Wade Ch 4 Wade Powerpoint/Lecture for Chapter 4

7/25/2019 Wade Ch 4 Wade Powerpoint/Lecture for Chapter 4

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Chapter 4

Organic Chemistry , 8th Edition

L. G. Wade, Jr.

The Study of

Chemical Reactions

• “ We cannot remember thousands

of specific organic reactions but wecan organize the reactions into

logical groups based on how the

reactions work and whatintermediates are involved!

• “ The most interesting and useful

aspect of organic chemistry is the

study of reactions!

"# Wade$ % th &d$ p '()

• “The amounts of reactants and

products present at e*uilibrium

depend on their relative stabilities!

• “ +to understand a reaction$ we

must+ know the mechanism$ the

step,by,step pathway from

reactants to products!

"# Wade$ % th &d$ p '()

To -etermine a Reaction.s

/echanism$ We Will "ook 0t+

1 Thermodynamic -ata

0 &*uilibrium constants

2 3ree energy change

C &nthalpy 756- &ntropy S56

& 2ond -issociation &nthalpy 2-&6

11 8inetic -ata

0 Reaction order

2 0ctivation &nergy

3or 9ur :ery 3irst Reaction$

We Will Study+• The 3ree Radical 7alogenation of

0lkanes$ starting with+

The Chlorination of /ethane

• Re*uires heat or light for initiation• The most effective wavelength is blue$ which is

absorbed by chlorine gas

• /any molecules of product are formed from absorptionof only one photon of light ∴

a chain reaction 6

7alogenation 9f 0lkanes 1s 0+• 3ree Radical Chain Reaction$ and a

• Substitution Reaction ; where one group takes the place of

another group in a molecule

Three Steps 1n a 3R Chain Reaction

1 1nitiation ; heat or light hν

6 generatesa free radical 3R6 intermediate

11 <ropagation ; 3R intermediate =

reactant > new 3R intermediate = product

111 Termination ; side r?ns consume @

destroy intermediates so r?n stops

1 1nitiationA 3ormation of a

Chlorine Radical

0 chlorine molecule splits

homolytically into ) chlorine atoms

free radicals6

"ewis Structures of 3ree

Radicals

• 3ree radicals have unpaired electrons

• 7alogens have B valence electrons so one electronwill be unpaired thus$ making halogen atom a freeradical6

11 3irst <ropagation StepA

3ormation of /ethyl Radical @ 7Cl

The chlorine atom 3R6 collides with a

methane molecule and abstracts removes6

a proton$ forming a methyl free radical and

one of the products 7Cl 6

11 Second <ropagation StepA

<roduct 3ormation

The methyl free radical collides with

another chlorine molecule$ producing theorganic product chloromethane 6 and

generating a new chlorine radical

1mportantA #eneration of a

ew Chlorine Radical /eansthe Reaction Will ContinueD

1n a free radical chain reaction$ the

reaction will continue until one of

the reactants is consumed

o more heat or light is necessary once

the reaction has startedD

The 9verall Reaction 1s The Sum

9f The Two <ropagation Steps

<rob 4,' -raw the structures for the

following free radicals

• &thyl radical radical

• tert,butyl radical

• 1sopropyl radical ),propyl6

• 1odine atom

<rob 4,) a6 Write the propagation

steps leading to the formation of

dichloromethane C7 ) Cl ) 6

b6 &?plain why free radical halogenation

usually gives a mi?ture of products

c6 7ow could an industrial plant controlthe proportions of methane and

chlorine to favor production of CCl 4F

C7 (ClF

<rob 4,( &ach of the following proposed

mechanisms for the chlorination of

methane is wrong &?plain how thee?perimental evidence disproves each

mechanism

a6 Cl ) = hν > Cl

) G G H an activated form of Cl

Cl ) G = C7 4 > C7 (Cl = 7Cl

b6 C7 4 = h

> IC7 ( = 7I

IC7 ( = Cl ) > C7 (Cl = ClI ClI = 7I > 7Cl

<rob 4,4 3ree,radical chlorination of

he?ane gives very poor yields of

',chlorohe?ane$ while cyclohe?ane can be

converted to chlorocyclohe?ane in good

a6 7ow do you account for this differenceF

b6 What ratio of reactants cyclohe?ane and

chlorine6 would you use for the synthesis of

chlorocyclohe?aneF

Lsing 8 e* values$ we can compare

stabilities of reactants and products

by comparing their energies

• We use the #ibbs 3ree &nergy$

#eneral RuleA 1f # 5 is more negative

than P ,') kENmol$ the reaction “goes

to completion$! ie O KKQ products

3actors -etermining # °

3ree energy change depends onA &nthalpy

75 H enthalpy of products6 , enthalpy of

reactants6

&ntropy

S5 H entropy of products6 , entropy of

reactants6

# ° H 7 ° , T

&nthalpy

• 7 5 H heat released or absorbed

during a chemical reaction at standard

conditions

• &?othermicA , 76 heat is released

• &ndothermicA =

76 heat is absorbed

• Reactions favor products with lowest

enthalpy strongest bonds6

&ntropy

• S 5 H change in randomness$

disorder$ or freedom of movement

• 1ncreasing heat$ volume$ or number

of particles increases entropy• Spontaneous reactions ma?imize

disorder and minimize enthalpy

• 1n the e*uation # 5 H 75 , T S 5 theentropy value is often small

<rob 4,B When ethene is mi?ed with

hydrogen in the presence of a platinum

catalyst$ hydrogen adds across the doublebond to form ethane 0t room temperature$

the reaction goes to completion <redict the

signs of 7 ° and S ° for this reaction

&?plain the signs in terms of bonding andfreedom of motion

2ond,-issociation &nthalpies 2-&6

• 2ond,dissociation re*uires energy =2-&6

• 2ond formation releases energy , 2-&6

• 2-& can be used to estimate7 for a reaction

2ond,-issociation &nthalpies 2-&6

• 2-& for homolytic cleavage of bonds in agaseous molecule

7omolytic cleavageA When the bond

breaks$ each atom gets one electron$eg 02 > 0I = 2I

7eterolytic cleavageA When the bond

breaks$ the most electronegative atomgets both electrons$ eg 02 > 0= = :2,

7omolytic and 7eterolytic Cleavage

We use homolytic cleavage 2-&.s for 75 calculations

&nthalpy Changes in Chlorination

C7 (,7 = Cl,Cl → C7 (,Cl = 7,Cl2onds 2roken 2-& per mole6 2onds 3ormed 2-& per /ole6

Cl,Cl )4) kM 7,Cl 4(' kM

C7 (,7 4( kM C7 (,Cl (' kM

T9T0"S BB kM T9T0" B%) kM

7 ° H Σ 2-& break , Σ 2-& make

7 ° H BB kM ; B%) kM H ,'J kM

<rob 4,K a6 <ropose a mechanism for

the free,radical chlorination of ethane

C7 (,C7 ( = Cl ) > C7 (,C7 ) Cl = 7Cl hν

1nitiationA Cl ) > )ClI

<ropagation 'A ClI = C7 (

> C7 (

I = 7Cl

<ropagation )A C7 (,C7 ) I = Cl ) > C7 (,C7 ) Cl = ClI

9verallA C7 (,C7 ( = Cl ) > C7 (,C7 ) Cl = 7Cl

<rob 4,K b6 Calculate 7 ° for each

step in this reaction

1) CH3-CH3 + Cl• → CH3CH2• + HCl

98 kcal 103 kcal ∆ H 1 = 98 – 103 = -5

CH3CH2• + Cl2 → CH3-CH2Cl + Cl• 58 kcal 81 kcal ∆ H 2 = 58 – 81 = -23

∆ H rxn = - 28 kcal/mol

c6 Calculate the overall value of 7 ° forthis reaction

8inetics

• 8inetics is the study of reactionrates

• Rate of the reaction is a measure

of how the concentration of the products increase while theconcentration of the reactants

decrease

8inetics

• 0 rate e*uation is also calledthe rate law and it gives therelationship between theconcentration of the reactantsand the reaction rate observed

Th R t "

The Rate "aw

• 3or the reaction 0 = 2 → C = -$

rate H k r 0U a2U b

a is the order with respect to 0

b is the order with respect to 2 a = b is the overall order k r is a temp,dependent rate constant

9rder is the number of molecules of that

reactant which are present in the rate, determining step of the mechanism

<rob 4,'' The reaction of tert,butyl

chloride with methanolA

is found to follow this rate e*uationA

C7 ( 6(C,Cl = C7 (,97 > C7 ( 6(C,9C7 ( = 7Cl

Rate H k r [C7 ( 6(C,Cl ]

a6 What is the kinetic order with respect to

t,butyl chlorideF

b6 What is the kinetic order with respect to

methanolF

c6 What is the kinetic order overallF

0 i i &

0ctivation &nergy • The value of k r depends on temperature as given by

the 0rrhenius e*uationA

RT E a Aek /

where 0 H constant fre*uency factor6 & a H activation energy

R H gas constant$ %('4 MNkelvin,mole

T H absolute temperature& a is the minimum kinetic energy

needed to react

The 0rrhenius e*n implies rate depends on the

fraction of molecules with kinetic energy V &a

• 0t higher temperatures$ more molecules have there*uired energy to react

• 0t higher temperatures$ all reactions go

faster$ even unwanted side reactionsD

&nergy -iagram of an

&?othermic Reaction

• The transition state is the highest point on the graph

• The activation energy is the energy difference betweenthe reactants and the transition state

∀ 75 is the energy difference between reactants and products

The Transition State

• The highest,energy state in a molecular

collision that leads to a reaction

• LnstableD Cannot be isolatedD

transition statereactant

intermediate intermediate product

The Transition State

• The “double dagger!6 symbol and

s*uare brackets are used to indicate a

transition state

• Transition state can go either direction$

forward to products$ or back to reactants

shows old bond

breaking

shows new bond

forming

shows charge or

radical e,

1ntermediates• either reactants nor products$

they e?ist for a short time

• Can sometimes be isolated

because they have some stability• 3ree radicals are Eust one of four

common types of intermediates in

organic chemistry

Th R t "i iti St

The Rate,"imiting Step

• Reaction intermediates are stable as long

as they don.t collide with another moleculeor atom$ but they are very reactive

• Transition states are at energy ma?imums

• 1ntermediates are at energy minimums

The reaction step with highest & a will be the

slowest$ therefore rate,determining for the

entire reaction

R ti & -i

Reaction &nergy -iagrams

This one,stepreaction has no

intermediates$ only a

transition state

R ti & -i f

Reaction &nergy -iagrams of

/ulti,Step Reactions+

e n e r g y

reaction coordinate >

The highest & a hill is the

rate,determining step

This two,step r?n has a

“valley! indicating that an

intermediate is involved

Reaction &nerg -iagram for

Reaction &nergy -iagram for

the Chlorination of /ethane

propagation

Rate & and Temperature

Rate$ & a$ and Temperature

X + CH4 HX + CH3

X Ea(per Mole) Rate at 27 ° Rate at 227 °

F ! "#$,$$$ %$$,$$$

Cl "7 "%$$ "8,$$$

Br 7! & ' "$8 $.$"!

I "#$ 2 ' "$"&

2 ' "$&

Conclusions 0bout /ethane

7alogenation Reactions

• With increasing & a$ rate decreases

• With increasing temperature$ rate

increases

• 3luorine reacts e?plosively

• Chlorine reacts at a moderate rate

• 2romine must be heated to react• 1odine does not react detectably6

0 t l t l th & b t d t ff t

0 catalyst lowers the & a$ but does not affect

the energies of reactants or products

& awith catalyst

S l d < b 4 4 Th i

Solved <rob 4,4 The reaction

C7 4 = ClI > IC7 ( = 7Cl

has an activation energy of ='B kMNmoland a 75 of =4 kMNmol -raw a reaction

energy diagram of this reaction

<rob 4 '4 -raw the reaction energy

<rob 4,'4 -raw the reaction energy

diagram for the reverse reactionA

IC7 ( = 7Cl > C7 (Cl = ClI

P b 4 1 "# b % t% & t# d

Prob 4-1! "#e bro$%nat%on o& $et#ane proceeds

t#rou'# t#e &ollo(%n' steps:

CH4 + Br

CH3 + HBr

CH3 + Br2 CH3Br + Br

HEa+ 46 kcal

46 kcal

+16 kcal

18 kcal

-24 kcal1 kcal

a) Draw the complete reaction-energy

diagram or thi! r"n#$) %a$el the rate-limiting !tep#

c) Draw the !tr&ct&re o each tran!ition!tate#d) Comp&te the o'erall 'al&e o H° or the$romination#

Chlorination of <ropane /echanism

Conclusion

• Some hydrogens are morereactive than others

2-&.s indicate that C,7 bondstrengths vary from '5 to )5 to

(5 carbons

<rimary '56$ Secondary )56$

and Tertiary (56 7ydrogens

2-&.s for the 3ormation of 3ree

2-& s for the 3ormation of 3ree

Radicals

Chlorination &nergy -iagram

• "ower & aH faster rate$ so most stable

intermediate is formed faster

2 i ti f <

2romination of <ropane

statistical yieldA BQ )Q

actual yieldA (Q KBQ

• 2romination of propane is much

more selective than chlorination

• WhyF

&nergy -iagramsA Chlorination

vs 2romination of <ropane

• Chlorination is e?othermic because bond energy

of 7,Cl bond is high 4(' kM6 while that of 7,2r is

( kM lower

:s 2romination of <ropane

• 7igher & a for bromination e?plains why

bromination is slower$ but not why more selective

Th l th & diff th

The larger the & a difference$ the

more selective the reaction will beD

The 7ammond <ostulate

•Related species that are similar inenergy are also similar in structure

• The structure of the transition stateresembles the structure of the closest

stable species• &ndothermic reactionA Transition state

is product,like

• &?othermic reactionA Transition state isreactant,like

1 & th i 3i t St

1n &?othermic 3irst Stepshorter line shows bond mostly still intact

longer line shows bond

mostly yet to form

transition state looks

mostly like reactantsA

C7 4 = ClI

1n &ndothermic 3irst Step

longer line shows bond mostly broken

shorter line shows bondmostly formed

transition statelooks mostly like

productsA

7 (CI = 72r

Radical 1nhibitors

• 9ften added to food to retard spoilageby radical chain reactions

• Without an inhibitor$ each initiationstep will cause a chain reaction so that

many molecules will react• 0n inhibitor combines with the free

radical to form a stable molecule

• :itamin & and vitamin C are thought to protect living cells from free radicals

3ree Radical 1nhibitors

• 0 radical chain reaction is fast and has manye?othermic steps that create more reactive radicals

• When an inhibitor reacts with the radical$ it creates

a stable intermediate$ and any further reactions willbe endothermic and slow

Reactive 1ntermediates

Reactive 1ntermediates• Short,lived species that are never present in high

concentrations because they react as soon asthey are formed

• Commonly contain carbon with only ) or ( bonds

instead of 4

Trivalent intermediatesCarbocation ='6

3ree radical J6

Carbanion ,'6

-ivalent intermediateA Carbene J6

Th / t C 1 t di t

The /ost Common 1ntermediates

Carbocation Structure

• Carbon has electronsX positively charged

• Carbon is sp)

hybridized$ ie trigonal planar geometry$ with vacant p orbital• Carbocations are STR9# &"&CTR9<71"&S

C b ti St bilit

Carbocation Stability

Same as free radical stability orderD

• Stabilized by alkyl

substituents in two waysA

' 1nductive effectA -onation

of electron density alongthe sigma bonds

) 7yperconEugationA

9verlap of sigma bonding

orbitals with empty porbital

<rob 4,)K Rank the following

<rob 4 )K Rank the following

carbocations in decreasing order of

stability Classify each as '5$ )5 or (5

• 1sopentyl cation$ C7 ( 6) C7C7 ) C7 ) =

• (,methyl,),butyl cation$ C7 (,C7 =,C7C7 ( 6)

• ),methyl,),butyl cation$ C7 (,C =,C7 ( 6C7 ) C7 (

3ree Radicals

• 0lso electron,deficient sp) species with ' unpaired e, $but zero chargeD

• Stabilized by alkyl substituents same as carbocations

• 9rder of stabilityA (° O ) ° O '° O methyl

Stability of 3ree Radicals

Same as carbocation stability orderD

"ater We Will See 7ow Resonance

&nhances The Stability of Radicals$

Carbocations and Carbanions

Carbanions

• &ight electrons on carbonA bonding plus one lone pair $∴

carbon is tetrahedral sp(

• Carbon has a negative

charge• -estabilized by alkyl

substituents

• Stability orderA

/ethyl O'° O ) ° O (°

Carbanions

• /ost common carbanions are stabilizedby resonance as in this e?ample from

ne?t semesterA

• 2ecause of their formal negative charge$carbanions are STR9# LC"&9<71"&SD

Carbenes

Ca be es

• Carbon has only ) bonds$ but has a lone pair

• sp) $∴

trigonal planar

• :acant p orbital$ so can be electrophilic

• "one pair of electrons$ so can be nucleophilic

The &nd

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