© E.V. Blackburn, 2012 Alkanes Nomenclature, Conformational Analysis, and an Introduction to...

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© E.V. Blackburn, 2012

Alkanes

Nomenclature, Conformational Analysis, and an Introduction to

Synthesis

© E.V. Blackburn, 2012

Alkanes

• saturated aliphatic hydrocarbons

• paraffins

• general formula CnH2n+2

• acyclic hydrocarbons

© E.V. Blackburn, 2012

Sources of methane

• major constituent of natural gas (97%)

• “firedamp” of coal mines

• “marsh gas”

• product of anaerobic plant decay

© E.V. Blackburn, 2012

Cycloalkanes

Single ring cycloalkanes have the general formula CnH2n thus they have two fewer hydrogen atoms than alkanes.

© E.V. Blackburn, 2012

Methane – its structure

tetrahedral

H

H HH

sp3

109.5o

© E.V. Blackburn, 2012

Methane – its structure

“Fischer Structure” “Lewis Structure”

H

HHH

H

HHH C: :

..

..

© E.V. Blackburn, 2012

Space-filling modelsSpace-filling models depict atoms as spheres and therefore show the volume occupied by atoms and molecules.

© E.V. Blackburn, 2012

Ethane - C2H6

sp31.10Å

1.53Å

A structural formula is a Lewis structure which shows the connectivity of its atoms - the order in which atoms are connected.

C CH

HH

H

HH

© E.V. Blackburn, 2012

What is ethane’s structure?

Or something in between?

H

H H

H

HH

staggered

H

H H

H

HH

eclipsed

H

H H

H

H H

© E.V. Blackburn, 2012

ConformationsConformations are structures that are interconvertible by rotation about single bonds.

This is the staggered conformation of ethane:

This is an example of a sawhorse formula.

H

H H

H

HH

© E.V. Blackburn, 2012

Newman projections

The nearest carbon is represented by the point where the three bonds meet.

The rear carbon is represented by the circle.

o

HH

H

H

HH

Look along the C-C bond. The nearest carbon masks the rear carbon but all six bonds to the two carbons are visible.

staggered

HH

H

eclipsed

© E.V. Blackburn, 2012

Space-filling model of ethane

staggered eclipsed

© E.V. Blackburn, 2012

Pot

entia

l ene

rgy

Stability of conformations

rotation

eclipsed staggered eclipsed

12 kJ/mol

60o 120o 180o

© E.V. Blackburn, 2012

Torsional strain

Torsional energy is the energy required to rotate the molecule about the C-C bond.

The relative instability of the eclipsed conformation is said to be due to torsional strain.

© E.V. Blackburn, 2012

Propane - C3H8

energy barrier = 14 kJ/mol

H C CH

H

H

HCH

HH

© E.V. Blackburn, 2012

Butane - C4H10

compound A Bbp -12 0 Cmp -159 -138solubility 1320 1813 mL/100mL C2H5OH

CH3CH2CH2CH3

(CH3)3CH

H3C CH3

CH3H

© E.V. Blackburn, 2012

Conformations

H3C

H

HH

H CH3

anti

All conformations are free of torsional strain.

H

H

CH3

H

H CH3

H

CH3

HH

H CH3

gauche

© E.V. Blackburn, 2012

The methyl groups in the gauche conformations are crowded together and steric repulsion results. These conformations are less stable due to steric strain.

Stability of conformations

H3C

H

HH

H CH3

anti

H

H

CH3

H

H CH3

H

CH3

HH

H CH3

gauche

© E.V. Blackburn, 2012

Stability of conformations

anti gauche

© E.V. Blackburn, 2012

Stability of conformationsP

oten

tial e

nerg

yH3C

CH3

anti

H3C

CH3

H3CCH3

gauche

H3C H3CH3C

16 kJ

3.8 kJ

19 kJ

© E.V. Blackburn, 2012

Nomenclature

C2H6 ethane

C3H8 propane

C4H10 butane

Subsequent alkanes are systematically named using a numeric prefix (Greek) (penta-, hexa-, etc.) and the suffix -ane.

CH4 methane

© E.V. Blackburn, 2012

Nomenclature

CH4 methane C7H16 heptane

C2H6 ethane C8H18 octane

C3H8 propane C9H20 nonane

C4H10 butane C10H22 decane

C5H12 pentane C11H24 undecane

C6H14 hexane C12H26 dodecane

C13H28 tridecane C14H30 tetradecane

C20H42 icosane C100H202 hectane

© E.V. Blackburn, 2012

n-ButaneCH3CH2CH2CH3

(CH3)3CH

H3C CH3

CH3H

n- - specifies a straight chain hydrocarbon, e.g. n-butane or normal butane

?

© E.V. Blackburn, 2012

Prefixes......

isobutane

H3C C CH3

H

CH3

iso-

iso- (CH3)2CH-

© E.V. Blackburn, 2012

Pentanen-pentane

isopentane

neopentane

CH3-CH2-CH2-CH2-CH3

CH3-CH-CH2-CH3

CH3

CH3-C-CH3

CH3

CH3 H3CCH3

CH3

neo

© E.V. Blackburn, 2012

Hexane

There are five alkane isomers of formula C6H14 ...

© E.V. Blackburn, 2012

n-hexane

CH3CH2CH2 CH2CH2CH3

© E.V. Blackburn, 2012

isohexane(CH3)2CHCH2CH2CH3

H3C CH

CH3

CH2-CH2-CH3

© E.V. Blackburn, 2012

Neohexane

(CH3)3CCH2CH3

H3C CCH3

CH3

CH2CH3

© E.V. Blackburn, 2012

and ........

CH3

CH3

CH3-CH-CH-CH3

CH3-CH2-CH-CH2-CH3

CH3

© E.V. Blackburn, 2012

Nomenclature

Why not name these more complex alkanes by first identifying and naming the longest carbon chain? – the parent chain.

Then consider the groups attached to the parent chain as substituents?

© E.V. Blackburn, 2012

Alkyl group substituents

These groups are named by replacing the -ane suffix of the corresponding alkane by -yl, hence “alkyl”.

• CH3- methyl (Me-)

• CH3CH2- ethyl (Et-)

• (CH3)2CH- isopropyl (i-Pr)

• CH3CH2CH2- propyl (Pr)

An alkyl group is the structure obtained when a hydrogen atom is removed from an alkane.

© E.V. Blackburn, 2012

Alkyl group substituants

• CH3CH2CH2CH2- butyl

• (CH3)2CHCH2- isobutyl

• but …. (CH3)3C- ?

• or CH3CHCH2CH3|

• CH3- methyl (Me-)

• CH3CH2- ethyl (Et-)

• (CH3)2CH- isopropyl (i-Pr)

• CH3CH2CH2- propyl (Pr)

© E.V. Blackburn, 2012

alkyl group classification

• a “secondary” carbon is bonded to two carbon atoms

• a “tertiary” carbon is bonded to three carbon atoms

sec-butyl tert-butyl

• a “primary” carbon is bonded to one other carbon

CH3-CH-CH2-CH3 CH3-C-CH3

CH3

© E.V. Blackburn, 2012

IUPAC nomenclature

• The longest continuous carbon chain forms the basic carbon skeleton.

C CC

CC

C

C

• If there are two of these chains, select the one with the greater number of branch points.

• The remaining alkyl groups are considered as substituents.

© E.V. Blackburn, 2012

Nomenclature

• The different substituent groups are assigned numbers based on their positions along this chain.

• Every substituent must have a number even if they are on the same carbon.

• If identical substituents are present use the prefixes di-, tri-, tetra- etc.

2,3-dimethylpentanenot

3,4-dimethylpentaneC C

CCC

C

C

• The carbon chain is then numbered from the end nearer the first branch point.

© E.V. Blackburn, 2012

Substituents

Substituents are named in alphabetical order.

C C C C C C CC C

C

4-ethyl-3-methylheptane

© E.V. Blackburn, 2012

Hexane

CH3

CH3

CH3-CH-CH-CH3CH3-CH2-CH-CH2-CH3

CH3

CH3-CH-CH2-CH2-CH3

CH3

CH3-CH2-C-CH3

CH3

CH3

CH3-CH2-CH2-CH2-CH2-CH3

© E.V. Blackburn, 2012

Nomenclature of branched alkyl groups

CH3-C-CH3

H

1-methylethyl or isopropyl

Numbering begins at the point where the group is attached to the main chain.

© E.V. Blackburn, 2012

Nomenclature of branched alkyl groups

CH3CH2-C-CH3

H

1-methylpropyl or sec-butyl

CH3

2-methylpropyl or isobutyl

CH3CHCH2-

CH3-C-CH3

CH3

1,1-dimethylethyl or tert-butyl

© E.V. Blackburn, 2012

Nomenclature of alkyl halides

H3C C CH3

H

Cl

(CH3)3CCl

© E.V. Blackburn, 2012

Nomenclature of alcohols

The OH group has a higher priority than a multiple C-C bond, a halogen, and an alkyl group in determining the carbon chain numbering.

Add the suffix ol to the name of longest, linear, carbon chain which includes the carbon bearing the OH and any double or triple C-C bond.

CH3CH2OH

ethanol

CH3CH2CH2CCH2OHH

CH2CH3

2-ethyl-1-pentanol

© E.V. Blackburn, 2012

Nomenclature of alcohols

CH3CH2CH2OH

1-propanol

CH2CH2OH

2-phenylethanol

H3C CCH3

HCOH

HCH3

3-methyl-2-butanol

phenyl

© E.V. Blackburn, 2012

Nomenclature of alcohols

H3C CH

CO

OH

OH

2-hydroxypropanoic acid

© E.V. Blackburn, 2012

Other nomenclature systems1. Name the alkyl group followed by the word alcohol:

2. Name alcohols as derivatives of carbinol, methanol:

ethyl alcohol

CH3CH2OHCH3CHCH3

OH

isopropyl alcohol

CH3OH

carbinolOH

triphenylcarbinol

© E.V. Blackburn, 2012

Vicinal glycols

Alcohols having two OH groups are called “glycols”:

HOCH2CH2OH is ethylene glycol or 1,2-ethanediol

“vicinal” means “adjacent” (vicinus, Latin for adjacent), “glycol” means “diol”

© E.V. Blackburn, 2012

Ethers

Structure:

R-O-R, Ar-O-R, or Ar-O-Ar

nomenclature

Name the two groups bonded to the oxygen and add the word ether.

CH3CH2OCH2CH3 - diethyl ether

© E.V. Blackburn, 2012

Odiphenyl ether

CH3OCH=CH2

OCH(CH3)2 isopropyl phenyl ether

CH3CH2CH2CHCH2CH3|OCH3

3-methoxyhexane

Nomenclature of ethers

methyl vinyl ether

© E.V. Blackburn, 2012

Nomenclature of cycloalkanes

cyclopropane

1,3-dibromocyclohexane

Cycloalkanes are named by adding the prefix cyclo to the name of the corresponding n-alkane.

Br

Br

Br

Cl

1-bromo-2-chlorocyclopentane

© E.V. Blackburn, 2012

Bicyclic compoundsUse the name of the alkane corresponding to the total number of carbons in the rings as the parent:

Seven carbons – a bicycloheptane.

Now determine the number of carbons in each bridge and place them in the name in order of decreasing length.

Bicylo[2.2.1]heptane!

© E.V. Blackburn, 2012

Bicyclic compounds

bicyclo[2.1.0]pentane bicylco[3.1.1]heptane

Number the carbons beginning at one bridgehead, along the longest bridge, then the next longest back to the original bridgehead, then along the shortest bridge.

H3C

1 2

345

6

77-methylbicyclo[2.2.1]heptane

© E.V. Blackburn, 2012

Bicyclic compounds

Cl

© E.V. Blackburn, 2012

Nomenclature of cyclic ethersUse the prefix oxa- to indicate that an O replaces a CH2 in the ring.

Ooxacyclopropane

ethylene oxide

Ooxacyclopentane

tetrahydrofuran

O

O1,4-dioxacyclohexane

1,4-dioxane

© E.V. Blackburn, 2012

Nomenclature of alkenes

1. To name alkenes, select the longest carbon chain which includes the carbons of the double bond. Remove the -ane suffix from the name of the alkane which corresponds to this chain. Add the suffix -ene.

C C C C C C C CCC

a derivative of heptene not octane

© E.V. Blackburn, 2012

Nomenclature of alkenes

2. Number this chain so that the first carbon of the double bond has the lowest number possible.

C C C C C C C CCC 1

2

3 4 5 6 7

3-propyl-1-heptene

© E.V. Blackburn, 2012

Nomenclature of alkenes

Cl

1

23

3-chlorocyclohexene

H

H

H H2C=CHCH2-vinyl allyl

H2C=CHCl H2C=CHCH2OH

© E.V. Blackburn, 2012

Butene - C4H8

The following are obviously butenes:

However there are four alkenes of formula C4H8!

compound bp mp A -7C -141C B -6C < -195C C +1C -106 D +4C -139C

1-butene

2-butene

methylpropene

CH3CH2CH=CH2

CH3CH=CHCH3

CH3C=CH2CH3

© E.V. Blackburn, 2012

The butenes - C4H8compound bp mp A -7C -141C B -6C < -195C C +1C -106 D +4C -139C

H2/Pt

H H

“A” must be methylpropene!

B, C, and DH2

Pt, Pd or Ni

CH3CH2CH2CH3

AH2

Pt, Pd or Ni

H3C-C-CH3

CH3

H

© E.V. Blackburn, 2012

The butenes - C4H8compound bp mp A -7C -141C B -6C < -195C C +1C -106 D +4C -139C

methylpropene

i. O3

ii. (CH3)2SO O

B1. O3

2. (CH3)2SH2C=O + CH3CH2C=O

H

“B” is 1-butene

© E.V. Blackburn, 2012

The butenes - C4H8compound bp mp A -7C -141C B -6C < -195C C +1C -106 D +4C -139C

methylpropene1-butene

C and D1. O3

2. (CH3)2SCH3C=O

H

C and D: CH3CH=CHCH3

© E.V. Blackburn, 2012

2-butene

C C C CH

H3C

CH3

H

H

H3C

H

CH3

trans cis

© E.V. Blackburn, 2012

NomenclatureReplace the -ane ending of the parent alkane with -yne. The numbering is analogous to that for alkenes.

1-butyne 2-butyne

4-methyl-2-pentyne

H C C C2H5 H3C C C CH3

H3C C C CH(CH3)2

© E.V. Blackburn, 2012

Nomenclature

“Enynes” are compounds containing both a double and a triple bond.

Numbering of the chain starts from the end nearer to the first multiple bond, be it double or triple.

HC CCH2CH2CH=CH2

1-hexen-5-yne

HC CCH2CHCH2CH2CH=CHCH3

4-methyl-7-nonen-1-yne

CH3

© E.V. Blackburn, 2012

Physical properties of alkanes and cycloalkanes

• low melting point (-183C for methane)

• low boiling point (-161.5C for methane)

• colorless

• insoluble in water

• soluble in non-polar solvents such as petrol, ether, etc.

• non-polar

© E.V. Blackburn, 2012

Cyclopropane

X

Y

Ni/H2

80o CH3CH2CH3

Br2/CCl4 CH2BrCH2CH2Br

H3O+

CH3CH2CH2OH

HI CH3CH2CH2I

© E.V. Blackburn, 2012

Cyclobutane

H2/Ni

200oCH3CH2CH2CH3

© E.V. Blackburn, 2012

Relative stabilities of cycloalkanes

Angle strain in cyclic compounds can be quantitatively evaluated by comparing heats of combustion for each -CH2- unit.

Baeyer (1885) proposed that rings smaller and larger than cyclopentane were unstable due to angle strain. How does this hypothesis fit the facts?

© E.V. Blackburn, 2012

Heats of combustion/CH2

Cyclane (CH2)n n H/n (kJ)

cyclopropane 3 697.0

cyclobutane 4 686.0

cyclopentane 5 664.0

cyclohexane 6 658.7

free of angle strain

free of angle strain!!! Why?

n-alkane 658.6

cycloheptane 7 662.4

cyclooctane 8 663.8

cyclopentadecane 15 659.0

© E.V. Blackburn, 2012

Cyclanes have puckered, not flat rings:

H H

HH

H

HH

H

HH

HH H

HH

H

H

H

cyclobutane

cyclohexanecyclopentane

HH

H

H

H

H

H

HH H

H

H

© E.V. Blackburn, 2012

Conformational analysis - angle strain

Any atom tends to have bond angles that match those of its bonding orbitals: 109.5o for sp3-hybridized carbons.

Any deviation from these normal bond angles is accompanied by angle strain.

© E.V. Blackburn, 2012

Any pair of sp3 carbons bonded to each other tend to have their bonds staggered. Any deviation from the staggered conformation is accompanied by torsional strain.

Conformational analysis - torsional strain

© E.V. Blackburn, 2012

Non-bonded atoms that just touch one another attract each other. If they are closer, they repel each other. Such crowding is accompanied by van der Waals strain (steric strain).

Conformational analysis - van der Waals strain

© E.V. Blackburn, 2012

Cyclohexane - the “chair” conformation

HH

H

H

H

H

H

HH H

H

H

© E.V. Blackburn, 2012

The “boat” conformation

This conformation is less stable (29.7 kJ/mol) than the chair conformation. It is situated at the top of a PE curve and is therefore a transition state between 2 conformational isomers.

1.83A

"flag pole"hydrogens

HH

H

HH

H

HH

H

H

HH

© E.V. Blackburn, 2012

Skew-boat conformations

H

HH

H

"boat" "skew-boat"

The skew-boat conformations are 23.0 kJ/mol less stable than the chair conformation.

© E.V. Blackburn, 2012

Conformations of cyclohexane

E45 kJ

23 kJ

6.7 kJ

© E.V. Blackburn, 2012

Axial and equatorial hydrogens

Ha

Ha

HaHa

Ha

Ha

Ha = axial

He

He

He

He

He

He

He= equatorial

© E.V. Blackburn, 2012

Axial and equatorial hydrogens

Ha

Ha

He

Ha

He

Ha

Ha

He

HeHa

He

He

Ha = axial He= equatorial

© E.V. Blackburn, 2012

Axial and equatorial hydrogens

axial

equatorial

© E.V. Blackburn, 2012

Methylcyclohexane - equatorial

HH

CH3

H

H

H

H

HH H

H

H

© E.V. Blackburn, 2012

Methylcyclohexane - axial

CH3

H

H

H

1,3 diaxialinteraction

1

3

3

© E.V. Blackburn, 2012

trans-1,2-dimethylcyclohexane

HH

CH3

H

CH3

H

H

HH H

H

H

HCH3

H

CH3

H

H

H

HH H

H

H

© E.V. Blackburn, 2012

cis-1,2-dimethylcyclohexane

HCH3

H

H

CH3

H

H

HH H

H

HH

H

CH3

CH3

H

H

H

HH H

H

H

© E.V. Blackburn, 2012

cis v trans

HH

CH3

H

CH3

H

H

HH H

H

H

HCH3

H

H

CH3

H

H

HH H

H

H

© E.V. Blackburn, 2012

cis-1,3-

cis

cis

© E.V. Blackburn, 2012

trans-1,3-

trans

© E.V. Blackburn, 2012

trans-1,4-

trans

?

© E.V. Blackburn, 2012

cis-1,4-

cis

© E.V. Blackburn, 2012

NomenclatureCH3

OH Br

Br CH3

Cl

I

Cl C(CH3)3

Br

© E.V. Blackburn, 2012

Synthesis of alkanes and cycloalkanes

© E.V. Blackburn, 2012

Hydrogenation of alkenes and alkynes

CnH2n C nH2n+2

H2

Pt, Pd or Ni

alkene alkane

H2/Ni

C2H5OH25o, 50 atm

(CH3)3CH

© E.V. Blackburn, 2012

Hydrogenation of alkenes and alkynes

Pt+ 2 H2

+ H2Pd

© E.V. Blackburn, 2012

Reduction of alkyl halides

RX + Bu 3SnH RH + Bu 3SnX

peroxide

Bu = CH3CH2CH2CH2-

Bu3SnH = tri-n-butylstannane

CH3Cl + Bu3SnD CH3D + Bu3SnCl

© E.V. Blackburn, 2012

Alkylation of terminal alkynes

An acetylenic hydrogen is weakly acidic:

C C HRNa

NH3

C CR-

Na+ + 1/2H2

a sodiumacetylide

(CH3)2CHC C HNaNH2

ether(CH3)2CHC C

- Na+

+ NH3

© E.V. Blackburn, 2012

Alkylation of terminal alkynes

The anion formed will react with a primary halide:

C C- Na+R + CH3X C CCH3 + NaXR

1. NaNH2

2. CH3Br

H2/Pt

© E.V. Blackburn, 2012

Corey – Posner – Whitesides - House Synthesis

R-X + 2Lidiethyl ether

RLi + LiX

alkyllithium1o, 2o,or 3o

2RLi + CuI R2CuLi + LiI lithium dialkylcupratea Gilman reagent

R2CuLi + R'X R-R' + RCu +LiX

1o alkyl or 2o

cycloalkyl halide

© E.V. Blackburn, 2012

Retrosynthetic analysis

targetmolecule

1st precursor

2nd precursor starting compound

Here is a target molecule. Plan a synthesis.

CH3CH2CHCH2CH2CH2CH2C

CH3

H3

© E.V. Blackburn, 2012

Retrosynthetic analysisCH3CH2CH

CH3

CH2CH2CH2CH2CH3

CH3CH2CH

CH3 2

CuLi BrCH2CH2CH2CH2CH3

CH3CH2CHBr

CH3

1. Li2. CuI

© E.V. Blackburn, 2012

CH3CH2CHCH2CH2CH2CH2C

CH3

H3

Retrosynthetic analysis

CH3CH2CHBr

CH3

1. Li2. CuI

(CH3CH2CH)2CuLi

CH3

BrCH2CH2CH2CH2CH3(CH3CH2CH)2CuLi

CH3

© E.V. Blackburn, 2012

Corey – Posner – Whitesides - House Synthesis

Muscalure is the sex pheromone of the common house fly. It is used to attract flies to traps containing insecticide. It can be synthesized by the Corey - House reaction. What lithium dialkylcuprate would you use?

H

(CH2)7CH2Br

H

H3C(H2C)7

(CH3(CH2)3CH2)2CuLi

H

(CH2)12CH3

H

H3C(H2C)7?

Muscalure

© E.V. Blackburn, 2012

Reactions of alkanes with halogens

C H + X2

250-400o

or hC X + HX

Reactivity:- X2 : F2 > Cl2 > Br2 (> I2)

H : 3o > 2o > 1o > H3C-H

© E.V. Blackburn, 2012

Chlorination - a substitution reaction

CH4 + Cl2 hor

CH3Cl + HCl

© E.V. Blackburn, 2012

Polychlorination

CH3Cl + Cl2 CH2Cl2 + HCl

CH2Cl2 + Cl2 CHCl3 + HCl

CHCl3 + Cl2 CCl4 + HCl

dichloromethanemethylene chloride

trichloromethane chloroform

tetrachloromethanecarbon tetrachloride

© E.V. Blackburn, 2012

A Problem?

Chlorination leads to the possible formation of four products - a mixture! How can we limit the reaction so that only one product is formed?

© E.V. Blackburn, 2012

Bromination

• bromomethane

• dibromomethane -methylene bromide

• tribromomethane - bromoform

• tetrabromomethane - carbon tetrabromide

Bromination takes place less readily than chlorination but it produces the four analogous brominated products:

© E.V. Blackburn, 2012

Iodination and fluorination

• iodine does not react

• fluorine reacts very readily

order of halogen reactivity:

F2 > Cl 2 > Br 2 (> I2)

© E.V. Blackburn, 2012

A Mechanism

• it must explain all experimental facts

• the mechanism should be tested by devising appropriate experiments - mechanistic predictions must be tested in the lab

• a detailed, step by step, description of the transformation of reagents into products

© E.V. Blackburn, 2012

Mechanism of the chlorination of methane

2. Reaction readily occurs, in the absence of light, at temperatures above 250C.

3. Reaction occurs at room temperature in the presence of light of a wavelength absorbed by chlorine.

1. No reaction occurs at room temperature in the absence of light.

The experimental facts

© E.V. Blackburn, 2012

The experimental facts

5. The presence of even a small quantity of oxygen slows down the reaction.

4. When the reaction is initiated by light, a large number of chloromethane molecules are produced for each photon of light absorbed by the system.

© E.V. Blackburn, 2012

The mechanism?

2, 3, 2, 3, 2 etc.

1. Cl Cl 2Clhor

Cl H CH3 CH3 + HCl2.

H3C Cl Cl CH3Cl + Cl3.

© E.V. Blackburn, 2012

Chain ReactionChain initiation:

Cl-Cl 2Cl

Chain propagation:

Cl + CH4

CH3 + Cl2

CH3 + HCl

CH3Cl + Cl

Chain termination:

2Cl 2CH3 Cl + CH3

Cl2C2H6 (ethane)

CH3Cl

© E.V. Blackburn, 2012

Inhibitors

A compound which slows down or stops a reaction, even when present in small quantities, is called an inhibitor.

CH3 + O2

a "peroxy" radical

CH3-O-O

© E.V. Blackburn, 2012

Lets test the mecanism

If tetraethyllead is heated at 140C......

F. Paneth and W. Hofeditz, Ber., 62, 1335 (1929)

(C2H5)4Pb

Pb + 4C2H5

© E.V. Blackburn, 2012

An alternative source of chlorine atoms.....

(C2H5)4Pb140 C

Pb + 4C2H5

C2H5 + Cl2 C2H5Cl + Cl

© E.V. Blackburn, 2012

The test(C2H5)4Pb

C2H5 + Cl2

Pb + 4C2H5

C2H5Cl + Cl

CH4 + Cl2 140C

0.02%

(C2H5)4Pb

CH3Cl + HCl

© E.V. Blackburn, 2012

Heat of reaction

H = + 438 + 243 - 351 - 432 = -102 kJ

H - CH3 + Cl - Cl Cl - CH3 + H - Cl

438 kJ 243 kJ

681 kJ

351 kJ 432 kJ

783 kJ

© E.V. Blackburn, 2012

Bromination

H - CH3 + Br - Br Br - CH3 + H - Br

438 kJ 193 kJ

631 kJ

293 kJ 366 kJ

659 kJ

H = + 438 + 193 - 293 - 366 = -28 kJ

© E.V. Blackburn, 2012

Iodination

ENDOTHERMIC!!!

H - CH3 + I - I I - CH3 + H - I

438 kJ 151 kJ

589 kJ

234 kJ 298 kJ

532 kJ

H = + 438 + 151 - 234 - 298 = +57 kJ

© E.V. Blackburn, 2012

Chlorination

H = - 102 kJ..........

Cl - Cl 2Cl

Cl + H - CH3 Cl - H + CH3

Cl - Cl + CH3 Cl + Cl - CH3

243 kJ

H

243 kJ

438 kJ 432 kJ + 6 kJ

243 kJ 351 kJ - 108 kJ

© E.V. Blackburn, 2012

How does Cl. react with CH4?

The H-Cl bond can only form if the two species come in contact.

A certain minimum energy must be provided by the collision in order for reaction to occur.

Why?????

In order for chlorination to occur, a Cl. and a CH4 must collide.

© E.V. Blackburn, 2012

Activation energy

Bond breaking and bond formation are not perfectly synchronous processes. Therefore energy liberated during bond formation is not completely available for bond breaking.

A collision must therefore provide a certain minimum amount of energy for reaction to occur. This is called the “activation energy”, Ea.

© E.V. Blackburn, 2012

Potential energy diagramsP

oten

tial

ene

rgy

Reaction coordinate

CH4 + Cl CH3 + HCl

CH3. + HCl

CH4 + Cl

H = +6 kJ

Ea = 16.7 kJ

© E.V. Blackburn, 2012

Pot

entia

l ene

rgy

Potential energy diagrams

© E.V. Blackburn, 2012

Reaction rates

rate = collisionfrequency

x xenergy factor

probability factor(orientation)

© E.V. Blackburn, 2012

Factors affecting collision frequency

• pressure

• molecular size

• momentum

• temperature

• concentration

© E.V. Blackburn, 2012

The probability factor

• depends on the nature of the reaction taking place

• depends on reactant geometry

© E.V. Blackburn, 2012

The energy factor

• depends on activation energy

• depends on temperature

© E.V. Blackburn, 2012

KE distribution among collisions

Num

ber

of c

ollis

ions

of p

artic

ular

ene

rgy

Energy

E1

E2

E2 > E1

© E.V. Blackburn, 2012

= e-Ea/RT

Fraction of collisions with E > Ea

© E.V. Blackburn, 2012

Relative rates of reaction

Cl + CH3-H

Br + CH3-H

H Ea

HCl + CH3

HBr + CH3

+6 16.7

+72 75.3

(kJ)(kJ)

rate = collisionfrequency

x xenergy factor

probability factor(orientation)

© E.V. Blackburn, 2012

Relative rates of reaction

At 275C, of every 15 million collisions, 375,000 are of sufficient energy to cause reaction when chlorine atoms are involved …

and only one is of sufficient energy when bromine atoms are involved.

Thus, solely due to Ea differences, the chlorine atom is 375,000 more reactive than the bromine atom.

© E.V. Blackburn, 2012

Relative reactivity of halogens

X2 2X

X + CH4 CH3 + HX

CH3 + X2 CH3X + X

X = F Cl Br I

H = +142 +243 +193 +151 kJ

-293 -108 -100 -83 kJ

-134 +6 +72 +140 kJ

© E.V. Blackburn, 2012

Obed Summit

© E.V. Blackburn, 2012

Rate determining step

Reaction coordinate

Cl +CH4

CH3Cl + Cl

Rate determiningstep

Pot

enti

al e

nerg

yObed Summit

© E.V. Blackburn, 2012

Transition state

Reaction coordinate

Ea

H

reagents

products

transition stateP

oten

tial

ene

rgy

© E.V. Blackburn, 2012

Transition state

CH

HH H + X H C

H

H+ HX

CH

HH H + X H C

H

HX H C

H

H+ HX

CH

HH H + X H C

H

HH X H C

H

H+ HX

© E.V. Blackburn, 2012

Transition state

CH

HH H + X H C

H

HH X

H C

H

H+ HX

CH

HH H + X H C

H

HH X

transition state

H CH

H+ HX

© E.V. Blackburn, 2012

+ HCl ?H

+ Cl-

Cl

1.

HCl

2.

HCl -

+

3.

Transition state

© E.V. Blackburn, 2012

Halogenation

CH3CH3 CH3CH2ClCl2

chloroethane

CH3CH3 CH3CH2BrBr2

hbromoethane

© E.V. Blackburn, 2012

Chlorination of propane

CH3CH2CH3 CH3CH2CH2Cl + Cl2

hCH3CHCH3

Cl

1-chloropropane 2-chloropropane

43% 57%

© E.V. Blackburn, 2012

Bromination of propane

CH3CH2CH3 CH3CH2CH2Br Br2

h+ CH3CHCH3

Br

1-bromopropane 2-bromopropane

3% 97%

© E.V. Blackburn, 2012

Halogenation of isobutane

CH3CHCH3

CH3Cl2

h(CH3)2CHCH2Cl + (CH3)3CCl

64% 36%

CH3CHCH3

CH3Br2

h(CH3)2CHCH2Br + (CH3)3CBr

trace >99%

Why this selectivity?

© E.V. Blackburn, 2012

Mechanism of the halogenation

1. X2 2X initiation250-400o

or h

2. X + RH HX + R

3. R + X2 RX + X

propagation

2, 3, 2, 3, 2, 3....etc.

© E.V. Blackburn, 2012

The intermediate alkyl radicalThe nature of the intermediate free radical determines the product:

CH4X CH3

methane methyl radical

X2 CH3X

halomethane

CH3CH3X

CH3CH2

ethane ethyl radical

X2CH3CH2X

haloethane

© E.V. Blackburn, 2012

The intermediate alkyl radical

CH3CH2CH3X

propane

X2CH3CH2CH2X

1-halopropane

X2CH3CHXCH3

2-halopropane

CH3CH2CH2

n-propyl radical

CH3CHCH3

isopropyl radical

© E.V. Blackburn, 2012

Orientation of halogenationabstraction of a primary hydrogen

abstraction of a secondary hydrogen

We have competing reactions and should review factors which influence reaction rates!

XHCHH

C CH

H

H

HCH

HH

HCHH

C CH

H

H

HCH

H

HCHH

C CH

H HCH

HH

© E.V. Blackburn, 2012

Reaction rates

rate = collisionfrequency

x xenergy factor

probability factor(orientation)

© E.V. Blackburn, 2012

Probability factor

The statistical product ratio for the chlorination of propane is 75% 1-chloropropane and 25% 2-chloropropane, a 3:1 mixture.

Why? There are three times as many primary hydrogens.

However:

CH3CH2CH3 CH3CH2CH2Cl + Cl2

hCH3CHCH3

Cl

43% 57%

© E.V. Blackburn, 2012

Relative reactivitiesLets look at the relative reactivities per hydrogen atom:

tertiary secondary primary

Chlorination: 5.0 : 3.8 : 1.0

Bromination: 1600 : 82 : 1

We need to look at activation energies and transition states!

© E.V. Blackburn, 2012

Transition state for rate determining step

C H + X C H X

C + HX

the carbon isdevelopingfree radicalcharacter

So let us look at the stability of free radicals….

© E.V. Blackburn, 2012

Free radical stability

Order of free radical stability is therefore

tertiary > secondary > primary > methyl

H3C-H CH3. + H. H = 438 kJ

CH3CH2-H CH3CH2. + H. H = 420 kJ

(CH3)2CH-H (CH3)2CH. + H. H = 401 kJ

H = 390 kJ(CH3)3C-H (CH3)3C. + H.

© E.V. Blackburn, 2012

Free radical stability - hyperconjugation

Using the concept of resonance:-

H C CH

H

H

HH C C

H

H

H

HH C C

H

H

H

H

A charged system is stabilized when the charge is dispersed or delocalized. Thus the order of free radical stability is tertiary > secondary > primary > methyl.

H C CHH

H H

© E.V. Blackburn, 2012

Free radical stability - hyperconjugation

The electrons are delocalised through overlap of a p orbital which is occupied by one, lone electron, and a orbital of the alkyl group:

ethyl radical

H

HH

HH

© E.V. Blackburn, 2012

Free radical stability - hyperconjugation

ethyl radical

H

HH

H

H

HH

isopropyl radicalH

HH

H

H

H

HH

tert-butyl radical

H

HH

HH

© E.V. Blackburn, 2012

Transition state for rate determining step

C H + X C H X

C + HX

the carbon isdevelopingfree radicalcharacter

Factors which stabilize free radicals will stabilize the transition state which is developing free radical character.

© E.V. Blackburn, 2012

E

Reaction coordinate

H3C CCH3

H

H3C CCH3

CH3H Br H Br

(CH3)3CH + Br (CH3)3CH + Br

(CH3)2CHCH2

+ HBr (CH3)3C+ HBr

CH2

Orientation of halogenationThis is determined by the stability of the transition state for the rate determining step.

Ea2Ea1

Ea1 > Ea2

© E.V. Blackburn, 2012

Reactivity and selectivity and the Hammond postulate

The postulate states that the transition state resembles the structure of the nearest stable species. Transition states for endothermic steps structurally resemble products whereas transition states for exothermic steps structurally resemble reactants.

Thus the later the transition state is attained in the reaction, the more it resembles the products.

In other words, the greater the Ea, the more the transition state resembles the products.

This will explain the greater selectivity of the bromine atom.

© E.V. Blackburn, 2012

Reactivity and selectivity

R H + Cl R Cl

R + HClH

This reaction has a low activation energy and so the transition state resembles the reactants - it has little radical character.

R H + Br R H Br

R + HBr

The activation energy for the bromination is far higher. The transition state has considerable radical character.

The free radical stabilizing factors are far more important in the bromination, hence the greater selectivity.

© E.V. Blackburn, 2012

Synthesis of alkanes• Hydrogenation of alkenes and alkynes

• Reduction of halides

• Corey - Posner, Whitesides – House Synthesis

H2/Pt, Pd or Ni

solvent, pressureHH

H2/Pt, Pd or Ni

solvent, pressureH

H

H

H

RXBu3SnH

peroxideRH

RX1. Li

2. CuIR2CuLi

R2CuLi R'X (1o) or ArX R-R or R-Ar

© E.V. Blackburn, 2012

Reactions of alkynes

• Alkylation of terminal alkynes1. NaNH2

2. R'X (1o)HR R'R

• Hydrogenation

H2/Pt, Pd or Ni

solvent, pressureH

H

H

H

© E.V. Blackburn, 2012

Reactions of alkanes

• Halogenation

RH RXX2/ or h