Post on 06-Apr-2018
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Chs. 7 & 8: Alkenes
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Alkenes (olefins): hydrocarbons containing C=C. Unsaturated less than 4 single (sigma) bonds to C.
Simplest alkene = ethene
C-C bond length = 1.34 average bond angle = ~120o
.
H
H
H
HC=C
ethene (ethylene)
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H
H
H
H
pi bond - sideways overlap of p orbitals
ethene (ethylene)
sigma bond - electron density along
bond axis
Orbital picture shows & bonding
electron density is along axis of bond and above and below planeof molecule in bond.
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I. Nomenclature : IUPAC rules
Alkenes
1. Select the longest carbon chain containing the double bond asthe parent. Name as ethene, propene, butene, pentene, etc.
(Alkane name minus ane plus ene)
2. # the C chain of parent so that C=C has the lowest possiblenumber (locant).
5 4 3 2 1Ex: CH
3CH
2CH=CHCH
3
If more than one way of numbering gives lowest # to C=C, choosethe way that gives the lowest #s to branch points.
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CH3CH3
CH3
CH3 CH3
CH3
123
4
56
NOT
65
4
3
21
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3. Write full name #ing substitutents & listing them alphabetically.
CH3CH3
CH3
CH3 CH3
2-pentene 2-methyl-3-hexene
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4. If two or more double bonds are present, choose as the parent thelongest C chain containing both and use the prefixes di, tri, tetra, etc.
before ene.
CH2CH3
H
H
H
H
H
CH3H
H
1,3-pentadiene 2-ethyl-1,3-butadiene
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Cycloalkenes
1. Give double bond lowest locant 1.
2. Number substituents thru the double bond so that they have thelowest possible locants.
3. List the substituents alphabetically.
4-methylcyclohexne
1-ethyl-4-methyl-1,3-cyclohexadiene
5,5-dimethyl-1,3-cyclohexadiene
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Common names
CH2=CH2 ethylene
CH3-CH=CH2 propylene
H2C=CH- vinyl (IUPAC: ethenyl)
H2C=CH-CH2- allyl (IUPAC: 2-propenyl)
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II. Cis/trans stereoisomerism
- Occurs in alkenes because there is no rotation about C=C
CH3H
H CH3
HCH3
H CH3
is not the same as
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Geometric isomers (cis/trans) exist for every alkene which hasnonidentical groups attached to both C atoms of double bond
CH3 CH3
CH3CH2 H
CH3 H
CH3CH2 H
H CH3
CH3CH2 H
-no cis/trans isomerism
cis-2 pentene trans 2-pentene
cis: substituents on same side of double bond
trans: substituents on opposite sides of double bond
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As molecules get larger and more complex, cis/trans nomenclaturegets complicated.
H CH3
CH2CH3CH3
CH3
For cpds like:
use Z,E nomenclature1. Assign priority to groups on each C of double bond
2. If high priority groups on both C's are on same side of the double bond,use Z (Zusammen - together).
3. If high priority groups on both C's are on opposite side of the double bond,use E (Entgenen -apart).
CC
high high
lowlow Z
CC
high low
highlow E
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Assignment of priority: Cahn-Ingold-Prelog Sequence Rules
1. Assign priorities of the atoms directly attached to C atoms of thedouble bond according to atomic #.higher atomic # = higher priority
C(6) > H(1)
N(7) > C(6)
O(8) > N(7)
D(2) > H(1) Isotopes = special case since they have the same atomic #Use atomic wt. # instead.
2. If priority cannot be established by rule #1, make a similar
comparison of the next atoms in each group.
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3. Multiply-bonded atoms are equivalent to the same # of singlybonded atoms
CH CH3
H
C
H
H
Ex:
= -C
H
C
C
C H
H
-C C - H = - C C H
C C
C C
=- C N
N
N
C
C
-C N:
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CH3
O
H
O
= -C
O
O
C
H
H
H
C
= -C
O
O
H
C
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Designate the following molecules as E or Z
H
Cl
Br
CH3
H
CH2OH
OHO
H
Br
H
CH3
CH3
Z Z E
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III. Stability of alkenes
a) cis vs trans
Generally, cis isomers are less stable than trans isomers
E = 2-3 kJ/mol
CH3
H
H
CH3
CH3
H
CH3
H
trans 2-butene cis 2-butene
-less stable: higher E due to stericstrain. The larger the groups on thedouble bond, the greater the Edifference between cis and trans.
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Energy difference is reflected in
1) interconversion equilibria and 2) heats ofhydrogenation
Interconversion equilibria:
G = -RTlnK = -(8.314J/K)(298K)ln(.24/.76) 2.8kJ
CH3
H
H
CH3CH3
H
CH3
H acidcat.
at 298K: 76% 24%
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heats of hydrogenation:
H2, cat H2, cat
trans 2-butane butane cis 2-butene
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Hhyd (cis)= -120 kJ/mol
Hhyd (trans) = -116 kJ/mol
Energy difference = 4 kJ/mol (close to 2.8 kJ, 4kJ is from H datawhile 2.8 kJ = G data)
b) stability of mono - vs polysubstituted alkenes
General trend:
R2C=CR2 > R2C=CHR > RCH=RCH R2C=CH2 > RCH=CH2
more stable less stable
lower Hhyd higher Hhyd
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Explanation of observed stability trend:
-hyperconjugation occurs when a pi orbital can overlap (partially)with a neighboring sp3 orbital. only happens when C=C issubstituted. In the figure, the pi orbital can interact with 3 sp3
orbitals.
-greater bond strength for sp2-sp3 bonds than for sp3-sp3 bonds.
Compare CH3-CH=CH-CH3 (disub.) and CH3-CH2-CH=CH2 (monosub.)
2 sp3-sp2 bonds vs 1 sp3-sp3 and 1 sp3-sp2
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Reactions of Alkenes 4 types well examine
1. electrophilic addition
2. hydrogenation (reduction)
3. oxidation4. radical addition
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Electrophilic Addition
I. Addition of acid: H-B -acid catalyzed
Y z
+ Y-Z
nucleophile electrophile
R2C=CHR RR
B HR
H+ H-B
H-B = HX (X=Cl, Br, I)H-HSO4
H-OH
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A) Mechanism
R2C=CHR
RR
B HR
H
BH R
R
H
R
H
B
R
R
H
R
HB
++
+
++
Rate determining step
1.
2.
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For H-Br + (CH3)2C=CH2
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Outline the mechanism:
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Rate law
from RDS: rate = k[alkene][H-B]
first order in alkene
first order in H-B
second order overall
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B) intermediate of rxn = carbocation
structure of R3C+:
Relative stabilities of carbocations effect of alkylgroups
R3C+ > R2CH+ > RCH2+ > CH3+most stable least stable
3o > 2o > 1o > CH3+
RR
R
120o
-sp2 hybridized-empty p orbital-6 valence electrons
+
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How do alkyl groups stabilize the positive charge?
1. Hyperconjugation
- Overlap between vacant p orbital and an adjacent C-H bond (sp3 orbital)
- Not possible in methyl cation
- The more R groups, the more hyperconjugative interactions are possible.
H
H
H
H
H
H+
+
=
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2. Inductive effects
- alkyl groups donate e- density by induction toward (+) charge.
- the more alkyl groups, the greater the stabilizing effect.
C) Orientation of addition
There are two possible products from the addition of H-B
Rxn is regioselective: only one of two possibledirections of addition is observed.
R2C=CHR R
R
B H
R
H
R2C=CHR R
R
H B
R
H
+ H-B
OR
+ H-B
observed product
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Markovnikovs Rule: For electrophilic addition of HB toan alkene, the acid H will become attached to the doublebond C already bonded to the greater # of Hs.
Predict the products: H-Br
ether
H-Cl
ether
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To understand:
Q: What does it mean to obtain only one of two possibleproducts?
A: Favored product is formed faster.RDS for two possibilities:
CH3
CH3
CH3
H
CH3
CH3
Br
CH3
H
H
CH3
CH3
H
CH3
H
BrH-Br
or
product
CH3
CH3
CH3
H
CH3
CH3CH
3
H
H
CH3
CH3
CH3
H CH3
CH3
H
CH3
H
H-Br+
+ Br-
H-Br
++ Br-
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Shown is the energy profile for the RDS
The rate is determined by the activation energy energydifference between reactant and transition state of RDS.
The rate is proportional to e (-Ea/RT). The greater the Ea,the slower the reaction.
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Since the Markovnikov product is the favored product, itmust be formed faster, that is, with a lower Ea. Thereforeits TS must be lower in energy than the TS of the non-
Markovnikov product.
Could this have been predicted?
Hammond Postulate:
For an endothermic reaction step, the TS resembles theproduct.
For an exothermic reaction step, the TS resembles thereactant.
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For the RDS in the addn of H-B, the TS resembles thecarbocation product of that step, so the more stable thecarbocation, the more stable the TS, the lower the E
a
,the faster the reaction.
So, compare the carbocations arising from theMarkovnikov and non-Markovnikov pathways.
CH3
CH3CH3
H
H
CH3
CH3
H
CH3
H
+
+
Markovnikovpathway
3o carbocation
2o carbocationNon-Markovnikovpathway
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stability of carbocations: 3o
> 2o
> 1o
E of the Mark. TS < E of non-Mark. TS
Ea (Mark. pathway) < Ea (non-Mark. pathway)
So Mark. pathway is faster than non-Mark pathway
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Predict the products:
Write the mechanism for the acid catalyzed reactionbetween 1-methylcyclohexene and water (the 3rd
reaction)
H-Cl
ether
H2SO4H2O
H2O, H+
All three reactions pass through the same intermediate carbocation.What is its structure?
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D. Rearrangement of carbocations
In some cases, more than one product may be obtainedas a result of rearrangement of the carbocationintermediate.
CH3
CH3
HH
H
H(CH3)2CH CH3
Cl
H
CH3
CH3
Cl
CH2CH3
H-Cl
ether
+
predicted Mark. product
rearranged product -major product
Example:
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Provide a mechanism for the formation of the second product.
CH3
CH3
H
H
H
H
CH3
CH3CH
2CH
3
CH
CH3
CH3CHCH3
CH3
CH3
H
CHCH3CH3
CH3CH2CH3
CH3
CH3
H
CHCH3CH3
CH3CH2CH3
Second Q: What carbocation is originally produced (product of Mark. addition of H+)?
H-Cl
First ask: What intermediate carbocation produced the rearranged product?
A.
A.
Third Q: How do you get from initial R+ to the rearranged R+?
A.
Justification: conversion of a 2o carbocation to a 3o carbocation with increase in stability
2o 3o
1,2 hydride shift
+
+
+
+
+
+
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Other possible rearrangement: 1,2-alkyl shift
H-Cl
H-Cl
Cl-
Example:
C
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II. Addition of X2 (X2 = Cl2, Br2; F2 is too reactive;
I2 is slow & not as useful)
A. Mechanism
R2C=CR2X2 R
X
R
X
R
R (a vicinal dihalide)
R
R
R
R
Br Br
Br
R RR R
Br
R R
R R
Br
Br
R
RBr
Br
R
R
++
cyclic bromonium ion
+
+
+
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B. Stereochemistry of addition = anti
due to cyclic intermediate: Br-
must add to side oppositeto first Br.
Br
R R
R RBr
R
RBr
Br
R
R
Br
R R
R RBr
RR
Br
Br
R
R
+ +
OR
+ +
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Predict the product and give a mechanism:
Br2
CCl4
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C. The Nu: in the 2nd step of the rxn can be any nucleophile.
R2
C=CR2 R2C CR2
BrBr
Cl
NO3-
Br
BrBr
ClBr
I
Br
NO3Br
OHhalohydrin
+
(NaCl)
I-
(NaI)
(NaNO3
)
H2O
formation
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Halohydrin
H2C=CH2 HO-CH2CH2-Br
Cl2, H2O
HO-CH2CH2-Cl
C C
X OH
2-bromoethanol- a bromohydrin
Br2, H2O
2-chloroethanol- a chlorohydrin
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For asymmetrical alkenes, 2 products are possible.
R2C=CHR
Br2, H2O
R
Br
R
OH
R
H R
OH
R
Br
R
Hor
observed product
-halogen becomes attached to the double bond C already bonded to the greater # of H atoms
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To understand:
Br
R R
R H
Br
R R
R H+
resembles 3o
carbocationhas more (+) character = point ofattack by nucleophile
resembles 2o carbocation
+
+ +
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When nucleoophile (H2O) attacks in 2nd step, it attacks the morehighly substituted C atom.
Br
H R
R ROH2
H
R
Br
O
R
R
H H
H
R
Br
O
R
R
H
+ +
+
-H+
..
:
:
: :
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Mechanism for halohydrin formation: noteregiochemistry and stereochemistry
Br2
OH2
OH
CH3Br
H
Br Br
H CH3Br
Br
H CH3Br
OH2
HCH3Br
O H
H
H CH3Br
O H
H BrOH
CH3
Br
H
1.
+
+
2.
+
+
3.
+
+ HBr
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III. Addition of H2O hydration 2 methods
A. oxymercuration - demercuration
-regiochemistry = Markovnikov
R2C=CHR
Hg(OAc)2, H2O1.
2.
- non-stereospecific
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Example: OH
CH3CH3
CH3CH2
H
H
HgOAc
HH
HgOAc OH2
OH
HH
HgOAc
H
OH
HH
HgOAc
H
OH
HH
NaBH4
Hg(OAc)2, H2O
AcO-Hg-OAc -OAc
-OAc
1.
2.
NaBH4
(exact mechanismnot known)
not stereospecific - syn & anti addition is observed
+
+
+
+
+
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B. Hydroboration oxidation non-Mark. Addition
- syn stereochemistry
BH3, THF H2O2, OH-
HBH2
OH
D
H
H H
H2O2, OH-
2.
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D CH2CH3
OH
D
H
CH2CH3
D CH2CH3
H
D
BH2
CH2CH3
H
D CH2CH3
H2B
Mechanism:BH3, THF H2O2, OH-
H-BH2
(- H and BH2 are added to the same side of the dbl bond - synstereochemistry- BH2 is added to the less substituted C atom - results in less steric
crowding in transition state - non-Mark. regiochemistry)
1.
-----
CH3
HCH
3
D
OH
CH2CH
3
CH3
D
CH3
CH2CH
3
CH3
D
CH3
CH2CH
3
BH3 CH
3
D
CH3
CH2CH3
H2B H
CH3
H
CH3
D
BH2
CH2CH
3
CH3
H
CH3
D
BH2
CH2CH
3
CH3
H
CH3
D
OH
CH2CH
3
BH3, THF H2O2, OH-
1. + -----
2.
-BH2 and -H add to the same side of the dbl bond - syn stereochemistry
-BH2 is added to the less substituted C atom and -H is added to the more highly substituted C atom
because that reduces crowding in the transition state. Since -BH2 is later replaced by -OH,
non-Markonikov regiochemistry results.
H2O2, OH-
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IV. Addition of carbene - R2C:
R2C:
CR2
:CCl2
I-CH2-Zn-I
+
cyclopropyl ringcarbene is generated"in situ"
Example:
KOH
or KOt-bu
(dichlorocarbene)
CH2I2Zn(Cu) (carbenoid - adds :CH2)
- Simmons-Smith reaction
CHCl3
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Reaction of carbenes with alkenes is stereospecific:
CHCl3
KOH
CH2I2, Zn(Cu)
CH3
H CH3
H CHCl3
KOH
CH2I2, Zn(Cu)
CH3
H H
CH3
H
CH3
H
CH3
CH3
H
H
CH3
Cl
Cl
H
CH3
CH3
H
H
CH3
H
CH3
Cl
Cl
cis
trans
cis
cis
trans
trans
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Hydrogenation (reduction)
H
R
R
H
R
R
R2C=CR2
catalyst =
H2
Addition is syn
Pt, PtO2, Pd/C, Ni
+
H2, cat.
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Under the relatively mild rxn conditions (Rm temp.)which reduce C=C, other multiple bonds are not
reduced.
Pd/C
phenyl groups:
carbonyl groups:
cyano groups:
H2
Pd/C
H2
PtO2
H2
PtO2
H2
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Oxidation
A. Epoxidation1. via halohydrin
CH3 Br2
CH3Br
OHH
CH3
O
CH3Br
OH
H
OH-
CH3Br
:O:-H
CH3
O
H2O
mechanism of epoxide formation:
..
via halohydrin
epoxide(oxirane)
::
::
epoxide(oxirane)
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2. via peroxyacid
CH3
CH3
O
R O
O
OH
epoxide
(oxirane)
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B. Hydroxylation glycol formation (glycol = cpd with vicinal hydroxygroups)
H
OHOH
H
O
HH
H
OHH
OH
OsO4, pyridine
NaHSO3 (aq)
RCOOOH
1.
2.
H3O+
cis
trans
To get cis glycol:
To get trans glycol:
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C. Oxidative Cleavage - 3 types
1. ozonolysis
2. permanganate oxidation
3. periodate oxidation of glycols
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R4
R3R1
R2 O
R1
R2 O
R3
R4Zn,H+
O3
+
1.
2.
1. ozonolysis of alkenes
R = alkyl, H
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H
HR
H
KMnO4, H+
KMnO4, neutral R OH
O
H
R3R1
R2
KMnO4, H+
KMnO4, neutral
O
R2
R1
R3OH
O
R = alkyl
or + CO2
or+
=CH2
=CHR
=CR2
CO2, carbon dioxideRCOOH, carboxylic acid
R2C=O, ketone
2. permanganate oxidation of alkenes
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R4
R3R1
R2
OsO4, pyr
NaHSO3 (aq) R4
OHOH
R2
R1 R3
HIO4
O
R1
R2
OR4
R3
1.
2.
3. periodate oxidation of glycols (result like ozonolysis)
alkene glycol
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Radical Polymerization use fish hook arrows in mechanism
nH2C=CH2
RO-OR
H
H
H
H
RO *n
RO OR
H2C=CH2
H2C=CH2
R R
3 steps in any radical reaction:
1) initiation: generation of a free radical 2RO.
2) propagation:
RO-CH2-CH2-CH2-CH2.
RO-CH2-CH2. +
Ex: Combination of 2 radicals
R-R
attack of alkene by a free radical generates another free radical
- generates another free radical
RO. + RO-CH2-CH2.
3) termination: ends the polymer and doesn't produce a radical
+
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Polymerization of unsymmetrical alkenes to predict product, needto take into account the stability of radicals.
observed product not observed
H2C=CHR
H2C=CHR CH2-CH-CH2-CHRRO
R
H2C=CHR
H2C=CHR CH-CH2-CH-CH2RO
R R
Mechanism for formation of first product:
1. RO - OR 2RO.
2. RO. + RO-CH2-CHR
.
RO-CH2-CHR
.
3. +
.
Mechanism for formation of second product:
1. RO - OR 2RO.
2. RO. + RO-CHR-CH2
.
RO-CHR-CH2
.
3. +.
H2C=CHRRO CH
2-CHR *n
RO CHR-CH
2 *n
RO-ORn or ??
Q
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H2C=CHR
RH
H HRO
HR
H H
RO
Why do radicals in step 2 & step 3 always add to the less hindered C atom?
A:
RO.
+or
. 2o radical
. 1o radical
Order of radical stability:
3o > 2o > 1o >.CH3
so the more stable radical is formed.
Q:
So the more highly substituted radical is formed