Reaction Mechanism Theory E

CHEMISTRY

REACTION MECHANISM 1.

NUCLEOPHILICITY & LEAVING GROUP ABILITY

1.1

Nucleophiles :

It is the electron rich species having atleast one lone pair of electrons. It can be

neutral or negativetely charged it is always a lewis base. e.g. CN , OH , Br , I , NH3 , H2O –

1.2

Basicity :

–

–

–

It is the tendency to donate electron pair to

ion.

(a)

(b)

1.3

Nucleophilicity :

The tendency to give electron pair to an electron deficient carbon atom is defined

as nucleophilicity.

Criteria for Nucleophilicity : – The factors which increases electron density at donor atom increases nucleophilicity. – The more polarisable donar atom is a better nucleophile. Therefore, large size of donor atom increases nucleophilicity.

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

Steric effects on nucleophilicity

Stronger base, yet weaker nucleophile cannot approach the carbon atom so easily.

Weaker base, yet stronger nucleophile

The effect of the solvent : – In polar protic solvent large nucleophiles are good, and the halide ions show the following order – – – – F < Cl < Br < I – In DMSO, the relative order of nucleophilicity of halide ions is – – – – F > Cl > Br > I

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CHEMISTRY – 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 :

Relative nucleophilicity in polar protic solvent >

e.g.

>

>

>

>

>

> H2 O

More extensive resonance.

e.g.

HO¯ > H2O

Anion are better nucleophile than their neutral molecule

e.g.

NH2¯ > NH3

e.g.

CF3SO3¯ < PhCOO¯ < PhO¯ < RO¯ Nucleophilicity

1.4

>

Basicity

Remarks

1.

CH3¯ > NH2¯ > OH¯ > F¯

CH3¯ > NH2¯ > OH¯ > F¯

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

2.

SiH3¯ > PH2¯ > SH¯ > Cl¯

SiH3¯ > PH2¯ > SH¯ > Cl¯

3.

F¯< Cl¯ < Br¯ < I¯

F¯> Cl¯ > Br¯ > I¯

4.

OH¯ < SH¯

OH¯ > SH¯

––//––

5.

RO¯ < RS¯

RO¯ > RS¯

––//––

6.

RO¯ > HO¯

RO¯ > HO¯

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

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

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. – The leaving group should have lower bond energy with carbon. – Negative charge should be more stable either by dispersal or delocalization. – The weaker bases are better leaving groups. Strongly basic ions rarely act as leaving group  (strong base / poor leaving group)

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CHEMISTRY (It is not a leaving group) Some other leaving groups are

1.5

e.g.

F¯ < Cl¯ < Br¯ < I¯

e.g.

CH3¯ < NH2¯ < OH¯ < F¯

e.g.

R–COO¯ > PhO¯ > HO¯ > RO¯

e.g.

SH¯ > OH¯

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

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CHEMISTRY

2.

Nucleophilic substitution reaction (SN) : Replacement (displacement) of an atom or group by an other atom or group in molecule is known as substitution reaction. If substitution reaction is brought about by a nucleophile then it is known as nucleophilic substitution reaction. Generally substitution takes place at sp3 carbon.  R – Nu + R–g +

2.1

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

2.1.1 SN1 Reaction of Alkyl halide Mechanism :

Ionisation of alkyl halide      Slow step (rds)

Characteristics of SN1 reactions : 1. It is unimolecular, two step process. 2. Carbocation intermediate is formed so rearrangement is possible in SN1 reaction. 3. It is first order reaction 4. Kinetics of the reaction Rate  [Alkyl halide] Rate of SN1 reaction is independent of concentration and reactivity of nucleophile. 5. Energetics of the SN1

Figure : Free energy diagram for the SN1 reaction. 6. Factors affecting the rates of SN1 (i) The structure of the substrate : The rds of the SN1 reaction is ionization step, a carbocation is formed 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 # Weak, neutral, mostly solvents (protic) itself functions as nucleophiles in SN1 reaction. So SN1 reaction are termed as solvolysis reaction. water  hydrolysis ; C2H5OH  ethanolysis  CH3COOH acetolysis ; NH3  ammonolysis

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CHEMISTRY (iii) Effect of the solvent : The ionising 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 ionisation of an alkyl halide in any SN1 reaction. It does this because solvation stabilises the transition state leading to the intermediate carbocation and halide ion more than it does the reactant, thus the energy of activation is lower. R – X

(Solvolysis)

Solvated ions

Table - : Dielectric constants () and ionisation rates of t-Butylchloride in common solvents Solvent



Relative rate

H2O

80

8000

CH3OH

33

1000

C2H5OH

24

200

(CH3)2CO

21

1

CH3CO2H

6

–

(iv) 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 stabilises the transition state and this lowers the free energy of activation and there by increases the rate of reaction. Leaving ability of halogen is F¯ < Cl¯ < Br¯ < I¯ 7. 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 racemisation. Mechanism of racemisation (SN1 

e.g.

e.g.

H2 O acetone   

CH OH / Ag

3    

+ HBr

(3° alkyl halide)

(carbocation rearrangement)

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CHEMISTRY 2.1.2 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. HBr  

e.g.

(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. Mechanism : R –OH

e.g.

e.g.

 RDS   R R  O H2  –H2O

Reactivity of

HX :

Reactivity of

ROH :

CH3CHCH3 | OH Isopropyl alcohol

CH3 | CH3  C  CH2 – OH | CH3

R – X

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

CH3CHCH3 | Br Isopropyl bromide CH3 | CH3  C  CH2 – CH3 | Br

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CHEMISTRY Lucas Reagent (i) A mixture of concentrated hydrochloric acid and zinc chloride is called the Lucas reagent. (ii) Whether an alcohol is primary, secondary or tertiary is identified 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. (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.

CH3  CH2  CH  CH2  CH3 | OH

anhy. ZnCl / HCl

2       

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

2.1.3 SN1 Reactions of Ethers (A)

Reaction with HX 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 are good nucleophiles for the substitution. If R or R’ is 3º then mechanism will be SN1 otherwise SN2 . Mechanism :

X  R` – X

   – ROH

R – O – R`



e.g.

(CH3)3COC(CH3)3

HCl  (CH3 )3 C O C(CH3 )3 | H



(CH3 )3 C

+ (CH3)3COH

 (CH3)3CCl

(CH3)3COH + HCl  (CH3)3CCl + H2O

2.2

Bimolecular nucleophilic substitution reaction (SN2) Nucleophilic substitution in which incoming group replaces leaving group in one step only.

2.2.1 SN2 Reaction of Alkyl halide : Mechanism :



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CHEMISTRY Characteristic of SN2 1. It is bimolecular, one step concerted process 2. It is second order reaction because in the rds both species are involved 3. Kinetics of the reaction  rate  [alkyl halide] [nucleophile] rate = k[alkyl halide] [nucleophile] If the concentration of alkyl halide in the reaction mixture is doubled, the rate of the nucleophilic substitution reaction is double. If the concentration of nucleophile is doubled the rate of reaction is also double. If the concentration of both are doubled then the rate of the reaction quadriples. 4. Energetics of the reaction :

Figure : A free energy diagrams for SN2 reaction 5. 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. 6. The 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 that is the target of nucleophilic attack. This inversion is also known as Walden inversion.

Inversion   

7. Factor's affecting the rate of SN2 reaction  Number of factors affect the relative rate of SN2 reaction, the most important factors are (i) Effect of the structure of the substrate 

SN2 reactivity 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. Table :

Relative rate of reactions of alkyl halide in SN2 reaction. Substituent

Compound

Relative rate

Methyl

CH3X

30

1°

CH3CH2X

1

2°

(CH3)2CHX

0.02

Neopentyl

(CH3)3CCH2X

0.00001

3°

(CH3)3CX

~0

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CHEMISTRY (ii) Concentration and reactivity of the nucleophile – As nucleophilicity of nucleophile increases rate of SN2 increases. – Anionic nucleophiles mostly give SN2 reaction – A stronger nucleophile attacks upon -carbon with faster rate than the rate of departing of leaving group. 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 stabilise the transition state and lowers its free energy of activation and there by increases the rate of the reaction. Order of leaving ability of halide ion F¯ < Cl¯ < Br¯ < I¯ Examples of SN2 reactions of alkyl halide  + R – X



Nucleophile

– R + X¯ Product

..

Class of Product

..



R – .I. :



R – O .. H



R – O .. R 



R – :S .. H



R – S . . R



R – N H3 X¯

R – X + ¯: N  N  N :¯



R – N  N  N :¯

azide

R – X + ¯:C  C – R



R – C  C – R

alkyne

R – X + ¯:C  N:



R – C  N:

nitrile

R – X + R – CO O . . :¯



R – COO – R

ester

R – X + :P(Ph)3



[R – PPh3]+ X

R – X +

¯ .I. :

R – X +

¯: .O . H

..

..

R – X + ¯: .O . R ..

R – X +

¯: S .. H

R – X +

¯: S . . R

..

R – X + : NH3 ..



..

..

alkyl halide

..

alcohol

..

ether (Williamson synthesis)

..

thiol (mercaptan)

..

thioether (sulphide)



..



amine ..

_

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CHEMISTRY KOH H2O

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

e.g.

e.g.

Na

 MeOH 

e.g.

PhCH Br

2    

2.2.2 SN2 Reaction of Alcohol (A)

Reaction w ith HX : The protonated  unbranched primary alcohol is well suited for the SN2 reaction. Mechanism :  X R  O H2  R – X + H2O

R –OH

e.g

(B)

CH3 CH2CH2 CH2 CH2Cl

CH3CH2CH2 CH2CH2OH

n  Pentyl chloride

n  Pentyl alcohol

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

Remarks The mechanism for the reaction involves attack of the alcohol group on the phosphorus atom, displacing a halide ion and forming a protonated alkyl dihalophosphite In second step a halide ion acts as nucleophile to displace HOPX2, a good leaving group due to the electronegative atoms bonded to the phosphorus.

e.g.

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

PBr

3   

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

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CHEMISTRY e.g.

P  I2   

CH3 CH2 OH Ethyl alcohol

CH3CH2 I Ethyl iodide

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 :

.. .. –H  R – O .. – S 



.. 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 SOCl 2     Py

e.g.

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 :

.. .. –H  R – O .. – S 



.. O:

+ HCl

Cl Chlorosulphite ester

 R – Cl + SO2

e.g.

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CHEMISTRY 2.2.3 (A)

SN2 Reaction of Ether Reaction with HX 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.

CH3CH2 – O – CH3

e.g.



H  

excess HBr

   

e.g.

2.3

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. In basic medium mechanism is SN2. Nucleophile atacks on less hindered carbon. Mechanism :

Nu Nu | |  H R — CH — CH2   R — CH — CH2 | | O OH

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



H  

e.g.

Nu | R — CH — CH2 | OH

H2O     H

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CHEMISTRY

3.

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. 

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

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

3.1

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

3.1.1 E1 Reaction of Alkyl halide : Mechanism : Step 1 :

Step 2 :

+ B – H

Characteristics of E1 reaction : (i) It is unimolecular, two step process. (ii) It is a first order reaction. (iii) Reaction intermediate is carbocation, so rearrangment is possible (iv) In the second step, a base abstracts a proton from the carbon atom adjacent to the carbocation, and forms alkene. (v) Kinetics 

Rate  [Alkylhalide] Rate = k [Alkylhalide]

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

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CHEMISTRY

CH3 | e.g. CH3  CH2  C  CH3 | Br



+



e.g.

3.1.2 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) H3 PO4/  (ii) H2SO4 / 160º

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

+ H2 O

(Rearrangement may occur)

Mechanism :

Step 1 :

Step 2 :

+

Step 3

+

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CHEMISTRY 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 molecule of water : 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°

e.g.

CH2

CH2OH e.g.

Conc.H2SO 4     

+

(I) Minor

e.g.



H  

Migration of bond

  



H  

3.1.3 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 / 

e.g.

3.2

Conc.H 2 SO 4 /       

E2 Reaction :

3.2.1 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. Reagent (i) Hot alcoholic solution of KOH or EtO¯ / EtOH

(ii) NaNH2

(iii) t-BuO¯ K



in t-BuOH

Mechanism :





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CHEMISTRY Characteristics of E2 reaction  1. This is a single step, bimolecular reaction 2. It is a second order reaction 3. Kinetics  Rate  [R – X] [Base] Rate = k [R – X] [ ] 4. Rearrangment is not possible 5. For the lower energy of activation, transition state must be stable 6. E2 follows a concerted mechanism 7. The orientation of proton & leaving group should be antiperiplanar. 8. Here  – H is eliminated by base hence called  elimination 9. 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 rule is called the saytzeff or zaitsev rule. Reactivity towards E2  R – I > R – Br > R – Cl > R – F For example : Dehydrohalogenation of 2-bromo-2-methylbutane can yield two products.

e.g.

+

CH3 e.g.

+ (major)

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CHEMISTRY e.g.

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 highly substituted product, called the Hoffmann product.

H CH3 | | CH3  C  C  CH2 | | | H Br H

+

+

Bulky base :

(CH3CH 2)3N: Triethylamine

Stereospecific E2 reactions The E2 is stereospecific because it normally goes through an anti periplanar transition state. The products are alkene. Base

Ph Ph

H

Ph



CH3

 H Br

CH3 H

Ph

H

H

Ph

CH3

Ph

= Ph

=

Ph

CH3

H

Br

3.3

E1 cB Reaction (Unimolecular conjugate base reaction) In the E1 cB, H leaves first and then the X. This is a two step process, the intermediate is a carbanion. Mechanism :

Step 1 :

Step 2 :

H | | CC X | |

(conjugate base)



–C  C – | |

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CHEMISTRY Remarks : First step consists of the removal of a proton, , by a base generating a carbanion In second step carbanion looses a leaving group to form alkene Condition : For the E1 cB, (i) substrate must be containing acidic hydrogens and (ii) poor leaving groups.

e.g.



e.g.

 X2C = CF2

e.g.

 R2C = O +

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CHEMISTRY

Miscellaneous Solved Examples Q.1

In the given reaction, CH3CH2 – X + CH3SNa  The fastest reaction occurs when ‘X’ is (A) – OH (B) – F (C) – OCOCF3

(D) OCOCH3

Ans. Sol.

C Leaving group ability  Stability of anion

Q.2

Give the solvolysis products expected when each compound is heated in ethanol

(a)

(b)

(c)

(d)

Ans.

(a)

(b)

(c)

(d)

Sol.

(a)

(b)

(c)

(d)

Q.3

Ans. Sol.

The rate of SN1 reaction is fastest with

(A)

(B)

(C)

(D)

(A) Rate of SN1 reaction  Stability of carbocation .

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CHEMISTRY Q.4.

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

Ans.

(a) II > I

Sol.

Rate of solvolysis  Stability of carbocation .

Q.5

CH3 | CH3  C  CH2 OH | CH3

Ans.

(b) II > I

H I 

?

(c) I > II

(d) II > I

Write the structure of major product

CH3 | CH3  C  CH2 – CH3 |

I

2 - Iodo - 2 - methylbutane

Sol.

CH3 | CH3  C  CH2 OH | CH3

CH3  H I  CH  C|  CH O  3 2 H2  | CH3

CH3 | CH3  C  CH2 – CH3 

CH3 | CH3  C  CH2 – CH3 |

I

2 - Iodo - 2 - methylbutane

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CHEMISTRY Q.6

3-Methyl-2-pentanol reacts with conc.HCl / ZnCl2 to gives 3-Chloro-3-methylpentane, explain the mechanism. Ans./Sol.

CH3 | CH3  CH2  CH  CH  CH3 | OH

CH3 CH3 | | H   CH3  CH2  CH  CH  CH3  CH3  CH2  C  CH2  CH3  H2O   

CH3 | CH3  CH2  C  CH2  CH3 | Cl

Q.7

Make distinction between following pairs of substances by using Lucas reagent.

Ans./Sol.

Q.8

Lucas (a)

(II) gives white turbidity immediately

(b)

(II) gives white turbidity after 5-7 minute

(c)

(II) gives white turbidity immediately

(d)

(I) gives white turbidity after 5-7 minute

What is the major product ?

+

HBr



+

HBr



A (Major)

Ans.

Sol.

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CHEMISTRY Q.9

Correct decreasing order of reactivity towards SN2 reaction (I) CH3CH2

(II) CH3

CH2Cl

CH2CH2Cl (III) CH3CH2CH2CH2Cl (IV) CH3CH2CH2

Ans. Sol.

(A) IV > I > II > III (B) III > II > I > IV (C) IV > I > III > II B As increases branching rate of SN2 reaction decreases.

Q.10

Write the structure of product ‘P’

KOH / DMSO

     

(D) II > I > IV > III

P

H2O

Ans.

Inversion of configuration is possible

Sol.

It is a SN2 reaction.

Q.11

Give the decreasing order of reactivity of alkyl halides with MeO– (Williamsons synthesis).

(CH3 )3 CCH2Br (A)

ClCH2CH  CH2 (B)

ClCH2 CH2 CH3 (C )

Ans. Sol.

B > D > C > A It is a SN2 reaction.

Q.12

Write mechanisms that account for the product of the following reaction :

BrCH2CH2CH3 (D)

NH2 – CH2 – CH2 – CH2 – CH2 – Br



Sol.

Q.13

products What is the product of the given reaction ?

Ans. Sol.

+ It is a SN2 reaction.

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CHEMISTRY Q.14

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

NaI / Acetone

(a)

(b)

    



Ans.

(a)

(b)

Sol.

It is a SN2 reaction.

Q.15

To prepare t-butyl ethyl ether, t-butyl bromide and sodium ethoxide is not taken ; instead ethyl bromide sodium t-butoxide is taken. Explain.

Sol.

CH CH O

3 2 (CH3)3CBr   E2

S 2

N

(CH3)3COCH2CH3

Q.16

CH3 CH2CHOH | CH3

Ans.

CH3 CH2CHOH | CH3

Sol.

It is a SN2 reaction.

Q.17

Explain why ArOR ethers are cleaved to give RI and ArOH rather than ArI and ROH ?

Sol.

A , Identify A

CH3CH2CHBr | CH3





SN2 attack on a carbon of a benzene ring does not occur nor does the

(phenyl) of high energy

form by an SN1 reaction. Hence, ArI can’t be form

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CHEMISTRY Q.18

Which one of the following does not give white precipitate with aqueous silver nitrate solution?

(A)

(B) CH2 = CH – Cl

(C) CH2 = CH – CH2 – Cl

(D)

Ans. Sol.

A B In (A) and (B) lone pair of Cl atom takes part in conjugation so partial double bond character between C–Cl bond.

Q.19

Me3CCH2OH

Ans.

A : Me2C = CHMe

+

, Identify A and B

B : CH2  CCH2Me | Me

Me2C = CHMe + CH2  CCH2Me | Me

Sol.

Me3CCH2OH

Me3C

Q.20

Write the intermediate species as indicated : 

H 3, 3 – dimethyl – 2 – butanol  .................. 

(A) oxonium ion

 H2 O Migration   ..................    ..................

(B) 20 carbocation

(C) 30 carbocation

-H+

-H+

.................... (E) Alkene

CH3

CH3 Sol.

(A) = CH3 C

CH CH3

(B) = CH3 C

CH3

CH3 OH2 +

.................... (D) Alkene

CH3 CH3

CH CH3 +

(C) = CH3 C +

CH CH3

CH3 CH3 CH3 (D) =

CH3 C

C

CH3

(E) =

CH3 C CH CH2 CH3

Major product is (D) Minor product is (E)

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CHEMISTRY Q.21

What are dehydration products of

Ans.

Sol.

Q.22

+

H SO , 

Write the major product of the following reaction

4 2   

Ans.

Sol.

Q.23

H SO , 

4 2   

Predict the elimination products of the following reactions (a) Sec. butyl bromide +

(b) 2-Bromo-3-ethylpentane + MeONa

(c) 1-Bromo-2-methylcyclohexane + EtONa

Ans.

(a) CH3 – CH = CH – CH3

(b)

(c)

Sol.

(a) It is an E2 reaction.

(b) It is an E2 reaction.

(c) It is an E2 reaction.

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CHEMISTRY Q.24

major + minor Write the structure of major and minor product.

Ans. Sol.

It is an E2 reaction.

Q.25

What are the various product due to loss of HBr from

CH3 CH3 Ans.

,

, (minor)

Sol.

It is an E2 reaction.

Q.26

What are the major products of dehydrohalogenation by alcoholic KOH of : (ii) CH3  CH2  CH2  CH  CH3 | Cl

(i) CH3 – CH2 – CH2 – CH2 – CH2 Cl

CH3 | (iv) CH3  CH2  C  CH3

(iii) CH3  CH2  CH  CH2  CH3 | Cl

| Cl

Cl | (vi) CH3  CH  C  CH3 | | CH3 CH3

(v) CH3  CH  CH  CH3 | | Cl

Ans.

CH3

(i) gives CH3CH2CH2CH = CH2,

(iv) & (v) Sol.

(ii) & (iii) CH3CH2CH = CHCH3

(vi)

It is an E2 reaction.

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