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HOMOLYTIC BOND FISSION

HOMOLYSIS

The bond cleavage in which each bonded atom gets theirown contribution

BAB–A

Free RadicalBA

or

• Cleavage takes place due to

HELP (H = Heat, E = Electricity,

L = light, P = Peroxide)

• Favoured when E.N. difference is less or zero.

• Cleavage favoured in non polar solvent.

2CHAPTER

GOC

HETROLYTIC BOND FISSION

C A

C

C

–

(Carbanion)

+ A–

(A is more electronegative)

(Carbocation or carbonium ion)

+ A+

(C is more electronegative)

+

• It is formed when the electronegativity differencebetween the bonded atoms is more

• formation is favoured by polar solvent

C

..............A–

H

Attraction

– O

|

H

–

+ve charge of the solvent attracts the –ve pole ofcompound and the –ve pole of the solvent attracts +vepole of compound and the bond breaks.

INTERMEDIATES OF ORGANIC COMPOUNDS

Free Radical Carbocation Carbanion

(1) Lone pair 0 0 1(2) Bond pair 3 3 3(3) Unpaired e– 1 × ×(4) Bond Angle 120º 120º 107º(5) Hybridisation sp2 sp2 sp3

(6) Shape Trigonal planer Trigonal planer Pyramidal(7) Magnetic property Paramagnetic Diamagnetic Diamagnetic(8) Stability order 3º > 2º > 1º 3º > 2º > 1º 1º > 2º > 3º

(As per inductive effect)(9) e– rich/deficient/poor ED(Deficient) E D ER(Rich)(10) Reactivity order 1º > 2º > 3º 1º > 2º > 3º 3º > 2º > 1º(11) +I/–I (stablized) +I +I –I

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2.2 Theory and Exercise Book


ELECTRONIC DISPLACEMENT EFFECTThe displacement of electrons within the same molecule isknown as electronic displacement. These effects affect thestability of a species or compound and it also affect theacidic & basic strength.

Electronic Displacement Effect is divided into twoparts:

(1)Permanent effect (2) Temporary effect(1)Permanent effect :

(i) Inductive effect(ii) Mesomeric (resonance) effect(iii) Hyperconjugation

(2)Temporary effect:(i) Electromeric effect (ii) Inductomeric effect

(i) Inductive effect:It is an effect in which permanent polarisation arisesdue to partial displacement of -electrons along carbonchain or partial displacement of sigma-bonded electrontoward more electronegative atom in carbon chain.

C – C – C – ClMagnitude of partial positive charge

– (net charge remainconstant in a molecule having inductive effect)Inductive effectIt is a permanent effect

C C C C C X5 4 3 2 1

– (–I effect of X)

if X i.e more electronegative(After carbon No. 3 the effect disappears)

(+ I effect of Y)

*H – N – H | H

R – N – R | R

+ +

< (– I effect order)

* O– < O < O+ (–I effect order)• It is a permanent effect• It is caused due to electronegative difference.• It operates via bonded electron.• It is distance dependent effect.• As distance increases, its effect decreases.• It can be neglected after third carbon.• It is a destablising effect.• It is divided into 2 parts. (On the basis of

electronegativity w.r.t. hydrogen atom)(1) +I effect (2) – I effect

If any atom or group having electronegativity greaterthan that of hydrogen. than it is considered as – I effectand vice-versa.+I effect

(i) e– releasing group(ii) EN less than H(iii) Those group which are showing + I effect, disperses

partial – ve charge on the C-chain

– I effect(i) e– accepting group(ii) EN greater than H(iii) Those group showing –I effect disperses + ve charge

on the C-chainEg. CH3 – CH2 – Cl (–I of Cl)Eg. CH3 – CH = CH2 (–I of –CH=CH2 & +I of –CH3)Eg. CH3 – CH2 – C CH (–I of –C CH & + I of –CH2–

CH3)Eg. I – Cl +I –I

Eg. CH2 = CH (–I of –ph)

Order of –I effect showing group:

– NF >– NR – NH –NO –CN

–C–OH –F –Cl –Br –I

3 2

> –C–R > > > > >

3 3 > > > >–C–H

O O

O

(–I order) – C CH > – CH = CH2

Order of + I effect showing group

– CH > – NH > – O > – CMe > – CHMe

> – CH Me > CH

2 3 2

2 3> CT > CD > T > D > H3 3

Bond Strength : CT3 > CD3 > CH3

(+ I of T > D > H)

Q. Why carbon - hydrogen bond is longer than C - T bondAns As the mass increases, vibration decreases as a result

of which the heavier isotope will be more closer to theC-atom for a longer time. Therefore C – T bond isstronger C – T > C – D > C – HWhich implies that C – H bond has longest bond

GOC 2.3

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APPLICATION OF INDUCTIVE EFFECTTo compare the stability of intermediates.

IntermediatesThese are real separable species having measurablestability formed during coversion of reactant to product.(After bond cleavage and before bond formation).6 types of intermediates:(i) Free radical (ii) Carbocation(iii) Carbanion (iv) Carbene(v) Nitrene (vi) BenzyneThey are formed by homolytical and heterolyticalcleavage.

MESOMERIC EFFECT(RESONANCE EFFECT)

Delocalisation of electrons in any conjugated systemis known as mesomeric effect

Types1 + M effect (+R)2 – M Effect (–R)

* Consider the following conjugated system

C = – C = C – C – C = C – C = +CH H H Y H H H H Y2

(+M effect of y)

H2

* Consider another conjugated system

C = C – C = C – N = O C – C = C – C = N +

O–

OO

(– M effect of NO )2

MESOMERIC EFFECT IN PHENOL(+ M EFFECT)

OH O–H O–H O–H O–H

–

–

–

+ + +

+M effect in aniline

NH2 NH2 NH2 NH2 NH2

–

–

–

+ + +

If the movement of e– is towards ring (+M effect)This effect increases the electron density over benzene ring.

* –M effect in Benzaldehyde

H – C – O–

+

+

H – C = O H – C = OH – C – O–

+

H – C – O–

SOLVED EXAMPLE

EXAMPLE 1

Idenfity the compound showing +M or –M seperately

(a)

OH – S = O

O

(b)

OH – C = O

(c)

SH

SOLUTION

(a) (–M) (b) (–M) (c) +M

* +M group increases electron density of ring while – Mdecreases the electron density of benzene ring.

* if NO2 is present on the ortho or para position thenalong with its –I effect, It will also show –M effect.

OH O–H O–H O–H

N = O N = O N = O N – O

O O O(–M)

O

+ + +

–

–

* Above compound have +M of –OH and –M of NO2group.

OH O–H O–H

N = O N = ON = O

OO O

+ +

–

–



as we can easily see that –NO2 at meta position is notattracting e– density towards it self and that's why itwill not show –M effect at m-position

RESONANCEDelocalisation of -electrons in conjugation is knownas resonance.

(Actual Structure)

(resonating structures) (Resonance hybrid)

in this form

CONDITION FOR SHOWING RESONANCE

1. Molecule should be planer, nearly planer or a part of itis planar

Q.1 Which are planer

(A) (B)

*(C) *(D)

Because all carbon atoms are sp2 hybridised.2. Molecule should have conjugated system.

Conjugated System :

Continuous unhybridised p-orbital parallel to each–other.

Types of Conjugated System:

(1) -bond alternate to -bondCH2 = CH – CH = CH2

(2) -bond alternate to + chargeCH2 = CH – CH2

+

+

Eg. CH = CH – CH = CH2 2

CH CH CH – CH2 2– =

CH – CH = CH – CH2 2R.S. CH2 CH CH2CH

+ _ +_

Eg. CH = CH – CH2 2

CH – CH = CH2 2 CH2 CH2CH

Eg. CH = CH – CH2 2

CH – CH = CH2 2 CH2 CH2CH

(4) CH = CH – NH2 2 CH CH NH2 2 – =

(5) CH = CH – CH2 2

(6) CH2 = CH – BH2B

H

H B

H

H

+

(7)

CH = CH 2 – C CH

CH – CH =2 C = CH

1. Resonance takes place due to delocalization of e–.(a) Resonance(b) Resonance absent

(c) Resonance

(d) Resonance

2. Position of the atoms remains the same, onlydelocalization of e– takes place.

Note:– CH – C – NH3 2 CH – C = NH3

O OH

[They are not resonating structure rather they are tautomer]

GOC 2.5


3. Bond pair get converted into lone pair and l.p. getconverted into b.p.

CH = C – NH2 2 CH C NH2 2– =

4. In Resonance No. of unpaired e– remains the sameC H 2 = C H – C H = C H 2

CH – CH = CH – CH2 2

(They are not resonating structure)

Resonating Structure :(1) Hypothetical strtucture exist on paper

(2) The energy difference b/w different resonatingstructure is very small.

(3) All R. S. contribute twoards the formation of resonancehybrid (Their contribution may different)

(4) A single R. S. Can't explain each & every property ofthat particular compound

Draw The Resonating Structures : –

Q.1 CH = CH – CH2 = CH – NH2

CH – CH = CH – CH = NH2 2

NO

O N N

N N

O O

O OO

O O

O

–m of NO2 group

Resonance Hybrid : –It is a real structure which explain all the properties ofa compound, formed by the contribution of different R.S.. It has got maximum stability as compared R. S.

Resonance Energy : –It is the diffrence b/w theoretical value of H.O.H &experimental value.

OrIt is the difference b/w more stable R.S. & R. H.

* More the resonance energy, more stable will be themolecule.

* Cyclohexane is thermodynamically more stable thanbenzene, even though resonance energy of benzene ismore.

* Resoance energy is a absolute term.

CONTRIBUTION OF DIFFERENT R. S.TOWARDS RESONANCE HYBRID

(1) Non-polar R. S. contribute more than polar R. S.(a) CH2 = CH – CH = CH2

(b) CH2–CH = CH – CH2

(c) CH2 – CH = CH – CH2

a > b = c stability

(2) Polar R. S. with complete octet will contribute more ascompared with the one with incomplete octetCH3 – CH+ – OCH3 CH3 – CH = :O+ – CH3

Incomplete octet Complete octet

(3) In polar R. S. The –ve charge should be on more electro– ve atom & +ve charge should be on more electro +ve atom

(a) CH – C – CH2 3 CH = C – CH2 3

O O

(more stable )

(b) CH – C – CH2 3 CH = C – CH2 3

O O

(4) Compound with more covalent bond will contribute more

(5) Unlike charges should be closer to each other whereaslike charges should be isotated.

(6) Extended conjugation contribute more than crossconjugation.

<

Cross conjugation < Extended conjugation

Fries Rule :–Compound with more benzenoid structure are morestable as their Resonance energy is greater than thosein which lesser no. of benzenoid structure are present.

R. E. is

<



* If double bond is participating in resonance then it willaquire partial single bond character as a result of whichbond length increase & bond strength decreases.If a single bond is involved in resonance then it willaquire partial doulbe bond character. As a result ofwhich bond length decreases & bond strength increase.

Q.1

OH O–H O–H

O–H O–H

–

–

–

+ +

+

(a) (b) (c)

(d) (e)

a = e > b = d > c

Q.2 (a) CH = N = N 2

(b) CH – N N 2

(c) CH – N = N 2 (incomplete)

(d) CH – N = N 2 (incomplete)

a > b > c > d

Q.3 H – CCl

ClCl H – C

F

FF

–H–

–H–

Stability CCl

ClCl < C

F

FF

(back bonding)

Q.4 (a)CH = CH – F2

(b) CH2 – CH = F+ a > b (stability)

*

N

H

N

H

N

H

N

H

N

H

N

H

sp -N2

Note:When lone pair as well as double bond is present onsome atom. Then only bond will participatingresonance. Where as lone pair remains sp2 hybridisedorbital.When an atom has two or more then two lone pairthen only one lone pair will participate in resonanceand the other one remains in sp2 hybridised orbital.

HYPER CONJUGATION

Permanent polarisation caused by delocalisation of-electrons into -molecular orbital is known ashyperconjugation

H – C – CH2 H – C CH2| |

|

H H

H H

Hyper conjugation is called No bond Resonance

* More C – H bond, more will be the no bondresonating structure (Hyper conjugation)

• More (C – H) bond, more will be the stability of freeradical.

(CH ) C > (CH ) CH > CH – CH > CH3 3 3 2 3 2 3

)H–C( 9 )H–C(6 )H–C(3 0

Stability order

Properties of Free Radical1. It is a neutral species.2. It has one upaired electron that's why paramagnetic in

nature.

Structure :

3HC methyl free Radical

23CHHC ethyl free radical

GOC 2.7


3. its hydridisation is sp2 and triangular planer shape.Note :

unpaired electron is not counted while calculating thehybridisation state.

H

H

H

(unpaired electron stay perpendicular to the plane)

Stability of free Radical :Its stability can be determined with the help ofhyperconjugation as well as Resonance effect

ALLYLIC FREE RADICAL

CH = CH – CH 2 2 CH – CH = CH 2 2(Homolysis)

(Free Radical is on next carbon to doubly bonded carbonatoms)

Effect of Resonance > Hyper conjugation

CH – CH = CH 2 2 > (CH ) C3 3

(stability)

BENZYLIC FREE RADICAL

CH2

CH2

CH2

CH2

CH2

(5 Resonating structure)

* More Resonating structure, more will be the stabilityof the free Radical.

CH

(di-benzylic free Radical)

No. of Resonating structure = 7

C

(Tri-benzylic free Radical)No. of Resonating structure = 10

Stability Order :

3 2 2 2 2Ph C Ph CH Ph CH CH CH – C H

3 3 3 2 3 2 3(CH ) C (CH ) CH CH – CH CH

EXAMPLE 2

Compare the stability of the following free Radical.

(a) 23 HC–CH

(b) HCCH2

(c) HCCH

SOLUTION

(a) 23 HC–CH

will be most stable due to hyper conjugation.

Between HCCH2

and

CCH

sp

more s-character more electronegative e– density maximum more repulsion less stable

Ans. a > b > c

* More repulsion, less stabilityEXAMPLE 3

Compare the stability of the following free Radicals

(a) Tendingto sp due to allylicstructure

(very unstable)

(b) HCCH

sp2

(c)

CCH

actual sp More repulsion less stability

(Therefore this resonating structure is not possible)

SOLUTION b > a > c



EXAMPLE 4

(a)

3

3

*

3

CH

|

CH–HC–CH (b) 323 CH–HC–CH

(c) 33 HC–CH

(d) 4HC

Compare the H–C

bond energy of the above compounds.SOLUTION

After forming free radical from the compound

3

33

CH

|

CH–C–CH

33 CH–HC–CH

23 HC–CH

3HC

(3°) (2°) (1°) methyl free radical(a) (b) (c) (d)

(most stable) therefore will have moretendency to come in this form And C – H bond will break very readily bond energies will be very less.a < b < c < d < (bond energies order)

* Radical free ofstability

1energy Bond

* Radical free ofstability length Bond

EXAMPLE 5

Compare the potential energy of the following compounds(above compounds)SOLUTION

If compound after being in free Radical form is very stable(i.e., less energy) it mean it would have possessed moreenergy initially i.e. it potential energy will be most

a < b < c < d* Potential energy stability of free Radical

EXAMPLE 6

Compare the bond energies of C – H bond(at a, b, c, d, e and f position)

CH – CH – 3 2 –CH – CH = CH – CH – 2 2 –CH3

(2° benzylic) (2° benzylic, allylic) (2° benzylic, allylic) (1° benzylic)

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

vinylic vinylic

b > e > a > f > c = d ?Stability order of free Radical that might be formed

after removal of H (Homolytically) from the givencarbon.

e < b < a < f < c = d(C – H bond energies)

• In the above compound while comparing 2° benzylicallylic stability at two given position

CH2 CH2

CH3 CH – CH2 3

and

while drawing the resonating structure of the

CH2 CH2

CH3 CH3

CH2

H–C–H

H

GOC 2.9


(Here inspite of Resonace three (C – H) bond areavailable for no bond Resonance.

Therefore extra stable than

CH2

CH CH2 3

which have

only two (C – H) bond for Hyper conjugation.Therefore 2° benzylic allylic corresponding to structure(a) is more stable than that of structure (b)

EXAMPLE 7

Compare the stability of the following free RadicalCH2

(a)(b)

(c)

CH3

(d)

CH = CH – CH2 2

SOLUTION

CH2

(a) (b)

1° allylic 2° allylic + 2H

(c)

CH3

(d)

CH = CH – CH2 2

3° allylic + 5H

(d)

CH – CH = CH2 2

(only resonance)

CH2

HH c > b > a > d

(Resonance + 2H)

EXAMPLE 8

Compare the potential energy of CH3 – CH3, CH2 = CH2

CH CH

SOLUTION

After making free Radical of the above compounds

)stablemost(

HC–CH 23

, HCCH2

)stableleast(

CCH

)c()b()a(

CCHHCCHHC–CH 223

a > b > c

CARBOCATION

3HC Methyl Carbocation

23 HC–CH Ethyl Carbocation

3

23

CH

|

HC–HC–CH

Isopropyl Carbocation

Properties of Carbocation :1. it is positivly charged species2. it has sixtet of electrons i.e. diamagnetic3. it is formed by heterolysis4. it is generally formed due to polar solvent

Structure :(sp2) Triangular planer

Stability :Its stability can be determined with the help of Inductiveeffect, Hyper conjugation and Resonance effect.

Stability of Carbocation :

,HC 3

CH CH3 2

+

(+ I effect)

Stability

1eargch

,HC 3

< CH CH3 2+

(stability order)

Stability of carbocation can also be determined byHyper conjugation (no bond Resonance)

H |H – C – CH | H

2

H |H – C = CH | H

2+

+



CH – C – CH | CH

3 3

3

CH – CH | CH

3

3

CH – CH > CH3 3

+ ++

> >+

9 C – H bond 6 C – H bond 3 C – H bond

ALLYLIC CARBOCATION

CH = CH – CH2 2 CH – CH = CH2 2

+ +

allylic carbocation Actual Resonance

BENZYLIC CARBOCATION

CH2 CH2 CH2 CH2

+

strucutresonatingRe7HCph2

strucutresonatingRe10Cph3

23323 HCphC)CH(HCphCph

EXAMPLE 9

Compare the stability of th following carbocation

(a) 23 HC–CH

(b) HCCH2

sp2

(c)

CCH

sp

more s charactor more electronegativity +ve charge on more electronegative element issymbol of unstability.

a > b > c

EXAMPLE 10

Compare the stability of the following compounds

(a) 32 CF–HC

(b) 32 CCl–HC

(c) 32 CBr–HC

(d) 3HC

SOLUTION

d > c > b > aF being most electron attracting group decreases thee– density from positively charged C-atom anddecreases the charge density and makes the carbocationless stable.

EXAMPLE 11

Compare the stability of the following carbocation :

(a) +CH – F2 (b)

+CH – Cl2

(c) –BrHC 2

(d) I–HC 2

SOLUTION

Due to greater size of Iodine, its L.P. will not be availablefor coordinate bond. Therefore L.P. would not stabilizecorbocation.In case of F due to its small size its lone pair can be

easily coordinated to +C making it most stable

a > b > c > d (Stability)* By coordination the carbocation completes its octet and

structure having complete octet of its atom is supposedto be most stable.

+ +CH – F 2 CH = F 2CH F 2

..

..

..

+

(Each atom has its full octet)

*+C A ph C3

+

(stability)

Note :

In Resonating Structure of Cph3, at least one C gets

sixtet of e– and hence less stable than coordinatedcompound.

EXAMPLE 12

Compare the stabilities of the following corbocation

(a) 22NHHC

(b) OH–HC 2

(c) F–HC 2

SOLUTION

N, O, F belongs to same period In period Electronegativity of the atom is deciding factor F being most electronegative, holds its e– pair very

firmly. Its L.P. will not be easily available for coordination. Stability by it will be minimum.

a > b > c

EXAMPLE 13

Compare the following corbocation in order of their stability.

(a) Cl–HC 2

(b) OH–HC 2

SOLUTION

If periods of atoms which have to donate their electronsfor coordination (for stability) is different then atomicsize will be deciding factor. The atom whose size isgreater will be unable to make its e– pair available forcoordination.

b > a

GOC 2.11


EXAMPLE 14

Compare the stability of the following compounds

(a) CH3 CH – CH2 2+

(b) )allylic(HC–CHCH 22

(c) ph CH – CH2 2+

SOLUTION

more s-character

more e.n.

attracts e–

reduces, stability

b > a > c

CARBANION

1. it is a –ve charged species

2. it has octet of electrons.

3. diamagnetic

Strucutre :

* if –ve charge is in Resonance then the hybridisation ofcarbanion is sp2 (Triangular planer shape)

* If –ve charge is not in Resonance then the hybridisationof carbanion is sp3 (pyramidal)

Stability :

Its stability can be determined with the help of

(1) Inductive effect

(2) Resonance effect

EXAMPLE 15

233 CHCH,CH– –

(a) (b)

a > b (stability)

* Stability of the carbanion is as follows

–P P Ph C > h CH > h CH > CH = CH – CH > > 3 2 2 2 2

CH = CH > CH > CH – CH > (CH ) CH > (CH ) C 2 3 3 2 3 2 3 3

CCH – –

–

– –

– – – –

EXAMPLE 16

Compare the stability of the following carbacation

(a)

SP2

+

(b) )sp(actual

CCH

(c)

2

2

sp

HCCH

SOLUTION

c > a > b

EXAMPLE 17

Compare the stability of the following carbanion

(a)

tending sp

–

(b)

–CCH

sp–ve charge is attracted bysp hybridised carbon (most electronegative)

(c) CH = CH2

–

sp2 become more stable

SOLUTION

b > a > c

EXAMPLE 18

Compare the stability of the following carbanion

(a) –CH – CF2 3

(b) –CH – CCl2 3

(c) –CH – CBr2 3

SOLUTION

a > b > c

EXAMPLE 19

Arrange the following anion order of their stability

(a) Cl–, (b) Br–

(c) F– (d) I– (maximum size)

maximum dispersion of –ve charge

max stability

SOLUTION

d > b > a > c



EXAMPLE 20

Compare the stability of the following

(a) CH3 (b) NH2 (c) OH (d) F

SOLUTION

Same period element (C, N, O, F)

Stability E.N. of the atom

d > c > b > a

EXAMPLE 21

Compare the acidic strength

(a) HCl (b) HF

(C) HBr (D) HI

SOLUTION

Acidic strength stability of the anion formed

(conjugate base)

as we know I– > Br– > Cl– > F–

H I > HBr > HCl > HF

EXAMPLE 22

Compare the Acidic strength of the following

(a) NH3 (b) PH3

(c) AsH3 (d) SbH3(e) BiH3

SOLUTIONAnion formed from there acids are

)Stability(

iHBbHSsHAHPHN 22222

acidic strength e > d > c > b > a

EXAMPLE 23

Compare the acidic strength of the following comounds

CH4, NH3, H2O, HF

SOLUTION

The conjugate base of the given acid is as follows

F,HO,HN,HC 23

––––

we have already proved that

F–

> HO–

> HN 2

– >

–CH3 (Stability)

HF > H2O > NH3 > CH4 (acidic strength)

EXAMPLE 24

Compare the stability of the following carbanion.

(a)

CH2

–

(b)

CH2

–

NO2

(–I)

(c)

CH2

–

NO2

(–M, –I)

(d)

CH2

–

NO2(–M, –I)

SOLUTION

d > c > b > a

* +M or –M is not distance dependent

EXAMPLE 25

compare the stability of the following carbocation

(a)

CH2+

(b)

CH2+

NO2

(–I)

(c)

CH2+

NO2

(–M, –I)

(d)

CH2+

NO2

(–M, –I)

SOLUTION

a > b > c > d

GOC 2.13


EXAMPLE 26

Compare the stability of the following carbocation.

(a)

CH2+

Cl(+M, –I)

but –I > +M for Cl

(b)

CH2+

NH2(+M)

(c)

CH2+

OH(+M)

(d)

CH2+

OCH3(+M)

SOLUTION

+M (OH) > +M (OCH3)b > c > d > a

EXAMPLE 27

Compare the stability of the following carbocation

(a)

CH2+

F (–I > +M)

(b)

CH2+

Cl

(c)

CH2+

SOLUTION

c > a > b

EXAMPLE 28

Compare order of dehydration of the following alcohols :

(a)

OH |C – C – C | C

(b) OH |C – C – C

(c) C – C – C – OHSOLUTION

After formation of carbocation

C – C – C , | C

+C – C – C

+ , C – C – C +

Since 3° carbocation is most stable therefore it willshow greatest tendency to lose water as after lose ofwater it comes in stable form.

TYPES OF REAGENT

1. Electrophilic reagent : All electron deficient atomor group of atoms is known as Electrophilic reagent, theelectrophile attacks at the electron rich centre.(a) all positively charged species are electrophile

H+, NO2+, Br+, Cl+, etc.

(b) The compound in which the octet of central atom isnot complete

BF3, AlCl3, ZnCl2, etc.(c) all the compound in which the central atom canexpand its octet

SnCl4, SiCl4, etc.(d) all polarising functional group are electrophile aswell as nuelophile

C = O, –C N, etc.

Nucleophile :

All electron rich compounds are nucleophile and attackat the electron deficient centre.(a) all negatively charqed species

H–, Cl–, NO2–, Br–, CH3

– etc.(b) the compound in which the central atom has lonepair of electron.

NH3, H2O, 2HNR

, HOR

etc.

(c) all organometallic compounds are nucleophileR – Mgx, RLi, R2Cd

(d) The compound having e– density, CH2 = CH2,

etc.

Nucleophilicity :

The power of nucleophile is known as nucleophilicity . The nucleophilicity of negative charge is greater than

the nucleophilicity of lone pair

OHHO 2

OHCHOCH 33

If lone pair or –ve charge is present on the differentatom then less electronegativity, more will be thenucleophilicity.

–3CH , –

2NH , –OH , –F



Nucleophilicity –––2

–3 FOHNHCH

NH3 < PH3 < AsH3 < SbH3 < BiH3 (Nucleophilicity) If –ve charge or lone pair of electron is present on the

same atom then the less stable –ve charge will be thebetter nucleophile

–33

– COOCH,OCH,OH

–3

––3 COOCHOHOCH (nucleophilicity)

ACTIVATOR & DEACTIVATORThe groups in benzene which show +M effect or +I effectIncreases the electron density on benzene it means theyactivate the ring towards electrophile and known as activator.

NH2

,

OH

, –CH3,–OR,–NHMe,

COO–

,

O–

The groups which shows –M or –I effect (resultant)decreases the e– density from benzene ring. It means theydeactivate the ring towards electrophile

CHO

,

NO2

,

COOH

,

SO H3

,

NO

,

CCl3

,

CN

,

NC

etc.

ORTHO, PARA & META DIRECTORThe groups which shows +I (resultant) or +M effect thennegative charge is developed at the ortho & para position.Therefore electrophile attack at the ortho & para positionand the groups are known as OP director.

G

+M Effect

–

G+

–

–

G+G+

• The groups which shows –M effect or – I effect(resultant) then +ve charge is developed at the ortho

& para position this means electron density is minimumat the ortho & para positions and electronphile willattack at the meta position the groups are known asmeta director.

CCl3

,

CN

,

NO2

,

NO

,

NC

,

COOH

,

CHO

,

SO H3

N = O

O

N – O–O

+

N – O–

O

+

N – O–

O

+

HEAT OF HYDROGENATION(H.O.H)It is the amount of energy realeased when one mole of H2is added to any unsaturated system.

CH2 = CH2 + H2 CH3 – CH3 + energyHOH is exothermic process H = – ve*HOH No. of -bonds in compoundIf no. of -bonds is same then

*HOH 1

stabilityof compound

In case of alkene

**1 1HOH

stability of compound No. of αH

Example

a + H2 + 29 kcal

b + 2 H2 + 58 kcal(expected) (actual)

55 kcal

c + 3 H 2 + 87 k cal 51 k cal (expected) (actual)

b > c > a

GOC 2.15


Energy

– 29H=H=

58 8751

36 k Cal

– 55

RE = 3 k Cal RE =

HOH

+H2 +2H2 +3H2

H=–

Aromaticity (Huckle Rule) :Cyclic, planar and completely conjugated system

with (4n+2) electrons (where n = 0, 1, 2, 3, ......) are knownas Aromatic compounds. These compounds gain extrastability which is known as aromaticity.

ExampleO

Anti aromatic compounds : Cyclic planar and completelyconjugated system, with 4n electrons (where n = 1, 2, 3,....)are known as Antiaromatic compounds. These compoundsare highly unstable & paramagnetic in nature due to presenceof unpaired electrons.

Ex.

(i) Cyclic (i) Cyclic (ii) Planar (ii) Planar (iii) Comp. conjugated (iii) Comp. conjugated (iv) 4n = 4, n = 1 (iv) 4n = 4, n = 1

Both the Compounds are anti aromatic.Note : Anti aromatic compounds have tendency to get

dimerise and leading to the formation of nonaromatic compounds.

Some examples of Arromatic(A), Non-arromatic(NA)and Anti-arromatic(AA)

(1) (A) (2) (AA)

(3) (AA) (4) (NA)

(5) (AA) (6) (A)

(7)

O O–H

H+

OH

(A)

(8) AgNO /3

Br

(A)

(9) (A) (2e) (10) (A) (6e)

(11) (NA) (12) (NA)

(13) (AA) (4e) (14) (A)

(15) (NA) (16)

O

H+

(A)

(17) AgNO3

Br

(A)

(18)

HH

NH2 (AA)

(19) O

(A)



(20) AgNO3

Br

(AA)

(21) H+

O

(AA)

(22) (A) (23)N

(A)

(24) N

H

(A) (25)S

(A)

(26) O

(A) (27)

O

N=N

(NA)

Acidity & Basicity

HA H + AAcid Conjugate base

Note :

More stable the conjugate base (i.e., A), more will be theforward reaction which results more acidic nature of HA.

EXAMPLE 29

Compare the acidic strength of the following acids.(a) C – C – C – COOH(b) C = C – C – COOH(c) C C – C – COOH

SOLUTION

The acid whose conjugate base is most stable will bemore acidic.After forming conjugate base from the above acids.

(a) C – C – C – COO–

sp3

(b) C = C – C – COO–

sp2

(c) C C – C – COO–

sp

It is clear that sp hybridised carbon being mostelectronegative will decrease e– density from O mosteffectively making the conjugate base most stable.

c > b > a (acidic strength)

EXAMPLE 30

Which is more acidic between the two(a) CHF3 (b) CHCl3

SOLUTION

CHF3 > CHCl3

If we consider the –I effect of F and Cl But this effectwill not be considered hereAfter the removal of proton

(a) C

F

F

F

– (b) C

Cl

Cl

Cl

–

(vacant d-orbital available where C will coordinate itselectron) (p – d bonding) a < b (acidic strength)

EXAMPLE 31

Compare the acidic strength of the following(a) CHF3

(b) CHCl3

(c) CHBr3 (p – d bonding in Br is not as much aseffective as in Cl due to large size of Br)

SOLUTION

CHCl3 > CHBr3 > CHF3

EXAMPLE 32

Compare the acidic strength of the following(a) CH (CN)3 (b) CH (NO2)3(c) CHCl3

SOLUTION

After removing H+

C–

NC

NC NC

(Resonance) In its resonating

structure, –ve charge will be on N)

C–

N = O

O

N = O

OO N = O

(Resonance) (– In its resonating

structure –ve charge will reside on O more effective Resonance

GOC 2.17


C–

Cl

Cl

Cl

(p – d)

b > a > c* –ve charge on O is more

stable than –ve charge on N as O is moreelectronegative than N.

* P – d Resonance < Actual Resonance

EXAMPLE 33

Compare the acidic strength of the following(a) CH CH (b) CH2 = CH2

(c) CH3 – CH3

SOLUTION

CCH

sp

> CHCH2

sp2 sp3

> 23 HCCH–

–

(Stability of the conjugate base) a > b > c (acidic strength)

EXAMPLE 34

Compare the acidic strength of the following :(a) CH3 – CH2 – CH2 – COOH(b) CH3 – CH – CH – COOH

| Cl

(c) CH3 – CH – CH – COOH|F

(d) CH3 – CH – CH – COOH|NO2

SOLUTION

d > c > b > a

EXAMPLE 35

Compare the acidic strength of the following :(a) H2O (b) H2S (c) H2Se(d) H2Te

SOLUTION

Conjugate base is in an stability order

TeHeSHSHHO 2

H2O < H2S < H2Se < H2Te (acidic strength)

EXAMPLE 36

Compare the acidic strength of the following compound

(a) (b)

CH3

(c)

CH Cl2

(d)

CH F2

SOLUTION

After forming conjugate base of the above

–CH2

–ve charge is not in resonance

CHCl

bonding due vacant d-orbital of Cl

d–p

(most stable)

CHF

C – I effect of F decrease e density from C making the carbanion stable

–

c > d > b > a

EXAMPLE 37

Compare the reactivity of the following compounds with1 mole of AgNO3

(a)

Cl

(b)

CH Cl2

(c)

CHCl2

(d)

CH Cl2

CH3



SOLUTION

After removing Cl–

+ CH2

+

(+ve charge is not on resonance

least stable)

CH – Cl+

(most stable as L.P. of Cl will be coordinated to +vecharge completing the octet of each atom and makingthe corbocation most stable)

CH2

+

CH3

(By hyper conjugation)

extent of +ve charge decreases stability increases

EXAMPLE 38

Compare the acidic strength

(a)

CH3

NO2

(b)

CH3

NO2

(c)

CH3

NO2(d)

CH3

SOLUTION

After making conjugate base

CH2

NO2(–I)

CH2

NO2

–

(–I, –M)

CH2

NO2

(–I, –M)

CH2

c > b > a > d

BASIC STRENGTH

A H+

A – H+

Basic strength directly depends on the availibility oflone pair for H+

EXAMPLE 39

Compare the basic strength of following

SOLUTION

(a) NH3

(b) PH3

(c) AsH3

(d) SbH3

(e) BiH3

Basic strength

EXAMPLE 40

Compare the basic strength of the following

(a) 3HC (b) 2HN

(c) HO (d) F

SOLUTION

3HC , 2HN , HO , FCH4 < NH3 < H2O < HF

(acidic strength)

CH > NH > OH > F3 2

– – –

(Basic strength)

–

GOC 2.19


* Strong Acids have weak conjugate base.

CH > NH > OH > F3 2

– – – –

(Nucleophilicity)

* For the same periodless electronegativity, more nucleophilicity as moreelectronegative element has less tendencey to give itselectron pair.

EXAMPLE 41

Which is more basic HO or SH ?

SOLUTION

HO > SH

Which is more basic NH3 or NH 2

–

forming conjugate acid

NH > NH (acidity)4 3+

NH < NH3 2

–(Basicity)

COMPARISON OF BASICITY OFAMMONIA AND ALKYL AMINES :

EXAMPLE 42

Compare the basic strength of the following NH3, CH3NH2,(CH3)2NH, (CH3)3NFactors which affect the basicity of Amines(1) steric effects(2) Inductive effect(3) solvation effect.

• The base whose conjugate acid is more stable will bemore acidic forming conjugate acid of the given base

4HN ,

33 HNCH ,

223 HN)CH( , HN)CH( 33

Stability order of conjugate acid

43322333 HNHNCHHN)CH(HN)CH(

(due to +I effect)Therefore basic strength(CH3)3N > (CH3)2NH > CH3NH2 > NH3

(vapor phase or gaseous is phase or in Non polarsolvent)In Aqueous solution or in polar solvent (CH3)2NH > CH3NH2 > (CH3)3N > NH3

• In aqueous solution, the conjugate acids form H-bonds(intermolecular) with water molecules and stabilise

them selves conjugat acid of 1° amine which has largestno. of H-atoms form maximum H-bond with water andis most stable. Consequently 1° amine is most basic.

• Due to steric effect 1° amine is considered more basicas compared to 3° amine as lone pair is hindered bythree alkyl group and less available for H+.Considering the combined effect of the three (Inductive,solvation and steric effect) we can conclude that

2° > 1° > 3° > NH3

• Aromatic amines are least basic as their lone pair is inconjugation and less avaibable for protonation.

EXAMPLE 43


(a)

N(no Resonance as ring will break if we draw the resonating structure)

(most basic)

(b)

NH2

(Resonance)

(c)

(if L.P. will be participate in Resonance, then moleculebecomes aromatic)

Hence L.P. will have a greater tendency to take part inResonance and will be less available for H+

This compound will be least basic.

EXAMPLE 44


22 NH–CH–CCH

sp

CH = CH – CH – NH2 2 2

CH – CH – CH – NH2 2 2 2

sp2

sp3

(a)

(b)

(c)

Common for all



SOLUTION

sp hybridised carbon being most electronegative willattract e– density from nitrogen and will make it lessavailable for H+. Hence basicity decreases.c > b > a

EXAMPLE 45

Compare the basic strength

(a)

H |N

O(–I of O at attracts e density from N making it less basic)

–

(b)

H |N

a < b

EXAMPLE 46

Compare the basicity of the following compounds

(a) CH3 – CH2 – CH = CH – 2HN

(b) CH – CH – CH – CH – NH3 2 2 2 2

sp3

(c) CH = CH – CH – CH – NH2 2 2 2

sp2

(d) 222 NH–CH–CH–CCH

sp

SOLUTION

In part (a) the lone pair of nitrogen in Resonancetherefore will be less available for H+ making it leastbasic among all followed by sp, sp2, sp3 hybridisedcarbon atoms.b > c > d > a

EXAMPLE 47

Compare the basicity of the numbered nitrogen atoms.

H – N

sp3

1 2

H |N N

3

N

sp2sp2

(As L.P. in Resonance)

as L.P. is not in Resonance (or in conjugation)

SOLUTION

The planerity of ring will be destroyed if L.P. will takepart in Resonance.Basicity order of Nitrogen follows the orderN(sp3) > N(sp2) > N(sp)

1 > 3 > 2

sp2 sp2

(In this sp2, l.p. is in Resonance with ring hence will beless available for H+ therefore it will be least basic)

EXAMPLE 48


(a)

NH2

NO2

(b)

NH2

NO2

(c)

NH2

SOLUTION

In part (a) NO2 is at p-position Hence will attract e–

density by both –M and –IIn part (b) NO2 is at m-position hence will attract e–

density by –I onlyThere is no such effect in part (c)

Availibity of L.P. on nitrogen in part (a) is minimumfollowed by b and then c.

c > b > a

Ortho effect :The ortho substituted aniline are less basic than anilineand ortho substituted benzoic acids are more acidic thanbenzoic acid.

• Ortho effect is valid only for benzoic acid and aniline.

e.g.

NH2

NO2

<

NH2

Also

NH2

CH3

<

NH2

GOC 2.21


EXAMPLE 49

Compare the basic strength of the following :

(a)

NH2

CH3

(+I, Hyperconjugation)

(b)

NH2

CH3

(+I)

(c)

NH2

CH3

(+I)

(d)

NH2

SOLUTION

a > b > d > c

* Due to ortho effect d > c

if c is less basic than d then it will be certainly less

basic than b as b is more basic than d.

EXAMPLE 50

Compare the basic strength of the following :

(a)

NH2

NO2

(b)

NH2

NO2

(c)

NH2

NO2(d)

NH2

SOLUTION Do your selves

S.I.P Steric inhibition of Protonation (ortho effect)

NH2 NH3

H+

+

GG

after protonation, repulsion increases therefore ortho

substituted aniline is less basic than aniline

S.I.R Steric inhibition of resonance

(a)

NH3

CH3

NO2

CH3

(Shows only –I effect)

(b)

NH2

NO2

CH3

(Shows –I as well as – M this means delocalisation of e is more) –



Basic Concept (Bonding)

1. Bond formation is:(A) always exothermic(B) always endothermic(C) neither exothermic nor endothermic(D) sometimes exothermic and sometimes endothermic

2. CH2 = CH – CN 3 2 1C1 - C2 bond of this molecules is formed by:(A) sp3-sp2 overlap (B) sp2-sp3 overlap(C) sp-sp2 overlap (D) sp2-sp2 overlap

3. Find out the hybridisation state of carbon atoms ingiven compounds from left to right.CH3 – CH = CH – CH = C = CH – C C – CH3(A) sp3 sp2 sp2 sp2 sp sp2 sp sp sp3

(B) sp3 sp2 sp2 sp sp sp sp sp sp3

(C) sp3 sp2 sp2 sp2 sp2 sp2 sp sp sp3

(D) sp3 sp sp sp2 sp sp2 sp sp sp3

4. Total number of and -bonds are in naphthalene is:(A) 5 and 18 (B) 6 and 19 (C) 5 and 19 (D) 7 and 26

5. In which of the following molecules resonance takesplace through out the entire system?

(A) (B)

O

(C) (D) COOCH|

3

COOCH3

(E)

NH

Inductive Effect6. The inductive effect -

(A) implies the atom’s ability to cause bond polarization(B) increases with increase of distance(C) implies the transfer of lone pair of electronsfrom more electronegative atom to the lesserelectronegative atom in a molecule(D) implies the transfer of lone pair of electronsfrom lesser electronegative atom to the moreelectronegative atom in a molecule

7. When – CH3, CH – CH – 3

CH3

and CH – C – 3

CH3

CH3

groups

are introduced on benzene ring then correct orderof their inductive effect is

(A) CH3 – > CH – CH – 3

CH3

> CH – C – 3

CH3

CH3

(B) CH – C – 3

CH3

CH3

> CH – CH – 3

CH3

> CH3 –

(C) CH – CH – 3

CH3

> CH3 > CH – C – 3

CH3

CH3

(D) CH – C – 3

CH3

CH3

> CH3 – > CH – CH – 3

CH3

8. Express in decreasing order of (+) -(a) CH3CH2 – CH2 – (b) CH3 –

(c) CH3 CH3–C–CH2–CH3

(d)

CH3 CH3–C–

CH3

(e) CH3–CH–CH2–

CH2CH3 Correct answer is -(A) (c) > (d) > (e) > (a) > (b)(B) (d) > (a) > (b) > (c) > (e)(C) (a) > (b) > (c) > (d) > (e)(D) (a) > (b) > (c) > (e) > (d)

9. Consider the following carbanions(i) CH – CH3 2

(ii) CH = CH2 (iii)

Correct order of stability of these carboanions indecreasing order is :(A) i > ii > iii (B) ii > i > iii(C) iii > ii > i (D) iii > i > ii

Exercise - 1 Objective Problems | JEE Main

GOC 2.23


Inductive & Acid, Base10. In which of the following compounds is hydroxylic

proton the most acidic ?

(A)

H

F

O (B) HOI

(C) H

F

O(D)

HF

O

11. Consider following acidClCH2COOH, CH3COOH, CH3CH2COOH I II IIICorrect order of their pH value is :(A) III < II < I (B) I < II < III(C) I < III < II (D) II < I < III

12. Which among the given acid has lowest pKa value -(A) Chloroacetic acid (B) Bromoacetic acid(C) Nitroacetic acid (D) Cyanoacetic acid

13. Arrange in decreasing pKa(a) F – CH2CH2 COOH

(b) Cl – CH – CH – COOH2

Cl(c) F – CH2 – COOH(d) Br – CH2 – CH2 – COOHCorrect answer is :(A) b > d > a > c (B) a > c > d > b(C) d > a > b > c (D) d > b > a > c

14. The correct order of increasing acid strength of thecompound is :(a) CH3CO2H (b) MeOCH2CO2H

(c) CF3CO2H (d)MeMe

CO H2

(A) d < a < c< d (B) d < a < b < c(C) a < d < c < b (D) b < d < a < c

15. The correct order of increasing basic nature of thebases NH3, CH3NH2 and (CH3)2NH is gas phase(A) NH3 < CH3NH2 < (CH3)2NH(B) CH3NH2 < (CH3)2NH < NH3(C) CH3NH2 < NH3 < (CH3)2NH(D) (CH3)2NH < NH3 < CH3NH2

16. Arrange basicity of the given compounds indecreasing order -(a) CH3 – CH2 – NH2 (b) CH2 = CH – NH2(c) CH C – NH2(A) a > b > c (B) a > c > b(C) c > b > a (D) b > c > a

17. Which one of the following is the strongest base inaqueous solution ?(A) Trimethylamine (B) Aniline(C) Dimethylamine (D) Methylamine

RESONANCE18. In which of the following molecules, all atoms are

not coplanar ?

(A)

OO

(B)

O

(C) (D)

O

O

19. (I) CH2 = CH – CH = CH2

(II) CH – CH = CH – CH2 2

(II) CH – CH = CH – CH2 2

Among these, which are canonical structures ?(A) I and II (B) I and III(C) II and III (D) all

20.

O||C

OHH

O|C

OHH

O|C

OHH

I II III

Among these canonical structures, the correct orderof stability is(A) I > II > III (B) III > II > I(C) I > III > II (D) II > I > III

21. CH = CH – CH = CH – OCH2 3(I)

CH CH CH CH OCH2 3– = – =

CH = CH – CH – CH – OCH2 3

CH = CH – CH – CH – OCH2 3

(II)

(III)

(IV)

Among these canonical structures which one is leaststable ?(A) I (B) II(C) III (D) IV



22. For phenol which of the following resonatingstructure is the most stable ?

(A)

OH

(B)

OH

(C)

OH

(D) All have equal stability

23. The most stable resonating structure of followingcompound is :

O = N N = O....

(A) O = N N = O..

(B) O – N N – O

(C) O = N N = O

(D) O – N N = O

24. N

N

N

N

I II III IV

N

VAmong these canonical structures of pyridine, thecorrect order of stability is :(A) (I = V) > (II = IV) > III(B) (II = IV) > (I = V) > III(C) (I = V) > III > (II = IV)(D) III > (II = IV) > (I = V)

25.N|H(I)

N|H(II)

N|H

(III)

N|H

(IV)

N|H

(V)(A) (III = IV) > (II = V) > I(B) I > (II = V) > (III = IV)(C) I > (III = IV) > (II = V)(D) (II = V) > (III = V) > I

26. ‘M’ effect is the resonance of -(A) electrons only(B) electrons only(C) and both(D) (+)ve and (–) charge.

27. Which of the following contain + M but - effect -(A) O = CH – (B) – NO2(C) – Cl (D) CH3 –

28. OH

om

p

om

In phenol, -electron density is maximum on(A) ortho and meta positions(B) ortho and para positions(C) meta and pera positions(D) none of these

29. Which of the following compounds has maximumelectron density in ring ?

(A)

NO2

(B)

OH

(C)

O

(D)

COO

30. In which of the following molecules -electrondensity in ring is minimum?

(A)

NO2

(B)

OCH3

(C)

NO2

H N2

(D) NO2

NO2

GOC 2.25


Stability of Intermediate1. Rank the following free radicals in order of

decreasing stability(I) C6H5 CHC6H5 (II) C6H5 – CH – CH = CH2

(III) CH3 – CH – CH3 (IV) C6H5 – CH – CH3

(V) CH3CH CHCH2CH2

(VI) CH3 – CH2 – C –

CH3

CH3

(A) I > II > IV > VI > III > V(B) VI > V > IV > III > II > I(C) I > II > III > IV > V > VI(D) I > IV > VI > V > II > III

2. Rank thefollowing radicals in order of decreasingstability

(I) (II)

(III) (IV)

(A) III > II > I > IV (B) III > IV > I > II(C) II > III > I > IV (D) IV > II > I > III

3. Select the most stable carboncation among thefollowing -

(A)

(B)

(C)

(D)

4. Write correct order of stability of followingcarbocations:

(I) MeMe

CMe3 (II)Me

MeCMe3

(III) MeCMe2 (IV)

Me

CMe3

Me

(A) I > II > III > IV (B) III > II > I > IV(C) III > I > II > IV (D) III > II > IV > I

5. Arrange the following carbocations in the increasingorder of their stability.

(I) (II)

(III)

(A) I > II > III (B) I > II = III(C) I > III > II (D) III > I > II

6. Which of the following carbocation will be moststable ?

(A) H C3 CH+

OCH3

(B) H C3 CH+

CH3

(C) H C3 CH+

CH3

(D) H C3 C

CH3

CH3

+

7. Statement-1: Me – CH2 is more stable than MeO

– CH2

Statement-2: Me is a +I group whereas MeO is a–I group.(A) Statement-1 is true, statement-2 is true andstatement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true andstatement-2 is NOT correct explanation forstatement-1.(C) Statement1 is false, statement-2 is true.(D) Statement1 is true, statement-2 is false.

8. Ease of ionization to produce carbocation andbromide ion under the treatment of Ag will bemaximum in which of the following compounds?

(A) O Br (B) O

Br

(C) N Br

Ph

(D) N Br

CH3

Exercise - 2 (Level-I) Objective Problems | JEE Main



9. In which of the following pairs, first species is morestable than second ?

(A) CH3CH2O– or CH CO3

–

O

(B) CH CCHCH CH3 2

O O

or CH CCHCH3 3

O O

(C) CH CHCH CCH3 2 3

O

or CH CH CHCCH3 2 3

O

(D) N–

O

O

or N–

O

10. The order of stability of the following carbanion is:

(I) CH CH3 2(II)

(III) (IV)

(A) I > II > III > IV (B) I > III > II > IV(C) IV > III > II > I (D) III > IV > I > II

11. Arrange the carbonions,

(CH ) C, CCl , (CH ) CH, C H CH3 3 3 3 2 6 5 2 in order of their

decreasing stability(A) (CH ) CH > CCl > C H CH > (CH ) C3 2 3 6 5 2 3 3

(B) CCl > C H CH > (CH ) C3 6 5 2 3 3(CH ) CH > 3 2

(C) (CH ) C > (CH ) CH > C H CH > 3 3 3 2 6 5 2 CCl3

(D) C H CH > CCl > (CH ) C > (CH ) CH6 5 2 3 3 3 3 2

12.

1

2

3

4

1

2

3

4

1

2

3

4

There are three canonical structures of napthalene.Examine them and find correct statement amongthe following:(A) All C–C bonds are of some length(B) C1-C2 bond is shorter than C2-C3 bond.(C) C1-C2 bond is longer than C2-C3 bond(D) None.

13. Which of the following has longest C – O bond:

(A) O

(B)

O

(C)

O

(D)

O

CH2

14. Among the following molecules, the correct orderof C - C bond length is :(A) C2H6 > C2H4 > C6H6 > C2H2

(B) C2H6 > C6H6 > C2H4 > C2H2 (C6H6 is benzene)(C) C2H4 > C2H6 > C2H2 > C6H6

(D) C2H6 > C2H4 > C2H2 > C6H6

AROMATICITY

15. In which of the following molecules -electrondensity in ring is maximum?

(A)

NO2

(B) O

(C)

NH2

(D)

OCH3

16. (I) (II) (III)

Which of these cyclopropene systems is aromatic?(A) I (B) II(C) III (D) all of these

17. (I) (II) (III)

Which of these species is anti-aromatic ?(A) I only (B) II only(C) III only (D) both II and III

GOC 2.27


18. Which of the following compouds is not aromatic?

(A) OO

(B)

O

O

(C)

O

(D)

O

O

19. N = N

The most stable canonical structure of this moleculeis :

(A) N = N (B) N = N

(C) N N (D) All are equally stable

20.O

The most stable canonical structure of this molecule is:

(A)

O

(B) O

(C)

O

(D)

O

21. (I) (II) (III)

The barrier for rotation about the indicated bondsWill be maximum in which of these three compounds ?(A) I (B) II(C) III (D) same in all

22. Identify the odd species out Which of the speciesamong the following is different from others ?

(A) (B)

(C) (D)

23. Which of the following heterocyclic compoundswould have aromatic character ?

(A) N

N(B)

NN – H

(C) N–H

N–H(D) N–H

24. Which one of the following carbonyl compoundwhen treated with dilute acid forms the more stablecarbocation ?

(A) CH3 –

O||C – CH3 (B)

O

(C)

CH3

OHHO

HO

O

(D) C6H5 –

O||C– C6H5



25. The order of the rate of formation of carbocationsfrom the following iodo compound is:

(I) IH

(II) IH

(III) IH

(A) I > II > III (B) I > III > II(C) III > II > I (D) II > III > I

26. Write correct order of reactivity of following halogenderivatives towards AgNO3.

(I)

Cl

(II) CH2 = CH – Cl

(III) Et3 C – Cl (IV) PhCH2Cl(V) Ph3C – Cl(A) I > V > IV > III > II(B) V > IV > I > III > II(C) V > I > IV > III > II(D) I > V > III > IV > II

27. Which of the following species is not aromatic ?

(A) O

(B) O

(C) (D)

28. (I) O

(II) N|H

(III) N

The aromatic character is maximum in which ofthese three compounds ?(A) I (B) II(C) III (D) Same in all

29. CH3COOH CH3COONa CH3CONH2 (I) (II) (III)Among these compounds, the correct order ofresonance energy is :(A) I > II > III (B) III > II > I(C) II > III > I (D) II > I > III

30. (I)

O

(II)

O

(III)

O

(IV)

O

Among these compounds, which one has maximumresonance energy ?(A) I (B) II(C) III (D) IV

31. (I) (II)

Which of the following orders is correct for theresonance energy of these two compounds ?(A) I > II(B) II > I(C) I = II(D) there is nothing like -electron energy

GOC 2.29


RESONANCE1. Which of the following statements is (are) true about

resonance.(a) Resonance is an intramolecular process.(b) Resonance involves delocalization of both and

electrons.(c) Resonance involves delocalization of electrons

and lone pair only.(d) Resonance decreases potential energy of a

molecule.(e) Resonance has no effect on the potential energy of

a molecule.(f) Resonance is the only way to increase molecular

stability.(g) Resonance is not the only way to increase molecular

stability.(h) Any resonating molecule is always more stable than

any nonresonating molecule.(i) The canonical structure explains all features of a

molecule.(j) The resonance hybrid explains all features of a

molecule.(k) Resonating structures are real and resonance hybrid

is imaginary.(l) Resonance hybrid is real and resonating structures

are imaginary.(m) Resonance hybrid is always more stable than all

canonical structures.

2. Resonance energy will be more if(a) canonical structures are equivalent than if canonical

structures are non-equivalent.(b) molecule is aromatic than if molecule is not aromatic.

3. A canonical structure will be more stable if(a) it has more number of bonds than if it has less

number of bonds.(b) the octate of all atoms are complete than if octate

of all atoms are not complete.(c) it involves cyclic delocalization of (4n + 2) –

electrons than if it involves acyclic delocalization of(4n + 2) – electrons.

(d) it involves cyclic delocalization (4n) – electronsthan if it involves acyclic delocalizationof (4n) –electrons.

(e) +ve charge is on more electronegative atom than if+ve charge is on less electronegative atoms.

(f) –ve charge is on more electronegative atom than if–ve charge is on less electronegative atom.

4. Consider structural formulas A, B and C:

(A) (B) (C)(a) Are A, B and C constitutional isomers, or are they

resonance forms?(b) Which structures have a negatively charged carbon?(c) Which structures have a positively charged carbon?(d) Which structures have a positively charged

nitrogen?(e) Which structures have a negatively charged

nitrogen?(f) What is the net charge on each structure?(g) Which is a more stable structure, A or B? Why?(h) Which is a more stable structure, B or C? Why?

5. How many of the following compounds give CO2 on

reaction with NaHCO3 ?

,HCl, , ,

, , HCOOH , C2H

5–OH , CH

3COOH ,

6. Identify more stable canonical structure in each ofthe following pairs :

(a) C CH HOH OH

..

O O

(b)

(c)

(d) OCHCHCHOCHCHHC 22

(e)

Exercise - 2 (Level-II) Multiple Correct | JEE Advanced



7. In the following sets of resonance forms, label themajor and minor contributors and state whichstructures would be of equal energy. Add anymissing resonance forms.

(a) CH3—CH–C N CH3—CH=C N

(b) CH3—C=CH–CH—CH3 CH3—C—CH=CH—CH3

O¯ O¯+

+

(c) CH3—C–CH–C—CH3 CH3—C=CH–C—CH3

O O¯O O

(d) [CH – CH – CH = CH – NO CH – CH = CH – CH – NO ]3 2 3 2

(e) CH3—C —C—NH2H2 CH3—C —C = NH2H2+

+

NH2 NH2

Resonance Energy8. Which of the following pairs has higher resonance

energy ?(a) CH3COOH and CH3COONa

(b) CH2 = CH –O and CH2 = CH – OH

(c) COO

and O

(d) and

(e) and CH2 = CH – CH = CH – CH = CH2

9. Which of the following pairs has less resonanceenergy ?(a) CO3

2– and HCOO–

(b) and CH2 = CH – CH2–

(c) and CH2 = CH – CH = CH2

(d) and CH2 = CH – CH2+

10. Which of the following pairs has higher resonanceenergy ?

(a) and

(b) and

(c) and

(d) CH2 = CH – OH and CH2 = CH – CH = CH – OH

(e) and

AROMATIC11. H – O – C N H – N = C = O

(Cyanic acid) (Isocyanic acid)Loss of proton from these two acids produces(A) same anion (B) different anions(C) same cation (D) different cations

12. Ease of ionization to produce carbocation andbromide ion under the treatment of Agwill bemaximum in whichof the following compounds ?

(A) Br (B) Br

(C) Br

OCH3

(D) Br

13.

Cl

Cl 2SbCl5 P will be

(A) 2– (B) 2+ 2SbCl6

(C) (D) mixture of (a) and (b)

14.

Pn-BuLi

KH

P will be

(A) (B)

(C) mixture of (A) & (B) (D) none of these

GOC 2.31


15. Which one of the following statements is True:

(1) (2)

(A) PhLi adds to both compound with equal ease(B) PhLi does not add to either of the compound(C) PhLi react readily with 1 but does not add to 2(D) PhLi react readility with 2 but does not add to 1

16. Correct order of rate of hydrolysis or rate of reactiontoward AgNO3 for following compounds is :

(I)

Br

(II)

Br

(III) Br

(IV) Br

(A) III > II > IV > I (B) I > II > III > IV(C) III > I > II > IV (D) III > II > I > IV

17.

O

HClO4 P will be

(A)

OH

ClO4

(B)

OH

H

(C)

O

H

ClO4(D) Mixture of (A) & (B)

18. Cl Ag ClO4 P

P will be :

(A) ClO4 (B) Ag

(C) Mixture of (A) & (B) (D) None of these

19. Aromatic compounds is/are:

(A) N

N

N(B)

O

(C) B|H

(D)

20. Which of the following reactions give aromaticcompound ?

(A) KH (B)

O

HBr

(C) HI (D)

O

HBr

Stability of Intermediate21. Write stability order of following intermediates:

(i) (a) 23 CHCH

(b) 33 CHCHCH

(c)

3

3

3

CH|CCH|

CH

(ii) (a)

(b)

(c)

(iii) (a)

(b)

(c)

(iv) (a) 23CH CH

(b) 33 CHCHCH

(c)

3

3

3

CH|

CH C

|

CH

(v) (a) (b) (c)



(vi) (a) (b) (c)

(vii) (a)

CHC (b) 33 CHCHCH

(c)

3

3

3

CH

|

CCH|

CH

(viii) (a) (b) (c)

(ix) (a) (b) (c)

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

(xi) (a) HC C (b)

CHCH2 (c)

23 CHCH

(xii) (a) HC C (b)

CHCH2 (c)

23 CHCH

22. Write stability order of following intermediates:

(i) (a)

CH2

N

O O

(b)

CH2

OMe

(c)

CH2

(ii) (a)

CH2

Cl

(b)

CH2

NO O

(c)

CH2CH2

CNº

(d)

CH2CH2

(iii) (a)

CH2

OH (b)

CH2

OH (c)

CH2

OH

(iv) (a)

CH2

F

(b)

CH2CH2

ClCl

(v) (a) CH2 CH

O

(b) 32 CHCH

(vi) (a) O O

(b) O

(c)

(vii) (a)

CH2

(b)

CH2–CH2

(c)

(viii) (a) (b) (c)

(ix) (a) (b)

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

(xi) (a) (b) (c)

(xii) (a)

CH2

C

HH

H

(b)

CH2

C

HH

H

(c)

CH2

C

HH

H

GOC 2.33


(xiii) (a)

CH2

CH3

(b)

CH2

CH2Me

(c)

CH2

CH Me2

(d)

CH2

CMe3

Bond Length

23. In which of the following pairs, indicated bond is ofgreater strength :

(a) BrCHCH 23 and

ClCHCH 23

(b) BrCHCHCH3 and

Br|

CHCHCH 33

(c) and ClCHCH 23

(d)

22 CHCHCHCH and

3222 CHCHCHCH

(e) and

24. In which of the following pairs, indicated bondhaving less bond dissociation energy :

(a) and 22 CHCH

(b) CHCCH3 and

CHHC

(c) and

(d) and

(e) and

(f) and

25. Compare the C–N bond-length in the followingspecies:

(i) (ii) (iii)

26. Which of the following statements would be trueabout this compound:

NO2

NO2NO2

Br

5

31

(A) All three C – N bonds are of same length.(B) C1 – N and C3 – N bonds are of same lengthbut shorter than C5 – N bond(C) C1 – N and C3 – N bonds are of same lengthbut longer than C5 – N bond(D) C1 – N and C3 – N bonds are of differentlength but bot are longer than C5 – N bond.

27. Choose the more stable alkene in each of thefollowing pairs. Explain your reasoning.

(a) 1-Methylcyclohexene or 3-methylcyclohexene(b) Isopropenylcyclopentane or allylcyclopentane

(c) or



28. Consider the given reaction:

+ 3H2 C/Pd

In the above reaction which one of the given ringwill undergo reduction?

Heat of Hydrogeneration & Combustion29. Compare heat of hydrogenation (Decreasing order)(a) heat of hydrogenation

(i)

(ii)

(b) and

(c) and

(d) and

(e) CH2 = CH – CH and

CH2 = C

30. (I) Stability order and (II) heat of hydrogenationorders.

(A) (i) (ii)

(iii) (iv)

(B) (i)

(ii)

(iii)

31. Among the following pairs identify the one whichgives higher heat of hydrogenation :

(a) and

(b) and

(c) CH3 – CH = CH – CH3 andCH3 – CH2 – CH = CH2

(d) and

32. Match each alkene with the appropriate heat ofcombustion:Heats of combustion (kJ/mol) : 5293 ; 4658; 4650;4638; 4632

(a) 1-Heptene(b) 2,4-Dimethyl-1-pentene(c) 2,4-Dimethyl-2-pentene(d) 4,4-Dimethyl-2-pentene(e) 2,4,4-Trimethyl-2-pentene

33. Write increasing order of heat of hydrogenation :

(i) (a) (b)

(ii) (a) (b)

(c) (d)

(iii) (a) (b)

(c) (d) (e)

GOC 2.35


(iv) (a) (b) (c)

(v) (a) (b) (c)

(HOH per bond)

(vi) (a) (b)

(c) (HOH per benzene ring)

(vii) (a) (b)

(viii) (a) (b)

34. Give decreasing order of heat of combustion (HOC):

(i) (a) (b) (c)

(ii) (a) (b) (c)

(d)

(iii) (a) (b)

(iv) (a) (b) (c)

35. Arrange in order of C–H bond energy

H

H-CH-CH-C-CH3

H H

H-CH2

CH2

Hf

e

d

c b

a

36. Use the following data to answer the questionsbelow:

H

Ni

2

H = – 28.6 Kcal mol–1

excess H2

(Ni)

H = – 116.2 Kcal mol–1 AnthraceneCalculate the resonance energy of anthracene inkcal/mol.

37. Arrange the given phenols in their decreasing orderof acidity:

(I) C6H5–OH (II) F OH

(III) Cl OH (IV) O2N OH

Select the correct answer from the given code:(A) IV > III > I > II (B) IV > II > III > I(C) IV > III > II > I (D) IV > I > III > II

38. Which one of the following is the most acidic?

(A) (B)

(C) (D) CH2=CH–CH3

Acid & Base39. Which one of the following phenols will show

highest acidity?

(A)

NO2

CH3

CH3

OH(B)

CH3

H C3

O N2

OH

(C)

H C3

H C3 NO2

OH

(D)

NO2

CH3

H C3

OH



40. Which of the following is weakest acid?

(A)

COOH

(B)

COOH

OH

(C)

COOH

OH

(D)

COOH

OH

41. The correct pKa order of the follwoing acids is :

OH OHOHO HO

(I) (III)(II)O OOO OO

(A) I > II > III (B) III > II > I(C) III > I > II (D) I > III > II

42. Arrange pH of the given compounds in decreasingorder:(1) Phenol (2) Ethyl alcohol(3) Formic acid (4) Benzoic acid(A) 1 > 2 > 3 > 4 (B) 2 > 1 > 4 > 3(C)3 > 2 > 4 > 1 (D) 4 > 3 > 1 > 2

43. Consider the following compound :

(A)

OHO

O OH

(B) OH

COOH

(C) CH CCOOH3

O(D)

OH

O N2 NO2

NO2

Which of the above compounds reacts withNaHCO3 giving CO2 ?

44. Match the column:Column I

(A)

(B)

(C) NH

NH

(D) H– N

Column II(P) Six electrons(Q) Four electrons(R) Aromatic Compounds(S) Anti-aromatic compound


(A)

(B)

(C)

(D)

Column II(P) Hybrid state of each atom sp2

(Q) Anti aromatic(R) Delocalisation of bond(S) Non aromatic(T) Obeys Huckel's Rule for aromaticity

GOC 2.37


46. Match the column :Column I

(A) NH

(B)

(C) O

(D)

Column II(P) Non aromatic(Q) Anti aromatic(R) Resonance(S) Aromatic


(A) CH OCH or CH NHCH3 2 2 2

+ +

(B) CH OCH or CH OCH3 2 3 2CH2

+ +

(C) + or +

(D) CHCH3

+

or CHCH3

+

Column II(P) First is more stable than second(Q) Second is more stable than first(R) Not resonating structure of each other(S) Resonance is present in both carbocation



ACIDS

1. Write the correct order of acidic strength offollowing compounds:

(i) (a) H–F (b) H–Cl(c) H–Br (d) H–I

(ii) (a) CH4 (b) NH3(c) H2O (d) H–F

(iii) (a) CH3–CH2–O–H (b)

3

3

CH|

HOCHCH

(c) CH3–C–O–H

CH3

CH3

(iv) (a) F–CH2–CH2–O–H (b) NO2–CH2–CH2–O–H(c) Br–CH2–CH2–O–H

(d) 3 2 2NH CH CH O H

2. Write the correct order of acidic strength offollowing compounds:

(i) (a) CH3COOH (b) CH3CH2OH(c) C6H5OH (d) C6H5SO3H

(ii) (a) COOH

(b) COOH

(c) COOH

(iii) (a) COOH|COOH

(b) CH2

COOH

COOH

(c) COOHCH

|COOHCH

2

2

3. Write correct order of acidic strength of followingcompounds:

(i) (a) Cl–CH –C–O2 –H

O

(b) Cl–CH –C–O2 –H

O

Cl

(c) Cl–C–C–O–H

O

Cl

Cl

(ii) (a) CH – –CH–C O3 CH – –H2

O

F

(b) CH – –CH –C O3 CH – –H2

O

F

(c) CH – –CH –C O2 CH – –H2 2

O

F

(iii) (a) HOCCHNO||O

22

(b) HOCCHF||

O

2

(c) HOCCHPh||O

2

(d) HOCCHCH||

O

23


(i) (a)

O–H

NO2

(b)

O–H

Cl

(c)

O–H

CH3

(ii) (a)

O–H

Cl (b)

O–H

Cl

(c)

O–H

Cl

(iii) (a)

O–H

(b)

O–H

(c)

O–H

(d)

O–H

Exercise - 3 | Subjective | JEE Advanced

GOC 2.39



(i) (a)

O–HN

O

O

(b)

O–H

NO

O

(c)

O–H

NO O

(d)

O–H

(ii) (a)

O–H

NO2

(b)

O–H

NO

O

(c)

O–H

NO2

NO2

(d)

O–H

NO2

NO2NO2


(i) (a)

C–O–H

O

(b)

C–O–H

O

CH3

(ii) (a)

COOH

Cl (b)

COOH

Br

(iii) (a)

C–O–H

OMe

O

(b)

C–O–H

OMe

O

(c)

C–O–HOMe

O

(iv)(a)

C–O–H

N

O

OO

(b)

C–O–H

NO2

O

(c)

C–O–H

NO2

O

7. Select the strongest acid in each of the followingsets :

(i) (a)

OH

CH3

(b)

OH

NO2

(c)

OH

Cl

(d)

OH

NH2

(ii) (a)

OH

NO2

(b)

OH

F

(c)

OH

CH3

(d)

OH

(iii) (a)

OHOMe

(b)

OH

OMe

(c)

OH

(d)

OH

OMe

8. Say which pka belong to which functional group incase of following amino acids :

(i) cysteine : NH2

COOHHS 1.8, 8.3 & 10.8

(ii) glutamic acid :

NH2

HO C2 COOH

: 2.19, 4.25, 9.67



9. Record the following sets of compounds accordingto increasing pKa ( = – log Ka)

(a)

OH

,

OH

, cyclohexane carboxylic acid.

(b) 1-butyne, 1-butene, butane(c) Propanoic acid, 3-bromopropanoic acid,

2-nitropropanoic acid(d) Phenol,o-nitrophenol, o-cresol(e) Hexylamine, aniline, methylamine

10. Explain which is a stronger acid.

(a) CH3CH3 BrCH2NO2

(b) CH –C–CH3 3

O

& CH –C–CH CN3 2

O

(c) CH3 – CHO CH3 – NO2

11. Explain which is a weaker acid.

(a)

OH

O=C–CH3

or

OH

(c)

OH

O=C–CH3

or

OH

CH3

(b)

SH

or

OH

12. Which of the following would you predict to be thestronger acid ?

(a)COOH

or NO

OOC – OH

(b) CH3 – CH2 – CH2 – OH or CH3 – CH = CH – OH(c) CH3 – CH = CH – CH2 – OH or CH3 – CH = CH – OH

Bases13. Write increasing order of basic strength of following:(i) (a) F (b) Cl (c) Br (d) I

(ii) (a) 3CH b)

2NH (c) OH (d) F

(iii) (a) R–NH2 (b) Ph–NH2 (c) R–C–NH2

O

(iv) (a) NH3 (b) MeNH2(c) Me2NH (d) Me3N (Gas phase)

(v) (a) NH3 (b) MeNH2(c) Me2NH (d) Me3N (in H2O)

14. Write increasing order of basic strength of following:

(i) (a) NH

O

(b) NH

(c) N

Me

(ii) (a)

NH2

(b)

NH2

(c) NH

(iii) (a) O N2

N

(b) Me

N

(c) F

N

(iv) (a)

NH2

NH3

(b)

NH2

Cl

(c)

NH2

CH3

(d)

NH2

H


(i) (a) CH3–CH2– 2HN (b) CH3 – CH = HN

(c) CH3 – C N

(ii) (a) CH –C–NH3 2

O

..(b) CH3 – CH2 –

2HN

(c) CH –C–NH3 2

NH

..

..

(d) NH –C–NH2 2

NH

..

..

(iii) (a) N

H

(b) N

(c)

NH2

GOC 2.41


(iv) (a)

NH – C – CH 3

O

(b)

NH2

(c)

NH–CH –CH2 3

(v) (a)

NH2

N

MeMe

O O

(b)

NH2

NMeMe

O O


(i) (a)

NO2

NH2

(b)

CN

NH2

(c)

OMe

NH2

(d)

NH2

NH2

(ii) (a)

NH2

C

H

H

H (b)

NH2

C

H

H

H

(c)

NH2

CH

HH

(iii) (a)

NO2

NH2

(b)

NH2

NO2

(c)

NH2

NO2

(iv) (a)

NH2

(b)

NH2

C

H

H

H (c)

NH2

(v) (a)

NMe2

OMe

(b)

NMe2

OMe

(c)

NOMe

Me Me

17. Select the strongest base in following compound :

(i) (a) N

(b)

N

H

(c) N

O

H

(d)

N

S

H

(ii) (a)

NH2

(b) NH

(c) N

(d) N

H

(iii) (a) N

N

H

(b) N

H

(c) N

CH3

(d) N

H

(iv) (a)

N Li¯ +

(b) N

H

(c) N

H

(d) N

Me



18. Arrange the following compound in decreasing orderof their basicity.

(i) (a) H2C = CHNa (b) CH3CH2Na(c) CH3CH2ONa (d) HC CNa

(ii) (a) NH2 (b) CH NH2 2 –

(c) NH2

NO2

(d) C NH2 –

O

(iii) (a) HO¯ (b) NH3 (c) H2O

19. Basicity order in following compound is :

N

N

NH

O CH3

CH3

CH2 – – – NH C CH3H2N CH2– C– b

d

a

c

CH3

CH3

(A) b > d > a > c (B) a > b > d > c(C) a > b > c > d (D) a > c > b > d

20. Consider the following bases:(I) o-nitroaniline(II) m-nitroaniline(III) p-nitroanilineThe decreasing order of basicity is:(A) II > III > I (B) II > I > III(C) I > II >III (D) I > III > II

21. Consider the basicity of the following aromaticamines:(I) aniline (II) p-nitroaniline(III) p-methoxyaniline (IV) p-methylanilineThe correct order of decreasing basicity is:(A) III > IV > I > II (B) III > IV > II > I(C) I > II > III > IV (D) IV > III > II > I

22. Which one of the following is least basic in character?

(A) N

H

(B) N N – H

(C) N

H

(D) N

H

23. In each of the following pair of compounds, whichis more basic in aqueous solution? Give anexplanation for your choice:

(a) CH3NH2 or CF3NH2

(b) CH3CONH2 or H2N

NH

NH2(c) n-PrNH2 or CH3CN(d) C6H5N(CH3)2 or 2,6-dimethyl-N,N-dimethylaniline(e) m-nitroaniline or p-nitroaniline

24. From the following pair, select the stronger base:(a) p-methoxy aniline or p-cyanoaniline(b) pyridine or pyrrole(c) CH3CN or CH3CH2NH2

25. Explain which compound is the weaker base.

(a)

NH2

or

NH2

NO2

(b) CH2 = CH – CH = CH – CH2– or CH2 = CH – CH2

–

(c) O – –C–O–

C H

O O

or H C HO– –C–O

O O

(d)

OH

CH2 or

OH

CF3

26. Rank the following amines in increasing basic nature.

(a)

NH2 NH2

CH3

(i) (ii)NH2

NO2

NH2

(iii) (iv)

(b)

NH2

CH3

NH2

CH3

(i) (ii)

NH2

CH3

(iii) (iv)

GOC 2.43


27. Arrange the basic strength of the following

compounds.

(a) OH– CH3COO– Cl–

(i) (ii) (iii)

(b) CH C– CH2 = CH– CH3CH2–

(i) (ii) (iii)

(c) (i) CH2 = CHCH2NH2

(ii) CH3CH2CH2NH2

(iii) CH C – CH2NH2

28. Arrange the basic strength of the following

compounds.

(a)

NH2

NH–C H6 5 NH2

(i) (ii) (iii)

(b)

NH2

Cl

NH2

Cl

NH2

Cl

(i) (ii) (iii)

(c)NH2

H C3

NH2 NH2

O N2

(i) (ii) (iii)

29. Arrange the following compounds in order of

increasing basicity.

(a) CH3NH2, CH33NH , CH3NH—

(b) CH3O—, CH3NH—, CH3

2CH

(c) CH3CH = CH—, CH3CH22CH , CH3CC—

30. Rank the amines in each set in order of increasing

basicity.

(a) NH

NH2 NH2

(b)

H

N NH N

(c) N NN NH HH

31. NN1

2 3

6

5

4

N

4 5

12

3N–H

NN

N N

56

1

2

3

49

8

7

H

Pyrimidine Imidazole PurineAmong the following which statement(s) is/are ture:(A) Both N of pyrimidine are of same basic strength(B) In imidazole protonation takes places on N–1.(C) Purine has 3 basic N.(D) Pyrimidine imidazole and purine all are aromatic



1. Amongst the following the most basic compound is -[AIEEE-2005]

(A) aniline (B) benzylamine(C) p–nitroaniline (D) acetanilide

2. The number and type of bonds between two carbonatoms in calcium carbide are [AIEEE 2005](A) two sigma, two pi (B) two sigma, one pi(C) one sigma, two pi (D) one sigma, one pi

3. Due to the presence of an unpaired electron, freeradicals are [AIEEE 2005](A) cations (B) anions(C) chemically inactive (D) chemically reactive

4. Among the following acids which has the lowestPKa value ? [AIEEE 2005](A) CH3CH2COOH (B) (CH2)2 CH – COOH(C) HCOOH (D) CH3COOH

5. The increasing order of stability of the followingfree radicals is - [AIEEE 2006]

(A) (C6H5)3C< (C6H5)2 HC

< (CH3)3

C < (CH3)2 HC

(B) (C6H5)2 HC < (C6H5)3

C < (CH3)3

C < (CH3)2 HC

(C) (CH3)2 HC < (CH3)3

C < (C6H5)3

C< (C6H5)2 HC

(D) (CH3)2 HC < (CH3)3

C < (C6H5)2 HC

< (C6H5)3C

6. The correct order of increasing acid strength of thecompound is : [AIEEE 2006](a) CH3CO2H (b) MeOCH2CO2H

(c) CF3CO2H (d) Me

Me CO2H is

(A) d < a < c < b (B) d < a < b < c(C) a < d < c < b (D) b < d < a < c

7. Which one of the following is the strongest base inaqueous solution ? [AIEEE-2007](A) Trimethylamine (B) Aniline(C) Dimethylamine (D) Methylamine

8. Presence of a nitro group in a benzene ring : [AIEEE 2007]

(A) activates the ring towards electrophilicsubstitution(B) renders the ring basic(C) deactivates the ring towards nucleophilicsubstitution(D) deactivates the ring towards electrophilicsubstitution

9. Arrange the carbanions, (CH3)3 C , 3ClC ,

(CH3)2 HC , C6H5 2HC , in order of their

decreasing stability - [AIEEE 2009]

(A) (CH3)2 HC > 3ClC > C6H5 2HC > (CH3)3 C

(B) 3ClC > C6H5 2HC > (CH3)2 HC > (CH3)3 C

(C) (CH3)3 C > (CH3)2 HC >C6H5 2HC > 3ClC

(D) C6H5 2HC > 3ClC > (CH3)3 C > (CH3)2 HC

10. The correct order of increasing basicity of the givenconjugate bases (R = CH3) is [AIEEE 2010]

(A) –2RCOO HC C R N H

(B) 2R HC C RCOO N H

(C) –2RCOO NH HC C R

(D) 2RCOO HC C N H R

11. The correct order of acid strength of the followingcompounds is : [AIEEE 2011]A. Phenol B. p-CresolC. m-Nitrophenol D. p-Nitrophenol(A) D > C > A >B (B) B > D > A > C(C) A > B > D >C (D) C > B > A > D

12. The non aromatic compound among the followingis - [AIEEE 2011]

(A)

S(B)

(C)

(D)

Exercise - 4 | Level-I Previous Year | JEE Main

GOC 2.45


13. The order of stability of the following carbocations: [JEE Main 2013]

(I) CH2= CH – CH2 (II) CH3–CH2– CH2

(III)

CH2

is :

(A) I > II > III (B) III > I > II(C) III > II > I (D) II > III > I

14. For which of the following molecule significant 0 ?

[JEE Adv. 2014]

(a)

Cl

Cl

(b)

CN

CN

(c)

OH

OH

(d)

SH

SH

(A) Only (c) (B) (c) and (d)

(C) Only (a) (D) (a) and (b)

15. Considering the basic strength of amines in aqueous

solution, which one has the smallest pKb value?

[JEE Adv. 2014]

(A) (CH3)3N (B) C6H5NH2

(C) (CH3)2NH (D) CH3NH2

16. Which of the following molecules is least

resonance stabilized ? [JEE Adv. 2017]

(A) O

(B) N

(C) O

(D)

17. The increasing order of basicity of the

following compounds is : [JEE Adv. 2018]

(a) NH2 (b) NH

(c) NH

NH2

(d) NHCH3

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

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

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

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

For 2019 & 2020 year questions you can visit

@ www.onlinetestseries.motion.ac.in



1. For 1-Methoxy-1,3-butadiene, which of the followingresonating structure is the least stable ?[IIT-2005]

(A) H C – CH – CH = CH – O – CH2 3

(B) H C – CH = CH – CH = O – CH2 3

(C) H C = CH = CH – CH – O – CH2 3

(D) H C = CH – CH – CH = O – CH2 3

2. Predict whether the following molecules are isostructural or not. Justify your answer. [IIT-2005](i) NMe3 (ii) N(SiMe3)3

3. When benzene sulfonic acid and p-nitrophenol aretreated with NaHCO 3, the gases releasedrespectively are [IIT-2006](A) SO2,NO2 (B) SO2,NO(C) SO2,CO2 (D) CO2,CO2

4. (I) 1, 2-dihydroxy beznene(II) 1, 3-dihydroxy benzene(III) 1, 4-dihydroxy benzene(IV) Hydroxy benzeneThe increasing order of boiling points of abovementioned alcohols is [IIT-2006](A) I < II < III < IV (B) I < II < IV < III(C) IV < I < II < III (D) IV < II < I < III

5. Among the following, the least stable resonancestructure is : [IIT-2007]

(A) N

O

O (B) N

O

O

(C) N

O

O

(D) N

O

O

6. Statement-1: p-Hydroxybenzoic acid has a lowerboiling point then o-hydroxybenzoic acid.Statement-2: o-Hydroxybenzoic acid has aintramoleculer hydrogen bonding. [IIT-2007](A) Statement-1 is true, statement-2 is true andstatement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true andstatement-2 is NOT correct explanation forstatement-1.(C) Statement1 is true, statement-2 is false.(D) Statement1 is false, statement-2 is true.

7. Hyperconjugation involves overlap of the followingorbitals [IIT-2008](A) – (B) – p(C) p – p (D) –

8. The correct stability order for the following species is [IIT-2008]

(I) O

(II)

(III) O

(IV)

(A) II > IV > I > III (B) I > II > III > IV(C) II > I > IV > III (D) I > III > II > IV

9. The correct acidity order of the following is [IIT-2009]

(I)

OH

(II)

OH

Cl

(III)

COOH

(IV)

COOH

CH3

(A) III > IV > II > I (B) IV > III > I > II(C) III > II > I > IV (D) II > III > IV > I

10. The correct stability order of the followingresonance structures is [IIT-2009]

(I) H C = N = N2

–+(II) H C – N = N2

–+

(III) H C – N = N2

– +(IV) H C – N = N2

– +

(A) I > II > IV > III (B) I > III > II > IV(C) II > I > III > IV (D) III > I > IV > II

Exercise - 4 | Level-II Previous Year | JEE Advanced

GOC 2.47


11. In the following carbocation; H/CH3 that is mostlikely to migrate to the positively charged carbon is

[IIT-2009]

H C – C – C – C –CH3 3

OH H CH3

HH1 2

3

+ 4 5

(A) CH3 at C-4 (B) H at C-4(C) CH3 at C-2 (D) H at C-2

12. The total number of basic group in the followingform of lysine is : [IIT-2010]

NH – CH3 2 – CH – CH – CH – CH – C2 2 2

NH2

O

O–

+

13. Amongst the following, the total number ofcompounds soluble in aqueous NaOH is

[IIT-2010]

N OHNO2

CH3H C3 COOH

NH C3 CH3

14. Among the following compounds, the most acidic is:(A) p-nitrophenol [IIT-2011](B) p-hydroxybenzoic acid(C) o-hydroxybenzoic acid(D) p-toluic acid

15. The total number of contrubting structure showinghyperconjugation (involving C-H bonds) for thefollowing carbocation is [JEE-2011]

CH CH2 3CH3

16. In Allen (C3H4), the type (s) of hybridisation of thecarbon atoms is (are) [JEE Adv. 2012](A) sp and sp3 (B) sp and sp2

(C) only sp2 (D) sp2 and sp3

17. Which of the following molecules in pure from is (are)unstable at room temperature [JEE Adv. 2012]

(A) (B) (C)

O

(D)

O

18. The compound that does NOT liberate CO2, ontreatment with aqueous solium bicarbonate solution is(A) Benzoic acid [JEE Adv. 2013](B) Benzenesulphonic acid(C) Salicylic acid(D) Carbolic acid (Phenol)

19. The hyperconjugative stabilities of tert-butyl cationand 2-butene, respectively, are due to

[JEE Adv. 2013](A) p (empty) and electrondelocalisations(B) and electron delocalisations(C) p (filled) and electrondelocalisations(D) p (filled) and electrondelocalisations

20. The number of resonance structures for N is : [JEE Adv. 2015]

OHNaOH

N

21. The correct order of acidity for the followingcompounds is : [JEE Adv. 2016]

(I)

CO H2

OHHO(II)

CO H2

OH

(III)

CO H2

OH (IV)

CO H2

OH

(A) I > II > III > IV (B) III > I > II > IV(C) III > IV > II > I (D) I > III > IV > II



22. Among the following the number of aromaticcompound(s) is : [JEE Adv. 2017]

23. The order of basicity among the followingcompounds is : [JEE Adv. 2017]

(I) NH

NH2H C3(II) N NH

(III) HN N (IV) NH2

NHH N2(A) IV > II > III > I (B) II > I > IV > III(C) I > IV > III > II (D) IV > I > II > III

24. The correct order of acid strength of thefollowing carboxylic acids is :

[JEE Adv. 2019]

I.

O

OH

H

II.OHH

O

H

III. O

OHMeO

IV. OH

OH C3

(A) II > I > IV > III (B) III > II > I > IV(C) I > II > III > IV (D) I > III > II > IV

GOC 2.49


Exercise - 1 Objective Problems | JEE Main

1. A 2. C 3. A 4. C 5. B 6. A 7. B

8. A 9. C 10. D 11. B 12. C 13. C 14. B

15. A 16. A 17. C 18. C 19. D 20. C 21. D

22. C 23. D 24. C 25. C 26. A 27. C 28. B

29. C 30. D

Exercise - 2 (Level-I) Objective Problems | JEE Main

1 A 2 A 3 C 4 B 5 A 6 A 7 C

8 D 9 D 10 D 11 B 12 B 13 B 14 B

15 B 16 C 17 A 18 D 19 C 20 B 21 B

22 B 23 D 24 C 25 C 26 A 27 B 28 C

29 C 30 C 31 B

Exercise - 2 (Level-II) Multiple Correct | JEE Advanced

1. (a), (c), (d), (g), (j), (l), (m) 2. (a), (b) 3. (a), (b), (c), (f)

4. a = Resonance form, b = A, c = C, d = A & B, e = B & C, f = 0, g = B, h = B

5. 6 (i, ii, iii, vi, vii, ix)

6. (a) I, (b) I, (c) I, (d) II, (e) II

7. (a) II, (b) II, (c) II, CH – CH3 3– C – CH – CH = C

O O

(d) II, CH3–CH=CH – CH = NO

O

(e) II, CH – CH – C – NH3 2 2

NH2

8. (a) II, (b) I, (c) I, (d) I, (e) I 9. (a) II, (b) I, (c) I, (d) II

10. (a) II, (b) I, (c) II, (d) II, (e) II 11. A 12. A13. B 14. B 15. C 16. A17. A 18. A 19. ABD 20. ABC21. (i) c > b > a (ii) c > b > a (iii) b > c > a (iv) c > b > a

(v) c > b > a (vi) b > c > a (vii) a > b > c (viii) a > b > c(ix) a > c > b (x) d > c > b > a (xi) a > b > c (xii) c > b > a

22. (i) b > c > a (ii) b > c > a > d (iii) c > a > b (iv) a < b(v) a > b (vi) a > b > c (vii) a > b > c (viii) a > b > c(ix) a > b (x) c > b > a > d (xi) a > c > b (xii) c > a > b (xiii) a > b > c > d



23. (a) II, (b) I, (c) I, (d) I, (e) II

24. (a) I, (b) I, (c) II (d) I (e) I (f) I

25. iii > ii > i 26. C 27. (a) i, (b) i, (c) ii 28. A

29. (a) (i) D > C > B > A (ii) E > C > D > B > A (b) 2 > 1 (c) 2 < 1 (d) 1 < 2 (e) 1 > 2

30. (A) (I) iv > iii > ii > i, (II) i > ii > iii > iv (B) (I) iii > ii > i (II) i > ii > iii

31. (a) I, (b) I, (c) II, (d) I

32. (a) 4658, (b) 4638, (c) 4632, (d) 4656 (e) 5293

33. (i) b > a (ii) a > b > c > d (iii) a > b > c > d > e (iv) b > c > a

(v) a > b > c (vi) a > b > c (vii) b > a (viii) b > a

34. (i) c > b > a (ii) a > b > c > d (iii) a > b (iv) c > b > a

35. d < f < b < c < a < e

On the basis of stability of free radical formed after removal of H .

36. Anthracene is 14 p e's system

i.e. there are 7 p bonds

Expected (theoretical) heat of hydrogen = – 28.6×7 = – 200.2 kcal/mol

Observed (experimental) heat of hydrogen = –116.2

R. E. = – 166.2 – (–200.2)

= 84 kcal/mol

37. C 38. B 39. C 40. B 41. B 42. B 43. ABCD

44. (A) P, R; (B) Q, S; (c) P, R; (D) P, R

45. (A) P, R, T; (B) P, R, S (C) S ; (D) P, Q, R

46. (A) P, R; (B) P, R; (C) R, S; (D) P, R

47. (A) Q, R, S; (B) Q, R; (C) P, R; (D) P, R

Exercise - 3 | Subjective | JEE Advanced

1. (i) d > c > b > a (ii) d > c > b > a (iii) a > b > c (iv) d > b > a > c

2. (i) d > a > c> b (ii) c > b > a (iii) a > b > c

3. (i) c > b > a (ii) a > b > c (iii) a > b > c > d

4. (i) a > b > c (ii) a > b > c (iii) d > b > c > a

5. (i) c > a > b > d (ii) d > c > a > b

6. (i) b > a (ii) b > a (iii) c > b > a (iv) c > a > b

7. (i) b (ii) a (iii) b

8. (i) cysteine : 1.88.3

10.8

COOH

NH2

HS (ii) glutamic acid : 2.19

9.67

4.25HO C2 COOH

NH2

GOC 2.51


9. (a) 3 < 2 < 1; (b) 1 < 2 < 3; (c) 3 < 2 < 1; (d) 2 < 1 < 3; (e) 2 < 3 < 1

10. (a) 2; (b) 2; (c) 2 11. (a) 2; (b) 2; (c) 2 12. (a) 2; (b) 2; (c) 2

13. (i) a > b > c > d (ii) a > b > c> d (iii) a > b > c (iv) a < b < c < d (v) c > b> d> a

14. (i) a < b < c (ii) c > a > b (iii) b > c > a (iv) c > d > b > a

15. (i) a > b > c (ii) d > c > b > a (iii) b > c > a (iv) c > b > a (v) b > a

16. (i) d > c > b > a (ii) c > b > a (iii) b > a > c (iv) a > b > c (v) c > a > b

17. (i) d (ii) b (iii) a (iv) a

18. (i) b > a > d > c (ii) b > a > c > d (iii) a > b > c

19. B 20. A 21. A 22. A

23. (a) i, (b) ii, (c) i, (d) ii, (e) i 24. (a) i, (b) i, (c) ii

25. (a) 2; (b) 1; (c) 2; (d) 2

26. (a) 3 < 2 < 1 < 4; (b) 1 < 2 < 3 < 4 27. (a) 1 > 2 > 3; (b) 1 < 2 < 3; (c) 3 < 1 < 2

28. (a) 2 < 1 < 3; (b) 1 < 2 < 3; (c) 2 > 1 > 3 29. (a) 2 < 1 < 3; (b) 1 < 2 < 3; (c) 3 < 1 < 2

30. (a) 2 > 1> 3, (b) 1 > 2 > 3, (c) 1 > 3 > 2, 31. A, B, C, D

Exercise - 4 | Level-I Previous Year | JEE Main

1. B 2. C 3. D 4. C 5. D 6. B 7. C8. D 9. B 10. D 11. A 12. D 13. B 14. B15. C 16. C 17. D

Exercise - 4 | Level-II Previous Year | JEE Advanced

1. C

2.N

MeMeMe

(Pyramidical)

NSiMe3SiMe3

SiMe3

Given componds are not isostrucutralDelocalised of l.p. of nitrogen in vacant d-orbital of silicon makes compound planar.

3. D 4. C 5. A 6. D 7. B 8. D 9. A10. B 11. D 12. 2 13. 4 14. C 15. 6 16. B17. B 18. D 19. A 20. 9 21. A 22. 5 23. D24. C

goc - motion...2.4 theory and exercise book: [email protected], url : , : 1800-212-179999,...

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