158 Essential Notes on General Organic Chemistry (G.O.C.) The IITian’s Prashikshan Kendra Pvt. Ltd. Topics Coverage :- • Substitution Reactions • Free Radical Reactions •

158 Essential Notes on General Organic Chemistry (G.O.C.)

The IITian’s Prashikshan Kendra Pvt. Ltd.

Topics Coverage :- • Substitution Reactions • Free Radical Reactions • SN1 Mechanism • SN2 Mechanism • SN1 v/s SN2 Mechanisms • E1 Mechanism • E2 Mechanism • E1 v/s E2 Mechanisms • Substitution v/s Elimination Reactions • Aromatic Substitution Electrophilic Mechanism • Aromatic Substitution Nucleophilic Mechanism • Addition Nucleophilic Mechanism • Addition Electrophilic Mechanism • Addition Free Radical Mechanism • RARE MECHANISMS : (SNi, E1cb, SN', NGP, Benzyne, Ipso) • Rearrangements Introduction : How exactly does an organic reaction take place is the subject of study in this chapter of Reaction Mechanisms. Just as detectives investigate a crime after it has taken place and arrive at a reasonable prediction about how the crime may have occurred, similarly organic chemists predict by a number of laboratory techniques using sophisticated instrumentation how an organic reaction may have taken place. Many of these techniques are based on the Physical Chemistry principles of Redox, Equilibria, Chemical Kinetics and Thermodynamics. The instruments used include Polarimeter, NMR Spectroscope, Electric Dipole measurement device, isotopic tracer, pH meter etc.

CLASSIFICATION OF ORGANIC REACTIONS Organic Reactions in JEE syllabi can be classified as under : 1. Substitution reactions - (i) Free radical (ii) Nucleophilic-SN1, SN2,SNi (iii) Electrophilic 2. Elimination reactions-E1, E2 & E1cb 3. Addition reactions - (i) Electrophilic (ii) Nucleophilic (iii) Free Radical 4. Rearrangements

4 CHAPTER

Reaction Mechanisms

Essential Notes on General Organic Chemistry (G.O.C.) 159

REACTION MECHANISMS.

1. SUBSTITUTION REACTIONS A reaction in which one group or atom is replaced by another is called a substitution reaction. The incoming group is bonded to the same carbon to which the leaving group was bonded. Substitution reaction has been classified according to the nature of the substituents involved. Thus-

A. Free radical substitution ••

+→+ BQAQBA ::

B. Nucleophilic substitution ΘΘ

+→+ BQAQBA ::::

C. Electrophilic substitution ⊕⊕

+→+ BQAQBA :: It will be seen that in all types of substitutions, the displaced species belong to the same class as the attacking species. A. Free Radical Substitution

Radical substitution reactions are initiated by radicals in the gas phase or in non-polar solvents.

Thus, methane and chlorine react in the presence of sunlight or heat to give methyl chloride. HClClCHClCH

C

orhv ++→+° 330024

Light energy or heat causes homolytic fission of chlorine producing chlorine radicals which attack methane to form methyl chloride.

••° + → ClClClCl Corhv 300: …….(i)

HClCHClHCH +→+••

33 : ……..(ii) ••

+→+ ClClCHClClCH :: 33 ……(iii)

B. Nucleophilic Substitution (Common in aliphatic compounds): Nucleophilic substitution is one of the two main reactions of alkyl halides. A nucleophile

replaces a leaving group on an sp3 hybridized carbon.

One σ bond is broken and one σ bond is formed. There are two possible mechanisms: SN1 and SN2

The electron pair in the C – Nu bond comes from the nucleophile.

R – X + −•• Nu R – Nu + −:X

nucleophile

leaving group



SN1 and SN2 Mechanisms Compared

SN2 Mechanism SN1 Mechanism [1] Mechanism One step Two steps [2] Alkyl halide Order of reactivity :

CXRCHXRXRCHXCH 2223 >>> Order of reactivity :

XCHXRCHCHXRCXR 3223 >>> [3] Rate equation Rate = ]][:[ NuRXk

Second – order kinetics Rate = ][RXk First – order kinetics

[4] Stereochemistry Backside attack of the nucleophile Inversion of configuration at a stereogenic center

Trigonal planar carbocation intermediate Racemization at a single stereogenic center

[5] Nucleophile Favored by stronger nucleophiles Favored by weaker nucleophiles [6] Leaving group Better leaving group → faster reaction Better leaving group faster reaction [7] Solvent Favored by polar aprotic solvents Favored by polar protic solvents [8] Reactivity structure of R-determining factor. Nature of X

3° > 2° > 1° > CH3 Stability of R + RI > RBr > RCl > RF

CH3 > 1° > 2° > 3° Steric hindrance in R group RI > RBr > RCl > RF

[9] Effect of nucleophile

Rate depends on nucleophilicity I − > Br − > Cl − ; RS > RO −

No effect on rate

[10] Catalysis Lewis acid, e.g. Ag + ,AlC3, ZnCl2 None of these [11]Competition reactions

Elimination, rearrangement Elimination

[12] TS of slow step

• Factors to decide whether SN1 or SN2 : Any factors that affects the energy of activation of a given type of reaction will affect the rate and / or mechanism. These factors are useful to decide whether SN1 or SN2 occurs : 1. Electronic effect 2. Steric effect 3. The nature of the leaving group. 4. The nature of the nucleophile. 5. Participation of neighbouring group. 6. Effect of solvent. 1. ELECTRONIC EFFECT :-

I-effect, R-effect, hyperconjugation etc. which can affect the stability of carbocation are collectively termed as electronic effect. Clearly if electronic effects stabilise the carbocation,

1NS will more probable than 2NS .

XCNu −− −δ −δ

X C +δ −δ

)1( °− XC

)2( °−XC

)3( °−XC

2NS

2NS or 1NS

1NS



2. STERIC EFFECT :- More steric restriction opposes 2NS mechanism. Simple alkyl halides show the following general order of reactivity in 2NS reactions: methyl > primary > secondary >>(tertiary)

Methyl halides react most react most rapidly and tertiary halides react so slowly as to be unreactive by the SN2 mechanism. 3. THE NATURE OF LEAVING GROUP :-

The ease of formation of carbocation supports SN1 and this formation depend upon nature of leaving group. The best leaving group is the weakest base. Leaving group ability increases left-to-right across a row and down a column of the periodic table. The ease of leaving is as follows :

The leaving power of some nucleophilic groups is given below in decreasing order :

4. THE NATURE OF THE NUCLEOPHILE :- Stronger nucleophile supports 2NS mechanism. Strength of nucleophile is as follows :

:3NMeClPhOBrOHOEtICNPhS >>>>>−>>> −−−−−−−− 5. Participation of Neighbouring Groups : Some neighbouring groups as :

,−COO −−−− PhCOOHXO||

,, etc. can also participate and determine the nucleophilic substitution.

6. Effect of solvent : As the dielectric constant (i.e., polarity) of solvent increases, the rate of 1NS reaction increases. Illustration : Indicate the effect on the rate of SNl and SN2 reactions of the following: (a) Doubling

the concentration of substrate (RL) or −Nu . (b) Using a mixture of ethanol and H2O or only acetone as solvent. (c) Increasing the number of R groups on the C bonded to the leaving group, L. (d) Using a strong −Nu .

Ans: (a) Doubling either [RL] or [ −Nu ] doubles the rate of the SN2 reaction. For SN1 reactions, the rate is doubled only by doubling [RL] and is not affected by any change in [ −Nu ].

CF3 – S – O

O

O

Θ > Br S – O

O

O

Θ > CH3 S – O

O

O

Θ > C6H5 – S – O

O

O

Θ

> CH3 – S – O

O

O

Θ > I > Br > CF3 – C – O Θ Θ

O

Θ > H2O > Cl > F

⊕ Θ Θ CH3 – C – O >

O



(b) A mixture of ethanol and H2O has a high dielectric constant and therefore enhances the rate of SN1 reactions. This usually has little effect on SN2 reactions. Acetone has a low dielectric constant and is aprotic and favors SN2 reactions.

(c) Increasing the number of R’s on the reaction site enhances SN1 reactivity through electron release and stabilization of R+. The effect is opposite in SN2 reactions

because bulky R’s sterically hinder formation of, and raise ++

∆H for, the transition state. (d) Strong nucleophiles favor SN2 reactions and do not affect SN1 reactions Illustration : List the following alkyl bromides in order of decreasing reactivity in indicated reactions.

(a) SN1 reactivity, (b) SN2 reactivity, (c) reactivity with alcoholic AgNO3. Ans. (a) Reactivity for the SN 1 mechanism is 3°(I) > 2°(III) > 1° (I). (b) The reverse reactivity for SN2 reactions gives 1° (II) > 2°(III) > 3°(1). (c) Ag+ catalyzes SN1 reactions and the reactivities are 3°(I) > 2°(III) > 1° (II). (C) Nucleophilic Substitution in Aromatic Compounds : Aryl halides are less reactive towards SN reaction due to the donation of lone pair electrons to the benzene ring via resonance. Aryl halides undergo SN reaction when a strong electron-withdrawing group (or activating group) is present at the ortho and para positions. ArSN reactions are of two types viz.

(i) Addition-Elimination reaction (ii) Elimination-Addition reaction via benzyne mechanism

(i) Addition-Elimination reaction (also called Bimolecular displacement mechanism)

••

••••Cl Nu Cl

H Θ

Nu Cl

H Θ

Nu Cl H

Θ ≡

Nu Cl

Θ

Fast

Z

+ Cl −

C

CH3

Br

CH3 CH2CH3

(I)

CH3CH2CH2CH2CH2Br

(II)

CH3CH2CHCH2CH3

(III)

Br



Any factor that stabilizes the carbanion will increase the rate of nucleophilic substitution reaction by dispersion the charge presence on resonating structures. An electron withdrawing group present at meta position does not activate the ring as much as it does from ortho and para position. This can be known by looking at following resonance structures.

(ii) Elimination-Addition Mechanism : (Please refer to RARE MECHANISMS at the end of this chapter)

(D) Electrophilic Substitution (Common in Aromatic compounds) : Electrophilic Substitution involves the attack by an electrophile. It is represented as SE. If the order of the reaction is 1, it is abbreviated as SE1 (unimolecular) and if the order is 2, it is SE2 (bimolecular).

(a) SE1 Mechanism : Electrophilic substitution in aliphatic compounds (SE1) are very rare. Some of the important examples are : (i) Replacement of metal atom in an organo-metallic compound by hydrogen

R---M + H+ R—H + M+ (ii) Isotopic exchange of hydrogen for deuterium or tritium

R---H + D+ R—D + H+ R---H + T+ R—T + H+

(b) SE2 Mechanism :

Electrophilic substitution (SE2) is every common in benzene nucleus (aromatic compounds) in which π electrons are highly delocalized and an electrophile can attack the region of high electron density. Aromatic electrophilic substitution reactions involve the following 3 steps mechanism : Step-1 : The formation of an electrophile. Step-2 : The electrophile attacks the aromatic ring to form carbocation (or arenium ion) which is stabilized by resonance. Step-3 : Carbocation loses the proton to form the substitution product. e.g. Bromination of Benzene, Nitration, Sulphonation & Friedel-Crafts reactions of benzene are classic examples of electrophilic substitution reactions.

NB: (i) The second step is found to be rate – determining step by isotopic studies. (ii) The intermediate carbocation is stabilized by resonance; a minimum of three resonance structures can be drawn. The positive charge is always located ortho or para to the new

C – W bond. Refer to Chapter 5 or lecture notes for more clarity on the SE2 mechanism.

Z Cl

NO2

Z Cl

NO2 Θ

Z Cl

NO2 (meta)

Z Cl

NO2

Z Cl

NO2

Z Cl

NO2

(para)



2. ELIMINATION REACTIONS

Elimination reactions of alkyl halides are important reactions that compete with substitution reactions. In an elimination reaction, the fragments of some molecule (YZ) are removed (eliminated) from adjacent atoms of the reactant. This elimination leads to the introduction of a multiple bond:

The reaction in which two atoms or groups are removed from a compound is referred to as an elimination reaction. The elimination reaction can be classified as shown below:

E1 & E2 Mechanisms Compared

E2 Mechanism E1 Mechanism Mechanism One step Two step Alkyl halide Rate : XRCHCHXRCXR 223 >> Rate : XRCHCHXRCXR 223 >> Rate equation Rate = :]][[ BRXk

Second – order kinetics Rate = ][RXk First – order kinetics

Stereochemistry Anti periplanar arrangement of H and X Trigonal planar carbocation intermediate

Base Favored by strong bases Favored by weak base Leaving group Better leaving group → faster reaction Better leaving group → faster

reaction Solvent Favored by polar aprotic solvents Favored by polar protic solvents Product More substituted alkene favored More substituted alkene favored

(Zaitsev rule (Satyzeff’s) When will the reaction be E1 and when will it be E2 ?

C C elimination (-YZ)

C = C

Y Z

1,1 Elimination

or α - elimination

1,2 Elimination

or β - elimination

1, 3 Elimination

1, 4 Elimination

Elimination Reaction

Syn Elimination

(Ei ) Elimination

Intra-molecular

Anti Elimination

E1

E2

E1CB



The strength of the base determines the mechanism of elimination. Strong bases favor E2 reactions Weak bases favor E1 reactions Illustration :

Regioselectivity : ORENTATION IN ELIMINATION REACTIONS :-

If substrate is unsymmetrical, then this will give more than one product. Major product of the reaction can be known by two empirical rules.

(1) Saytzeff’s Rule (also spelled as Zaitsev’s Rule by some books) :- According to this rule, major product is the most substituted alkene, i.e., major product is obtained

by elimination of ⊕H from that β - carbon which has the least number of hydrogens. Product of the

reaction in this case is known as Saytzeff product.

product Saytzeff

3|3/.

3

|

|3

31

2

3

CHCHCCHCHCHCHCHCH

KOHalc

Cl

CH

−=− →−−− ∆

βα

β

(2) Hofmann Rule :- According to this rule, major product is always least substituted alkene, i.e., major product is formed from −β carbon which has maximum number of hydrogens. Product of the reaction in this case is known as Hofmann product.

ProductHofmann

22

|

|3/.

3

|

2

|

|3

3

312

3

3

CHCHCHCCHCHCHCHCCHCH

CH

KOHalc

BrCH

CH

=−−− →−−−− ∆

βαβ

Note :- (i) In 1E reactions, product formation always takes place by Saytzeff rule. (ii) In 1E cb reactions, product formation always takes place by Hofmann rule. (iii) In 2E reaction, product formation takes place by Saytzeff as well as Hofmann rule. In

almost all E2 reactions, product formation takes place by Saytzeff rule. Only in four cases, product formation takes place by Hofmann rule.

4 CASES OF HOFFMANN ELIMINATION REACTIONS :-

CH3– C – CH3.

CH3

Br

- OH E2

H2O

Weak base E1

Same Product, Different Mechanism

C = CH2 + H2O+ + Br-

CH3

CH3

Strong base

C = CH2 + H2O + Br-

CH3

CH3



(1) Dehydrohalogenation of alkyl halides when leaving group is very poor, i.e., alkyl fluoride

(primary or secondary).

30% product Saytzeff 70% productHofmann

3232223/.

3

|

2231

2CHCHCHCHCHCHCHCHCHCHCHCHCHCHCH KOHalc

F

−=−−+=−−− →−−−− ∆

βαβ

(2) Primary and sec – alkyl halides give Hoffmann elimination when the size of the base is bulky, i.e., sodium or potassium salt of tert butoxide.

(3) Primary and sec – alkyl halides having quarternary −γ carbon gives Hofmann elimination.

product) (Saytzeff product)(Hofmann productMinor product Major

3

|

|322

|

|3/.

3

|

2

|

|3

3

3

3

312

3

3

CHCHCHCCHCHCHCHCCHCHCHCHCCHCH

CH

CH

CH

KOHalc

BrCH

CH

−=−−+=−− →−−− ∆

βαβ

γ

(4) If leaving group is bulky, then compound gives Hofmann elimination reaction. The most common type of lager, bulky leaving group which leads to Hofmann products has positively

charged nitrogen )( 3

⊕NR or positively charged sulphur )( 2

⊕SR

Illustration : Q. Which of the following will give Hofmann product in E2 reaction?

(a) Θ⊕

−−−− OHCHNCHCHCHCHCH

33|223 )(3

(b) Θ⊕

−−−− OHCHSCHCHCHCHCH

23|223 )(3

(c) 3

|

23 CHCHCHCHF

−−− (d) All of these Ans. (d) Illustration : Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with +−KOHC 52

to give mainly 2-pentene (the Saytzeff product), with +−KCOMe3 it gives mainly l-pentene (the anti-Saytzeff, Hofmann, product). Ans :

12

3

|

23βαβ

CHCHCHCHBr

−−−

∆−−⊕Θ

/|

|3

3

3

OKCCHCH

CH

20% 80% product) (Saytzeff product)(Hofmann

33223 CHCHCHCHCHCHCHCH −=−+=−−



Since −COMe3 is a bulky base, its attack is more sterically hindered at the 2° H than at the

1° H. With −COMe3 ,

hinderedless

BrH

productmajorCHCHCHCHCH ←= − 1

3222 32|||2

21

CHCHCHCHCHHBrH

−−

hinderedmore

BrH →− 2

productorCHCHCHCHCH

min323 =

Elimination v/s Substitution : All nucleophiles are potential bases and all bases are potential nucleophiles. This is because the reactive part of both nucleophiles and bases is an unshared electron pair. It should not be surprising, then, that nucleophilic substitution reactions and elimination reactions often compete with each other. Thus, elimination reactions are usually accompanied by substitution reactions. When the reagent is a good base, it accepts protons to yield elimination products (alkenes) and if it is a good nucleophile, then it attacks the carbon to give substitution products. The proportion of elimination and substitution depends upon the following: (i) Structure of the Substrate (ii) Nature of the Base (iii) Nature of Solvent (iv) Effect of Temperature In general, the proportion of elimination increases on using a strong base of high concentration and a solvent of low polarity. On the other hand, the proportion of substitution increases by using a weak base of low concentration and a solvent of high polarity A Comparison Between Nucleophilic Substitution and β Elimination : Nucleophilic substitution – A nucleophile attacks a carbon atom

β Elimination – A base attacks a proton

Similarities Differences

- C – C -

H

X

−Nu:

- C – C -

H

X

Nu

+

−:X

substitution product

good leaving group

- C – C -

H

X

:B

elimination

good leaving group

C = C + H – B+ + −:X



In both reaction RX acts as an electrophile, reacting with an electron – rich reagent.

In substitution, a nucleophile attacks a single carbon atom.

Both reaction require a good leaving group −:X that can accept the electron density in the C – X bond.

In elimination, a Bronsted – Lowry base removes a proton to form a π bond, and two carbons are involved in the reaction .

Summary Chart on the Four Mechanisms : ,,2,1 1ESS NN or 2E

Alkyl halide type Conditions Mechanism

XRCH 21° strong nucleophile → 2NS

strong bulky base → 2E

CHXR22° strong base and nucleophile → 22 ESN +

strong bulky base → 2E

weak base and nucleophile → 11 ESN +

CXR33° weak base and nucleophile → 11 ESN +

strong base → 2E

Mechanism Review : Substitution versus Elimination

SN2 SN1 & E1 Primary substrate Back-side attack of Nu: with respect to LG Strong/polarizable unhindered nucleophile

Tertiary substrate. Carbocation intermediate Weak nucleophile/base (e.g., solvent)

Bimolecular in rate-determining step concerted bond forming/bond breaking. Inversion of stereochemistry. Favored by polar aprotic solvent.

Unimolecular in rate-determining step Racemization if SN1 Removal of β -hydrogen if E1 Protic solvent assists ionization of LG Low temperature (SN1) / high temperature (E1)

SN2 and E2

E2

Secondary or primary substrate Strong unhindered base/nucleophile leads to SN2 Strong hindered base/nucleophile leads to E2 Low temperature (SN2) / high temperature (E2)

Tertiary or secondary substrate. Concerted anti-coplanar TS Bimolecular in rate-determining step. Strong hindered base. High temperature

Detailed Discussion on SN2 v/s E2 :

−δLG

H

−δBNu /



SN2 and E2 reactions are both favored by a high concentration of a strong nucleophile or base. When the nucleophile (base) attacks a β hydrogen atom, elimination occurs. When the nucleophile attacks the carbon atom bearing the leaving group, substitution results:

• Consider the following examples with small (unhindered) nucleophiles and alkyl halides of

different classes.

Primary Substrate :- When the substrate is a primary halide and the base is unhindered, like ethoxide ion, substitution is highly favored because the base can easily approach the carbon bearing the leaving group:

223223)(

55232352 CHCHCHOCHCHCHBrCHCHNaOCHCH

NaBrC

OHHC =+ →+−°

+−

2NS E2 (90%) (10%)

Secondary Substrate :- With secondary halides, however, a strong base favors elimina-tion because steric hindrance in the substrate makes substitution more difficult:

Tertiary Substrate :- With tertiary halides, steric hindrance in the substrate is severe and an SN2 reaction cannot take place. Elimination is highly favored, especially when the reaction is carried out at higher temperatures. Any substitution that occurs must take place through an SN1 mechanism:

(a)

Elimination E2

(b)

substitution 2NS

(a)

H – C

(b)

– C X

:−Nu

C || C

+ Nu – H + −X:

H – C

Nu – C

+ −:X

3323 CHCHCHNaOCHCH ++−

Br

OHHC 52

C°55

)( NaBr−

3233 CHCHCHCHCHCH =+

OCH2CH2

2NS

(21%)

2E (79%)

(9%)

3323 CCHCHNaOCHCH ++−

Br

CH3

C2H5OH C°55

(-NaBr)

3233 CCHCHCCHCH =+

CH3

32CHOCH

CH3

1NS Mainly E2

(91%)

3323 CCHCHNaOCHCH ++−

Br

CH3

C2H5OH C°55

(-NaBr)

OHCHCHCCHCH 2333 +=

CH3

(100%)

E2 + E1



• Size of the Base/Nucleophile :- Increasing the reaction temperature is one way of favorably influencing an elimination reaction of an alkyl halide. Another way is to use a strong sterically hindered base such as the tert-butoxide ion. The bulky methyl groups of the tert-butoxide ion inhibit its reacting by substitution, allowing elimination reactions to take precedence. We can see an example of this effect in the following two reactions. The relatively unhindered methoxide ion reacts with octadecyl bromide primarily by substitution; the bulky tert-butoxide ion gives mainly elimination.

Unhindered (small) Base / Nucleophile

Hindered Base / Nucleophile

Basicity and Polarizability Another factor that affects the relative rates of E2 and SN2 reactions is the relative basicity and polarizability of the base/nucleophile. Use of a strong, slightly polarizable base such as amide ion ( −

2NH ) or alkoxide ion (especially a hindered one) tends to increase the likelihood of elimination (E2). Use of a weakly basic ion such as a chloride ion ( −Cl ) or an acetate ion ( −

23COCH ) or a weakly basic and highly polarizable one such as Br-, 1-, or RS- increases the likelihood of substitution (SN2). Acetate ion, for example, reacts with isopropyl bromide almost exclusively by the SN2 path:

−− +−− →−+− BrCHCHOCCHBrCHCHOCCHCHO

S

CHO

N 3

|||

3%)100(~

2

|

3

||

3

33

The more strongly basic ethoxide ion reacts with the same compound mainly by an E2 mechanism.

• Temperature :- Increasing the reaction temperature favors elimination (El and E2) over substitution. Elimination reactions have greater free energies of activation than substitution reactions because more bonding changes occur during elimination. When higher temperature is used, the proportion of molecules able to surmount the energy of activation barrier for elimination increases more than the proportion of molecules able to undergo substitution, although the rate of both substitution and elimination will be increased.

Furthermore, elimination reactions are entropically favored over substitution because the products of an elimination reaction are greater in number than the reactants. Additionally, because temperature. is the coefficient of the entropy term in the Gibbs free-energy equa-tion °∆−°∆=°∆ STHG , an increase in temperature further enhances the entropy effect.

BrCHCHCHCHOCH −+−2215233 )(

CH3OH C°65 322152321523 )()( OCHCHCHCHCHCHCHCHCH +=

E2 (1%)

2NS (99%)

CH3

C

BrCHCHCHCHO −+−

221523 )(

CH3

CH3

COHCH 33 )(

C°40

322152321523 )()( CHCOCHCHCHCHCHCHCHCH −−+=

CH3

CH3

E2

(85%) 2NS

(15%)



Detailed Discussion on Tertiary Halides SN1 v/s E1 Because E1 and SN1 reactions proceed through the formation of a common intermediate, the two types respond in similar ways to factors affecting reactivities. El reactions are favored with substrates that can form stable carbocations (i.e., tertiary halides); they are also favored by the use of poor nucleophiles (weak bases) and they are generally favored by the use of polar solvents. It is usually difficult to influence the relative partition between SN1 and E1 products because the free energy of activation for either reaction proceeding from the carbocation (loss of a proton or combination with a molecule of the solvent) is very small. In most unimolecular reactions the SN1 reaction is favored over the E1 reaction, especially at lower temperatures. In general, however, substitution reactions of tertiary halides do not find wide use as synthetic methods. Such halides undergo eliminations much too easily.

Increasing the temperature of the reaction favors reaction by the E1 mechanism at the expense of the SN1 mechanism. If the elimination product is desired, however, it is more convenient to add a strong base and force an E2 reaction to take place instead. Illustration : Given the product (or products) that you would expect to be formed in each of the following reactions. In each case give the mechanism ,1,2,1( ESS NN or E2) by which the product is formed and predict the relative amount of each (i.e., would the product be the only product, the major product, or a minor product?).

Answer :

(a) The substrate is a °1 halide. The base/nucleophile is ,3−OCH a strong (but not a

hindered one) and a good nucleophile. According to Table, we should expect an 2NS reaction mainly, and the major product should be 3223 OCHCHCHCH . A minor product might be 23 CHCHCH = by an E2 pathway. (b) Again the substrate is a 1° halide, but the base/nucleophile, −COCH 33 )( , is a strong

hindered base. We should expect, therefore, the major product to be CH3CH = CH2 by an E2 pathway and a minor product to be CH3CH2CH2OC(CH3)3 by an SN2 pathway.

(a) OHCH

COCHBrCHCHCH3

503223 →+ °−

(b) −+ COCHBrCHCHCH 33223 )(COHCH

C

33 )(

50 → °

(c) OHCH

CHSBrC3

50 →+− °− CH3

CH3CH2

H

(d) OHCH

COHCBrCHCH3

50323 )( →+ °−

(e) OHCH

CCBrCHCH3

25323 )( → °



(c) The reactant is (S)-2-bromobutane, a 2° halide and one in which the leaving group is attached to a stereogenic center. The base/nucleophile is −HS , a strong nucleophile but a weak base. We should expect mainly an SN2 reaction, causing an inversion of

configuration at the stereogenic center and producing the (R) stereoisomer:

(d) The base/nucleophile is OH-, a strong base and a strong nucleophile. However, the substrate is a 3° halide; therefore, we should not expect an SN2 reaction. The major product should be CH3CH = C(CH2CH3)2 via an E2 reaction. At this higher temperature and in the presence of a strong base; we should not expect an appreciable amount of the SN1 product, CH3OC(CH2CH3)3. (e) This is solvolysis; the only base/nucleophile is the solvent, CH3OH, which is a weak base (therefore, no E2 reaction) and a poor nucleophile. The substrate is tertiary (therefore, no SN2 reaction). At this lower temperature we should expect mainly an SN1 pathway leading to CH3OC(CH2CH3)3. A minor product, by an E1 pathway, would be CH3CH = C(CH2CH3)2.

Illustration : Indicate the products of the following reactions and point out the mechanism as : 1,2,1 ESS NN or .2E

(a) 4223 LiAlHBrCHCHCH + (source of −H: ) (b) ,)( 5233 OHHCCBrCH + heat at C°60 (c) 23 NaNHCHClCHCH += (d) MgBrCHBrCH +22 (ether) (e) MgBrCHCHBrCH +222 (ether) (f) 333 NaOCHCHBrCHCH + in OHCH 3 Answer : (a) ;323 CHCHCH an 2NS reaction, −H: of −

4AlH replaces .−Br

(c) )2(3323 ENaClNHCHCCHNaNHCHClCHCH ++≡→+= Vinyl halides are quite inert toward 2NS reactions.

(d) 22222 MgBrCHCHMgBrCHBrCH +=→+ This is an E2 type of β - elimination via an alkyl magnesium iodide.

(e) This reaction resembles that in (d) and is an internal 2NS reaction

CBrCH 33 )( RX°3

+− + CCHBr 33 )( OHCHCH 23 +− H 233233 )( CHCCHCHCOCHCH =+

3CH

major ; SN1 very minor ; E1 (b)

H3C CH2Br

CH2Br + Mg H3C

−2CH

CH2 – Br

+MgBr 2NS H3C

CH2

CH2 + MgBr2

Mg + BrCH2CH2Br +

−MgBr CH2 – CH2 – Br MgBr2 +H2C = CH2

HS

CH3

CH3CH2

H

C



(f) This °2 RBr undergoes both 2E and 2NS reactions to form propylene and isopropyl methyl ether.

3

|33233333 )(

OCH

CHCHCHCHCHCHOHCHCHONaCHBrCHCH +=→++

Illustration: Will the following reactions be primarily substitution or elimination? (a) →+ −IClCHCHCH 223 (b) −+CNCBrCH 33 )( (ethanol) → (c) →+ − )( 333 OHOHCHBrCHCH (d) −+OHCHBrCHCH 33 (ethanol) → (e) →+ OHCBrCH 233 )( Answer : (a) 2NS substitution. −I is a good nucleophile, and a poor base. (b) E2 elimination. A °3 halide and a a fairly strong base. (c) Mainly 2NS substitution. (d) Mainly E2 elimination. A less polar solvent than that in (c) favors E2. (e) 1NS substitution. OH 2 is not basic enough to remove a proton to give elimination. 3. ADDITION REACTIONS : (1) Addition reactions are given by those compound which have at least one −π bond, i.e.,

).,,,(||

NCCCCCCO

≡−−−≡−<=> (2) In this reaction, there is loss of one pi bond and gain of two sigma bonds. Thus product of the reaction is generally more stable than reactant. (3) The Addition reaction is a spontaneous reaction. Types of Addition reactions : Addition reaction can be classified into three categories on the basis of the nature of initiating species :

(1) Electrophilic Addition (2) Nucleophilic Addition, and (3) Free radical Addition

• Electrophilic Addition Reactions (1) This reaction is mainly given by alkenes and alkynes (2) Electrophilic addition reactions of alkenes and alkynes are generally two – step reactions. (3) Alkenes and alkynes give electrophilic addition with those reagents which, on dissociation, give electrophile as well as nucleophile.

(4) If the reagent is a weak acid then electrophilic adition is catalysed by strong acids

(generally H2SO4) (5) Unsymmetrical alkenes and alkynes give addition reaction with unsymmetrical regents according to Markovnikov’s rule.



Regio-selective Study : (Markovnikov’s & Anti-Markovnikov’s Rules) Addition Reaction of Alkenes : Markovnikov’s Rule According to Markovnikov’s rule, the negative part of the reagent adds on that doubly bonded carbon if the alkene which has least number of hydrogen (s). Conditions for the use of Markovnikov’s Rule : This rule can be used only in those alkenes which fulfill the following conditions: (a) Alkenes should be unsymmetrical (b) Substituent/substituents present on doubly bonded carbon /(s) should only be + I group. (c) If phenyl group is present on doubly bonded carbon, then both doubly bonded carbons should be substituted by phenyl groups. For example, the following alkenes will give addition according to Markovnikov’s rule:

Following alkenes will not give addition reaction according to Markovnikov’s rule.

Illustration : Q. Markovnikov’s rule can be used in which of these?

Ans. (d) All alkenes are unsymmetrical. Substituted Alkenes which give Addition Reaction according to Anti – Markovnikov’s Rule : Unsymmetrical alkenes having the following general structure give addition reaction according to anti – Markovnikov’s rule. GCHCH −=2 where, G is a strong – I group such as :

),,,(,,,,,, 2

||

23 NHOROHClZZCCOOHCORCHOCNNOCXO

=−−−−−−−− Some examples are : ,22 NOCHCH −= ,2 CHOCHCH −= ,2 CNCHCH −=

,||

2 RCCHCHO

−−= 32 CHCHCH −= etc.

C = C , R

R CH2 = CH2, R – CH = CH – R

R

R C = C

C6H5

C6H5

C6H5

C6H5

CH2 = CH – CF3 ; CH2 = CH – NO2

C = CH2

H3C

H3C (c)

CH3 – CH = CH2 (a) C6H5 – CH = CH2 (b)

all of these (d)

C = CH2, H3C

CH3 – CH = CH2, H3C

C = CH – CH3, H3C

H3C C = CH – C6H5,

H3C

H3C

C6H5 – CH = CH2



Illustration : Q. Which of these will give addition reaction according to anti – Markovnikov’s rule? (a) 32 CFCCH −= (b) 22 NOCHCH −= (c) 32 CClCHCH −= (d) all of these Ans. (d) Stereo-selective Study (Syn & Anti Addition) An addition that places the parts of the adding reagent on the same side (or face) of the reactant is called syn addition. We have just seen that the platinum-catalyzed addition of hydrogen (X = Y = H) is a syn addition

The opposite of a syn addition is an anti addition. An anti addition places the parts of the adding reagent on opposite faces of the reactant.

• Nucleophilic Addition Reactions Addition of nucleophile in the first step and electrophile in the second step is called nucleophilic addition reaction. Thus this reaction is two – step reaction. In this reaction, addition of nucleophile is rate determining step. Nucleophilic addition reaction is given by : (1) Alkenes (2) Alkynes (3) Carbonyl compounds and (4) Nitriles. A reaction in which the substrate and the reagent add up to form a product is called addition reaction. The reaction occurs at the site of unsaturation in a molecule. Thus, compounds having multiple bonds such as :

undergo addition reactions. The reactivity of these compounds is due to the more exposed and easily available π electrons to the electron-seeking (electrophilic) reagent.

C = C −≡− CC, C = O, ,NC ≡− etc.

C + X – Y C – C

X

Y

C An anti addition

C + X – Y C – C

X Y

C A syn addition



Free Radical Addition Radicals, like cations are electron-deficient species. Thus radicals will attack the pi-system of a double bond of an alkene because pi-electrons of an alkene can provide the electron needed to complete the outer shell of a radical. The net result of this usually is the addition of some species across the double bond. These reactions involve radical intermediate. Free radical addition reactions occur either in the gas phase or in inert non-polar solvents in the presence of UV light or sunlight, heat or catalytic amount of radical initiators such as organic peroxides, labile azo compounds like azoisobutyronitrile (AIBN), etc. The mode of the addition reaction involves the' general steps for the radical reactions, i.e., initiation, chain propagation and termination. The most important reaction of this category is hydrobromination. Hydrobromination :- The radical addition of HBr to a carbon-carbon double bond occurs in the presence of UV light or a small amount of initiators such as dibenzoyl peroxide. HF, HCl and HI do not, give this addition. When propene reacts with hydrogen bromide in the presence of dibenzoyl peroxide, the product is 1-bromopropane.

BrCHCHCHHBrCHCHCH −−− →+=− 223 peroxide dibenzoyl

23 In this addition the hydrogen that is being added ends up on the carbon bearing the smaller number of hydrogens. Therefore, this is an anti-Markonikov addition. In general, unsymmetrical alkenes add, HBr, in the presence of a radical initiator, in an anti-Markonikov manner.

BrCHCHRHBrCHCHR −−− →+=− 22source

radical2

The addition of HBr to alkenes in the presence of light or peroxide is a free radical reaction and is trans-addition at low temperature but at room temperature two isomers give the same mixture of diastereomers, e.g., I.

Meso + DL HBr + Peroxide

at room temp. (trans – and cis - addition )

C = C Me Me

Br H

HBr + Peroxide + hv at - C°78 (Br) trans – Addition

cis – 2 – Bromo – but – 2 – ene (or E)

Br | C | C | H

Me H

Me Br (meso)

HBr

Br | C | C

Me H

Me Br (meso)



II.

At room temperature, the less stable radical (from cis butene) equilibrates with the more stable radical (from trans butene) by internal rotation which is now faster than the hydrogen abstraction. The result is that both butenes give the same mixture of diastereomers. In both cases, the addition of hydrogen is trans with respect to the bromine atom. Free radical addition to alkenes is trans; no rotation is now possible, e.g.,

NB: Free Radical Addition Reactions are exhibited by electron-rich alkenes when they react with electron-deficient radicals. Radicals react with alkenes via a radical chain mechanism that consists of initiation, propagation and termination steps analogous to Free Radical substitution. (4) REARRANGEMENT REACTIONS Rearrangement refers to structural changes within a species. Rearrangements, in general are of two types: (a) Shifting of a group from one atom to the other (b) Rearrangement because of resonance (A) SHIFTING OF A GROUP FROM ONE ATOM TO THE OTHER ATOM Although shifting of a group can occur from an atom to any other atom yet most of the shifts occur to adjacent atoms only. The main reason for this shifting is presence of six valence electrons and a vacant orbital on an atom. Such an atom has tendency to complete its octet for which the group from the adjacent atom migrates with the bonding electrons. There are two different conditions around an atom with six valence electrons.

(a) When atom with six valence electrons and a vacant orbital:

- A – B - - A – B -

G G

Θ ⊕

Six valence electrons around B

Six valence electrons around A

Br Br Br

H

Br

Br

H

Br

H

+ Br

Meso + DL At room

temp. C C

Me

Me

Me

H

(trans or Z)

at C°− 78

(± ) Product



In this case when a group is shifted from atom S, the octet of S becomes incomplete which is not favourble condition. So we can say that this shift will occur only if it increases stability.

33/

223 CHCHCHCHCHCH ShiftH −− →−−⊕

Ring expansion and ring contraction can also result because of adjacent shift.

If migration of two (or more) different groups from adjacent atom can take place, major migration depends on the migrating tendency of the migrating groups (migration of group With higher migrating tendency is favoured) and the stability of resultant species (more stable resultant species is more favoured). Migrating tendency of a group is generally higher if it is more electron rich i.e. more electron donating. The general order of migrating tendency (or migratory aptitude) is as given below:

In several cases migrating tendency and the stability of resultant species favours the same migration. In such cases it will be easier to predict the major migration product.

CH3 – C – CH2

CH3

CH3

⊕ CH3 shift CH3 – C – CH2 - CH3

CH3

⊕

CH2 ⊕ H shift

CH3

⊕

alkyl shift Ring expansion

⊕

CH3 ⊕

CH3 shift

CH3

⊕

Ring Contraction

CH3

CH3 C ⊕

CH3

CH3

< <

Hydride < Alkyl < Aryl < Hydride

(If not involved in hyperconjugation with atom having 6 valence electrons)

RRR 000 321 << (If involved in hyperconjugation with atom having 6 valence electron )

E.W.G E.D.G



However, if migrating tendency and stability of resultant species oppose each other, following logics can be applied to predict the major product. (i) If both resultant species are very highly stabilized (stabilized by resonance of/one pair or

stabilized by formation of additional bond), small amount of stability does not matter a lot. Therefore, major migration occurs according to migrating aptitude.

CH3 – C - CH2

CH3

H

Hydride shift (higher migrating tendency )

CH3 – C - CH2

CH3

+

more stable (Major)

CH3 – C - CH2

CH3

H

Hydride shift (higher migrating tendency )

CH3 – C - CH2 - CH3

H

+

less stable

+

+

CH3 – C – CH2 +

H

hydride shift Highest migrating tendency)

CH3 – C – CH2

+

most stable (Major)

CH3 – C – CH2 +

H

phenyl shift (higher migrating tendency)

CH3 – C – CH3 +

least stable

H

CH3 – C – CH2 +

H

methyl shift

(least migrating tendency)

C – CH2 – CH3 +

more stable

H



(ii) If both resultant species are not very highly stabilized (not stabilized by resonance of lone

pair or not stabilized by formation of additional bond), small amount of stability also matters a lot. Therefore, major migration occurs according to stability of resultant species.

(iii) If the difference of stability is large, stability will dominate over the migrating tendency.

Therefore, groups containing lone pairs migrate rarely.

CH3 – C – CH2 +

CH3


CH3 – C – CH2 +

least stable

CH3

CH3 – C – CH2 +

CH3

methyl shift


C – CH2 – CH3 +

more stable (Major)

CH3

(Both stabilized by resonance of lone pair)

CH3 – C – CH2 +

OH


CH3 – C – CH2 +

least stable (Major)

OH

CH3 – C – CH2 +

O H

methyl shift


C – CH2 – CH3 +

more stable

OH

(Both stabilized by resonance of lone pair)



(b) When atom with six valence electrons and a vacant also has lone pair

In this case when a group from atom B is shifted, octet of B becomes incomplete which can be completed (stabilized) by lone pairs present on A atom. Hence under these conditions stability increases always and these rearrangements are, in general, bound to take place in the given species. Moreover, in such cases formation of additional bond or stabilization by resonance of lone pair will be generally observed, therefore migrating tendency generally dominates.

(B)REARRANGEMENT BECAUSE OF RESONANCE (ALLYLIC REARRANGEMENT) In these cases, actually rearrangements do not occur but more-than one product is formed because of the resonance in the intermediate. These rearrangements are also called pseudo rearrangements. These rearrangements are possible when the intermediates like carbocation, carbanion or carbon free radical etc. are resonance stabilized and more than one different resonating structure are possible. e.g.

2323

⊕⊕−=−→←=−− CHCHCHCHCHCHCHCH

−=−→←=−−⊕

CHCHCHCHCHCHCH 323 2⊕

CH

2323

⊕⊕−=−→←=−− CHCHCHCHCHCHCHCH

- A = B -

G

- A = B -

G

CH3 – C – H H Shift CH2 = CH2

CH3 – C – C – H H Shift CH3 - C = C – H

CH3

CH3 CH3

CH3

R – C – N

O

O = C = N – R



Following reactions illustrate allylic rearrangements:

In IIT-JEE syllabi, we will have to study few Rearrangement reactions :

Benzilic Acid Rearrangement Curtius Rearrangement Schmidt Rearrangement Claisen Rearrangement Pinacol – Pinacolone Rearrangement Fries Rearrangement Hoffmann’s Brommamide Reaction / Degradation Beckmann’s Rearrangement

NB: Refer Lecture Notes for the same.

CH3 – CH – CH = CH2

OH

CH3 – CH – CH = CH2 + CH3 – CH = CH – CH2

Cl

CH3 – CH2 – CH = CH2 Cl/hv High temp

CH3 – CH – CH = CH2 + CH3 – CH = CH – CH2

Cl Cl


OH

⊕H CH3 – CH = CH – CH2

OH


Cl

(i) Mg/dry ether

(ii) H2O CH3 – CH2 – CH = CH2 + CH3 – CH = CH – CH3



RARE MECHANISMS (NOT mentioned in JEE ADVANCED syllabi) (But a few questions asked in JEE in the past 5 years needed a background of these mechanisms) 1. E1 cb Mechanism It may be argued that a second-order elimination reaction may as well

proceed in two steps as in E1 reaction. The first step involves a fast and reversible removal of a proton from the β -carbon with the formation of a carbanion which then loses the leaving group in the second slow rate-determining step.

The overall rate of this reaction is thus dependent on the concentration of the conjugate base of the substrate (carbanion). Hence, this mechanism has been designated as E1cB (Elimination, Unimolecular from conjugate base).

2. SNi Mechanism : Most common example of this type of mechanism is the mechanism of the reaction between alcohol and SOCl2 It is actually substitution nucleophilic internal.

In this type of mechanism the configuration is retained. i.e., R form of alcohol gives R type of halide and 5 form gives 5 type of alkyl halide. One reaction in which this has been shown to occur is in the replacement of OH by Cl through the use of thionyl chloride, SOCl2:

The reaction has been shown to follow a second order rate equation, Rate = ],][[ 22 SOClROHk but clearly cannot proceed by the simple SN2 mode for this would lead to inversion of configuration in the product, which is not observed.

R – O H + Cl – S – Cl ClSORHCl −−− →−

O O

S = O Cl

R – O S = O

O

Cl

R R – Cl + SO2

C OH

H Ph

Me

SOCl2 C – Cl + SO2 + HCl

H Ph

Me

– C – C - + EtO

H

Fast …………(i) – C – C - + EtOH

Br Br

Θ Θ

– C – C - …………(ii)

Br

Θ Slow – C= C - + Br Θ



3. Neighbouring Group Participation (N.G.P.) A number of nucleophilic substitution reactions are known which occur with complete retention (not inversion or racemisation) of configuration and with unexpectedly greater rate of reaction. In these cases usually there is atom or group with an unshared electron pair β to the leaving group (or sometimes farther away). The mechanism operating in such cases is called neighbouring group mechanism or neighbouring participation. It consists of two consecutive SN2 substitutions with inversion of configuration, thus, the net result is retention of configuration. In the first step of this reaction the neighbouring group acts as a nucleophile pushing out the leaving group. In the second step the external nucleophile pushes out the neighbouring group. A common feature of all neighbouring group mechanisms is the formation of a cyclic intermediate.

(a). Neighbouring carboxylate anion

C – C

Z:

L

Step.1 Slow Intramolecular SN2 (inversion)

Θ− L

C C

⊕Z

Intermolecular SN2 Step 2. Fast (inversion)

ΘNu

C – C

Z

Nu Net result is retention of configuration

O

C

C

Me

H

Br

O

i. Θ Slow Br + O – C

O

C H

Me Θ

OH

Me

C OH

H C

O

O Θ

Protection

Retention

ΘCOO

Me

H NH2

D(-) alanine

ii. HNO2

ΘCOO

Me

H OH HNO2

O

CH

C

O

Me

H2O

COOH

Me

H OH

D(-) Lactic acid (Retention)



(b). Neighbouring halogen atoms : Reaction of HBr on (-) threo – 3 – bromo – butan – 2- ol

If no N.G.P. of Br occurred and the reaction was SN2, complete inversion would have occurred only at C1. If the reaction was SNl, C1 would have been a classical carbonium ion (flat), so inversion and retention (racemisation) would have occurred only at C1. Since either retention or inversion occurs at both C1 and C2, the results are explained by the N.G.P. of bromine atom. Similarly, optically active erythro -3-bromobutane 2 - ol with fuming HBr gives meso-2-3-dibromo-butane.

Br

iii. Me – CH – CH2

C = O ΘO

H2O Θ− Br

Me

O O

Br

iv. Me – CH – CH2 – CH2

C = O ΘO

H2O Θ− Br

O

Me O

OH

+H C1

Me H

C2

H Me Br (-) form

+ OH

C1

Me H

C2

H Me Br

Inversion at C1

OH 2− C1

Me H

C2

H Me

⊕ Br

Θ+ Br (from HBr)

Br

+H C1

Me H

C2

H Me Br (+) form

Br

C1

Me H

C2

H Me Br (-) form

Reacemate



(4) Elimination - Addition Mechanism (Benzyne Mechanism)

In the absence of an electron withdrawing group, nucleophilic substitution takes place in presence of very strong bases, but the mechanism is entirely different from that what we have seen in bimolecular nucleophilic substitution reactions. These reactions proceed by a mechanism called benzyne mechanism the positions. .

Benzyne is a symmetrical intermediate and can be attacked by nucleophile at both.

Isotopic labelling confirmed that there is an equal chance of abstraction from both carbons. An equal halide which does not contain alpha hydrogen with respect to halogen does not undergo this reaction. In the reactions involving/ benzyne intermediates, two factors affect the position of incoming group, the first one is direction of aryne formation. When there are groups ortho or para to the leaving group, then, the following intermediates should be formed.

When a meta group is present, aryne can form in two ways, in such cases.

More acidic hydrogen is removed, i.e. an electron attacking ‘Z’ favours removal of para hydrogen.

X

Θ•• 3NH+

Benzyne

2HN+ NH2

Θ••

NH2

2HN+

••

••

••X

2HN+

••

••

••X

3NH+ Θ••

H – NH2

Z

X → 2HN

Z

or

Z

Z X

→ 2HN

Z

Intermediate

Z

X

→ 2HN

Z

Cl 14 → 2HN

NH2 14

NH2

14



(5) SN’ MECHANISM : Allylic substrates undergo nucleophilic substitution reactions rapidly, and are usually accompanied by a rearrangement known as an allylic rearrangement or an allylic shift. When allylic substrates are treated with nuc1eophiles under SN1 conditions, two products are usually formed the normal product and a rearranged product:

The formation of two products can be easily explained because the allyl cation is resonance hybrid of two structures:

23 CHCHCHRCHCHCHR =−−→←−=−⊕⊕

Thus, C - 1 and C - 2 each carry a partial positive charge, and both are attacked by ΘY (nucleophile) resulting in the formation of two products. This mechanism is called SN1' mechanism. In SN1' reactions at equilibrium more stable product (thermodynamically controlled product) is formed in greater amount, whereas if equilibrium is not reached, less stable product (kinetically controlled) is formed in greater amount. If the double bond in one isomer is in conjugation with an aromatic ring, a triple bond, another double bond or a carbonyl group, then that isomer is more stable and predominates at equilibrium, e.g.,

100% of the I is formed at equilibrium because it is more stable due to conjugation of its double bond with the phenyl ring. If equilibrium is not reached, then II is the major product. Nucleophilic substitution at an allylic carbon may also take place by SN2 mechanism without allylic rearrangement. However, allylic rearrangement can also take place under SN2 conditions through the following mechanism in which the nuc1eophile attacks at the γ -carbon instead of at the usual position. This mechanism is called SN2' mechanism and is an allylic rearrangement:

SN2' mechanism takes place under SN2 conditions where a substitution sterically retards the normal SN2 mechanism.

R – CH = CH – CH2 X ΘY R – CH = CH – CH2Y + R – CH – CH = CH2

Y Normal product

Rearranged product

R – C C – C – X

R

R R

R

ΘY

R – C = C

R Y

R R

R

Ph – CH = CH – CH2 – X OHΘ

Ph – CH = CH – CH2 – OH + Ph – CH – CH = CH2

OH 100% at equilibrium I

major product if equilibrium not reached

II



(6) IPSO reaction : A position which is already occupied by a non-hydrogen substituent in an aromatic ring is called ipso position (Latin: ipso, on itself), the attack on this position is called ipso attack (or ipso addition), and the aromatic substitution in which a substituent already present is replaced is called ipso substitution. For example, protodealkylation of an alkylbenzene (reverse of Friedel-Crafts alkylation). In this reaction, tertiary alkyl groups are most easily removed, since they depart as stabler carbocations. Thus, t-butyl group is used to protect the most reactive position in a compound to effect reaction elsewhere. The mechanism is as follows :

⊕

⊕+=→ HCHCCHCCH 22333 )()(

CONCEPT TESTING QUESTIONS :

Details :- ,/,,,,,,, 21

21 ESSNEESEArSNArSNESNSN free radical mechanism. AdN, AdE, Ad free radical, rare Mechanism etc. Note :-

1. In all the questions below, ‘product’ means major product only 2. Both students and teachers should write the detailed mechanism where ever possible to

understand the answers. 1. Substitution Reactions Q.1 How many transition states are involved in 2SN reaction ? (a) 1 (b) 2 (c) 3 (d) 0 Ans.1 (a) It is one step second-order reaction & hence only one T.S. Q.2 How many intermediates are involved in 1SN mechanism? (a) 0 (b) 1 (c) 2 (d) 3 Ans.2 (b) It is two-step mechanism with one intermediate as carbocation. Q.3 Among the following nucleophiles, which of the following order is correct for their nucleophilicity in water ?

(a) ΘΘΘΘ

>>> FOHNHCH 23 (b) Θ

ΘΘΘ

>>> 32 CHNHOHF

(c) Θ

ΘΘ

Θ >>> 32 CHFNHOH (d) ΘΘ

ΘΘ >>> OHNHFCH 23 Ans.3 (a) From left to right in periodic table, nucleophilicity decreases.

C(CH3)3

⊕H

ipso attack

C(CH3)3 H

+

Electrophilic ipso substitution

33 )(CHC⊕

+



Q.4 Give correct order of leave ability of the anions : −−−− MeCOOROSOCNMeO ,,, 2 (a) −−−− <<< MeCOOROSOCNMeO 2 (b) −−− <<< MeCOOROSOCNMeO 2

2 (c) −−−− <<< 2ROSOMeCOOCNMeO (d) −−−− <<< MeOCNROSOMeCOO 2 Ans.4 (c) Good L – G are conjugate bases of strong acids. Q.5 There are two reactions (i) ROH + NaBr (ii) ROH + HBr. Which is not possible? (a) (ii) (b) (i) (c) both possible (d) both not possible Ans.5 (b) −Br V – weak base, −OH bad L – G – reaction (i) not possible – In reaction (ii) ROH is converted to ROH2 & hence RBr possible Q.6 List the following nucleophiles in order of decreasing order of 2SN : (i) −−− OMeCHiiiMeOiiCOMe 23 )(,)(, and

(a) iv > iii, > ii, > i (b) iv > ii, iii > i (b) i > iii > iv > ii (d) ii > iii > iv > i

Ans.6 (d) Bulkier ,ΘNu steric hindrance and hence 2SN reactivity decreases Q.7 Arrange the following ΘNu in correct order of nucleophilic strength : (a) −−− >>> OHCCHOCOOCHOH 5633 (b) −−−− <<< OHOCHOHCCOOCH 3563 (c) OHCOCOCHOCHHO 5633 >>> (d) OCHHOOHCOCOCH 3563 <<< Ans.7 (d) Acetic acid CH3COOH & Phenol C6H5OH are acidic. Anions are stabilized by resonance. 3CHO and −OH are not having resonance. Q.8 The order of decreasing nucleophilicity of the following species is : (a) OHCHCOOCHOCHSCH 3333 >>> ΘΘ

(b) OHCHOCHSCHCOOCH 3323 >>>Θ

ΘΘ (c) ΘΘΘ >>> OCHCOOCHSCHOHCH 3333 (d) ΘΘΘ >>> SCHCOOCHOHCHOCH 3333

(iv) −O



Ans.8 (a) Anion better ΘNu than neutral molecule. Nucleophilicity increases as electronegativity of donor atom decreases.

Q.9

Ans.9 (a) Q.10 Arrange the following as increasing order of their ability as L – G (leaving group) )(&)()(),( 333 SFRCFQOCHPSCH −−−− (a) PRQS <<< (b) SRQP <<< (c) SPQR <<< (d) PQSR <<<

Ans.10(d) Conjugate base of strong acid is L – G (d) CHF3 will be very strong acid & hence Conjugate base will be very weak.

Q.11 Which of the following will react with EtONa to give only substitution & no elimination product?

Ans.11 (d) no H−β for elimination

Q.12 Which of the following halide is almost inert in SN or E reaction?

(a)

Br

(b)

Br

(c)

I

(d)

Br

(a)

Br

(b) Br

(c) Br (d)

Br

⊕

rearranges to

⊕ (b) (a)

⊕

(c)

⊕

(d) ⊕



Ans.12 (c) Positive charge at bridge-head carbon is least likely & hence inert. It is C°3 also Q.13 Give correct order of nucleophilicity of )(),( QClPF −Θ and )(RBr − in .DMSO (a) QRP >> (b) RQP >> (c) RPQ >> (d) PRQ >> Ans.13 (b) No solvation by polar aprotic solvent for anions. Basicity parallels nucleophilicity. Q.14 For stability of carbocation, which is the correct order out of the following? (a) −2,1 Ph shift > 1, 2 – Me shift > 1, 2 hydride shift (b) 1, 2 Ph shift > 1, 2 ΘH shift > 1, 2 - <e shift (c) 1, 2 H shift > 1, 2 Me shift > Ph2,1 Shift (d) 1, 2 Me Shift > 1, 2 Ph shift > 1, 2 −H shift Ans.14 (a) Q.15 Get correct order of nucleophilicity (a) 3NrBHOHS >>> (b) rBNHOHS >>> 3 (c) HOrBNHS >>> 3 (d) HOrBHSN >>>3 Ans.15 (b) Less EN, higher nucleophilicity and larger size, higher nucleophilicity. Q.16 Arrange the following in decreasing order of nucleophilicity in ethanol −−−−

)()()()( ,,, SRQP IBrClF (a) SRQP >>> (b) SPQR >>> (c) PQRS >>> (d) PSRQ >>> Ans.16 (c)

Ethanol is polar protic solvent and hence nucleophilicity is proportional to size of atom. In polar aprotic solvents, larger size ions are better solvated & hence nucleophilicity decreases with increase of size.

Q.17 Which of the following is most likely to undergo a favourable hydride shift ?

Ans.17 (a) °3 carbocation produced Q.18 In decarboxylation of acid, intermediate is ____ (a) carbocation (b) carbanion (c) carbonium ion (d) carbene Ans.18 (b)

CR −

O

O

Θ CO2

⊕

(a) (b) ⊕ (c) ⊕

⊕

(d)



Q.19 What is the decreasing order of strength of bases ?,,, 232ΘΘ−− ≡ CHCHCCHNHOH

(a) ΘΘΘΘ >≡>> OHCCHNHCHCH 223 (b) ΘΘΘΘ >>>≡ OHNHCHCHCCH 223 (c) ΘΘΘ− >≡>> 232 CHCHCCHNHOH (d) ΘΘΘΘ >>≡> 232 CHCHOHCCHNH Ans.19 (a) Conjugate base of strong acid is weak. Acidic strength HOH > CH ≡CH > NH3 > C2H6 Q.20 ΘΘ +→+ BrNuCHNuBrCH 33 . The decreasing order of rate of above reaction

with ΘNu A to D is ____ [ −−−−Θ ==== MeODOHCAcOBPhOANu ,, ] (a) A > B > C > D (b) B > D > C > A (c) D > C > A > B (d) D > C > B > A Ans.20 (c) Stronger acid, weaker conjugate base. Q.21 In the compound given below, the correct order of acidity of positions x, y and z is

(a) z > x > y (b) x > y > z (c) x > z > y (d) y > x > z Ans.21 (b) Electron withdrawing group (EWG) will increase positive charge of nitrogen. 2. Free Radical Reactions Q.22 The no. of possible enantiomers (pairs) can be produced during monochlorination of

2 – methyl butane is ___ (a) 3 (b) 4 (c) 2 (d) 1 Ans.22 (c)

COOH X

⊕yNH 3 z

HN 3

+

Cl

* &

Cl

*



Q.23

How many monobrominated products will be obtained by above reaction? (a) 6 (b) 4 (c) 5 (d) 3 Ans.23 (d) 13,12,11 →°→°→° HHH and compound is symmetrical. Q.24

rh

Br→ 2 Products

How many products (all isomers) are possible? (a) 1 (b) 6 (c) 4 (d) 3 Ans.24(b) 2 products optically active (d, l) 2 products non active – Total products = 2 + 2 + 2 = 6. Q.25

Ans.25 (b) °3 carbon free radical is more stable than °1

CH3 CH3

H

H

H H

rh

Br→ 2 Product. It can be

rh

Br→ 2 Products

C CH2Br

CH3

H

CH3 (a) C CH3

CH3

Br

CH3 (b)

(c)

Br

(d)

Br



Q.26 How many total products will be obtained by monobromination of 2 – methyl butane and how many can be separated by fractional distillation ?

(a) 6, 4 (b) 5, 4 (c) 6, 2 (d) 4, 2 Ans.26 (a)

Q.27 Consider the following reaction :

Identify X among the following

Ans.27 (b) °3 carbon free radical is most stable. Q.28 The relative reactivity of H°°° 3&2,1 bromination has been found to be

1600:82:1 respectively.

% of A & B can be ____ (a) 99% 1% (b) 50%, 50% (c) 1%, 99% (d) 80 %, 20%

Br Br2

λh + *

Br

+ *

Br

Br

+

*

Br

& hv

+ Br X + HBr .

D

D

2CH •

(a)

D

• (b)

D

CH 2 •

(c)

D

• (d)

+ Br2 λh

Br A +

Br B



Ans.28(a) Reactivity of 9911,16003 =×⇒°=° H There are 9H as °1 and 1H as .3°

∴% of A = %4.9910016091600

=×

Q.29 The relative reactivity of °° 2,1 and H°3 in chlorination has been found to be 1: 3.8 : 5

The ratio of amount of products A, B, C & D will be ____ (a) 1 : 3.8 : 5 : 1 (b) 3 : 7 – 6 : 5 : 6 (c) 3 : 7. 6 : 5 : 3 (d) 1 : 7.6 : 5 : 1

Ans.29(b) Reactivity 8.322,91 ×=°=° HH 53 ≡°H % of 56.79

1003++

×=A and hence ans is (b)

Because for % C = 56.79

1005++

× and % 56.79

1006++

×=D

Q.30 How many isomers (including stereo) can be obtained by chlorination of

2, 2, 4 – trimethyl pentane? (a) 3 (b) 4 (c) 5 (d) 6 Ans.30 (d)

2 optically active (4) 2 optically inactive Total 6 Q.31 How many dichlorocyclohexanes would be obtained on chlorination of chlorocyclohexane (including stereo)? (a) 4 (b) 6 (c) 8 (d) 9 Ans.31 (d)

Cl

Cl

1 Cl

Cl

2 +1

3 Cl

Cl

Cl Cl +

(2) (geo + opt) (2 geo + 1 opt) (geo) = total 9 (d)

h v 2Cl Cl +

(A) Cl (B)

+

Cl

(C)

+ Cl

(D)



Q.32

How many monohalogenated products are possible? (a) 3 (b) 4 (c) 5 (d) 7 Ans.32 (c)

at a, b & d possible 3 at c optically active 2 total 5 Q.33 How many monochlorinated product may be obtained when alkane shown below

reacts in presence of Cl2 and uv ?

(a) 1 (b) 2 (c) 0 (d) 3 Ans.33 (a) All ‘H’ are °1 only. Q.34

Bromination will not occur at (a) 1 (b) 4 (c) 3 (d) 2 and 5 Ans.34 (d) Benzylic carbon undergoes free radical bromination with NBS.

2

3

1

4

5 CTC NBS, hγ

CH

H3C

H3C CH2 – CH3

a

c b d

CH

H3C

H3C CH2 – CH3

X2

hλ



Q.35

Ans.35 (a) Allylic free radical bromination

Q.36 4 – methyl – 2 – pentene is brominated with NBS. The major product is (a) Br CH2CH = CH – CH – Me2 (b) 22 CHCHBrCHMeCH = (c) 22 CHCHBrCHMeCH = (d) 23 CMeCHBrCHCH = Ans.36 (d) More substituted product and allylic C°3 involved. Q.37 3 – methyl methyl enecyclohexane is brominated by NBS. Total products will be ___ (a) 6 (b) 4 (c) 5 (d) 3 Ans.37 (a)

At α and β position rearrangement of π bond is possible along with bromination.

Cis and trans products are also possible. Total products = 3 + 3 = 6

Q.38 Arrange bonding energy of C – H bonds in increasing order for following compound:

(a) b < d < c < a (b) b < c < d < a (c) b < a < c < d (d) c < a < b < d

H

H

d

H

H b

a

c

(a)

Br

(b)

(c) (d)

NBS hλ

?

Br

Br

Br

Br

CH2

β

α



Ans.38 (b) adcb <<< °3 °2 °2 Vinyllic allylic allylic Q.39

→NBS A major product. Bromination takes place at

(a) a (b) b (c) c (d) d Ans.39 (a) Allyic bromination (free radical mechism). And C°3 also

Q.40 BrCHCHCHHBrCHCHCH OH22323

22 →+=− The mechanism is ____

(a) electrophilic addition (b) free radical substitution (c) free radical addition (d) electrophilic substitution Ans.40 (c) Anti – Markownikov’s addition. Q.41 When 22 CHCHCHR =− reacts with NBS, the mechanism is ____ (a) AdE (b) free radical substitution (c) AdN (d) Elimination Ans.41 (b)

gives Br and it shows allylic substitution.

Q.42 The compound is

FEDCBACHCHCHCHCHCHCH 23223 )(−−−=−

Arrange them in decreasing order of reactivity towards radical substitution : (a) C > A > E > D > F > B (b) F > B > A > C > D > E (c) B > C > A > F > D > E (d) A > B > C < D > E > F Ans.42 (a) Allylic substitution

O

O

NBr

NBS

CH3 H a

b

d c



3. SN1 Mechanism Q.43

is subjected to solvolysis. In which solvent, the racemization will be maximum?

(a) 80 % acetone & rest water (b) 100 % water (c) 100 % acetone (d) 50 % acetone 50% water Ans.43 (b) Solvolysis is .1SN Polar protic solvent is required. Acetone is polar aprotic Q.44 ΘΘ +→+ XNuGGXNu . This mechanism can be elimination-addition. In organic displacement reaction, the mechanism is called as (a) 1SN (b) 2SN (c) 1E (d) 2E Ans.44 (a) It is 1SN because bond breaking is RDS (energy of activation larger). In rate equation

only GX will be involved Q.45 1SN reaction mechanism is favoured by (a) high temperature (b) Powerful base (c) solvent with higher dielectric constant (d) substrate with bad L – G Ans.45 (c) 1SN required ionization of covalent bond which is encouraged by solvent of higher dielectric constant. Q.46 Correct order of reactivity towards 1SN in case of BrPh − (P), BrCHCHPhQBrCHPh 222 ),( −−− (R) )()( 3 SCHBrCHPh − is (a) PRQS >>> (b) QPRS >>> (c) SRQP >>> (d) PQRS >>> Ans.46 (a) ,2° benzyllic (s) where as P can’t have stable carbocation. Q has °1 benzyllic carbocation Q.47 Arrange the following in decreasing order of 1SN

(a) SRQP >>> (b) RQPS >>> (c) SPQR >>> (d) SQPR >>> Ans.47 (c)

I

P

I

Q R

I

S

I

H

Br



Q.48 Arrange the following in decreasing order of 1SN :

(a) P > Q > R > S (b) Q > R > S > P (c) Q > P > R > S (d) Q > R > P > S Ans.48 (d) Q is °2 allylic where as S is vinyllic carbocation which is highly stable R is °1 allylic Q.49 Arrange the following in decreasing order of

),(),(

2|3

231

QClCHNCHPClOCHCHSN

H

(a) SRQP >>> (b) SPQR >>> (c) SRPQ >>> (d) RSPQ >>> Ans.49 (c) Electron-donating groups increase .1SN

Lone pairs can be donated to stabilized by carbocation.

Q.50 In the following compound, arrange the reactivity of different Br atoms towards NaSH in decreasing order :

(a) SRQP >>> (b) RPQS >>> (c) RPSQ >>> (d) RQSP >>> Ans.50 (c) No 2SN at bridge head. Less sterically-hindered & °2 will show .2SN Q.51 1SN reaction as hydrolysis is catalysed by (a) NaOH (b) H2O (c) AgNO3 (d) Pt Ans.51 (c) Precipitation of AgCl facilitates formation of carbocation

P

Br

Q

Br

R

Br

S Br

),(2 RClCHPh −

Cl (S)

Br (P)

O

Br (Q) Br (S)

Br (R)

O



Q.52 1SN reaction occurs most rapidly in which of the following cases?

Ans.52 (c) −I is better LG and °3 Carbocation is formed. Q.53 Which of the following is most reactive alkyl halide for ?1SN

Ans.53 (d) °3 resonance stabilized carbocation Q.54 Arrange the following in increasing order of their reactivity in 1SN

(a) PSQR <<< (b) PQSR <<< (c) RPSQ <<< (d) RSQP <<< Ans.54 (a) −I better LG & carbocation is stabilised by lone pair of oxygen. Q.55

Ans.55 (a) Nu is not powerful - +°C2 can be stabilised by −H shift

Cl

HOH Major product will be

(a) OH

(b)

OH

(c)

OH

(d)

I

O P)

Cl

O Q)

I

O

R)

Br

O S)

(a) Br (b) I (c)

I

CH3

(d)

Br

CH3

Cl

(a) Cl (b) (c) (d) Cl

Cl



Q.56 Which is correct?

Ans.56 (c) Actually (a) will show inversion and (b) will show racemisation & for (c) 1SN Q.57 Give correct order of rate of 1SN for

(a) SRQP >>> (b) SPQR >>> (c) SQRP >>> (d) SQPR >>> Ans.57 (a) Stability of carbocation. Q.58

Ans.58 (b) Benzyllic carbocation by rearrangement

OH

HCl P. Product P can be

Cl

(a)

Cl

(b)

Cl

(c) (d)

H

Br

Et

D

−EtO Racemisation (a)

(b)

Br

Me C2H5OH Walden inversion

(c)

Et

Me

Ph

Br EtOH Racemisation

(d) All are correct

Cl

P

Q

Cl

CHCl Ph

Ph R

Me2CHCl S



Q.59

Ans.59 (c) 1SN mechanism 4. SN2 Mechanism Q.60 ΘΘ +→+ XNuRNuRX . For this reaction following data is available.

[RX] [ ΘNu ] Rate 0.1 0.1 4102.1 −× 0.2 0.1 4104.2 −× 0.1 0.2 4104.2 −× 0.2 0.2 41040 −×

In this case, we can have (i) stability of carbocation (ii) polar aprotic solvent (iii) complete inversion (iv) rearrangement

Select correct options. (a) i, ii, iiii, iv (b) ii, iii (c) i, ii, iv (d) ii, iv Ans.60 (b) The kinetics indicates 2SN Q.61 2SN rate is – (a) directly proportional to LG (b) directly proportional to ΘNU (c) inversely proportional to steric hindrance (d) directly proportional to dielectric constant Ans.61 (b) [αr substrate] ][ ΘNu

Br

CH3OH C°25

Product will be

(a)

OMe

(b) C

CH3

(c) (d)

OMe



Q.62 Arrange the following in decreasing order of .2SN

(a) RQPS >>> (b) PRSQ >>> (c) SPRQ >>> (d) SPQR >>> Ans.62 (d) °1 prefers .2SN Steric hindrance will decrease .2SN Q.63 Arrange the following in decreasing order of :2SN

(a) QPRS >>> (b) PQRS >>> (c) PQSR >>> (d) QRPS >>> Ans.63 (a) °3 will not prefer °1,2SN will always prefer 2SN due to steric problems Q.64

This can be described as - (a) 1SN with racemisation (b) 2SN intermolecular with inversion (c) 2SN intramolecular with retention (d) 2SN Intramolecular with inversion Ans.64 (d) rB as good L- G. Lone pair of O atom makes cyclic compound of 5 atoms by removal of proton due to heating. It is Walden inversion. Q.65 3 – Bromo – 2 – methoxy butane undergoes 2SN reaction with .MeO Can we have product ___ ? (a) optically active (b) meso (c) enantiomer (d) homomer Ans.65 (b)

O

∆ NaOH

Br OH

Cl (P), Cl

(Q) Cl

(R) Cl (S)

Br (P),

Br

(Q), CH3Br (R), CH2 = CH - Br (s)



∴meso form (optically inactive) Q.66 Which of the two iodines will be more reactive in 1SN and 2SN reaction?

(a) A will be faster in 1SN but slower in 2SN (b) A will be faster. Both in 1SN and 2SN (c) A & B will be equally reactive. (d) B will be faster in 1SN and 2SN Ans.66 (b) Allylic halide has greater reactivity in both 1SN and 2SN Q.67 Which of these compounds could be used to establish stereo-specificity of ?2SN

Ans.67 (c) It has a chiral carbon Q.68 Give correct decreasing order of 2SN for following

(a) ADBEC >>>> (b) ADEBC >>>> (c) BADEC >>>> (d) DEBAC >>>> Ans.68 (b) Good L – G, °1 & steric factor to be counted.

Cl (A) Cl (B) Br (C)

Cl

(D) Cl (E)

(a)

Cl

(b)

Cl

(c)

Cl

(d) a & b

I (A)

I (B)

OMe

H3C H

H

Br

CH3

OMe

H3C H

CH3

OMe

H



Q.69 Which of the following is most reactive towards ?2SN

Ans.69 (d) (a), (b), (c) can show 1SN due to stability of carbocation (benzylic) but there is no chance of stability of carbocation in (d) due to (–R) effect. Q.70 (s) – 2 – bromobutane is obtained from (s) – 2 – chlorobutane when (a) 1SN and 2SN are used (b) 12SN steps are used (c) 1SN and iSN are used (d) 22SN steps are used Ans.70 (d) Cl can not be replaced by Br in one step Q.71 Consider the following for reaction with HBr :

The order of decreasing reactivity will be ____ (a) QSPR >>> (b) SQPR >>> (c) QSRP >>> (d) SQRP >>> Ans.71 (a) Stabilising of carbocation due to +R effect. Q.72 Dehydration of excess alcohol at high temperature in presence of acid gives the

product by which mechanism ? (a) 2SN (b) 1E (c) 2E (d) AdE Ans.72 (a) Ether product by substitution.

(P)

CH2OH CH2OH

NO2

(Q)

CH2OH

OMe

(R)

CH2OH

Br

(S)

Cl

(a)

Cl

(b)

Cl

(c) Cl

Cl

(d) NO2



Q.73

The reaction is ____ (a) AdN (b) AdE (c) SN (d) E Ans.73 (c) Q.74 Which of the following is more reactive towards AgNO3?

Ans.74 (b) Carbocation stability after −Br is leaving Q.75 Which of the following will not react with Na metal?

Ans.75 (c) Stability of anion is not possible after reaction due to anti-aromatic nature. Q.76 2SN reaction at an asymmetric carbon of a compound always gives (a) an enantiomer of substrate (b) a product with opposite optical rotation (c) a mixture of diastereomers (d) a single stereoisomer Ans.76 (d) In 2SN , there is inversion. ΘLG is replaced by different ΘNu . Q.77 The product produced will have

(a) SN2 mechanism (b) SNAr mechanism due to activator (c) SEAr mechanism due to activator (d) SN1 mechanism Ans.77 (a) Nucleophile is strong and solvent is polar aprotic.

F

Br Me

NO2

ΘPhS DMF

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

OH

(c)

Br

(a)

O Br

(b)

O

Br

(d)

Br

O CH3 HBr OH + CH3Br



5. SN1 v/s SN2 Mechanisms Q.78

where, hydrolysed in aq. acetone will give (i) substitution by −OH at position 2 (ii) substitution by −OH at position 1 (iii) substitution by −OH at position 3 (a) mixture of (i) and (ii) (b) mixture of (i) and (iii) (c) mixture of (ii) and (iii) (d) only (i) Ans.78 (a) SN mechanism, having OCH3 as activator to stabilize carbocation

H3CO 2 1 3

NO2

CH3 CH3 CH3

H Cl CH3



6. E1 Mechanism Q.79 Find number of possible 1E products from following reaction : (a) 3 (b) 4 (c) 5 (d) 2 Ans.79 (b)

Q.80 Which of the following compounds will give E1 reaction?

Ans.80 (e) All can form stable carbocation Q.81

EtOH

tONaE →

How many total alkenes can be made? (a) 4 (b) 5 (c) 3 (d) 6

Br

C Et

Br

(a) (b) Ph – CH – CH3

Br

(c) H2C = CH – CH – CH3

Cl

(d) Cl

(e) All

CH3OH ∆

Br

, (2) (2)



Ans.81 (b)

Q.82 Which product is formed in E1 reaction of 1 – bromo -2, 3 – dimethyl cyclohexane? (a) 3, 4 – dimethyl Cyclohexane (b) 2, 3 – dimethyl Cyclohexane (c) 4, 5 – dimethyl cyclohexene (d) 1, 2 – dimethyl cylohexene Ans.82 (d)

Q.83 How many products can be formed by E1 reaction of 1 – chloro – 2, 2 – diphenyl propane ? (a) 2 (b) 1 (c) 3 (d) 4 Ans.83 (d)

Q.84 At higher temperature, 1E chances are more than 1SN . This is due to (a) T. S. are increased (b) Endothermic reaction favourable at high temperature. (c) Nucleophile is converted to base (d) Carbocation is stabilised Ans.84 (b) Elimination is bond breaking & hence endothermic.

E1

Br ⊕

−H Shift

Ph

Ph

Cl

Ph Ph

(2)

Ph Ph

(1)

Ph

Ph (1)

(1) (2)

(2)



Q.85 For the reaction,

Ans.85 (c)

Q.86 The rate of hydrolysis of following compounds by 1SN is :

(a) 2 > 1 > 3 > 4 (b) 3 > 4 > 1 > 2 (c) 4 > 2 > 1 > 3 (d) 2 > 1 > 4 > 3 Ans.86 (c) Q.87

cannot undergo the reaction : (a) 1SN (b) 2SN (c) 1E (d) ES Ans.87 (c) There is no β - H on ring for elimination. Q.88 Which alkyl halide would you expect to be least reactive in E1 reaction ? (a) 2|2 CHCHCHCH

Br

−−= (b) 3|CHCHPh

Br

−−

(c) 322 CHCHCCHBr|

−−= (d) BrCHCHPh 22 −−

Ans.88 (c) Vinylic carbocation is most unstable and hence no chance of E1.

Cl

CH2Cl

(I) (II)

CH2Cl

(II) CH2Cl

OMe

(IV) OMe

CH2Cl

Cl

CH2OH C – H2SO4 ∆

the major product is __

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

22 HOCH+

2HC+

C ⊕



7. E2 Mechanism Q.89 Which of the following is expected to give more than 2 products in E2 reaction?

Ans.89 (d) & (a) (a) 3→ (b) pair of enantiomer (d) 4 (Count cis & trans isomers also) Q.90 Which of the following cannot undergo E2 reaction?

Ans.90 (c) There is no H atom at β position Q.91 2 – bromopentane + →∆

−EtO The total products are (a) 2 (b) 4 (c) 3 (d) 1 Ans.91 (c) Trans & cis – 2 – pentene & 1 – pentene Q.92 E2 dehydrobromination of −− 3,2),( RR dibromobutane can make (a) E product (b) Z product

(c) both E & Z (d) none of these Ans.92 (b)

(a) Br (b)

Br

(c) Br

(d) None

H

H

CH3

CH3

Br

Br

≡

CH3

H

H

Br

Br

CH3

≡

CH3

H

H

Br Br

CH3 :ΘB

Br

Me CH3

H anti

2 – bromo – 2 – butane

(z)

Cl

(a) (b) Cl Cl

Cl

(c)

Cl

(d)



Q.93 Meso – (R, S) – 2, 3 – dibromobutane is subjected to E2 dehydrobromination. The product can be

(a) E (b) trans (c) Z (d) none of these

Ans.93 (a)

Q.94

In the above reaction, maximum Saytzeff product will be obtained when x is ____ (a) – I (b) – Cl (c) – Br (d) – F Ans.94(a) −I is powerful base °2 carbon, anti-elimination product possible since X is simple.

L > G.

Q.95

Ans.95 (b) trans-product as major

H

H

CH3

Br

Br

CH3

H

Br

H

CH3

CH3

≡

Br

H

Me

Me (E)

meso (R,S)

Br

X

E2 +

Saytzeff product

Hofmann product

CH3

CH3

Ph

Br

H

Ph

alc. KOH

Product can be

Ph Ph

CH3 H3C

(a)

Ph CH3

Ph H3C

(b) C = C

Ph CH3

CH3 Ph

(c)

Ph CH3

CH2 Ph

(d) C C



Q.96

The product obtained when it is subjected to E2 reaction will be

Ans.96 (a) No carbocation, no rearrangement. Q.97

is subjected to E2. The product will be

Ans.97 (b) Anti-elimination (- HBr) Q.98

is subjected to E2 reaction Major product will be

Ans.98 (c) No anti-elimination @ 1, 2 position. Q.99 Which reaction is steroselective and regioselective too ? (a) 2SN (b) 1E (c) 2E (d) AdN Ans.99 (c) More substituted alkene and trans-product predominantly.

(a) (b) (c) None (d)

Br CH3

CH3

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

Br

CH3

CH3

(a) (b) (c) (d) All

Br



Q.100

DMF

PhSNa → The reaction mechanism is

(a) SNAr (b) 2SN by inversion (c) 1SN (d) 1E Ans.100 (b) Polar aprotic solvent & rB as LG Q.101

→NaOEt Major product will be

Ans.101 (a) E2 shows anti elimination. Apply wedge & dashed line structure concept. Q.102 Major product obtained in reaction of 1 – phenyl – 2- bromobutane with NaOMe is : (a) (E) – 1 – phenyl but - 1 – ene (b) (E) – 1 – phenyl but - 2 – ene (c) 1 – phenyl – 2 – ethoxybutane (d) (z) – 1 – phenyl but – 2- ene

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

CH3

Cl

F

Br

NO2



Ans.102 (a)

Q.103 Rank the following in order of decreasing rate of an E2 reaction

(a) P > Q > R (b) R > P > Q (c) R > Q > P (d) Q > P > R Ans.103 (b) Anti elimination Q.104

Ans.104 (b) The given compound is cis & hence

Anti coplanar elimination will take place and hence product (b).

If the compound is trans, then the answer will be (a)

Q.105 A highly hindered base gives (a) E1 (b) E2 (c) SN1 (d) SN2 at high temperature irrespective of nature of substrate Ans.105 (b) E2 only

Br

H H

CH3

E2 The major product is

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

H

H CH3

H H

Br

CH3

Br

(P)

Br

(Q)

Br

(R)

CH3 CH3

Br

Ph E Ph



Q.106 Which of the following can be decarboxylated most readily on being heated?

Ans.106 (d) −β keto acid. Hence, easily decarboxylated by heating 8. E1 v/s E2 Mechanisms Q.107 Which one of the following will most readily be dehydrated in acidic medium?

Ans.107 (a) Conjugation

Q.108

Ans.108 (b) Hoffmann’s elimination

CH3I AgOH, ∆

CH3I AgOH, ∆

y. Here Y can be ____

N |

CH3

(a)

N Me

Me

(b)

(c)

3)(MeN⊕

(d)

(a)

O OH

(b)

OH

(c)

O

OH

(d)

O

OH

O

OH O

(a)

O OH

HO

(b)

O

O HO

O

(c) O

HO

(d)

O O OH



Q.109 Hofmann’s elimination is possible when _____ (a) base is bulky (b) L.G. is very good (c) L.G. is very bad (d) alkyl halide contains one or more double bonds.

Find incorrect one Ans.110 (b) Q.111

How many isomers of P are possible? (a) 2 (b) 5 (c) 6 (d) Ans.111 (b) Dehydration gives 3 products. AdE of these products will give 4 optically active products and one meso product. 9. Substitution v/s Elimination Reactions Q.112

→+ DMSOOCH 3 Which of following pathways is predominantly followed?

(a) 1SN (b) 2SN (c) 11 /ESN (d) 22 /ESN Ans.112 (d) Polar aprotic solvent is used Q.113 t – butyl chloride can react with alcoholic ONaCH 3 . Select the correct statement : (a) It must be 2SN reaction to make ether (b) It must be 2E reaction to form isobutene (c) It must be 2E reaction at low temperature (d) It must be 2SN reaction at high temperature Ans.113 (b) No steric hindrance in 2E .

Br

CH3 OH

CH3 +H

)( 2OH

2Br

CTC 284 BrHC (P)



Q.114

→EtOH Give total products (including stereoisomers) of 11 & ESN (a) 5 (b) 6 (c) 7 (d) 4 Ans.114 (b)

Q.115 The order of reactivity of alkyl halide towards elimination is ____ (a) °>°>° 123 (b) °>°>° 312 (c) °>°>° 213 (d) °>°>° 321 Ans.115 (a) No steric hindrance in E & more substituted alkene as major product. Q.116 The correct stability order of the following species is :

(a) II > IV > I > III (b) I > II > III > IV (c) II > I > IV > III (d) I > III > II > IV Ans.116 (d) Resonance stabilization due to lone pair of oxygen.

⊕ O I

⊕

II

⊕

III

O ⊕

IV

Br

OEt (1)

OEt (2)

(1)

(1)

(1)



10. Aromatic Substitution Electrophilic Mechanism Q.117

Ans.117 (b) More activating group (– CH3) & hence ortho product para position is already blocked. Q.118

.3 AAlCl

∆ → A can be ____

Ans.118 (a) It is a Friedal-Crafts alkylation (intramolecular at ortho position) (a)

OMe

(a)

OCH3

(b)

(c)

OCH3

(d)

C – HNO3 + H2SO4 Major product can be

(a)

NO2

(b)

O2N

(c)

NO2

(d)

NO2

OCH3

Cl



Q.119

P can be ______

Sol.119 (c) F/C acylation as EAS. The group (–CH2) will be activator & hence ⊕E attack at ortho position. Para position blocked. Q.120

Nitration is possible at (a) A (b) B (c) C and (d) D Ans.120 (d) 3OCH− activator for EAS & attack of ⊕E possible at ortho position Q.121 Which of the following cannot be electrophilic substitution? (a) nitration (b) sulphonation (c) bromination (d) ammonolysis Ans.121 (d) Electrophile is not possible from NH3

.33 / PHOH

AlClCoClCH →

C

O

(a)

C

O

(b)

C

O

(c)

C O

(d)

A B D C

OCH3

O



Q.122

(i) CF3 is activator (ii) CF3 is deactivator (iii) CF3 is m – directing (iv) CF3 is o/p directing Select the correct options : (a) (i) and (iv) (b) (ii) and (iii) (c) (i) and (iii) (d) (ii) and (iv) Ans.122 (b) Electron withdrawing effect Q.123 In the following reaction,

The major product X will be produced by +

2NO when substituted at (a) position 1 (b) position 2 (c) position 3 (d) position 4 Ans.123 (d) p – position for activator

3 NH

O 1

2

C – HNO3

C – H2SO4 X.

4

CF3

⊕E Product



11. Aromatic Substitution Nucleophilic Mechanism Q.124

P can be

Ans.124 (a) NO2 is activator for SNAr and para position only suitable here because L – G is only at para. 12. Addition Nucleophilic Mechanism Q.125 Addition reaction is not shown by (a) alkene (b) alkyne (c) alcohol (d) aldehyde Ans.125 (c) Alcohol does not have any π bond Q.126 Which of the following will react with HCN/NaCN? (a) Alkane (b) alkene (c) Alkyne (d) benzene Ans.126 (c) Alkyne shows AdN due to −CN . Q.127 Alkynes are more reactive than alkenes towards AdN because of stability of (a) vinyl carbanion (b) °2 alkyl free radical (c) vinyl carbocation (d) °2 carbocation Ans.127 (a)

NUCCNuCC −=−→+−≡−ΘΘ |

Q.128 For −−=−=−=−⊕

|1

|

2

|

3

|

4

|

5

|

6

|

7CCCCCCC , thermodynamically controlled product due to ΘNu is

possible at (a) 51 ,CC (b) 71 ,CC (c) 31 ,CC (d) 42 ,CC

F

Cl

NO2

.PKSH

∆→

Cl

SH

(a)

NO2

(b)

F

SH

NO2

(c)

SH

SH

NO2

(d)

Cl

SH

NO2



Ans.128 (b) conjugated triene will be produced Q.129 Grignard reagent when reacted with carbonyl compound works as (a) ΘNu (b) ⊕E (c) free radical (d) cation Ans.129 (a) XRM9 makes ΘR . The reaction mechanism will be AdN. Q.130

The relation between 1K and 2K is (a) 21 KK = (b) 21 KK > (c) 21 KK < (d) 121 == KK Ans.130 (b) Reverse reactivity for AdN. Aldehydes are more reactive than ketones for AdN. Q.131 Aldol condensation reaction of aldehyde or ketone can be explained as ____ (a) AdN (b) AdE (c) SN2 (d) E1 Ans.131 (a) Carbanion produced by NaOH works as ΘNU for other molecule of carbonyl compound. Q.132 Tautomerism is not shown by

Ans.132 (b) No −α H on CSP 3

(a) Ph – CH = CHOH (b) O O

(c) O

O

(d) O

O

OH

CH3 – C - CN

CH3

O

+ HCN K1

OH

CH3 – C - CN

H

O

+ HCN K2 H



Q.133

Ans.133 (c) ΘNu attacks at less substituted carbon atom (basic). Q.134 In case of epoxide reaction with ,ΘNu in presence of acid the attack of ΘNu is on (a) less substituted ‘C’ (b) any ‘C’ (c) more substituted ‘C’ (d) α - H atom Ans.134 (c) Stability of carbocation Q.135 Arrange the following compounds in decreasing order of :

IIIIII

COCHCHCHOCHHCHO 333

(a) IIIIIVII >>> (b) IVIIIIII >>> (c) IIIIIIIV >>> (d) IIVIIIII >>> Ans.135 (b) Aldehydes are more reactive than ketones for AdN Q.136 Arrange the following compounds in decreasing order of AdN reaction :

IVIIIIII

COOHCHCONHCHCOClCHCOCHCH 323333

VCHCOOCOCH 33 −−−−

(a) IIIIVIVII >>>> (b) IIIVIIVIII >>>> (c) IVIIIVIII >>>> (d) IIIVIIIIV >>>> Ans.136 (a) Greater unstability of carbonyl carbocation

- C – CH3

O

IV

O −CN

H2O Product can be

CN

OH

(a)

OH

(b) OH (c)

CN

OH

(d)

CN



Q.137 Which carbonyl group of the given compound is most reactive for nucleophilic addition reaction?

(a) 1 (b) 2 (c) 3 (d) All have rqual reactivity Ans.137 (b) At 1 and 3 carbonyl carbocation are stabilized and hence less reactive Q.138 AdN reaction rate is increased by – (a) EWG at −α position (b) ERG at α - position (c) EWG at −β position (d) EWG on carbonyl carbon Ans.138 (a) Positive nature of carbonyl carbon is increased Q.139 When MgBrCH 3 reacts with CHCCH ≡3 , the product will be ____ (a) 23 CHCHCH =− (b) 33 CHCCCH −≡− (c) 2|3

3

CHCCHCH

=− (d) CH4

Ans.139 (d) Terminal alkyne is acidic ( +H ) and hence produce will be 4CH .

Q.140 CHCHOCHHCCHCOHCHHC OH = →≡+

56563)(

In above reaction, the steps involved are _____ (a) 4 (b) 5 (c) 6 (d) 7 Ans.140 (c) →≡−−→≡− −

⊕

+ OH

OH

H CHCCHHCCHCCHOHHC 2

2

|5656

→==→≡−⊕⊕

OHCHCCHHCCHCCHHC 25656

CHOCHCHHCCHCCHHCCHCCHHCOH

H

OH

−=→==→==+−

⊕

56|56|56

2

O

O

O

1

2

3



Q.141 The electrophile, ⊕E attacks the benzene ring to generate intermediate complex. Of the following, which σ - complex has lower energy?

Ans.141 (b) NO2 is deactivator for ⊕E attack. Stability more and hence energy lower compared to complex b. 13. Addition Electrophilic Mechanism Q.142

How many transition states and intermediates will be formed during this reaction? (a) 3 T. S. and 3 intermediates (b) 4 T. S and 3 intermediates (c) 3 T. S. and 2 intermediates (d) 5 T. S. and 4 intermediates Ans.142 (b) AdE

+

E H

NO2

(a) + E

H

(b) + E

H NO2

(c) +

E

H

NO2

(d)

OH

H2O +H

H2O

⊕

⊕

⊕ OH2

⊕−H

OH

4 steps in reaction



Q.143

Ans.143 (c) Ring expansion for stability of carbocation (AdE) Q.144 cis – isomer of 2- butene when reacted with Br2 in presence if CCl4 we get (a) d l pair (b) meso form (c) only d - (d) only l product. Ans.144 (a)

Q.145 Which of the following compound is most reactive towards electrophilic addition

reaction? (a) 22 CHCH = (b) 223 )( CHCCF = (c) CHOCHCHNC −=−

Ans.145 (a) Electrophilic addition as RDS occus at ΘC where +C should not be reactive. Stability of

(d)

NO2

CH = CH2 NO2

Nl2

C = C

CH3 CH3

H

H C is

Br2

Me H

Br

Me H

Br

Br

Br

CH3

CH3 H

H

d l pair as products

HCl Major product is

Cl

(a)

Cl

(b)

(c) Cl

(d) Cl



+C is essential in such a step. Q.146 When 1 – methyl cyclohexene is converted to alcohol, which reagent can make syn addition only? (a) acid catalysed hydration (b) hydroboration – oxidation (c) oxymercuration – demercuration (d) base catalysed hydration Ans.146 (b) Anti – Markovnikov’s addition Q.147 Two moles of aldehydes are produced by using alkene. The reaction is (a) reductive ozonolysis (b) oxidative ozonolysis (c) reductive oxidation (d) oxidative hydration Ans.147 (a) Zn/HOAc is used along with O3 Q.148 In addition reaction for ,CC = T.S. in energy profile diagram are / is (a) 1 (b) 2 (c) 0 (d) 3 Ans.148 (b) One intermediate, 2 steps in reaction Q.149 Peroxy benzoic acid reacts with 2 – butene to form oxirane. This mechanism can be (a) reduction (b) redox reaction (c) oxidation (d) electrophilic addition Ans.149 (c) & (d). OH of peroxyacid can be considered as E+. However −RCOO removes H form OH as O bonds to C = C, giving epoxide Q.150 When 1 – pentene – 4 – yne reacts with one equivalent of HBr, the product will be (a) CHCCHCHBrCH ≡− 23 (b) 222 CBrCHCHCHCH −= (c) CHCCHBrCHCH ≡222 (d) 222 CHCBrCHCHCH == Ans.150 (a) More stable carbocation to be taken. Alkenes react faster than alkynes for AdE Q.151 Propyne reacts with water in presence of 42SOH and .4HgSO The intermediate is (a) carbocation (b) carbanion (c) ketol (d) enol Ans.151 (d) Enol. It is Markovnikov’s addition of H2O.



Q.152 The compound 1010HC gives only one organic compound after oxidative cleavage.

The product is acid. It must be ____ (a) monobasic (b) aibasic (c) tribasic (d) tetrabasic acid Ans.152 (c) CHCCHCHCCHHC

CHCCH

≡≡≡ .

|22

2

Q.153

Ans.153 (c) AdE with ring expansion Q.154

Ans.154 (a)

+ HBr P. ‘P’ can be -

Br (a) Br

(b) Br (c) (d)

⊕ ⊕

⊕ P

HBr CCl4

P. Product P is

Br

(a)

Br

(b) (c)

Br

Br

(d)





Q.155

Ans.155 (b) Acid-catalysed dehydration, followed by carbocation stabilization to get °3 electrophile. Finally, addition-elimination product. Q.156 How many alkenes on catalytic hydrogenation can produce isopentane? (a) 2 (b) 3 (c) 4 (d) 5 Ans.156 (b) Q.157

Ans.157 (a) Mechanism can be given as

⊕ ⊕

⊕

Product

OH

∆

+H Product can be

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

OH

+H ∆

P. ‘P’ can be

(a) (b)

(c) (b)



Q.158

Ans.158 (a)

Q.159 In a reaction

How many transition states and intermediates are involved? (a) 4 & 3 (b) 3 & 3 (c) 2 & 2 (d) 3 & 4 Ans.159 (a) Q.160 Acid-catalysed hydration of alkene except ethene leads to formation of (a) mixture of °2 and °3 alcohol (b) mixture of °1 and °2 alcohol (c) mixture of °1 amd °3 alcohol (d) primary alcohol only Ans.160 (a)

2CHCHR =− and |

||2

R

CH

CR − will produce mixture of °2 and °3 alcohol.

Br HBr

OH

O

⊕

OH

⊕

O

O

O

⊕−H

OH

O

+H Product can be

O

O

(a) (b)

O

O (c) O

O

(d)

O

O



14. Addition : Free Radical Mechanism Q.161 Both HCl and HI can’t show peroxide effect in case of alkenes. The reason may be (a) both are strong acids (b) both radicals Cl and I are unstable (c) both have propagation reactions energetically unfavorable (d) both can’t make electrophiles Ans.161 (c)

There are 2 steps of propagation and in both cases, ∆H is – ve for HBr only. So, addition is possible to C = C by free radical mechanism in presence of peroxides.

Q.162 Anti – Markovnikov’s rule is not observed in ____ (a) propene (b) 1 – butane

(c) 2 – butane (d) 2 – pentene Ans.162 (c) Symmetrical alkene 15. RARE MECHANISMS : SN i, E1cb, SN', NGP, Benzyne, Ipso Q.163 For 1E cb mechanism, select the incorrect statement from following : (a) poor LG required (b) substrate mostly have acidic hydrogen (c) conjugate base is resonance stabilized (d) less substituted product as major product Ans.163 (d) Q.164 R – 2 – bromopropanoate is treated with −OH in presence of Ag2O. The product will

be (a) R – 2 – hydroxyl propanoate (b) S – 2 – hydroxypropanoate (c) R – 2 – hydroxypropanol (d) S – 2 – hydroxyl propanol. Ans.164 (a) It is double inversion due to NGP and hence retention OC

O

=−−

|is NGP working as

internal nucleophile. Q.165

The mechanism is _____ (a) 2SN (b) 1SN (c) 1SN (d) 1E

OH +H

ΘBrHOH ,



Ans.165 (c) It is 1SN with rearrangement. The product is

Nucleophilic substitution at −γ carbon. Q.166 Give decreasing order of heat evolved due to catalytic hydrogenation for

(a) Q > R > P > S (b) S > P > R > Q (c) S > R > P > Q (d) R > Q > S > P Ans.166 (b) All are is isomers. Greater the stability, lesser the heat of hydrogenation Q.167

Ans.167 (d) Bridge head carbocation (unstable) Q.168 The products will be

(a) ketone and alkyl halide (b) alcohol and alkyl halide (c) alcohol and vinyl halide (d) alcohol and alkene Ans.168 (a) Enol produced and then it is converts to ketone.

O

CH3 cone. H I

O HBr ∆

OH (a) Br (b)

Br (c) No reaction (b)

Product can be

Q R S P

Br or

Br



Q.169

Ans.169 (b) Acidic medium and ΘNu attack, accordingly Q.170 Carboxylic acid when reduced by LAH the product is ___ (a) °2 alcohol (b) aldehyde (c) °1 alcohol (d) alkene Ans.170 (c) Q.171 Which of the following is strongest base?

Ans.171 (a) °2 amine. All others are aromatic Q.172 Find the correct order for energy required for heterolytic cleavage of C – Br bond forming carbocation.

(a) Z > Y > X (b) Z > X > Y (c) X > Y > Z (d) Y > X > Z Ans.172 (a) More stability of C+, lesser energy required for cleavage.

CH2 - Br N | H

X

CH2 - Br O

Y

CH2 - Br

O

Z

N | H

(a) N

(b) (c)

NH2

(d) N | H

O

HI Product can be

I OH

(a)

I OH

(b)

I

OH (c) (b)

I

O



Q.173 Get the correct order for energy required for heterolytic cleavage of C – Cl bond

forming carbocation.

(a) x > y > z (b) z > y > x (c) z > x > y (d) y > x > z Ans.173 (b) After cleavage, carbocation produced is much stabilised by resonance.

Vinyllic carbocation is least stable. Q.174 Consider the following carbanions

Correct decreasing order of stability is (a) II > III > IV > I (b) III > IV > I > II (c) IV > I > II > III (d) I > II > III > IV Ans.174 (a) Resonance effect -R will stabilize carbanion and +R will destabilize. Q.175 In which of the following compounds the hydroxylic proton is most acidic?

Ans.175 (d) F is EWG and charge on O atom is stabilized

Cl

NH2

X

Cl

NH2

y

Cl

NH2

z

(I) H3CO Θ

2CH (II) NO2 Θ

2CH

(III)

Θ

2CH (IV) CH 3 Θ

2CH

(a) O

H F (b)

O

H

F

(c) O

H

F

(d) O

H

F



Q.176 The abstraction of proton will be fastest in which carbon of the following

compound?

(a) x (b) y (c) z (d) p Ans.176 (a) Conjugation and hence stabilization after loss of .+H Q.177 Arrange the following in increasing order of their heat of hydrogenation

(a) SRQP <<< (b) PQRS <<< (c) QPRS <<< (d) RSQP <<< Ans.177 (a) Highly stabilized compound will have lesser heat of hydrogenation.

Hydrogenation is exothermic. P is most stable and S is least stable. Q.178 Which of the following can give E1cb?

Ans.178 (a) & (b).

O

CH3 Z CH3 X

P

Y

P Q R

S

(a)

O F

(b)

O CN

(c)

NO2

(d)

CN Br



Q.179 Carbanion is produced by heterolytic cleavage of a covalent bond from (a) terminal alkyne (b) carbonyl compound (c) active methylene compound (d) all the above compounds Ans.179 (d)

and – CH2 (methylene) group connected with EWG groups Q.180

when reacted with ___ΘA (a) substitution occurs at 1, when working as nucleophile (b) degree of unsaturation increased when working as base at high temperature (c) carbocation is stabilized and hence both the reaction will be first order (d) substitution can occur at position 3.

Select the incorrect statement.

Ans.180 (d)

Q.181 Among the following, which one is not soluble in aq. NaOH ?

Ans.181 (c) All others are acidic

Q.182

(a) 1SN (b) Redox (c) Acid – base neutralization (d) E1

Ans.182 (c) All are acidic. Two of them will react with Θ+

2NHNa

OH

NO2

COOH OH

CH

This reaction is ____ 2 moles of NaNH2

OH

(a)

COOH

(b) (c)

NO2

(d)

OH

NO2

Br

3 2

O

1

Θ≡− CCR , CCH −Θ

2

H

O



Q.183 Select the correct statement from following : (a) In Tollen’s reagent, silver is oxidised (b) In LAH, oxidation number of hydrogen is + 1 (c) In Fehling’s solution test, copper is oxidised. (d) In nitration of benzene, HNO3 works like base Ans.183 (d)

HO – NO2 + H2 SO4 422 HSOOHNO Θ+

++→ Q.184 When

reacts with NaNH2/NH3, the mechanism is _____ (a) addition (b) elimination – addition (c) elimination – substitution (d) elimination Ans.184 (b) It is benzyne mechanism giving

The intermediate is called benzyne. It is Q.185 In benzyne mechanism, entering group occupies a position other than vacated by

leaving group. The reaction is _____ (a) syn – substitution (b) cine substitution (c) syn – addition (d) anti – substitution Ans.185 (b) The entering group does not occupy the position vacated by leaving group. The reaction of this type is cine – substitution.

OCH3

and

OCH3

NH2

Products

OCH3

NH2

OCH3 Br



Q.186

This substitution is called as (a) Ipso (b) pseudo (c) syn (d) Nucleophilic Ans.186 (a) Ipso (on itself). Ipso substitution would ease formation of potential leaving group like

⊕⊕⊕⊕ BrSiMeHSOR ,,, 33 etc. It is a rare reaction mechanism 16. Rearrangements Q.187 The number of stereoisomers obtained by bromination of trans – 2 – butane is (a) 1 (b) 2 (c) 3 (d) 4 Ans.187 (a) meso

SO3H

OMe

Br2

Br

OMe

Documents

158 Essential Notes on General Organic Chemistry (G.O.C.) The IITian’s Prashikshan Kendra Pvt. Ltd. Topics Coverage :- • Substitution Reactions • Free Radical Reactions •