JWCL234-10 JWCL234-Solomons-v1 December 11, …JWCL234-10 JWCL234-Solomons-v1 December 11, 2009 16:44 CONFIRMING PAGES RADICAL REACTIONS 189 (fandg) Inadditiontothe(2S,4S)-2,4-dichloropentaneand(2R,4S)-2,4-dichloropentane

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Bind Rite

JWCL234-10 JWCL234-Solomons-v1 December 11, 2009 16:44

CONFIRMING PAGES

10RADICAL REACTIONS

SOLUTIONS TO PROBLEMS

10.1 (a) H H

(DH° = 436)

2 H F

2(DH° = 570)

F F

(DH° = 159)

+

+595 kJ mol−1

is required forbond cleavage

−1140 kJ mol−1

is evolved information of thebonds in 2 mol

of HF�H ◦ = +595 − 1140

= −545 kJ mol−1

(b) CH3 H CH3 F

(DH° = 440) (DH° = 461)

F F

(DH° = 159)

+ H F

(DH° = 570)

+

+599 kJ mol−1


−1031 kJ mol−1

is evolved inbond formation

�H ◦ = +599 − 1031

= −432 kJ mol−1

(c) CH3 H CH3 Cl

(DH° = 440) (DH° = 352)

Cl Cl

(DH° = 243)

+ H Cl

(DH° = 432)

+

+683 kJ mol−1


−784 kJ mol−1


�H ◦ = +683 − 784

= −101 kJ mol−1

(d) CH3 H CH3 Br

(DH° = 440) (DH° = 293)

Br Br

(DH° = 193)

+ H Br

(DH° = 366)

+

+633 kJ mol−1


−659 kJ mol−1


�H ◦ = +633 − 659

= −26 kJ mol−1

182



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RADICAL REACTIONS 183

(e) CH3 H

(DH° = 440)

CH3 I

(DH° = 240)

I I

(DH° = 151)

+ H I

(DH° = 298)

+

+591 kJ mol−1


−538 kJ mol−1


�H ◦ = +591 − 538

= +53 kJ mol−1

(f) CH3CH2 H

(DH° = 421)

CH3CH2 Cl

(DH° = 353)

Cl Cl

(DH° = 243)

+ H Cl

(DH° = 432)

+

+664 kJ mol−1


−785 kJ mol−1


�H ◦ = +664 − 785

= −121 kJ mol−1

(g)

Cl

Cl Cl

(DH°= 243)

(DH°= 413) (DH°= 355)

+ H Cl

(DH° = 432)

+

H

+656 kJ mol−1


−787 kJ mol−1


�H ◦ = +656 − 787

= −131 kJ mol−1

(h) ClClH

(DH° = 349)

Cl ClH

(DH° = 243)(DH° = 400)(DH° = 432)

+ +

+643 kJ mol−1


−781 kJ mol−1


�H ◦ = +643 − 781

= −138 kJ mol−1

10.2

3° 2° 1° Methyl

> > CH3>

> > >

10.3 The compounds all have different boiling points. They could, therefore, be separated bycareful fractional distillation. Or, because the compounds have different vapor pressures,they could easily be separated by gas chromatography. GC/MS (gas chromatography/massspectrometry) could be used to separate the compounds as well as provide structural infor-mation from their mass spectra.



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184 RADICAL REACTIONS

10.4 Their mass spectra would show contributions from the naturally occurring 35Cl and 37Clisotopes. The natural abundance of 35Cl is approximately 75% and that of 37Cl is ap-proximately 25%. Thus, for CH3C1, containing only one chlorine atom, there will bean M .+ peak and an M .++2 peak in roughly a 3 : 1 (0.75 : 0.25) ratio of intensities. ForCH2Cl2 there will be M .+, M .++2, and M .++4 peaks in roughly a 9 : 6 : 1 ratio, respectively.[The probability of a molecular ion M .+ with both chlorine atoms as 35Cl is (.75)(.75) =.56, the probability of an M .++2 ion from one 35Cl and one 37Cl is 2(.75)(.25) = .38,and the probability of an M .++4 ion peak from both chlorine atoms as 37Cl is(.25)(.25) = 0.06; thus, their ratio is 9 : 6 : 1.] For CHCl3 there will be M .+, M .++2,and M .++4, and M .++6 peaks in approximately a 27 : 27 : 9 : 1 ratio, respectively (basedon a calculation by the same method). (This calculation does not take into account thecontribution of 13C, 2H, and other isotopes, but these are much less abundant.)

10.5 The use of a large excess of chlorine allows all of the chlorinated methanes (CH3Cl, CH2Cl2,and CHCl3) to react further with chlorine.

10.6 Chain Initiation

F F

(DH° = 159)

+159 kJ mol−12 FStep 1 ΔH° =

Chain Propagation

CH3 CH3H FH

(DH° = 440) (DH° = 570)

−130 kJ mol−1FStep 2 ΔH° =+ +

CH3 F

(DH° = 159) (DH° = 461)

−302 kJ mol−1FStep 3 ΔH° =+CH3 FF+

Chain Termination

(DH° = 461)

−461 kJ mol−1CH3 F ΔH° =+ CH3 F

(DH° = 159)

−159 kJ mol−1FF ΔH° =+ F F

(DH° = 378)

−378 kJ mol−1CH3 CH3 ΔH° =+ CH3 CH3

10.7 FH −130 kJ mol−1CH3F ΔH° =+ +CH3 H

FH −432 kJ mol−1ΔH° =+ +CH3 H

FF

FF

−302 kJ mol−1CH3 F ΔH° =+ +CH3 F

CH3 F

10.8 (a) Reactions (3), (5), and (6) should have Eact = 0 because these are gas-phase reactionsin which small radicals combine to form molecules.

(b) Reactions (1), (2), and (4) should have Eact > 0 because in them covalent bonds arebroken.



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(c) Reactions (1) and (2) should have Eact = �H ◦ because in them bonds are brokenhomolytically but no bonds are formed.

10.9 (a)

ΔH° = −130 kJ mol−1

Eact = +5.0 kJ mol−1

F H CH3

CH4

δ δ

F +

CH3 HF

Step 2

PE

Reaction coordinate

+

ΔH° = −302 kJ mol−1


F2

FFCH3δ δ

+

CH3F

CH3

FStep 3

PE

Reaction coordinate

+

(b)

FF

ΔH° = Eact = +159 kJ mol−1

2 F

PE

Reaction coordinate



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FCH3

ΔH° = −461 kJ mol−1

Eact = 0

PE

Reaction coordinate

F+CH3

(c) Notice that this is the reverse of Step (2) in part (a)

ΔH° = +130 kJ mol−1

Eact = +135 kJ mol−1

FH+CH3

PE

Reaction coordinate

HCH3 F+

10.10 (a)

(DH° = 421) (DH° = 432)

Cl+CH3CH2 CH3CH2H + ClH

�H ◦ = −432 + 421 = −11 kJ mol−1

ΔH° = −11 kJ mol−1


Cl H CH2CH3

Cl

δ δ

CH3CH3 +PE

Reaction coordinate

Transitionstate

ClHCH3CH2 +



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(b) The hydrogen abstraction step for ethane,

HCl (Eact = 4.2 kJ mol−1)Cl+ +CH3CH2 CH3CH2H

has a much lower energy of activation than the corresponding step for methane:

HCl (Eact = 16 kJ mol−1)Cl+ +CH3 CH3H

Therefore, ethyl radicals form much more rapidly in the mixture than methyl radicals,and this leads to the more rapid formation of ethyl chloride.

10.11hv

or heatCl2 2 Cl

H

Cl

ClCl + +Step 2a

HCl Cl+ +Step 2b

Cl

1,1-dichloroethane

Cl

ClCl

+ +Step 3a

1,2-dichloroethane

Cl++Step 3b

H

Cl Cl

Cl

Cl

ClH

ClCl

Cl

Cl

Cl

10.12 (a) There is a total of eight hydrogen atoms in propane. There are six equivalent 1◦ hydrogenatoms, replacement of any one of which leads to propyl chloride, and there are twoequivalent 2◦ hydrogen atoms, replacement of any one of which leads to isopropylchloride.

Cl2+ +Cl

Cl

If all the hydrogen atoms were equally reactive, we would expect to obtain 75% propylchloride and 25% isopropyl chloride:

% Propyl chloride = 6/8 × 100 = 75%

% Isopropyl chloride = 2/8 × 100 = 25%

(b) Reasoning in the same way as in part (a), we would expect 90% isobutyl chloride and10% tert-butyl chloride, if the hydrogen atoms were equally reactive.

Cl2+ +Cl

Cl

% Isobutyl chloride = 9/10 × 100 = 90%

% tert-Butyl chloride = 1/10 × 100 = 10%



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(c) In the case of propane (see Section 10.6), we actually get more than twice as muchisopropyl chloride (55%) than we would expect if the 1◦ and 2◦ hydrogen atoms wereequally reactive (25%). Clearly, then, 2◦ hydrogen atoms are more than twice as reactiveas 1◦ hydrogen atoms.

In the case of isobutane, we get almost four times as much tert-butyl chloride(37%) as we would get (10%) if the 1◦ and 3◦ hydrogen atoms were equally reactive.The order of reactivity of the hydrogens then must be

3◦ > 2◦ > 1◦

10.13 The hydrogen atoms of these molecules are all equivalent. Replacing any one of them yieldsthe same product.

Cl2 (+ more highly chlorinated products)

Cl

+

Cl2 (+ more highly chlorinated products)

Cl

+

hv

hv

We can minimize the amounts of more highly chlorinated products formed by using alarge excess of the cyclopropane or cyclobutane. (And we can recover the unreactedcyclopropane or cyclobutane after the reaction is over.)

10.14 (a) (b)

10.15 (a)Cl2

light+

(S )-2-Chloropentane (2S,4S )-2,4-Dichloro-pentane

(2R,4S )-2,4-Dichloro-pentane

Cl ClCl ClCl

(b) They are diastereomers. (They are stereoisomers, but they are not mirror images of eachother.)

(c) No, (2R,4S)-2,4-dichloropentane is achiral because it is a meso compound. (It has aplane of symmetry passing through C3.)

(d) No, the achiral meso compound would not be optically active.

(e) Yes, by fractional distillation or by gas chromatography. (Diastereomers have dif-ferent physical properties. Therefore, the two isomers would have different vaporpressures.)



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(f and g) In addition to the (2S,4S)-2,4-dichloropentane and (2R,4S)-2,4-dichloropentaneisomers described previously, we would also get (2S,3S)-2,3-dichloropentane, (2S,3R)-2,3-dichloropentane and the following:

++

(optically active) (optically inactive) (optically active)

Cl

Cl

Cl

Cl

Cl Cl

10.16 (a) The only fractions that would contain chiral molecules (as enantiomers) would bethose containing 1-chloro-2-methylbutane and the two diastereomers of 2-chloro-3-methylbutane. These fractions would not show optical activity, however, because theywould contain racemic forms of the enantiomers.

(b) Yes, the fractions containing 1-chloro-2-methylbutane and the two containing the2-chloro-3-methylbutane diastereomers.

(c) Yes, each fraction from the distillation could be identified on the basis of 1H NMRspectroscopy. The signals related to the carbons where the chlorine atom is bondedwould be sufficient to distinguish them. The protons at C1 of 1-chloro-2-methylbutanewould be a doublet due to splitting from the single hydrogen at C2. There would be noproton signal for C2 of 2-chloro-2-methylbutane since there are no hydrogens bondedat C2 in this compound; however there would be a strong singlet for the six hydrogens ofthe geminal methyl groups. The proton signal at C2 of 2-chloro-3-methylbutane wouldapproximately be a quintet, due to combined splitting from the three hydrogens at C1and the single hydrogen at C3. The protons at C1 of 1-chloro-3-methylbutane would bea triplet due to splitting by the two hydrogens at C2.

10.17 Head-to-tail polymerization leads to a more stable radical on the growing polymer chain. Inhead-to-tail coupling, the radical is 2◦ (actually 2◦ benzylic, and as we shall see in Section15.12A this makes it even more stable). In head-to-head coupling, the radical is 1◦.

10.18 (a)

n( (

R

(frominitiator)

Monomer

OCH3

OCH3OCH3 OCH3

OCH3 OCH3 OCH3

OCH3 OCH3

+OCH3

OCH3 OCH3R R

R

OCH3

repetition



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(b)

n( (

R

(frominitiator)

Monomer

Cl

Cl

+ Cl

Cl

Cl

Cl

Cl

Cl

R

Cl Cl

Cl Cl Cl Cl

Cl Cl Cl Cl Cl Cl

Cl Cl ClCl

Cl ClR

R

repetition

10.19 In the cationic polymerization of isobutylene (see text), the growing polymer chain has astable 3◦ carbocation at the end. In the cationic polymerization of ethene, for example, theintermediates would be much less stable 1◦ cations.

H

H

1∞ Carbocation

H

H

A H CH3CH2etc.

H

H

H

H

+ + +

With vinyl chloride and acrylonitrile, the cations at the end of the growing chain would bedestabilized by electron-withdrawing groups.

Cl etc.++A H

Cl

+Cl

Cl

Cl

CN etc.++A H

CN

+CN

CN

CN

Radical Mechanisms and Properties

10.20 (1) Br2 2 Brhv

(2) Br HBr+ +



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(3)

Br2

BrBr++

then 2,3 2,3 etc.

10.21 For formation of

Br

, the rate-determining step is:

H

H

BrA� A

�

Br

δ

δ+

For formation of Br, the rate-determining step is:

H

H

Br

B� B

�

Br + δ

δ

Eact

Eact

P.E.

Reaction coordinate

Br+

B HBr+

A HBr+

B�

A�

A· is a 3◦ alkyl radical and more stable than B·, a 1◦ alkyl radical, a difference anticipatedby the relative energies of the transition states.

As indicated by the potential energy diagram, the activation energy for the formation ofA|–– is less than that for the formation of B |––. The lower energy of A|–– means that a greater

fraction of bromine atom-alkane collisions will lead to A· rather than to B·.Note there is a statistical advantage to the formation of B· (6:1) but this is outweighed

by the inherently greater reactivity of a 3◦ hydrogen.

10.22Cl

and I are the only monosubstitution products which can be formed from

cyclopentane and ethane, respectively. (However, direct iodination of alkane is not a feasiblereaction, practically speaking.)



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Br

and

Br

would be formed in amounts much larger than the isomeric alterna-

tives due to the highly selective nature of bromine.

Formation of and

Br

Cl

would be acompanied by considerable

amounts of isomeric byproducts in each case.

10.23 Chain-Initiating Step

Cl2heat, hv

light2 Cl

Chain-Propagating Steps

Cl

Cl2

Cl2 +

+ HCl

+ HCl Cl

+ Cl

Cl

Cl

10.24 (a) Three

+ +

I II III

Cl2

Enantiomers as aracemic form

ClCl H ClH

(b) Only two: one fraction containing I, and another fraction containing the enantiomersII and III as a racemic form. (The enantiomers, having the same boiling points, woulddistill in the same fraction.)

(c) Both of them.

(d) The fraction containing the enantiomers.

(e) In the 1H spectrum for 1-chlorobutane the signal furthest downfield would be thatfor CH2C1; it would be a triplet. The corresponding signal for CHC

Cl

in either

enantiomer of 2-chlorobutane (also furthest downfield) would be an approximate sextet.



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The DEPT spectra for 1-chlorobutane would identify one CH3 group and three CH2

groups; for 2-chlorobutane, two (non equivalent) CH3 groups, one CH2 group, and oneCH group would be specified.

(f) Molecular ions from both 1-chlorobutane and the 2-chlorobutane enantiomers wouldbe present (but probably of different intensities). M .+ +2 peaks would also be present.Both compounds would likely undergo C Cl bond cleavage to yield C4H9

+ cations.The mass spectrum of 1-chlorobutane would probably show loss of a propyl radical byC C bond cleavage adjacent to the chlorine, resulting in an m/z 49 peak for CH2Cl+

(and m/z 51 from 37C1). Similar fragmentation in 2-chlorobutane would produce an m/z63 peak for CH3CHCl+ (and m/z 65).

10.25 (a) Five

Cl2

(R)-2-Chlorobutane

+ + +

+

A B C D

H Cl HClH Cl H ClH Cl H Cl

E

H Cl

Cl

Cl Cl

Cl

(b) Five. None of the fractions would be a racemic form.

(c) The fractions containing A, D, and E. The fractions containing B and C would beoptically inactive. (B contains no chirality center and C is a meso compound.)

10.26 (a) Oxygen-oxygen bonds are especially weak, that is,

DH° = 214 kJ mol−1HO OH

DH° = 184 kJ mol−1

O OCH3

This means that a peroxide will dissociate into radicals at a relatively low temperature.

RO OR100-200°C

2 RO

Oxygen-hydrogen single bonds, on the other hand, are very strong. (For HO H,DH ◦ = 499 kJ mol−1.) This means that reactions like the following will be highlyexothermic.

RO R+ +R H RO H

(b)Chain

Initiation

2Step 1

Cl Cl Cl

R

R

R

+ +

+ +

heat

Step 2

ChainPropagation

Step 3

+ +ClStep 4

R H

R H

R Cl

H Cl

O

O OH

O O



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10.27

(3°)

>

(2°)

>

(1°) (1°)

∼

10.28 R CH3 bond dissociation to form a primary radical:

BDE[CH3CH2CH2CH2 CH3] = 80.9 + 147 − (−146.8) = 375 kJ/mol

R CH3 bond dissociation to form a secondary radical:

BDE[CH3CH2CH(CH3) CH3] = 69 + 147 − (−153.7) = 370 kJ/mol

R CH3 bond dissociation to form a tertiary radical:

BDE[(CH3)3C CH3] = 48 + 147 − (−167.9) = 363 kJ/mol

These calculations are consistent with the relative stability of radicals presented in Sec-tion 10.2B based on C H bond dissociation energy comparisons.

10.29 Single-barbed arrows show conversion of the enediyne system to a 1,4-benzenoid diradicalvia the Bergman cycloaromatization. Each alkyne contributes one electron from a pi bondto the new sigma bond. The remaining electrons in each of the pi bonds become the unpairedelectrons of the 1,4-benzenoid diradical. The diradical is a highly reactive intermediate thatreacts further to abstract hydrogen atoms from the DNA sugar-phosphate backbone. Thenew radicals formed on the DNA lead to bond fragmentation along the backbone and todouble-stranded cleavage of the DNA.

OO

O

S

HO

NH

OMe

Sugar

Bergman

cycloaromatization

Calicheamicin enediyne intermediate

O

O

S

HO

NH

Sugar

Hydrogen

abstraction

from DNA

1,4-Benzenoid diradical

O

OS

HO

NH H

H

Sugar

O OMe O OMe



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10.30 DH◦(kJ mol−1)CH2 = CH H 465(CH3)2CH H 413CH2 = CHCH2 H 369

(a) It is relatively difficult to effect radical halogenation at Ha because of the large DH◦ fordissociation of a bond to a vinylic hydrogen.

(b) Substitution of Hb occurs more readily than Hc because DH◦ for the generation of anallylic radical is significantly smaller than that for the formation of a simple 2◦ radical.

10.31

O

OBr

O

Oheat Br+(1)

O

CO2+(2)O

then 2,3 2,3 etc.

O

O

O

OBr Br

++(3)

Synthesis

10.32

Br2 Br

heat,

light

heat (E2)

(c)

Br2 Br Iheat,

light

NaI

(SN2)(a)

Br

O

NaOH OH

[from part (a)]

Br O− Na+NaH

(SN2) (−H2)(b)

CH3CH3

OK

OH



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Br

Br

HBr,

ROOR,

heat, light

Br2

heat,

light heat

(E2)

(by anti- Markovnikov

addition)Br2

heat,

light

2) CH3Br

1) NaNH2

CH3BrCH4

H Hliq. NH3

2) CH3Br

1) NaNH2

liq. NH3

(e)

(d)

H

ONa

OH,

1) NaNH2

liq. NH3

2)

[from part (a)]

[from part (a)]

H2, Ni2B

or

H2, Lindlar’s

catalyst

(f )

(g)

OH

HA, H2O

Br + Na+ N−N N − NN N −SN2

++

Br

H H H

10.33 Br2

hv

H2O

H2SO4(a)

Br OHONa

OH

HBr

ROOR

[from (a)]

(b) Br

[from (a)]

(1) BH3 : THF

(2) H2O2, OH�(c) OH



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[from (a)]

(d) OCH3(2) NaBH4, OH�

(1) THF-CH3OH

O�

F 2

2�

FHgF

O

(e)Br2

hv

Br2Br

Br

Br

enantiomer

ONa

OH

�

(f )Br2

hv

BrONa

OH

� enantiomer

Br2

H2O

OH

O

Br

NaOH

H2O

10.34 (1)

N2+

AIBN

NN

NN N 260° C

(2) NH

+ +N

H

(3) O2+O

O

(4)H

then 3,4 3,4 etc.

+

+

O

O

O

OH



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10.35 Besides direct H· abstraction from C5 there would be many H· abstractions from the threemethyl groups, leading to:

Any of these radicals could then, besides directly attacking chlorine, intramolecularly ab-stract H· from C5 (analogous to the “back biting” that explains branching during alkeneradical polymerization).

10.36 (1) Ph

O O

O

Ph

O

OO

O Ph

heat2

(2) Ph Ph

O

O

CO2+

(3) HPhHPh + +

O O

(4)CO+

O

(5) H

then 4,5 4,5 etc.O

+O

+

10.37 HOHO HOH

HO

X

+ +•

HO OHdimerization

10.38 (1) Bu3SnH Bu3Sn

(from AIBN)

N

+ +N

H

(2)I

Bu3Sn Bu3SnI+ +



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(3) Bu3SnBu3SnH+ +

then 2,3 2,3 etc.

10.39O O

O

Ph2

O

Ph O

O

Ph

O

PhPh CO2+

O

PhHPh + Bu3 SnH Bu3 Sn+

++ Br Bu3SnBrBu3Sn

Bu3SnH+ Bu3Sn+

Bu3SnH+ Bu3Sn+

10.40 Unpaired electron density in the methyl radical is localized solely above and below the car-bon atom, in the region corresponding to the p orbital of the approximately sp2-hybridizedcarbon atom. The ethyl radical shows some unpaired electron density at the adjacent hy-drogen atoms, especially the hydrogen atom that in the conformation shown has its H Csigma bond aligned parallel to the unpaired electron density of the p orbital of the radical.The larger size of the spin density lobe of the hydrogen with its H C bond parallel tothe p orbital of the radical indicates hyperconjugation with the radical. This effect is evenmore pronounced in the tert-butyl radical, where three hydrogen atoms with H C sigmabonds parallel to the radical p orbital (two hydrogens above the carbon plane and one belowin the conformation shown) have larger unpaired electron density volumes than the otherhydrogen atoms.

10.41 The sequence of molecular orbitals in O2 is σ1s (HOMO-7), σ ∗1s (HOMO-6), σ2s(HOMO-5), σ ∗2s (HOMO-4), π2py (HOMO-3), π2pz (HOMO-2), σ2px (HOMO-1),π∗2py (HOMO), π∗2pz (LUMO). Therefore (a) HOMO-3 and HOMO-2 represent bond-ing pi molecular orbitals, (b) HOMO-1 is a bonding sigma molecular orbital comprisedof overlap of the px orbitals on each oxygen, and (c) the HOMO and LUMO representthe antibonding pi molecular orbital counterparts to the bonding pi molecular orbitals



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represented by HOMO-3 and HOMO-2. Note that the orbitals in s and p orbitals in O2 arenot hybridized. A diagram of the orbitals and their respective energy levels is shown below.

HOMO (π∗2py)

HOMO-1 (σ2px)

HOMO-3 (π2py) HOMO-2 (π2pz)

HOMO-4 (σ∗2s)

LUMO (π∗2pz)

HOMO-6 (σ∗1s)

HOMO-7 (σ1s)

HOMO-5 (σ2s)



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QUIZ Use the single-bond dissociation energies of Table 10.1 (page 202):

10.1 On the basis of Table 10.1, what is the order of decreasing stability of the radicals,

?

> >

HC C

HC C(a)

CHCH2CH2 CH

CH2 CH

CH2

CHCH2CH2

> >HC C(b) CH2 CH

CH2 CH

CHCH2CH2

>> HC C

HC C

(c) CHCH2CH2

CH2 CH> >(d) CHCH2CH2

HC CCH2 CH > >(e) CHCH2CH2

10.2 In the radical chlorination of methane, one propagation step is shown as

Cl CH4 HCl CH3+ +

Why do we eliminate the possibility that this step goes as shown below?

Cl CH4 CH3Cl H+ +

(a) Because in the next propagation step, H· would have to react with Cl2 to form Cl· andHCl; this reaction is not feasible.

(b) Because this alternative step has a more endothermic �H ◦ than the first.

(c) Because free hydrogen atoms cannot exist.

(d) Because this alternative step is not consistent with the high photochemical efficiencyof this reaction.

10.3 Pure (S)-CH3CH2CHBrCH3 is subjected to monobromination to form several isomers ofC4H8Br2. Which of the following is not produced?

(a) (b) (c)

(d) (e)

Br

Br H

H

Br Br

Br

H Br

HH

Br H

Br

H BrBr

10.4 Using the data of Table 10.1, calculate the heat of reaction, �H ◦, of the reaction,

+ +CH3CH3 HBrBr2 Br

(a) 47 kJ mol−1 (b) −47 kJ mol−1 (c) 1275 kJ mol−1

(d) −1275 kJ mol−1 (e) −157 kJ mol−1



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Table 10.1 Single-bond homolytic dissociation energies DH◦ at 25◦C

A : B A + B

Compound kJ mol−1 Compound kJ mol−1

H H 436 (CH3)2CH Br 298D D 443 (CH3)2CH I 222F F 159 (CH3)2CH OH 402Cl Cl 243 (CH3)2CH OCH3 359Br Br 193 (CH3)2CHCH2 H 422I I 151 (CH3)3C H 400H F 570 (CH3)3C Cl 349H Cl 432 (CH3)3C Br 292H Br 366 (CH3)3C I 227H I 298 (CH3)3C OH 400CH3 H 440 (CH3)3C OCH3 348CH3 F 461 C6H5CH2 H 375CH3 Cl 352 CH2 CHCH2 H 369CH3 Br 293 CH2 CH H 465CH3 I 240 C6H5 H 474CH3 OH 387 HC C H 547CH3 OCH3 348 CH3 CH3 378CH3CH2 H 421 CH3CH2 CH3 371CH3CH2 F 444 CH3CH2CH2 CH3 374CH3CH2 Cl 353 CH3CH2 CH2CH3 343CH3CH2 Br 295 (CH3)2CH CH3 371CH3CH2 I 233 (CH3)3C CH3 363CH3CH2 OH 393 HO H 499CH3CH2 OCH3 352 HOO H 356CH3CH2CH2 H 423 HO OH 214CH3CH2CH2 F 444 (CH3)3CO OC(CH3)3 157CH3CH2CH2 ClCH3CH2CH2 BrCH3CH2CH2 ICH3CH2CH2 OHCH3CH2CH2 OCH3

(CH3)2CH H(CH3)2CH F(CH3)2CH Cl

354294239395355413439355

C6H5CO OCC6H5

O O

139

CH3CH2O OCH3 184CH3CH2O H 431

CH3C–H

O

364



CONFIRMING PAGES


10.5 Which gas-phase reaction would have Eact = 0?

+ +(a) CH4

+CH4 CH3CH2

CH3CH2 CH3CH2

H Cl ClH Br

+

+

+ +

(b)

(c)

Br(d)

H I IH Br+ +Br(e)

CH3CH3CH3

CH3

10.6 What is the most stable radical that would be formed in the following reaction?

Cl HCl+ +

10.7 The reaction of 2-methylbutane with chlorine would yield a total of differentmonochloro products (including stereoisomers).

10.8 For which reaction would the transition state most resemble the products?

CH4 CH3 HF(a) F+ +

CH4 CH3 HCl(b) Cl+ +

CH4 CH3 HBr(c) Br+ +

CH4 CH3 HI(d) I+ +

Documents

JWCL234-10 JWCL234-Solomons-v1 December 11, …JWCL234-10 JWCL234-Solomons-v1 December 11, 2009 16:44 CONFIRMING PAGES RADICAL REACTIONS 189 (fandg) Inadditiontothe(2S,4S)-2,4-dichloropentaneand(2R,4S)-2,4-dichloropentane