Reaction Mechanism

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2 10 Br elimination

( iii) CHCHCHCH → 3 2 3 + CHONa 2 5 2° Halide

CH3 1.

( i)

+ CH3 CH2 C HCH3

( ii)

H 3C — C — C H2C H3

Strong base

+

CHCH = CHCH3 3 ci ss- and t ra ran ss- (Major)

(3°)

H +

( iii)

2.

H  → →

+

2.

+

H  →

OH

→

(– HO) 2

2, 2-Dimethylcyclohexanol

⊕

+

H

+

2° Carbocation

H

2° carbocation

+

+

–H →

+ 3° Carbocation

rearranges to

→

3° carbocation      →Cl –  –  Cl      Cl →

Cl

α

+

: O :H ( a)

H →

(i)

CH3

OH2 +

3.

1, 2-Dimethylcyclohexene

+

→

α

(– HO) 2

Carbocation 2-Chloro 2-Ch loro-3, -3, 3-di 3-dimeth methylb ylbutan utane e

2-Chloro 2-Ch loro-2, -2, 3-dim 3-dimeth ethylb ylbutan utane e

O 3.

( i)

O

OH

+

electrophilic + HCN →

+ CN

addition

1, 2-addition (A)

( ii) ( iii) ( iv )

CH3 CH = CH 2 + CCl4

Carbocation can undergo elimination to form 1-methylcyclohexene and methylenecyclohexane. : :OH2 H

peroxide

   → (C)

1, 4-addition (B)) (B

+

→  →

H

(ii)

+ : OH3

CN

(Major)

:

:OH2

CH2 —H

Cl  CH3 — CH CH2 CCl3

CH2

+

+

→  →

(iii)

+ : OH3

peroxide

CH3 CH = CH2 + CBrCl3  CCl3   → CH3CH(Br)CH2 CCl (D) CH3 CH = CH 2 + CHCl3

peroxide     →

(E)

(Minor)

:

: OH

CH3 CH2 CH2 CCl3

+

+

( b)

H →

(i)

α

+ H2 O

α Carbocation +

+

(ii) :

Br 1.

( i)

+ C2H 5O K

H 2O :

elimination

    →

2° Halide

H

Br

H

+

Substitution

CH 3CHCH 2CH 3 + CHOH CH3OH → CH 3CHCH 2CH3 2° Halide

(Minor)

H

: H2O :

Strong base

( ii)

+ + : OH3

H →

Weak base

EXERCISE 4.1

OCH OC H3

(iii)

+ + : OH3 0

→ (Minor)

















2 11 1.

2.

3.

( i)

Ag + has positive charge, hence electron deficient, in H 2 C: , , C has only 6 electrons ; in SiF 4 , Si Si can acquire acquire more than 8 electrons by utilizing its d -orbitals. -orbitals. In the reaction, A and C are reacta nts, B and E are intermediates; and D and F are products. This can be observed by addition of the three steps to get the following net reaction 2A + C   →D + 2F The reaction occurs in two steps, in the first step CH 2 = CH 2 is nucleophile, while in the second step Br – is nucleophile

δ + δ− H 2 C = CH 2

δ+ δ – + Br—Br   → CH 2CH 2Br

+

Nucleophile1

( ii)

+ CH 2 CH 2 Br Electrophile1

4.

Electrophile2

  →

Br–

+

15.

Nucleophile2

CH2 BrCH2 Br

Nucleophile2

Let us study the free radical formed by the removal of three H .’s .

Least stable because H. is to be removed from sp2 C

More stable because H. is to be removed from sp3 C

Isobutane has two types of carbon atoms (1° and 3°) ; hence it will form two products

CH3

CH 3

CH 3

17. 18.

: → CH 3C H2C H C H 3 + C H — CHCHCH 3 3 + CH2 3 C— CH 3 Isobutane

Insertion on 1°C

CH 3

19.

Insertion on 3°C

6.

In the formation of carbocation (as well as free radical) sp 3 hybridised carbon atom changes to sp2 . This is of special importance in formation of tert -carbocation (or free radical) where steric relief is observed. In the parent compound the three bulkyl groups are pushed together due to tetrahedral nature of the carbon atom having bond angle of 109.5° (steric strain) which is greatly relieved in formation of carbocation (or free radical) d ue to planar structure (120°) of the product, where the three bulky groups are separ ated from each other by an angle of 120° (steric relief).

7.

8.

In car banion, banion, sp3 hybridised carbon atom has three pairs of bonding electrons and one pair of non-bonding electrons. Thus here two types of repulsions are observed : lp-bp and bp-bp ; since lp-bp > bp-bp repulsions, angle between two bonding pairs is slightly reduced than the normal tetrahedral value of 109.5°. Thus the geometry of simple carbanions is is not exactly tetr ahedral, but pyramidal. Carbocations Carbocations ar e sp 2 hybridised, simple (unconjugated) carbanions carbanions ar e sp3 hybridised but conjugated carbanions are sp2 hybridised because here delocalization of electrons results in the formation of a double bond which requires all involved atoms to lie in the same plane (coplanar), i.e. the molecule becomes flat.

CH3 CH3 – : : HC—C = O ←→ HC = C—O 2 2 – : : ←→ HC = CH—CH HC—CH = CH 2 2 2 2 –

–

9. 10.

or

–δ

H2C

CH

–δ

CH2

Carbanion from CH3 CHO and cyclopentadiene is stabilised due to resonance and aromatic sextet respectively. Carbanion is formed as an intermediate which being flat can be attacked by Br2 on either face forming racemic mixture.

CH—C—OH 3 CH3 -Butanol tert -Butanol

.

Most stable due to allylic nature

Rate of a multi-step reac tion is determined by the slowest step. Thus transition state of such step will require more energy of activation, and thus the cor responding transition state will have highest enthalpy. In multistep reactions, the step with the highest enthalpy transition state (i.e. with highest ∆H≠ ) is the slowest (rate determining) step. The nucleophilicity of species whose nucleophilic atoms neither lie in the same period nor in the same group of the periodic table should be compared by considering their basic strength or the relative acidic character of their conjugate conjugate ac ids. The decreasing basic character of the species is HS– > RCOO– > ROH > RCOOH Conjugate acids H2 S < RCOOH < ROH2 + < RCOO COOH2 +

CH3 16.

.

5.

14.

Br–

+

Electrophile1

13.

20.

H

+

→

CH3 CH—C 3

⊕

CH3 -Butyl carbocation tert -Butyl

tert -Butyl carbocation is formed as an intermediate, which being stable, stable, does not rearr ange to the less stable stable 2° or 1° car bocation. bocation. Oxygen, being an electronegative element, can best accommodate negative charge. More the difference between the electronegativies of two concer ned atoms, higher will be the chance for dissociation into ions. More the stability of a species greater is the ease of its formation and hence lesser w ill be be the dissociation energy of the bond to be cleaved. Benzyl radical is highly stable stable due to delocalisation of its odd electron while methyl radical is quite unstable. Free ra dical substitution substitution leads to complete racemization because the intermediate free radical is coplanar and can be attac ked on either of the face equally forming two enantiomers in equal amounts (complete racemization). SN2 reaction involves the formation of transition state in which attack of nucleophile and removal of the leaving gr oup take place simultaneously, simultaneously, hence the nucleophile can attack back-side from that the leaving group. Consequently, compound having inverted configuration is formed (100% inversion of configuration). configuration). In S 1 reaction, the N

nucleophile attacks the ion pair (R+X– ), hence back-side attack of the nucleophile predominates although attack from the side of the leaving group (front-side attack) also takes place. Thus the product produc t will be having both enantiomers, and the enantiomer with the inverted configuration predominates. In short, SN 1 reaction leads to par tial racemization along with some amount of inverted configuration. 21. Discussed in the above answer. along with some 22. Since SN1 reac tion yields racemic modification along amount of the enantiomer of inverted configuration ; as far as specific specific rotation rotat ion is concerned, the product prod uct will be having opposite specific rotation. Further, its value will be less because optical pu rit y of the product will be less than that of reactant. 23. The weaker the base, better will be the leaving group. Further, basic character of the groups can be easily judged by acidic character of its conjugate acid (recall that weak bases have stronger conjugate acids). Thus – OH – OSO CF Base F– OH2 – OTs 2 3

















2 12

25.

+

+

–

OH CHCH CHC CHCH H 2Br → [CHCH = CHCH2 ←→ CH3 CH CH—C —CH H = CH 2] → CH3CH = CHCHO CHCHOH = C H2 3 H = CHC 3 2 H + CHCHCH 3 1-Bromobutene-2

Allylic cation

OH Normal Norm al produ product ct (SN1)

26.

Since Since allylic cation stabilise stabilise easily easily due to resonance, its formation is easy and hence, the compound undergoes S N1 reaction ; the product formed by rearranged carbocation is known as SN1, product (unimolecular nucleophilic substitution with rearrangement). It is an example of allylic rearrangement . All are primary alkyl bromides and undergo S N2 reactions in which nucleophile attacks on the carbon from back side. Thus more the branching branching on the alkyl group atta ched to C having Br, lesser will be be its reactivity towards SN2 reactions. Thus

> (CH)CH—CHBr > n -CH—CHBr 3 7 2 3 2 2 (I)

27. 28. 29.

(II)

CH3

CH—CHBr > (CH)C—CHBr 2 3 3 2

C2H5

(III)

(IV)

Addition of halogens to alkenes, involve formation of cyclic halogenonium ion, and not a carbonium ion. Stronger a base, poor will be the leaving group. In nucleophilic additions on carbonyl groups, nucleophile adds on the electron deficient carbon. Thus any factor which can increase electron deficiency of carbonyl carbon will increase rate of nucleophilic addition on carbonyl compounds. Acids perform this function by protonating carbonyl oxygen which thus becomes more electronegative.

R

30. 31.

Rearranged product pro duct (SN1’)

: + C = O: + H

→

R

+

C = OH :

→  

R′ R′ Reaction takes place in presence of a base ( :B) which removes hydrogen as proton. F– is a strong base and causes ca uses elimination reaction with tert -alkyl halides halides ; hence the expect expected ed product produc t in both should should be isobutene. However, in presence of H2 O, F– ion is solvated vi a hydrogen bonding and thus can’t exert its influence, the weak base H2O causes substitution reaction. On the other hand, F – is not solvated in presence of (CH 3 ) 2 SO (DMSO (DMSO)) ca using elimination reaction to form isobutene. +

32. 33. 34. 35.

(CH3 ) 3 COH   →(CH3 ) 3 C . Acid ( SbCl SbCl5)–catalysed r eactions generally involve carbocation as intermediates which here is further confirmed by rac emization, since carbocations are flat and can be attacked on either side of the face forming both enantiomers. Electron-withdrawing group like —NO2 enhances positive charge on the carbon, thus destabilises the carbonium ion, while electronpushing group like —OH, —OCH 3 disperses positive charge, hence stabilises the carbonium ion. More the stability of the product, faster is debromination of the parent compound.

EXERCISE 4.2 >1 CORRECT OPTION

1

(b,c)

2

(a, b)

3

(b,c)

PASSA G E 1

7

11

(c) (c) (b)

9

10

(d) (b) (b)

8

PASSA G E 2

12

(b) (a) (b)

PASSA G E 3

13

MATCHING TYPE QUESTIONS

16

A /R

22

T R U E / FA LS LS E

27

17 18

14

15

(A) - a, d ; (B) - b ; (C) - c, (D) - b (A)-a, c ; (B)-d ; (C)-a, c ; (D)-b, c (A) – b,c; (B) – b; (C) – a; (D) – d 23 24 (a) (a) (c) 28 29 False; False; False;

4

19 20 21 25

(b, d)

5

(a, b, c, d)

6

(a, b)

(A)-a, b ; (B)-c, d ; (C)-a, b ; (D)-a, d (A) – c; (B) – a; (C) – d; (D) – b (A) – b; (B) – a, d; ( C) – c, d; ( D) – b 26 (a) (a)

EXERCISE 4.3 1.

(i) (ii) (iii) (iv ) (v ) (v i)

NH2 – and NH3 . Here nucleophilic site is same (N), hence nucleophilicity and basicity should be of same order. Since the conjugate acid NH4 + is stronger than NH3 , therefore therefore N H 3 s h o u l d b e w e a k e r b a s e a n d w e a k e r n u c l e op op h i l e t h a n N H2 – . – – OH is a stronger base than SH because the O—H bond is stronger than the S—H bond ; while SH – is a stronger nucleophile than OH– because S is less electronegative than O. H2 O is better nucleophile and stronger base than H 3 O+ (Explanation as that of ( i)). On the the same same line line as that that of ( i), CH 3 CH2 O– is a stronger base and better nucleophile than CH 3 COO– . Since the bond bond dissoci dissociation ation energy of of the the H—Br H—Br is less less than than that that of H—Cl, H—Cl, Br Br – will be a weaker base than Cl– . However, – – electronegativity of Cl is more than Br, Cl will be weaker nucleophile than Br . In OH– and F – , the two nucleophilic nucleophilic atoms belong to same period, hence the basicity basicity and nucleophilicity nucleophilicity both decrease with the

















2 13 3.

Species having two different nucleophilic sites are known as a m b i d e n t n u c le le o p h i l e s , e.g. NO2 – and CN–

:– : O :

:N

–

: O :

: C ≡ N:

(N and O are nucleophilic)

R—NO2

R—O—N = O

R—C ≡ N

R—N == C

Nitrites

Nitriles

Isonitriles

Nitro co c ompounds

4.

5.

(C and N are nucleophilic)

( a)

When the the nucleophil nucleophilic ic site site is the same atom (here O), O), nucleophili nucleophilicity city parallels parallels basicity. basicity. Therefore, Therefore, – – CH3 O > OH > CH3 COO– > H2 O ( b) When the the attacking attacking (reacting) atoms are different different but in the the same periodic family, family, the the one with the largest atomic weight is is the most reactive. Therefore, PH3 > NH3 . This order is the reverse of basicity. We know that weaker a base better will be the leaving group. Further we know that weaker a base, stronger will be its conjugate acid and hence lower will be its p K a value. Thus acid strength of the three conjugate acids is C6 H5 OH < CH3 C OOH < C6 H5 SO3 H pK a v alue 1 0 .0 4 .5 2 .6 Hence, C6 H5 SO3 – , CH3 COO– , C6 H5 O– Basiccharacterincreases

→ Leavingpropertydecreases –)

6.

7.

Thus best leaving group is the weakest base (C 6 H5 SO3 ; and the poorest leaving group is the strongest base (C 6 H5 O– ). Greater the dispersal of the positive charge, higher will be the stability of the carbocation. Electron-releasing inductive effect and hyperconjugation is maximum in (CH 3 )3 C+ and minimum in CH 3 +. Further stability of these carbocations can be explained on the basis of steric acceleration. Bulkier the group on the carbon bearing positive charge ( sp2 hybridised), more will be steric acceleration (steric relief) in their formation due to conversion of bond angle from 109° present in parent compound (sp3 hybridised) to 120° in carbocation (steric relief). In short, Me3 C—Br > Me2 CH—Br > MeCH2 — Br > CH3 —Br Steric strain and Ma x imum Minimum hence unstability + + + Me3 C+ > Me2 C H > Me C H2 > C H3 Steric relief and Ma x imum Minimum hence formation ( a) Due Due to stron strong g elect electron ron-wi -withd thdrawi rawing ng flu fluori orine nes, s, a δ+ develops on the atom adjacent to C +. Due to positive charges on adjacent atoms, the species is destabilised. +

( b)

8.

Unshared Unshared elect electron ron pair on fluorin fluorine e can be shifte shifted d to vacant p -orbital of C ( p-p -overlap). Hence positive charge is dispersed leading to stability of the carbocation. ( c) Same ame exp expllanat anatiion as in (b). ( d ) Positive Positive charge charge is present present on the the two adjacent adjacent atoms, atoms, leadin leading g to destabil destability ity of the the carbocati carbocation. on. Remember Remember a weak base favours substitution, substitution, while a strong base favours elimination. elimination. ( i) HS– is a powerful nucleophile which reacts rapidly with the alkyl halide to form mercaptan CH3 CHBrCH3 + HS–

( ii)

I–

CH 3OH →  →

CH3 CHSHCH3

is a powerful nucleophile but weaker base and reacts by S

group than

Br– , it reacts w ith the nucleophilic nucleophilic

N

1

mechanism to form Me 3 CI, but again iodide is a better leaving

solvent HCOOH to form formate

O

(CH3 ) 3 CBr + I– ( iii) ( iv )

 → – (– Br )



HCOOH

(CH3 ) 3 CI    → (CH3 )3C—O— C —H

AgCN is an ambi ambident dent nucleop nucleophil hile e and hence, hence, it will will form form two products. products. CH3 CH2 Br + AgCN   →CH3 CH2 CN + CH 3 CH2 NC Since Since S provides provides a more powerful nucleophi nucleophilic lic site site than O, alkyl alkyl group will be be linked linked to S. S.

O CH3CH2Br +

—

S—S —S—O —O —

2–

O

O –

(v )

MeCO

–

–

CH 2= CH 2

—

CH—S—S—O —O →  → CH3 CH—S—S 2

– Cl : 3 CHCl3 → [: CCl 3] → CCl2 → H 2C—– —–C CH2

—

O

–

















2 14 9.

(a)

S

N

2

Reactivity follows the order

CH3 (CH2 ) 3 CH2 Br > 1° (b)

(CH3 ) 2 C(Br)C2 H 5 3°

Ag+ catalyzes S N 1 reactivity and thus the order is (CH3 ) 2 C(Br)C 2 H5 > 3°

10.

(C2 H5 ) 2 CHBr > 2° (C 2 H5 ) 2 CHBr > 2°

CH3 (CH 2 ) 3 CH2 Br 1°

+

(i)

Br– is an extremely w eak Bronsted base, hence it can’t displace the strong base OH– . However, in presence of acid, RO H2 is first

(ii)

formed. Now Br– displaces H 2 O which is a very weak base and a good leaving group. The bulky (CH3 ) 3 C group sterically hinders backside backside attack by a nucleophile. nucleophile.

(iii)

Solvo olvoly lysi siss goe goess by by an SN1 mechanism. Thus relative rates of different reactants in S N 1 reactions depend on the stabilities of the +

+

intermediate carbonium ions. CH 2 = CH 2 CH2 Cl is more reactive because CH 2 = CH— C H2 is more stable than (CH 3 ) 3 C . (iv )

CH3 CH2 Br and (CH 3 ) 2CHBr react by SN2 pathway in which reactivity of the latter halide is very less because of steric hindrance. +

However, (CH3 )3 CBr reacts by S N1 pathway which involves formation formation of carbocation, (CH 3 ) 3 C formation is very rapid because of steric acceleration as well as inductive effect. (v )

KI,fast

slow

ROH + Cl– ← RCl + H2 O → ROH + Cl– I– is a powerful nucleophile which reacts rapidly with RCl to form RI. Further, I – is also a better leaving group than Cl – , and RI is therefore hydrolysed rapidly to form ROH ROH and r egenerate I– , which recycles recycles in the the reaction Slow RCl + HO ROH → 2 –

I fast

RI

HO, 2 fast – –I

(v i)

This This is an E2 reaction which which involves involves the cleavage of C—H (or C—D) bond bond in the rate determining determining step. Since Since C—H bond bond is broken broken at a higher rate than the stronger C—D bond, formation of ethylene is easy in CH 3 CH2I than in CD3 CH2 I. This ratio of the rate constants, KH / KD is called isotope effect. effect. (vii ) Since SN1 react ions involve involve the formation of carbocations and primary carbocations (from primary a lkyl halides) halides) are least stable, 1° RCl are least reactive towards SN1 solvolysis. However, in case of CH 3 CH2 OCH2 Cl, the carbocation formed has a lone pair of electrons on an atom (O) adjacent to C + , hence delocalisation delocalisation (by p-p overlap) of the positive charge stabilises the carbocation. Therefore, Therefore, the compound shows SN1 reactivity : + C2H 5O H CHCHOCHCl CHCH—O—CH → – 3 2 2 3 2 2 → C H3C H2— O — C H 2 OC 2 H5 : (– Cl) Delocalisation of positive charge possible

11.

(a) (b) (c) (d ) (e)

SN2 displacement, I– is a good nucleophile and a poor base. E2 elim elimin inati ation on,, a 3° halid halide e and a fair fairly ly stron strong g base. base. Mainly SN2 displacement. Mainl ainly y E2. A les lesss pola polarr solv solven entt than than H 2 O (in c) favours E2. SN1 displacement ; H 2 O is not basic enough to remove a proton to ca use elimination. elimination.

12.

(a)

(CH3 ) 3 CBr   →(CH3 ) 3 C

(b)

+

CH3 CH = CHCl

C 2 H5 OH  →

(CH 3 ) 3 COC2 H 5 + Ma jo jo r( r( SN 1 )

(CH 3 ) 2 C = CH 2

Veryminor(E1)in absenceofastrongbase

NaNH 2   → CH 3C ≡ CH E2

Vinyl halides are quite inert toward SN2 reactions. (CH3 ) 3 CI + H 2 O   →(CH3 ) 3 COH + HI

(c)

(SN 2)

In presence of a nucleophilic solvent and in absence of a strong base, 3° RX undergoes S (d )

(CH3 ) 3 CI +

OH–   →(CH

3)2C

= CH2 + H 2

O + I–

(E2)

In presence of a strong base (OH– ), 3° RX undergoes mainly E2 reaction.

CH 2OH 13.

CH 2OH >

CH 2OH >

CH 2OH >

N

1

solvolysis.



















2 15 14.

There are seven isomeric pentyl alcohols, C 5 H11 OH, of which four four are pr imary, two secondary a nd one tertiary. Primary alcohols follow follow SN2 pathway and hence isomer having bulky alkyl group will be less reactive. Thus

CHCHCHCH—CHOH > (CH)CHCH—CHOH > 3 2 2 2 2 3 2 2 2

15.

CH3

CH—C CH —CH H 2OH > (CH)C—CHOH 3 3 2 C2H 5 More the stability of the carbocation, higher will be SN1 reactivity of the parent compound. Br

Br Br

Br

Br

Corresponding cations are +

CH2

+

+

+ +

Benzyl carbocation

16.

3°

2°

Vinyl

Stabili Stability ty of car bocation (Intermediate) determines the reactivity of alkene towards addition of HCl ; more stable the carbocation more will be the reactivity of alkene. Here carbocations of both alkenes are given along with their stability. ⊕ ⊕ ( i) CH—CH and an d 3 2 (1° Car bo bocation)

⊕ ( ii) ( iii)

(2° Ca Carbocation) Mo More st stable

⊕

and

2° Carbocation

3° Carbocation (More stable)

⊕

⊕

and

(2° Carbocation) 2° Car bocation, with conjugation (More (More stable)

( iv )

⊕

⊕

and an d

(2° Carbocation)

2° Carbocation, having conjugated double bond (More stable) +

17.

1°

CH2

CH

( i)

–

H →

Cl +

Cl ⊕

+

H

→

( ii)

–

Cl

→

Cl

3° Carbocation +

( iii)

–

H→

Cl →

+

Cl

3° Carbocation

C6H5 ( iv )

H

C6H5

+

→

+ Benzylicc carbocation Benzyli

Cl

–

→  →

C6H5 Cl

















2 16 20.

(i)

Prima Primary ry alco alcoho hols ls main mainly ly unde undergo rgo S N 2 reactions vi a the formation of a transition state which can form two products. –

Cl H

+

OH

CH2

–

→

Cl

→

+ OH2

+

OH2

–

—Cl — H2 O

—H 2 O

–

Cl

ClCH2 (ii)

+

→  

Cl

CH2

Secon econdar dary y alco alcoho holls unde undergo rgo SN1 mechanism. +

H →

OH OCH OC H3 Br

NH2

Cl

→

2° Carbocation

OCH OC H3

NaNH 2 → NH3

(iii)

→

OH2

OCH OC H3

–

+

– H2 O

+

Cl

OCH OC H3 NH2

–

→

+ NH2 Benzyne

21.

(a)

(b)

IV

conjugated allylic

>

III

3ºal 3ºally lyli lic, c,hi high ghly ly conjugated

..

(c) (d)

II

allylic

>

> III > VI V I > VI V II > 3º

II

2º

2ºal 2ºally lyli lic, c,hi high ghlly conjugated

..

1º

>

I

> V

vinylic

I

2ºall allylic ylic

..

(C 6 H5 )2 C > C6 H5 C H > C H 3 C H > : CH 2 Yes, more is the the stabili stability ty of a free radical, radical, weaker will will be the the parent bond. bond. Thus write write down the structures structures of the correspondi corresponding ng free radical formed and observe the relative stability of the free radical.

H

H

.

. Radical formed by a type C–H bond (I)

H

Radical formed by b type C–H bond (II)

II

Thus bond energy of the a , b and c C – H bonds :

> II III > 2º

I

.

Radical formed by c type C–H bond (III)

Relative stability of the free radical : allylic

H

vinylic

Reaction Mechanism

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