ and thus more product-like (more similar to
the radical). The reaction shows selectivity that reflects the stability of the radical.
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Problem 10.3, p. 323
Draw and name all monochloro products you would expect to obtain from radical
chlorination of 2-methylpentane. Which, if any, are chiral?
Problem 10.4, p. 323
Taking the 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.)
10.5
Allylic Bromination of Alkenes
Find the allylic positions on cyclohexene.
Reaction: Allyic Bromination of an Alkene
NBS
h, CCl4

Cyclohexene
3-Bromocyclohexene
The presence of h causes the Br2 to form Br radicals.
Br2 --hv
2 Br .
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Mechanism:
What type of intermediate is formed? ________________________________
What step of the mechanism is shown above? __________________________
What step of the mechanism is occurring when the Br radical forms (shown above the
“mechanism”)? ________________________________
What contributes to the stability of the allylic radical? _____________________
Why does bromination occur exclusively at the allylic position?
Compare bond dissociation energies at the other positions – alkyl, allylic, and vinylic.
The bond dissociation energy is less because the radical formed is more stable.
--------------------------_________________ Radical Stability----------
10.6
Stability of the Allyl Radical: Resonance Revisited
Why are allylic radicals so stable?
Draw the resonance structures for allyl radical.
Delocalization of electrons via resonance gives increased stability.
The greater the # of resonance structures, the greater the stability of the molecule.
Example: Propyl radical
Electron is localized on Carbon – it is _________ stable.
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Bromination of unsymmetrical alkenes yields an unequal mixture of products.
Reaction at less hindered end is more stable.
1-octene
NBS
h, CCl4

3-Bromo-1-octene (17%)
1-Bromo-2-octene (83%) (53/47- trans/cis)
Mechanism:
Useful Reaction:
NBS
h, CCl4
Cyclohexene

KOH 
3-Bromocyclohexene
1,3-Cyclohexadiene
What type of reaction occurs when KOH is added? ____________________
Problem 10.5, p. 327
Draw three resonance forms for the cyclohexadienyl radical.
Problem 10.6, p. 327
The major product of the reaction of methylenecyclohexane with N-bromosuccinimide is
1-(bromomethyl)cyclohexene. Explain.
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Problem 10.7 b., p. 327
What products would you expect from reaction of the following alkenes with NBS? If
more than one product is formed, show the structures of all.
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10.7
Preparing Alkyl Halides from Alcohols
Most general method of preparation of alkyl halides.
Reaction:
Alcohol
Hydrohalic acid
Alkyl Halide
Water
where X = Cl, Br, or I
Works best for 3o Alcohols.
1o and 2o alcohols react very slowly and require very high temperatures – generally not
practical.
Reactivity of Alcohols with Hydrohalic acids
Methyl
Primary
Secondary
Tertiary
------_______________ reactivity -------
Example:
HCl(aq), 25oC
t-butyl alcohol
+ H2O
t-butyl chloride
Primary and Secondary alcohols are best converted to alkyl halides by reaction with
Thionyl Chloride (SOCl2) or Phosphorus Tribromide (PBr3).
Examples:
PBr3
Ether, 35 oC

2-Butanol
SOCl2

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Pyridine
2-Butanol
Why does this reaction work better for primary and secondary alcohols? These reagents
are less acidic and less likely to cause acid catalyzed rearrangements.
(Mechanisms are covered in Chapter 11.)
Problem 10.8, p. 369
How would you prepare the following alkyl halides from the appropriate alcohols?
a) 2-chloro-2-methylpropane
b) 1-bromo-5-methylhexane
10.8
Grignard Reagents
Grignard reagents are organometallic reagents, RMgX
Preparation of Grignard Reagents:
R-X
Alkyl halide
+
Mg
Ether or THF
R-Mg-X
Grignard Reagent
where R = 1o, 2o, or 3o alkyl, aryl, or alkenyl (all work equally well – best to use THF
with aryl and alkenyl)
X = Cl, Br, or I (Cl is less reactive than Br or I, organofluorides rarely react with Mg)
Examples:
Bromobenzene
Phenylmagnesium bromide
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2-chlorobutane
sec-butylmagnesium chloride
Polarity of C-Mg bond:
Think of the C as partial negative or even negative like a carbanion. These species do act
like bases and react with acids (proton donors) such as H2O, ROH, RCOOH, and RNH2
to yield hydrocarbons.
Example:
1-bromobutane
butane
Problem 10.10, p. 330
How might you replace a halogen substituent with a deuterium atom if you wanted to
prepare a deuterated compound?
10.9
Organometallic Coupling Reactions
Preparation of Organometallic Reagents
2 Li 
pentane
1-bromobutane
Alkyllithium
(Butyllithium)
Lithium Bromide
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Alkyllithiums are basic and act as nucleophiles. They are similar to RMgX (Grignard
Reagents).
One of the most valuable reactions of alkyllithiums is the preparation of diorganocopper
compounds or Gilman reagents.
Preparation of Gilman Reagents
+
Methyllithium
ether
Copper(I) Iodide
Lithium Dimethylcopper
(Gilman Reagent)
Lithium Iodide
Reaction of Gilman Reagent with Alkyl Halides (Cl, Br, or I) to Produce Alkanes
Ether, 0oC
Lithium Dimethylcopper
(Gilman Reagent)
Ethyl Iodide
Propane
This organometallic coupling reaction is very versatile. It also works with vinylic halides
and aryl halides (not just alkyl halides). See p. 331.
Example:
Problem 10.11, p. 332
How would you prepare the following compounds using an organocopper coupling
reaction? More than one step is required in each case.
a) 3-methylcylohexene from cyclohexene
c) Decane from 1-pentene
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10.10 Oxidation and Reduction in Organic Chemistry
Inorganic Definitions:
Oxidation - ____________________ of electrons
Reduction - ____________________ of electrons
Still true in organic, but the definition is different.
Oxidation is a gain of _______________________onto C and/or a loss of __________
onto C.
Reduction is a gain of _______________________onto C and/or a loss of __________
onto C.
Examples on p. 333
Methane + Cl2
 Chloromethane _________________ Why? __________________
Chloromethane 1) Mg, ether/ 2) H3O+
Why?____________________________
Methane _______________________
More Examples on p. 333
See Figure 10.5 on p. 334.
--------------------------_________________ Oxidation Level -----------------
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Problem 10.12, p. 334
Rank each of the following series of compounds in order of increasing oxidation level:
(Strategy: Compounds that have the same number of carbon atoms can be compared by
adding the number of C-O, C-N, and C-X bonds in each and then subtracting the number
of C-H bonds. The larger the resultant value, the higher the oxidation level.)
a)
b) CH3CN, CH3CH2NH2, H2NCH2CH2NH2
Problem 10.13, p. 334
Tell whether each of the following reactions is an oxidation, a reduction, or neither.
Explain your answers.
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