RESONANCE Haloalkanes & Haloarenes

LECTURE NOTES Session : 2012-13

ORGANIC CHEMISTRY TOPIC :

HALOALKANES & HALOARENES

CONTENTS :

1 2. 3.

Nucleophile Leaving Group Nucleophilic substitution (SN1) of RX. ROH, ROR

Refer sheet :- Reaction mechanism JEE Syllabus [2009] Alkyl halides : Rearrangement reactions of alkyl carbocation, nucleophilic substitution reactions; preparation of alkenes by elimination reactions; Alcohols : Dehydration reaction, reaction with phosphorus halides, ZnCl2/conc HCl, Ethers : Preparation by Williamson’s Synthesis;

1. NUCLEOPHILE & NUCLEOPHILICITY 1.

NUCLEOPHILE : It is the e– rich species having atleast one lone pair of electrons. It can be neutral or –vely charged it is always a lewis base. BASICITY : It is the tendency to donate e– pair to ion. All bases are nucleophiles, but all nucleophlies are not base all the time.

(i)

(ii)

1.1

NUCLEOPHILICITY : The tendency to give e– pair to an electron deficient carbon atom is defined as nucleophilicity. (i) Criteria for Nucleophilicity : 1. The factors which increases e– density at ‘donor atom increases nucleophilicity’. 2. The more polarisable donar atom is a better nucleophile. Therefore, large size of donor atom increases nucleophilicity. (ii) Periodicity : Nucleophilicity decreases from left to right in a period.

 In a group, nucleophilicity increases from top to bottom due to increases in size of donor atom, but basicity decreases from top to bottom. Acid strength : HI > HBr > HCl > HF Basic strength : F¯ > Cl– > Br– > I– Nucleophilicity : F¯ < Cl– < Br– < I– (iii) Steric effects on nucleophilicity

Stronger base, yet weaker nucleophile cannot approach the carbon atom so easily. (iv) The effect of the solvent  In polar protic solvent large nucleophiles are good, and the halide ions show the following order

In DMSO, the relative order of reactivity of halide ions is >

>

>

Page # 2

This effect is related to the strength of the interaction between nucleophile and solvent molecules of polar protic solvent forms hydrogen bond to nucleophiles in the following manner :

(v) Relative nucleophilicity in polar protic solvent >

(vi)

>

>

>

>

>

> H2 O

Comparative chart Nucleophilicity

Basicity

Remarks

1.

CH3¯ > NH2¯ > OH¯ > F¯

CH3¯ > NH2¯ > OH¯ > F¯

2.

SiH3¯ > PH2¯ > SH¯ > Cl¯

SiH3¯ > PH2¯ > SH¯ > Cl¯

3.

F¯< Cl¯ < Br¯ < I¯

F¯> Cl¯ > Br¯ > I¯

In a group nucleophilicity increases while basicity decreases. on moving top to bottom.

4.

OH¯ < SH¯

OH¯ > SH¯

,,

5.

RO¯ < RS¯

RO¯ > RS¯

,,

6.

RO¯ > HO¯

Same

If donor atom is same, nucleophilicity and basicity have same order

7.

RCOO– < PhO¯< HO¯ < RO¯

Same

,,

Same

,,

8.

>

If donor atomis belong to same period, nuclephilicity and basicity order is same

Tetrahedral (more extensive resonance) [d–d bonds] 9.

HO¯ > H2O NH2¯ > NH3

Same

CF3SO3¯ < PhCOO¯ < PhO¯ < RO¯ (Nucleophilicity/Basicity)

Page # 3

2. LEAVING GROUP ABILITY 2.

Leaving group Ability/Nucleofugality : The best leaving groups are those that become the most stable ion after they leave, because leaving group generally leave as a negative ion, so those leaving group are good, which stabilise negative charge most effectively and weak base do this best, so weaker bases are always good leaving groups. A good leaving group always stabilize the transition state and lowers its free energy of activation and there by increases the rate of the reaction. (a) Order of leaving ability of halide ion

>

>

>

(b) Other good leaving groups are

(c) Strongly basic ions rarely act as leaving group  (strong base / poor leaving group) (It is not a leaving group)

(d) The weaker bases are better leaving groups. (e) The leaving group should have lower bond energy with carbon. (f) Negative charge should be more stable either by dispersal or declocalization. (g) Leaving group ability : 1.

CH3¯ < NH2¯ < OH¯ < F¯

2.

R–COO¯ > PhO¯ > HO¯ > RO¯

3.

SH¯ > OH¯

4.

>

Page # 4

2.1.

Types of solvents (a) Non polar

(b) Polar (These solvents are of two type - polar protic and polar aprotic) Solvents

Polar

Protic

Aprotic

1.

H2O

2.

CH3OH

3.

CH3CH2OH

4.

H–COOH

5.

CH3–COOH

6.

NH3

     

     

– – – – – –

7.



×



8.



×



DMSO – Dimethyl sulphoxide

9.



×



DMF

– Dimethyl formamide

10.



×



DMA

– Dimethyl acetamide

11.



×



12.



×





×



13.

C–C–C–C–C–C

Page # 5

3. NUCLEOPHILIC SUBSTITUTION REACTION 3.

Nucleophilic substitution reaction. General reaction (RX + Nu¯  RNu + X¯)

Unimolecular nucleophilic substitution reaction (SN1) : Nucleophilic substitution which involves two step process (a) First step : - Slow step involves ionisation (to form carbocation) R–g  R+ + g – (b) Second step : - Fast attack of nucleophile on carbocation to result into product . R+ + Nu –  R–Nu

3.1

SN1 Reaction of alkyl halide Mechanism : Ionisation of alkyl halide      Slow step (rds)

# Carbocation intermediate is formed so rearrangement is possible in SN1 reaction.

Kinetics: Rate  [Alkyl halide] # It is unimolecular, two step process. # It is first order reaction. # Rate of SN1 reaction is independent of the concentration and reactivity of nucleophile.

Energetics of the SN1 :

Figure : Free energy diagram for the SN1 reaction.

Stereochemistry of SN1 reactions : In the SN1 mechanism, the carbocation intermediate is sp2 hybridized and planar, A nucleophile can attack on the carbocation from either face, if reactant is chiral than after attack of nucleophile from both faces gives both enantiomers as the product, which is called racemization. Mechanism of racemization (SN1  

Ion-pair concept : - % inversion > % retention So 100 % racemisation is not occur Page # 6

Factor's affecting the rates of SN1 / Reactivity order : (i) The structure of the substrate : The Rds of the SN1 reaction is ionization step, a carbocation is form in this step. This ionisation is strongly endothermic process, rate of SN1 reaction depends strongly on carbocation stability because carbocation is the intermediate of SN1 reaction which determines the energy of activation of the reaction. SN1 reactivity : 3° > 2° > 1° > CH3 – X (ii) Concentration and reactivity of the nucleophile  The rate of SN1 reactions are unaffected by the concentration and nature of the nucleophile #

(iii) Nature of nucleophile:- Weak, neutral, protic solvents In SN1 reaction, mostly the solvent functions as nucleophile with oxygen & nitrogen as donor atom. So SN1 reaction are termed as solvolysis reaction. H2O  hydrolysis C2H5OH  ethanolysis CH3COOH  acetolysis NH3  ammonolysis SN1 reaction is disfavoured by strong, anionic nucleophiles which attacks faster before ionisation takes place leading to SN2 mechanism.

–

+ – H  R – OH + HX R – O – H   H (iv) Effect of the solvent : the ionizing ability of the solvent : Because to solvate cations and anions so effectively the use of a polar protic solvent will greatly increase the rate of ionization of an alkyl halide in any SN1 reaction. It does this because solvation stabilizes the transition state leading to the intermediate carbocation and halide ion more than it does the reactant, thus the energy of activation is lower.

OH2 R – X + H2O  R+

R – X

(Solvolysis)

Solvated ions (v) The nature of the leaving group  In the SN1 reaction the leaving group begins to acquire a negative charge as the transition state is reached stabilisation of this developing negative charge at the leaving group stabilizes the transition state and ; this lowers the free energy of activation and thereby increases the rate of reaction. leaving ability of halogen is

>

>

> >

Examples : 1.

2.

3.

H2 O acetone   

H O / Ag

2   

CH OH / Ag

3    

+ HBr

(3° alkyl halide hence no SN2 reaction)

+ AgI 

(carbocation rearrangement)

Page # 7

4. 5.

Flourine exchange reaction : (Swart’s reaction) AgF/ H2O (Major) + C2H5OH (Minor)

S 1 + F N R – X + Ag–F  s)  R – F + AgX  ppt + (R – OH + R–OC2H5 minor products)   R  Note : Only AgF is soluble among all silver halides  To prepare alkyl flouride 

6

7.

Br2 / h AgF / H2 O      

H O  acetone

2    

H shift

 

8.

9.

10.

11.

12.



MeOH / Ag   

[Diastereomers](Not recemic mix []  0)

Page # 8

13.

Q.1

Give the solvolysis products when each compound is heated in ethanol

(a)

(b)

(c)

(d)

Sol.

(a)

(b)

(c)

(d)

Q.2

Predict the compound in each pair that will undergo solvolysis (in aqueous ethanol) more rapidly.

Sol.

(a) II > I (d) II > I

Q.3

The rate of SN1 reaction is fastest with

Ans.

(b) II > I (e) II > I

(c) I > II

(A)

(B)

(C)

(D)

A>B>D>C

Page # 9

SN1’

(’ - Prime)

The substitution of alkylic system under SN1 conditions proceeds with rearrangement. 

–

–X R–CH=CH–CH2–X   R –CH=CH– C H2



R– C H –CH=CH2

R–CH=CH–CH2–Y

3.2

SN1 Reaction of Alcohols

(A)

Reaction with hydrogen halides A common method is to treat the alcohol with a hydrohalic acid, usually HI or HBr. These acids are used to convert alcohols to the corresponding alkyl halides. (i) In acidic solution, an alcohol is in equilibrium with its protonated form. Protonation converts the hydroxy group from a poor leaving group to a good leaving group (H2O). If the alcohol is protonated all the usual substitution and elimination reactions are feasible, depending on the structure (1°, 2°, 3°) of the alcohol. (ii) Halides are anions of strong acids, so they are weak bases. Solutions of HBr and HI contain nucleophilic

and

ions.

(iii) Concentrated hydrobromic acid rapidly converts t-Butyl alcohol to t-Butyl bromide. The strong acid protonates the hydroxyl group, converting it to a good leaving group. The hindered tertiary carbon atom cannot undergo SN2 displacement, but it can ionise to a tertiary carbocation. Attack by bromide ion gives the alkyl bromide. The mechanism is similar to other SN1 mechanism. (iv) 1-Butanol reacts with sodium bromide in concentrated sulfuric acid to give 1-Bromobutane by an SN2 displacement.

CH3 (CH2 )2  CH2 OH 1  bu tan ol

NaBr, H SO

2 4       

CH3 (CH2 )2  CH2Br 1  bromobutane (90%)

Protonation converts the hydroxy group to a good leaving group, but ionization to a primary carbocation is unfavourable. The protonated unbranched primary alcohol is well suited for the SN2 displacement. (v) Secondary alcohols also react with HBr to form alkyl bromides usually by the SN1 mechanism.

e.g.

HBr  

(vi) HCl (Hydrochloric acid) reacts with alcohols in much the same way that as the hydrobromic acid. (vii) Chloride ion is a weaker nucleophlile than bromide ion because it is smaller and less polarizable. Lewis acid, such as ZnCl2, is sometimes necessary to promote the reaction of HCl with primary and secondary alcohols.

Lucas Reagent (i) A mixture of concentrated hydrochloric acid and anhydrous zinc chloride is called the Lucas reagent. (ii) Whether an alcohol is primary, secondary or tertiary i.e. identify by the Lucas test, which is based upon the difference in reactivity of the three classes of alcohol towards hydrogen halides. (iii) Alcohol (of not more than six carbons in their molecule) are soluble in the Lucas reagent. The corresponding alkyl chlorides are insoluble.

Page # 10

(iv) Formation of a chloride from an alcohol is indicated by the cloudiness that appears when the chloride separates from the solution hence, the time required for cloudiness to appear is a measure of the reactivity of the alcohol. (v) A tertiary alcohol reacts immediately with the Lucas reagent, a secondary alcohol reacts within five minutes and a primary alcohol does not react appreciably at room temperature. e.g.

anhy . ZnCl / HCl

2       

CH3  CH2  CH  CH2  CH3 | OH

CH3  CH2  CH  CH2  CH3 | Cl 3  Chloropentane

Mechanism : R –OH

e.g.

e.g.



Reactivity of

HX :

Reactivity of

ROH :

CH3 CHCH3 | OH Isopropyl alcohol

CH3 | CH3  C  CH2 – OH | CH3



R  O H2   R –H2O

R – X

HI > HBr > HCl allyl  benzyl > 3° > 2° > 1°

CH3 CHCH3 | Br Isopropyl bromide

CH3 | CH3  C  CH2 – CH3 | Br

Example :

Ex.1

Ex.2

Page # 11

Ex.3

CH3 – CH = CH – CH2OH

HBr

Ex.4



HI

Ex.5

 

Ex.6

Why is ZnCl2 (Lewis acid) required with HCl in its reaction with 1° and 2° alcohol. Where as reaction with HBr proceeds without any Lewis acid.

Sol.

Because the chloride ion is a weaker nucleophile than bromide or iodide ions. HCl does not react with 1° or 2° alcohols unless zinc chloride or some similar Lewis acid is added to the reaction mixture as well. ZnCl2 is a good Lewis acid, forms a complex with alcohol through association with an unshared pair of electrons on the oxygen atom. This provides a better leaving group for the reaction than H2O.

Ex.7

Make distinction between following pairs of substances by using Lucas reagent

Ans.

(a) II > I

3.3

SN1 Reactions of Ethers

(A)

Reaction with HX

(b) II > I

(c) II > I

Ethers are unreactive towards most bases, but they can react under acidic conditions. A protonated ether can undergo substitution or elimination with the expulsion of an alcohol. Ethers react with conc. HBr and HI because these reagents are sufficiently acidic to protonate the ether, while bromide iodide

Page # 12

are good nucleophiles for the substitution. If R or R’ is 3º then mechanism will be SN1 otherwise SN2 . Mechanism : R – O – R`

X (3º)  R` – X

   – ROH



e.g.

HCl (CH3)3COC(CH3)3  (CH3 )3 C O C(CH3 )3 | H



(CH3 )3 C

+ (CH3)3COH

(CH3)3CCl



(CH3)3COH + HCl  (CH3)3CCl + H2O

HBr (1eq.)

e.g.



HBr (1eq.)

e.g.

(B)



Reaction with H3O+ 

H3O R–O–R`   R–OH + R`–OH



H R–O–R`  

e.g.

H O

3

e.g.

H O

3

Page # 13

LECTURE NOTES Session - 2009-10

ORGANIC CHEMISTRY TOPIC :

REACTION MECHANISM

CONTENTS :

4. 5. 6. 7.

Nucleophilic substitution (SN2) /RX. ROH, ROR Elimination Reaction (E1)/ RX. ROH, ROR Elimination Reaction (E2)/ RX. ROH, ROR Elimination Reaction (E1cB)

Refer sheet Reaction mechanism JEE Syllabus [2009] Alkyl halides : Rearrangement reactions of alkyl carbocation, nucleophilic substitution reactions; preparation of alkenes by elimination reactions; Alcohols : Dehydration reaction, reaction with phosphorus halides, ZnCl2/conc HCl, Ethers : Preparation by Williamson’s Synthesis;

4. NUCLEOPHILIC SUBSTITUTION REACTION (SN2) 4.1

SN2 Reaction of alkyl halide : Mechanism :



+

# No intermediates are formed in the SN2 reaction, the reaction proceeds through the formation of an unstable arrangment of atoms or group called transition state. Kinetics : rate  [alkyl halide] [nucleophile] rate = k[alkyl halide] [nucleophile] # It is bimolecular, one step concerted process # It is second order reaction because in the RDS both species are involved

Energetics of the reaction :

Figure : A free energy diagrams for SN2 reaction

Stereochemistry of SN2 reactions : As we seen earlier, in an SN2 mechanism the nucleophile attacks from the back side, that is from the side directly opposite to the leaving group. This mode of attack causes an inversion of configuration at the carbon atom. This inversion is also known as Walden inversion.

Inversion   

Factor's affecting the rate of SN2 reaction/Reactivity order   (i) Effect of the structure of the substrate  Order of reactivity in SN2 reaction : – CH3 > 1° > 2° >> 3° (unreactive) The important reason behind this order of reactivity is a steric effect. Very large and bulky groups can often hinder the formation of the required transition state and crowding raises the energy of the transition state and slows down reaction. (ii) Concentration and reactivity of the nucleophile As nucleophilicity of nucleophile increases rate of SN2 increases. – Anionic nucleophiles mostly give SN2 reaction Page # 15

– A stronger nucleophile attacks upon -carbon with faster rate than the rate of departing of leaving group. R– > NH2– > OH– > F– RO– > ROH NaOH > H2O NH3 > H2O

Table :

(iii) The effect of the solvent: Polar aprotic solvent have crowded positive centre, so they do not solvate the anion appreciably therefore the rate of SN2 reactions increased when they are carried out in polar aprotic solvent. (iv) The nature of the leaving group  Weaker bases are good leaving groups. A good leaving group always stabilize the transition state and lowers its free energy of activation and thereby increases the rate of the reaction. Order of leaving ability of halide ion F¯ < Cl¯ < Br¯ < I¯

Examples : aq.KOH  

Ex.1

CH3–Cl

Ex.2

– KOH DMSO CH3–CH–Cl    CH3–CH2–OH + Cl

Ex.3

– NaOH DMF CH3–CH2–CH2–Br     CH3–CH2–CH2–OH + Br

Ex.4

 (CH3)2CHCH2CH2 – OH (CH3)2CHCH2CH2 – Br  

Ex.5

 MeOH 



(:OH ) strong nucleophil e

CH3–OH + Cl–

KOH H2O

Na

PhCH Br

2    

CH2 – O – CH3 Benzylmethyl ether

Ex.6

Page # 16

KOH DMSO   

Ex.7.

2° RX, [D] []  0° = x° (Assumed)

2° ROH, [L] []  0  –x° (optically pure only one enantiomer is obtained)

KOH DMSO   

Ex.8.

KCN  

Ex.9.

Only one optically active product (Walden inversion at carbon 2) Replace nucleophile with halogen directly and make any 1 (only1) change on the -chiral carbon.

Halogen exchange reactions : (Finkelstein Reaction) General reaction

Acetone R – Cl + NaI   R – I + NaCl  ppt in acetone

R – Br + NaI Note:

Acetone   R – I + NaBr  ppt in acetone

In acetone, NaI is soluble / ionised but NaCl/NaBr are insoluble  For I – exchange NaI/acetone is used so that chlorides and bromides get precipitate out CH3

Ex.10. H

CH3 I + NaI*

C6H13 (reactant) [] = xº



Transition state

I



H + NaI C6H13 (product) [] = –xº

Polarimeter can detect difference between H and D, but not isotopes of higher elements. For polarimeter reactants and products are mirror image in the above example. (1) In this reaction, the nucleophile and the LG are the same species (I–). (2) The reaction is an equilibrium reaction and at equilibrium rate of forward reaction = rate of backward reaction. (3) In this reaction the optically pure reactant loses its optical activity and net result is a racemic mixture. (4) Since Nu:- and LG are same species, so the reactant and product are enantiomers. Although it is an SN2 reaction, but racemisation shows there is complete inversion of configuration in forward reaction.

Page # 17

Q.1 Sol.

What will be the retationship between rate of I* exchange and rate of racemisation. 2 × Rate of I* exchange = Rate of racemisation or, Rate of Racemisation = 2 × Rate of reaction.

Q.2

Complete the following reactions with mechanism ?

Na  

Sol.

Q.3.

+ CH3I



KOH

   ? 1 eq.

Sol.

Q.4.

+ Ph – CH2Cl

Sol. Q.5.

is present in excess and is stronger nucleophile than

so product is Ph – CH2 – OEt

CH3 CH2  I Na  X   CH3 – C  CH     Y

Sol.

 CH3  C  C  CH2  CH3

Y

Q.6.

+



salt

Sol. Q.7.

Ans. Q.8.

Ans.

In the given reaction, CH3CH2 – X + CH3SNa  The fastest reaction occurs when ‘X’ is (A) – OH (B) – F (C) – OCOCF3 C Correct decreasing order of reactivity towards SN2 reaction

(D) OCOCH3

CH3CH2

CH3CH2CH2CH2Cl

CH3CH2CH2

(C) IV > I > III > II

(D) II > I > IV > III

CH2Cl

(A) IV > I > II > III B

CH3

CH2CH2Cl

(B) III > II > I > IV

Page # 18

Q.9.

Draw a fischer projection for the product of the following SN2 reaction

(a)

Sol.

NaI / Acetone

(b)

    

(a)



(b)

Intramolecular SN2 reactions rate of intramolecular reaction > intermolecular reaction

NaOH   

(i)

(ii)

Na  

(Wurtz reaction)

ether

int ramolecula r     S 2 reaction N

(iii)

Na  

(iv)

Na  

(v)

Mg / ether   

(vi) Write mechanisms that account for the product of the following reactions : NH2 – CH2 – CH2 – CH2 – CH2 – Br

Sol.



Page # 19

SN2 Mechanism Nucleophilic substitution on allylic system may proceed also via SN2 mechanism, in which the rearrangement does not take place. By the same concepts allylic rearrangement may proceed via an SN2 like mechanism called the SN2’ mechanism. Conditions for SN2’ (1) A nucleophilic attack on the -carbon (2) A Concerted movement of three e– pair occurs.

The reaction is particularly susceptible to steric hinderance especially at the -C. Substrates with C=C–CH2–X shows a strong preference for SN2 while sustrates of the type –C=C–CR2–X favours SN2’ mechanism. OH 

Ex.

  

SH

Ex.

 

The SN21 mechanism has also been demonstrated in propargyl systems, which results in an allene. Ph–CC–CH2–OTs + CH3MgBr 

.

Comparision between SN1 / SN2 reaction : S N2

Characteristics

S N1

1.

Energetic

2.

Kinetics

r  [RX] [NU:]

r  [RX]

3.

Stereochemistry

inversion

racemisation

4.

Rearrangement

not possible

possible

5.

Activation energy

Ea S

less

6.

Nature of R – X

Me – X > R CH2 X > R2 CH X > R3CX

R3CX > R2CHX > RCH2X > CH3X

7.

Nucleophile

strong anionic

Weak neutral

N2

> Ea SN1

R > NH2 > OR > OH –

–

–

–

H2O > MeOH > EtOH > NH3

8.

Leaving group

– > Br– > Cl– > F–

(same)

9.

Solvent

Polar aprotic

Polar protic

10.

Temperature

High

Low

Page # 20

Example : +

1.



2.

3.

4.

4.2

SN2 Reaction of Alcohol

(A)

Reaction with HX : The protonated  unbranched primary alcohol is well suited for the SN2 reaction. Mechanism : R –OH

e.g

 X R  O H2  R – X + H2O

CH3 CH2 CH2 CH2CH2OH n  Pentyl alcohol

CH3 CH2 CH2 CH2 CH2Cl n  Pentyl chloride

Page # 21

(B)

Reaction with phosphorus trihalides Several phosphorus halides are useful for converting alcohols to alkyl halides. PBr3, PCl3, & PCl5 work well and are commercially available. Phosphorus halides produce good yields of most primary and secondary alkyl halides, but none works well with tertiary. alcohols. The two phosphorus halides used most often are PBr3 and the P4/I2 combination. 3R – OH + PX3

3R – X + H3PO3

Mechanism : Step : 1

Step : 2

RCH2X+ HOPX2

e.g.

CH3 | CH3 – CH2 – CH – CH2 – OH 2  Methyl  1  bu tan ol

e.g.

CH3 CH2 OH Ethyl alcohol

P  I2   

PBr

3   

CH3 | CH3 – CH2 – CH – CH2 – Br 2  Methyl  1  bromobutane

CH3 CH2 I Ethyl iodide

e.g.

e.g.

(C)

Reaction with thionyl chloride in presence of pyridine Thionyl chloride (SOCl2) is often the best reagent for converting an alcohol to an alkyl chloride. The by products (gaseous SO2 and HCl) leave the reaction mixture and ensure that there can be no reverse reaction.

O || R – OH + Cl  S  Cl

Heat

   Pyridine

R – Cl + SO2 + HCl

Mechanism :



.. ..  R – O .. – S – H

.. O:

+ HCl

Cl Chlorosulphite ester

 R – Cl + SO2 In the first step, the nonbonding electrons of the hydroxy oxygen atom attack the electrophilic sulphur atom of thionyl chloride. A chloride ion is expelled a proton and gives test of chloro sulphite ester. Second step is an SN2 mechanism

Page # 22

SOCl 2     Py

e.g.

(D)

Reaction with thionyl chloride ROH + SOCl2  RCl + SO2 + HCl In this mechanism an internal nucleophile attacks from the same side of leaving group , means retension of configuration . It is an SNi mechanism , where i means internal Mechanism :

.. ..  R – O .. – S – H



.. O:

+ HCl

Cl Chlorosulphite ester

 R – Cl + SO2

Cl

H C CH3(CH2)4CH2

e.g.

CH3

(R) 2-Chlorooctane 84%

4.3

SN2 Reaction of Ether

(A)

Reaction with HX Ethers are unreactive toward most bases, but they can react under acidic conditions. A protonated ether can undergo substitution reaction. Ether react with conc. HBr and HI because these reagents are sufficiently acidic to protonate the ether. If R or R’ is 3º then mechanism will be SN1 otherwise SN2. Mechanism : +



X R

+

HX

 X – R + X – R

alkyl halide

e.g.

HBr (i) CH3CH2OCH2CH3  2CH3 – CH2Br excess

Mech. Br |  CH3  C  H | H

+

ethyl bromide

CH3CH2Br Ex.1

Ex.2

CH3CH2 – O – CH3

H   

excess HBr

   

Page # 23

Ex.3

Ex.4

Ex.5

Ph

– –

CH3 HBr 2 CH3 – C – O – CH2 – CH3   CH3 – C – OH + CH3 – CH2 – Br (S N ) major H H Me Me Me Me

– –

CH3

HBr H   1 (SN )

O D

Et 2

3

Ex.6 1

4.4

O

4

D (Retention) 2

HI   1 (S N ) I

H + Ph

OH

1

Me

Br + Br

Et (Racemisation)

Ph Et

4 3

OH

SN Reaction of Epoxide

Epoxides are much more reactive than ether because of angle strain in three membered ring therefore epoxide readily undergo nucleophilic substitution reaction. (a) In basic medium mechanism is SN2. Nucleophile atacks on less hindered carbon. Mechanism : Nu | R — CH — CH2 | O

Nu |   R — CH — CH2 | OH H

e.g. (b) In acidic medium mechanism is SN1 type. Nucleophilic attacks on more substituted carbon. Mechanism : 

H  

e.g.

Nu | R — CH — CH2 | OH

H O18

2   H

e.g.

Page # 24

5. ELIMINATION REACTIONS In an elimination reaction two atoms or groups (YZ) are removed from the substrate and generally resulting into formation of  bond.

| | CC | | Y Z

E lim ination      YZ

-elimination : When two groups are lost from the same carbon atom to give a carbene (or nitrene). This is also called 1–1 elimination.  C–C: – XY

-elimination : When two groups are lost from adjacent atoms so that a  bond is formed. This is also called 1–2 elimination.  C=C

–elimination : It is also called 1–3 elimination, In this a three membered ring is formed. 

E1 Reaction : Proton and leaving group depart in two different step. (a) First step : - Slow step involves ionisation to form carbocation (b) Second step : Abstraction of proton

5.1

E1 Reaction of alkyl halide : Mechanism : Step 1 : Formation of the carbocation (RDS)

Step 2 : Base (

) abstracts a proton (fast) + B – H

# In the second step, a base abstracts a proton from the carbon atom adjacent to the carbocation, and forms alkene. # Reaction intermediate is carbocation, so rearrangment is possible

Kinetics  Rate  [Alkylhalide] Rate = k [Alkylhalide] # It is unimolecular, two step process. # It is a first order reaction.

Page # 25

Energetics  The free energy diagram for the E1 reaction is similar to that for the SN1 reaction.

Reactivity Order : # Similar to SN1 becuase Carbocation Intermediate is formed in the rds step.

SN1 vs E1 : In case of alkyl halide SN1 product is always more than E1 product

Example :

Ex.1

CH3 | CH3  CH2  C  CH3 | Br

Ex.2

+

and

% yield I > II > III

[JEE 2000]

Note : The minor product formed before the carbocation rearrangement is not consider. Ex.3

Propose mechanisms to account for the four products shown below

CH OH, 

+

3    

Sol.

CH OH

3   

+

+



or

Page # 26

In the above example : SN1 > E1

# #

Et – OH is not a stronge base, not strong enough to break C-H bond and hence attack at positive charge site. In such reactions, SN1 is major and E1 minor so, alkenes are not prepared by this method. water or MeOH do not cause E1 (very weak base)

Ex.4

(X) + (Y)

(W) + (Z)

+ve iodoform test given by both W & Z. Sol.

Rearranged Carbocation :

Y  (d/) mixture of

X 

and

5.2

E1 Reaction of Alcohol Dehydration requires an acidic catalyst to protonate the hydroxyl group of the alcohol and convert it to a good leaving group. Loss of water, followed by loss of proton, gives the alkene. An equilibrium is established between reactants and products. For E1 mechanism reagents are (i) H3PO4/  (ii) H2SO4 / 160º

| | | | acid  C  C     C  C  | | H OH Mechanism :

+ H2 O

(Rearrangement may occur)

Step 1 :

Step 2 :

Step 3

+

+

Page # 27

Remarks In first step, an acid-base reaction a proton is rapidly transferred from the acid to one of the unshared electon pairs of the alcohol. In second step the carbon oxygen bond breaks. The leaving group is a water molecule : Finally,in third step the carbocation transfers a proton to a molecule of water. The result is the formation of a hydronium ion and an alkene.

Reactivity of ROH : 3° > 2° > 1° Examples

CH2

CH2OH

Ex.1

Conc.H2SO 4     

+

(I) Minor

Ex.2 Ex.3

Ex.4

Ex.5

+

+

Ex.6

Ex.7

Ex.8

Ex.9

H SO , 

4 2   

Ex.10

Page # 28

Pinacol-Pinacolone rearrangement

Mechanism

Examples : H SO / 

4 2 

Ex.1

Ex.3

5.3

H SO / 

4 2 

Ex.2

+

Ex.4

E1 Reaction of Ether : Elimination is not a favourable reaction for ether, but however few reactions have been observed. E1 Elimination takes place via formation of stable carbocation. Ether undergoes dehydration reaction in the presence of conc. H2SO4 /  Conc.H 2 SO 4 /       

e.g.

Ethers dissolve in concentrated solutions of strong inorganic acids to from oxonium salts, i.e. ether behave as bronsted Lowry bases. +

-



(R – O – R) HSO4  R – OH + R  O  SO 2 OH alkyl hydrogen | sulphate H

R2O + H2SO4

H

 Ex.1 CH3 CH2 – O – CH2 CH3 

2

3



H SO 

4 2  

Ex.2 1

O

4

Page # 29

6. ELIMINATION REACTION (E2) 6.1

E2 Reaction of alkyl halide : Dehydrohalogenation is the elimination of a hydrogen and a halogen from an alkyl halide to form an alkene. Dehydrohalogenation can take place by E1 and E2 mechanism. Reagents : (i) Hot alcoholic solution of KOH or EtO¯ / EtOH

(ii) NaNH2

(iii) t-BuO¯ K



in t-BuOH

Mechanism :





+ BH

Kinetics : Rate  [R – X] [Base] ; Rate = k [R – X] [ ] # This is a single step, bimolecular reaction # It is a second order reaction # Rearrangment is not possible # For the lower energy of activation, transition state must be stable # E2 follows a concerted mechanism # The orientation of proton & leaving group should be antiperiplanar. # Here  – H is eliminated by base hence called  elimination

Saytzeff rule (Positional orientation of elimination) : In most E1 and E2 eliminations where there are two or more possible elimination products, the product with the most highly substituted double bond will predominate. This is called the saytzeff or zaitsev rule.

Energetics : P.E. diagram for dehydrohalogenation of 2-bromo-2-methylbutane.

Reactivity order: R – I > R – Br > R – Cl > R – F Page # 30

Examples :

e.g.

+

CH3 | e.g. + H2 O + CH3  C  Br +  | CH3 Formation of the Hoffmann product Bulky bases can also accomplish dehydrohalogenations that do not follow the saytzeff rule. Due to steric hindrance, a bulky base abstracts a less hindered proton, often the one that leads to formation of the least substituted product, called the Hoffmann product. H CH3 | | CH3  C  C  CH2 | | | H Br H

+

+

Stereochemistry of E2 Reaction : 1. It is a stereospecific reaction. 2. In the T.S., the L.G. and  hydrogen are anti-parallel (or, anti-periplanar) to each other. 3. In the anti conformation, the T.S. is most stable due to minimum electronic repulsion. 4. So the product alkene of E2 reaction has one of the two possible orientations (i.e., either cis or trans) 5. Stereospecificity is observed only when one stereoisomer of reactant R – X is used. 6. Stereospecific reaction means the one stereoisomer gives one stereoisomer as a product 7. In the T.S., 5 atoms lie in the same plane

(

,  – H,  – C, -C, g.)

Plane containing 5 atoms. T.S. (E2 reaction) Base

Ph Ph

H

Ph

H

Ph

CH3

H Br

CH3 H

Br

CH3 Ph

H

Ph

Ph CH3

Trans

Ph

H

Examples : Page # 31

Ex.1

Ex.2

Ex.3

Alc.KOH  No. reaction

Ex.4

+

Ex.5

Ex.7

Ex.6

Alc.KOH



Alc.KOH



Ex.8

H

Ex.9

Page # 32

Ex.10

Comparison of E1 and E2 elimination: Promoting factors

E1

E2

(i) Base

W eak base

Strong base required

(ii) Solvent

Good ionizing solvent

W ide variety of solvent

(iii) Substrate

3° > 2°

3° > 2° > 1°

(iii) Leaving group

Better one required

Better one required

Characteristics (i) Kinetics

K [R – X] Ist order

K [R – X] [Base] IIst order

(ii) Orientation

Saytzeff alkene

Saytzeff alkene

(iii) Stereochemistry

No special geometry is required

transition state must be co-planar



6.2

E1:

 X:  

E2:



1 rds

1



H   R–CH=CH 2

2

 R–CH=CH2

Elimination of alcohol : Reagent : Al2O3, P2O5, ThO2 Cyclic Transition state is formed and rearrangement is not possible

Ex.1

Ex.2

Page # 33

7. E1cB Reaction E1cB Reaction (Unimolecular conjugate base reaction) In the E1cB, H leaves first and then the X. This is a two step process, the intermediate is a carbanion.

Mechanism : 

Step - I : The removal of a proton (H) by a base, generating a carbanion

H | | (Conjugate base) CC X | | Step - II : Carbanion looses a leaving group to form alkene –C  C – | |



Remark : (i) The strong base abstracts B – H (Significantly acidic) in 1st fast rev. step. As a result a carbanion intermediate (cb) is formed. (ii) In the 2nd slow step the leaving group is eliminated. (iii) molecularity = 1, order = 1. (iv) Elcb is unimol . 1st order wrt. cb. Overall order = 2 Kinetics : Rate  Evidence : A:

R:

When the rxn is conducted in and a sample of rxn mixture is tested after some time interval, D - exchange is observed at  – c. (C – D is found) It indicates that a carbanion intermediate is formed and first step is rev.

Factor which make  – H significantly acidic (a)

When the leaving group is strongly electronegative (i.e. exerts strong -  effect on  – H) F,

Note:

(b)

Presence of EWG (– m) at  – c

(c)

poor l.g. favours E1cB

if EWG is attached to  – c, 



then X can be Cl, Br, I, F, OH, OR, N R3 , S R 2 Regioselectivity

in E1cB

(Hofmann elimination)

Ex.1

Page # 34

Ex.2



Ex. 3

 R2C=O

Ex.4

Note :

Hofmann Elimination in Quaternary Ammonium Salts : In Hoffmann degradation of quaternary ammonium salts, Hofmann alkene [less substiuted alkene] is formed as major product

Ex.5

CH2 = CH2 + CH3

1. CH I excess

Ex.6

3 

Ex.7

3 

2. AgOH / 

1. CH I excess 2. AgOH / 

Ex. 8

Ex.9

Page # 35

Faculty Remark Your suggessions / deficiencies are invite regarding this lecture note in detail. please submit to the R & D coordinator (Organic chemistry) S.No. 1.

2.

3.

4.

5.

NOTE : Please submit your valuable suggestions on previous lecture notes

Page # 36