chapter 10. introduction alkyl halides –alkyl halides – are compounds containing halogen bonded...

Chapter 10

IntroductionIntroduction

• Alkyl halides – Alkyl halides – are compounds containing halogen bonded to a saturated, sp3-hybridized carbon atom

– – are haloalkaneshaloalkanes

Properties and some usesProperties and some uses

It is a pain reliever It is more potent than morphine

A.A. Naming Alkyl HalidesNaming Alkyl Halides

A.A. Naming Alkyl HalidesNaming Alkyl Halides

• Alkyl halides are named according to the system of nomenclature devised by the IInternational nternational UUnion of nion of PPure and ure and AApplied pplied CChemistryhemistry (IUPACIUPAC):

Steps to naming alkyl halidesSteps to naming alkyl halides11

1.1. Find the parent hydrocarbonFind the parent hydrocarbon • Include the double or triple bond if presentInclude the double or triple bond if present

2.2. Number the atoms in the main chainNumber the atoms in the main chain

a.a. The correct sequence is when the substituents have the The correct sequence is when the substituents have the lowest possible numberlowest possible number

b.b. Use multiplicity prefixes di-, tri-, tetra-, …Use multiplicity prefixes di-, tri-, tetra-, …

c.c. Put the substituents in alphabetical order (Do not Put the substituents in alphabetical order (Do not consider multiplier prefixes)consider multiplier prefixes)

Steps to naming alkyl halidesSteps to naming alkyl halides22

3.3. If the parent chain can be properly numbered from If the parent chain can be properly numbered from either end by step 2, begin at the end nearer the either end by step 2, begin at the end nearer the substituent that has alphabetical precedencesubstituent that has alphabetical precedence

2,4,5 3,4,6 2,4,5 3,4,6

2,3,4 3,4,5

1,3,4 2,3,5

Practice ProblemPractice Problem: Give IUPAC names for the following alkyl : Give IUPAC names for the following alkyl halides: halides:

Practice ProblemPractice Problem: Draw structures corresponding to the : Draw structures corresponding to the following IUPAC names: 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.B. Structure of Alkyl HalidesStructure of Alkyl Halides

B.B. Structure of Alkyl HalidesStructure of Alkyl Halides

• As you go down the periodic table,– C-X bondC-X bond is longerlonger– C-X bondC-X bond is weakerweaker

• Due to EN, C-X bond is polarized with:

– slight positiveslight positive on carboncarbon and – slight negativeslight negative on halogenhalogen

• The C-X carbonC-X carbon atom can behave as an electrophileelectrophile

C.C. Preparing Alkyl HalidesPreparing Alkyl Halides

C.C. Preparing Alkyl HalidesPreparing Alkyl Halides

• Alkyl halideAlkyl halide -- is synthesized from addition of HCl, HBr, HI to alkenes to give MarkovnikovMarkovnikov product

• Alkyl dihalideAlkyl dihalide -- is synthesized from antianti addition of Br2 or Cl2

Reaction of Alkanes with HalogensReaction 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 methaneMechanism of the radical chlorination of methane

D.D. Radical Halogenation of AlkanesRadical Halogenation of Alkanes

D.D. Radical Halogenation of AlkanesRadical 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’sPrimary H’s Secondary H’sSecondary H’s

35% / 1 = 35% 65% / 9 = 7.2%

Tertiary H’sTertiary H’s Primary H’sPrimary H’s

Relative ReactivityRelative 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 ProblemPractice Problem: Draw and name all monochloro products you : Draw and name all monochloro products you would expect to obtain from radical chlorination would expect to obtain from radical chlorination of 2-methylpentane. Which, if any, are chiral? of 2-methylpentane. Which, if any, are chiral?

Practice ProblemPractice Problem: Taking the relative reactivities of 1: Taking the relative reactivities of 1oo, 2, 2oo, and 3, and 3oo hydrogen atoms into account, what product(s) hydrogen atoms into account, what product(s) would you expect to obtain from would you expect to obtain from monochlorination of 2-methylbutane? What monochlorination of 2-methylbutane? What would the approximate percentage of each would the approximate percentage of each product be? (Don’t forget to take into account product be? (Don’t forget to take into account the number of each type of hydrogen) the number of each type of hydrogen)

E.E. Allylic Bromination of AlkenesAllylic Bromination of Alkenes

E.E. Allylic Bromination of AlkenesAllylic Bromination of Alkenes

• N-bromosuccinimideN-bromosuccinimide (NBS) selectively brominates allylic positions

– It requires light for activation– It is a source of dilute bromine atoms

NBS allylic brominationNBS 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 Bromination with NBS occurs exclusively at an allylic position because:allylic position because:

– an allylic radical is more stable than a typical alkyl an allylic radical is more stable than a typical alkyl radicalradical by about 40 kJ/mol

Stability OrderStability Order

• Allylic radical is more stable than tertiary alkyl radical

F.F. Stability of the Allyl RadicalStability of the Allyl Radical

F.F. Stability of the Allyl RadicalStability of the Allyl Radical

• Allyl radical is resonance-stabilizedresonance-stabilized

– Allyl radicalAllyl radical has two resonance formstwo resonance forms

– Alkyl radical has only one structure

The greater the number of resonance forms, the greater the stability

• Allyl radical is delocalizeddelocalized

– The unpaired electron is spread out over an extended 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 ProblemPractice Problem: Draw three resonance forms for the : Draw three resonance forms for the cyclohexadienyl radical cyclohexadienyl radical

Practice ProblemPractice Problem: The major product of the reaction of : The major product of the reaction of methylenecyclohexane with methylenecyclohexane with NN-bromo -bromo succinimide is 1-(bromomethyl)cyclohexene. succinimide is 1-(bromomethyl)cyclohexene. Explain. Explain.

Practice ProblemPractice Problem: What products would you expect from the : What products would you expect from the reaction of the following alkenes with NBS? If reaction of the following alkenes with NBS? If more than one product is formed, show the more than one product is formed, show the structures of all. structures of all.

G.G. Preparing Alkyl halides fromPreparing Alkyl halides from

AlcoholsAlcohols

G.G. Preparing Alkyl halides fromPreparing Alkyl halides from

AlcoholsAlcohols

• 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 effectiveReaction 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 AlcoholsPreparation of Alkyl Halides from Primary and Secondary Alcohols

• PrimaryPrimary and secondary alkyl halidessecondary alkyl halides are normally prepared from alcohols using either

– thionyl chloridethionyl chloride (SOCl(SOCl22))

SOClSOCl22 : ROH RCl

– phosphorus tribromidephosphorus tribromide (PBr(PBr33))

PBr3 : ROH RBr

– These reactions use mild conditions – less acidic and less likely to cause rearrangements

Practice ProblemPractice Problem: How would you prepare the following alkyl : How would you prepare the following alkyl halides from the corresponding alcohols? halides from the corresponding alcohols?

H.H. Reaction of Organohalides:Reaction of Organohalides:

Grignard ReagentsGrignard Reagents

H.H. Reaction of Organohalides:Reaction of Organohalides:

Grignard ReagentsGrignard Reagents

• Alkyl halides react with magnesium in ether solution to form organomagnesium halides, Grignard reagents Grignard reagents (RMgX)(RMgX)

• Reaction of RX with Mg in ether or THF

• Product is RMgXRMgX – an organometallic compound (alkyl-metal bond)

– RR is alkyl 1°, 2°, 3°, aryl, alkenyl– XX = Cl, Br, I

• The carbon-magnesium bond is polarized

The carboncarbon atom is thus both nucleophilicnucleophilic and basicbasic

Reactions of Grignard ReagentsReactions of Grignard Reagents

• Many useful reactions– RMgXRMgX behaves as RR--

– RR-- (carbon anions) are very strong basesvery strong bases– RMgXRMgX reactreact with such weak acids as Hweak acids as H22OO,, ROH ROH,,

RCORCO22H H and RNH RNH2 2 to abstract H+

– RMgX + H3O+ R-H

Practice ProblemPractice Problem: Just how strong a base would you expect a : Just how strong a base would you expect a Grignard reagent be? Look at Table 8.1, and Grignard reagent be? Look at Table 8.1, and then predict whether the following reactions then predict whether the following reactions will occur as written. (The pK will occur as written. (The pKaa of NH of NH33 is 35) is 35)

a) CH3MgBr + H-CΞC-H CH4 + H-CΞC-MgBr

b) CH3MgBr + NH3 CH4 + H2N-MgBr

Practice ProblemPractice Problem: How might you replace a halogen substituent : How might you replace a halogen substituent by a deuterium atom if you wanted to prepare by a deuterium atom if you wanted to prepare a deuterated compound? a deuterated compound?

I.I. Organometallic Coupling ReactionsOrganometallic Coupling Reactions

I.I. Organometallic Coupling ReactionsOrganometallic Coupling Reactions

1.1. Alkyllithium (RLi) forms from RBr and Li metalAlkyllithium (RLi) forms from RBr and Li metal

2.2. RLi reacts with copper iodide to give lithium RLi reacts with copper iodide to give lithium dialkylcopper (Gilman reagents)dialkylcopper (Gilman reagents)

3.3. Lithium dialkylcopper reagents react with alkyl Lithium dialkylcopper reagents react with alkyl halides to give alkaneshalides to give alkanes

1.1. Alkyllithium (RLi) forms from RBr and Li metalAlkyllithium (RLi) forms from RBr and Li metal

Alkyllithiums are both nucleophiles and bases

2.2. RLi reacts with copper iodide to give lithium RLi reacts with copper iodide to give lithium dialkylcopper (Gilman reagents)dialkylcopper (Gilman reagents)

3.3. Lithium dialkylcopper reagents react with alkyl Lithium dialkylcopper reagents react with alkyl halides to give alkaneshalides to give alkanes

Gilman reagents undergo organometallic coupling reactionswith chlorides, bromides and iodides (but not fluorides)

Utility of Organometallic Coupling in SynthesisUtility 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 ProblemPractice Problem: How would you carry out the following : How would you carry out the following transformations using an organocopper transformations using an organocopper coupling reaction? More than one step is coupling reaction? More than one step is required in each case. required in each case.

J.J. Oxidation and Reduction in Organic Oxidation and Reduction in Organic

ChemistryChemistry

J.J. Oxidation and Reduction in Organic Oxidation and Reduction in Organic

ChemistryChemistry

• OxidationOxidation is a reaction that results in loss of electron density at carbon either by

• bond formationbond formation between carbon carbon and a more electronegative more electronegative atom (oxygen, nitrogen, or halogen)

• bond breakingbond breaking between carbon carbon and a less electronegative less electronegative atom (usually hydrogen)

– Not necessarily defined as loss of electrons by an atom as in inorganic chemistry

OxidationOxidation: break C-H (or C-C) and

: form C-O, C-N, C-X

ReductionReduction: form C-H (or C-C) and

break C-O, C-N, C-X

• Organic ReductionOrganic Reduction is the opposite of oxidation. It results in increase of electron density at carbon either by

• bond breakingbond breaking between carbon carbon and a more electronegative more electronegative atom (oxygen, nitrogen, or halogen)

• bond formation bond formation between carbon carbon and a less electronegative less electronegative atom (usually hydrogen)

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 ProblemPractice Problem: Rank each of the following series of : Rank each of the following series of compounds in order of increasing oxidation compounds in order of increasing oxidation level: level:

Practice ProblemPractice Problem: Tell whether each of the following reactions is : Tell whether each of the following reactions is an oxidation, a reduction, or neither. Explain an oxidation, a reduction, or neither. Explain your answers. your answers.

Chapter 10