Contents · Contents Syllabus ELECTRONIC EFFECTS ... Resonance generally occurs when there is a...

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Theory 01 - 03

Exercise - 1 04 - 20




Answer Key 42 - 45

Contents

Syllabus

ELECTRONIC EFFECTS

Resonance and hyperconjugation, Polarity and inductive effects in alkyl halides; Reactive

intermediates produced during homolytic and heterolytic bond cleavage; Formation, structure

and stability of carbocations, carbanions and free radicals, acidic strength & basic strength.

Name : ____________________________ Contact No. __________________

ELECTRONIC EFFECT # 1

A-479 Indra Vihar, KotaPh. - 9982433693 (NV Sir) 9462729791(VKP Sir)

Physical & Inorganic By

NV SirB.Tech. IIT Delhi

Organic Chemistry By

VKP SirM.Sc. IT-BHU

ELECTRONIC EFFECTS AND THEIR APPLICATIONSELECTRONIC EFFECTS

Permanent Electronic Effects : Permanent electronic effects are mainly divided into following categories:(1) Inductive Effect (2) Mesomeric(3) Hyperconjugation (4) Reverse hyperconjugationInductive Effect (Polar nature of covalent bonds)The displacement of an electron (shared) pair along the carbon chain due to the presence of anelectron withdrawing or electron releasing groups in the carbon chain is known as Inductive Effect(–effect).

C----

C ----

C -- G (G : functional group)

C ----–

C

----–

C

-- G

But the effect is insignificant beyond third carbon atom. This effect is transmitted through the chain of bonds and diminishes. with increasing chainlength.– I group :

The group which withdraws electrons is known as – I group and its effect is known as – I effect.– I power of various group & cation.

> – > – > – NO2 > – SO2R > – CN > – CHO > – COOH > – F > – Cl > – Br >– I > – OR > – OH > – C CH > – NH2 > – C6H5 > – CH = CH2 > – H.

– I power of groups in decreasing order with respect to the reference H+ I group :The group which donates or gives electron is known as + I group and effect is known as + I effect.

Alkyl groups, O ,

OCO are + I groups.

(a) + I power of different type groups and anions :

> > – C(CH3)3 > – CH (CH3)2 > – CH2 – CH3 > – CH3 > – D > – H+ I power in decreasing order with reference to H(b) + I power of same type of alkyl groups :

+ I power number of C’s in same type of alkyl group

For example

ResonanceWhen a molecule or ion can be represented by two or more structures which have the samearrangement of atomic nuclei but differ in distribution of electrons, the phenomenon is called asResonance. The various structures are called contributing or resonating structures. None of thesestructures truly represents all the properties of that molecule or ion. The actual structure is a resonancehybrid of several contributing structures.






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In the molecule if resonance is present then mesomeric effect may be present or may not be present.Resonance generally occurs when there is a conjugation of -bonds (alternate double-single-doublebond arrangement) known as conjugated system.

Conjugate positions of the molecule :Type of Conjugates(i) , conjugation (ii) Positive charge, conjugation(iii) Negative charge, conjugation (iv) Odd electron, conjugation(v) Lone pair, conjugation

Mesomeric Effects : This is of following types :

+ M effect (+R) :

C or

When flow of e– pair (movement) starts from the group (G). It takes place when G has a lone pairor an extra electron (in ion).

Groups which shows '+ m' effect are :

– > – NH2 > – NHR > – NR2 > – OH > – OR > –

O||

RCNH > –

O||

RCO > –Ph, – CH3 > – F >

– Cl > – Br > – I, – N = O.

– M effect (– R) :

– C C G or

When e– pair movement takes place towards G from the molecule.

Groups which can show – m effect are

– NO2 > – C N > – SO3H > – CHO >

O||

RC >

O||

OHC

HyperconjugationWhen (C – H) sigma electrons are in conjugation to pi bond, this conjugation is known as (C–H), conjugation, excessive conjugation or hyperconjugatioon.Structural requirement for hyperconjugation:

(A) Compound should have at least one 2sp -hybrid carbon of either alkene alkyl carbocation or alkyl

free radical.

(B) -carbon with respect to 2sp hybrid carbon should have at least one hydrogen.

If both these conditions are fulfilled then hyperconjugation will take place in the moelucle.(C) Hyperconjugation is of three types :

(i) (C – H), conjugation : This type of conjugation occurs in alkenes.

(ii) (C–H), positive charge conjugation : This type of conjugation occurs in alkyl carbocation.(iii) (C–H), odd electron conjugation : This type of conjugation occurs in alkyl free radicals-






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Reverse Hyperconjugation :

The phenomenon of hyperconjugation is also observed in the system given below :

– |C|X

– C = C where X - halogen

Application of Electronic Effects :

Stability of Alkenes

Carbon-carbon double bond length in alkenesStability of Alkyl CarbocationsStability of Alkyl free radicals :Electron releasing (or donating) power of R in alkyl benzene :

Aromaticity :

Aromatic Character : [The Huckel (4n + 2) rule]

The following three rules are useful in predicting whether a particular compound is aromatic or non–aromatic.

1. Aromatic compounds are cyclic and planar.

2. Each atom in an aromatic ring is sp2 or sp hybridised.

3. The cyclic molecular orbital (formed by overlap of p-orbitals) must contain (4n + 2) electrons, i.e., 2,6, 10, 14 ........ electrons. Where n = an integer 0, 1, 2, 3,..............


A-479 Indra vihar, kotaPh. - 9982433693 (NV Sir) 9462729791(VKP Sir)




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PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.Section (A) : Inductive Effect

A-1. The inductive effect :(A) implies the atom's and group’s ability to cause bond polarization(B) increases with increase of distance(C) implies the transfer of lone pair of electrons from more electronegative atom to the lesser

electronegative atom in a molecule(D) implies the transfer of lone pair of electrons from lesser electronegative atom to the more

electronegative atom in a molecule

A-2. Which of the following orbital overlaps is involved in the formation of the carbon–carbon single bond inthe molecule HC C – CH = CH2(A) sp3 - sp3 (B) sp2 - sp3 (C) sp - sp2 (D) sp3 - sp

A-3. Inductive effect involves :(A) Delocalisation of - electrons (B) Partial displacement of - electrons(C) Delocalisation of - electron (D) Displacement of lone pair electrons.

A-4. Which of the following has incorrect direction of Inductive effect.

(A)

CH3

CH3CH3CH3CH3CH3CH3CH3CH3CH3CH3CH3CH3CH3CH3CH3

(B)

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

O

C – OH

(C)

-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O-O

C O

(D) CH3

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C

O

C CH2 CH3

A-5. Decreasing –I effect of given groups is -(a) –CN (b) –NO2 (c) – NH2 (d) –F(A) b > a > d > c (B) b > c > d > a (C) c > b > d > a (D) c > b > a > d

Section : (B) Resonance and Mesomeric EffectB-1. Which of the following resonating structure contributes equally to the resonance hybrid.

(A) 3

3

3 CH–CHCH–CHC–CH|

3

3

3 CH–HC–CH

CH

C–CH|

(B) 23 NH–

O

C–CH||

CH – C = NH3 2|

O -

+

(C) CH – C3

O

O- CH – C3

O

O

-

(D) CH2 = CH – CH = O CH – CH = CH – O2

-+






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B-2. Which of the following compounds have delocalisation of -electrons ?

(A) (B)

(C) CH3CH2NH CH2 CH = CH2 (D) CH2 = CH – CH2 – CH = CH2

B-3. Which of the following statements is not true about the resonance contributing structures of a resonancehybrid -(A) Contributing structures contribute to the resonance hybrid in proportion of their energies(B) Equivalent contributing structures make the resonance very important(C) Contributing structures represent hypothetical molecules having no real existence(D) Contributing structures are less stable than the resonance hybrid

B-4. Which of the following is not a pair of resonating structures ?

(A) HNCH;HNCH..

(B) CH3CH = CHCH3; CH3CH2CH = CH2

(C) ; (D) ;

B-5. In which of the following delocalisation of -electron is possible.(A) CH2 = CH – CH2 – CHO (B) CH2 = CH – CH = O

(C)

OH|

CH–CH–CH 33 (D) CH2 = CH – CH2 – CH = CH2

B-6. Which of the following species have conjugated system ?

(A) (B) (C) CH = C = CH – NH2 2

: (D) All of these

B-7. The type of hybrid orbitals used by C and N in the formation of C–N bond in pyrrole NH is :

(A) sp2 and sp3 (B) sp2 and sp (C) sp2 and sp2 (D) sp2 and s

B-8. Which type of orbitals C and O use in the formation of C–O bond in furan, O

(A) sp2 and sp3 (B) sp2 and sp (C) sp2 and sp2 (D) sp2 and p






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B-9. Which is the most stable resonance form ?

(A) (B)

(C) (D)

B-10. I 2 2CH CH CH CH CH CH

II 2 2CH CH C H CH CH C H

III 2 2C H CH CH CH CH C H

Among these three canonical structures (through more are possible) what would be their relativecontribution in the hybrid :(A) I > II > III (B) III > II > I (C) I > III > II (D) III > I > II

B-11. I II III

The stability order of these canonical structures is :(A) I > II > III (B) II > III > I (C) I > III > II (D) III > I > II

B-12. Which of the following statements is NOT true ?(A) Resonance forms must obey normal valency rules(B) Resonance forms are in equilibrium(C) The hybrid is more stable than the individual resonance forms(D) Resonance forms differ only in the distribution of electrons

Section (C) : Hyperconjugation and Application of Electronic EffectsC-1. The correct order of decreasing hyperconjugative effect of the following compounds is -

(A) CH3CH=CH2 > CH3CH2CH=CH2 > (CH3)2 CHCH=CH2(B) (CH3)2 CHCH=CH2 > CH3CH=CH2 > CH3CH2CH=CH2(C) (CH3)2 CHCH=CH2 > CH3CH2CH=CH2 > CH3CH=CH2(D) CH3CH2CH=CH2 > CH3CH=CH2 > (CH3)2 CHCH=CH2

C-2. Which of the following molecule has longest C=C bond length ?

(A) CH2 = C = CH2 (B) CH3 — CH = CH2 (C)

3

23

3

CH|

CHCHCCH|CH

(D)

3

23

CH|

CHCCH

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C-3. In hyperconjugation there is overlap between :(A) p- and -orbitals (B) 2 - orbitals (C) d-and--orbtial (D) -and - orbitals

C-4. Hyperconjugation phenomenon is possible in :

(A) (B) CH2 = CH2

(C) C6H5 – CH = CH2 (D) CH3 – CH2 – CH = CH2

C-5. The decreasing order of bond length of C = C bond in the following compounds is:

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

C-6. Which one of the following compounds maximum electron density on aromatic ring :

(A) (B) (C) (D)

C-7. Which of the following statements is true about this molecule .

(A) C1-C2 and C3-C4 bonds are of same length.(B) C1-C2 bond is shorter than C3-C4 bond.(C) C1-C2 bond is longer than C3-C4 bond.(D) C1-C2 and C2-C3 bonds are of same length.

C-8. Which of the following statements is true about this molecule ?

(A) C5-N bond is longer than C1-N and C3-N bonds.(B) C5-N bond is shorter than C1-N bond and latter is shorter than C3-N bond.(C) All three C-N bonds are of same length.(D) C1-N and C3-N bonds are of same length but they are longer than C5-N bond.

C-9. Which of the C–C single bond is strongest in H C CH CH CH CH CHa b c

3 3 :

(A) a (B) b (C) c (D) All equal

C-10. Predict the relative bond lengths a and b, in napthalene a

b

(A) a > b (B) a < b (C) a = b (D) None of these






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C-11. The C–O bond lengths in the following three compounds is :

CH3–OH

O

CH C OH3 | |

O

CH C H3 | |

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

C-12. In which of the following molecules – electron density in ring is minimum :

(A) (B) (C) (D)

C-13. In which of the following molecules – electron density in ring is maximum :

(A) (B) (C) (D)

Section (D) : AromaticityD-1. Which has the most aromatic nature ?

(A) (B)

(C) (D) All having same aromatic character

D-2. Which of the following is aromatic species ?

(A) (B)

N

H

(C) (-)

(D) All

D-3. Which of the following is anti-aromatic species ?

fuEu esa l s i zfr , j ksesfVd i zt kfr gS \

(A)

N

(B)

S

(C) (D)

+

D-4. Which is the most aromatic in character ?

(A)N

H

(B)

S

(C)

O

(D) All having same aromatic character






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D-5. Which is the most aromatic in character ?

(A)

N

H

N

H

(B)

S(C)

O(D)

N

D-6. Identify the aromatic compound ?

(A) (B) (C) (D)

Section (E) : Reaction Intermediate

E-1. The species 33 CHCHCH is less stable than -

(A) (CH3)3 C+ (B)

223 CHCHCH (C) CH3

2CH (D) 3CH

E-2. Homolytic fission of a hydrocarbon will liberate :(A) Carbonium ions (B) carbanions (C) free radicals (D) carbenes

E-3. Heterolytic fission of carbon-chlorine bond produces :(A) two free radicals (B) two carbonium ions(C) two carbonions (D) one cation and one anion

E-4. Which carbocation is least likely to form as an intermediate?

(A) C)HC( 356 (B) (C) (D) HCCH2

E-5. Which allylic carbocation is the most stable carbocation :

(A) CH3—CH CH— (B) CH3—CH CH—

(C) C6H5 —CH CH— (D) All have same stability

E-6. Arrange stability of the given carbocations in decreasing order :

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






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E-7. Which of the following carbocation is most stable?

(A) CH3

C+ CH3

CH3

(B) CH3

C+ CH3

CH3

OH

(C) OH

C+ CH3

CH3

(D)CH3

C+

CH3

OH

E-8. The most unstable carbocation is

(A) CH3 2HC

(B) Cl – CH2 – 2HC

(C) 2HC

– CHO (D) 2HC

– O – CH3

E-9. Nonplanar cation is :

(A) (B) (C) CH3 (D) (CH3)3C+SbF6

–

E-10. The stability order of following carbocation is :

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

E-11. Most stable radical among following.

(A) (B) (C) (D)

E-12. Arrange the following radicals in decreasing order of their stability.

(I) (II) (III) (IV)

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

E-13. Consider the following carbanions -

(a) CH3 – 2HC

(b) CH2 = HC

(c) CH C

Correct order of stability of these carbanions in decreasing order is -(A) a > b > c (B) b > a > c (C) c > b > a (D) c > a > b

E-14. Arrange the following carbanions in decreasing order of stability :

I II III IV

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

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E-15. Arrange the following radicals in decreasing order of their stability.

(I) (II) (III) (IV)

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

E-16. Rank the followIng radIcals in order of Decreasing stability

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

Section (F) : AcidF-1. Consider following acid

ClCH2COOH, CH3COOH, CH3CH2COOH I II IIICorrect order of theIr pH value.(A) III < II < I (B) I < II < III (C) I < III < II (D) II < I < III

F-2. Which one of the following is most acidic ?(A) Phenol (B) 2-chlorophenol (C) 3-chlorophenol (D) 4-chlorophenol

F-3. Which of the following has the most acidic hydrogen ?(A) 3-hexanone (B) 2, 4-hexanedione(C) 2, 5-hexanedione (D) 2, 3-hexanedione.

F-4. In following acid, the correct increasing order of acidic strength acidI. HCOOH II. CH3COOH III. C6H5COOH(A) I < II < III (B) II < I < III (C) III < II < I (D) II < III < I

F-5.

I II IIIArrange following phenol in increasing order of Pka value(A) I < II < III (B) III PhSO3H > PhCH2OH > PhOH(B) PhSO3H > PhOH > PhCH2OH > PhCH2OH(C) PhCO2H > PhOH > PhCH2OH > PhSO3H(D) PhSO3H > PhCO2H > PhOH > PhCH2OH

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F-7. Give the correct order of increasing acidity of the following compounds -

(I) (II)

(III) (IV)

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

F-8. Among the following compounds, the strongest acid is :(A) HC CH (B) C6H6 (C) C2H6 (D) CH3OH

F-9. Amongest the following compounds the strongest acid is :

(A) (B) (C) (D)

F-10. Which one of the hydrocarbon is the most acidic ?

(A) (B)

(C)

|

|

|

| | |

||

CH3CH3

CH3CH3H

H C3H C3

H C3

(D)

|

|

|

| |

|

||

CH3CH3

CH3H

H C3H C3

H C3H C3

F-11. Arrange Increasing order of their Ka value.oxalic acid, succinic acid,

I IImalonic acid, adipic acid III IV (all dibasic)(A) III < II I > IV (C) I > III > II > IV (D) II > I > III < IV

F-12. (CH3)3COH is less acidic than (H3C)3SiOH because :(A) Carbon is more electronegative than Si (B) (CH3)3SiO is stabilized due to p- d-(C) Bigger size of silicon than carbon (D) None of these

F-13. Arrange pH of the given compounds in decreasing order :(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

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F-14. Find the order of reactivity of labelled hydrogen with NaNH2.

(A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 > 1 > 2 (D) 2 > 3 > 1

F-15. Use resonance arguments to rank the following in order of increasing acidity .

(A) Hc < Ha < Hb (B) Ha < Hb < Hc (C) Hb < Hc < Ha (D) Hc < Hb < Ha

F-16. 1,3-cyclopentadiene has a pKa = 15 whereas simple cycloalkanes have pKa > 50 with respect totheir conjugate bases . This is because :(A) 1, 3-cyclopentadiene is aromatic.

(B) 1, 3-cyclopentadiene is anti-aromatic.(C) 1, 3-cyclopentadine is non-aromatic.(D) 1, 3-cyclopentadiene is non-aromatic with an aromatic conjugate base.

F-17. The most acidic ‘H ’

(W) HO OH(Z)

CH OH(X)2

H

OHOO(Y)

Vitamin-C

is :

(A) X (B) Y (C) Z (D) W

F-18. H–O–CN 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






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F-19. ANaNHmoles2 2 . The product (A) will be :

(A) (B)

(C) (D)

Section (G) : Base

G-1. What is true about the following equilibrium ?

|K

H + H3C–C || CH |

HH + H C–C3

|| C K |

(A) It will be almost completely shifted to the left(B) It will be almost completely shifted to the right(C) The equilibrium constant is very close to one(D) The equilibrium constant is zero

G-2. Select the decreasing order of relative basic strengths of following species :

OH¯, NH2¯, H – C C¯, CH3 – CH2¯

(A) CH3 – CH2¯ > NH2¯ > H – C C¯ > OH¯ (B) H – C C¯ > CH3– CH2¯ > NH2¯ > OH¯

(C) OH¯ > NH2¯ > H – C C¯ > CH3 – CH2¯ (D) NH2¯ > H – C C¯ > OH¯ > CH3 – CH2¯

G-3. Select the decreasing order of relative basic strengths of following species :

(I) (CH3)3 (II) (CH3)2 (III) (CH3)3 (IV)

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

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G-4. Which is maximum basic in nature ?

(A) (B) (C) (D)

G-5. Which of the following shows the correct order of decreasing basicity in aqueous medium ?(A) (CH3)3N > (CH3)2NH > CH3NH2 > NH3 (B) (CH3)2NH > (CH3)3N > CH3NH2 > NH3(C) (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (D) (CH3)2NH > CH3NH2 > NH3 > (CH3)3N

G-6. Give the correct order of decreasing basicity of the following compounds ?C6H5NH2 (C6H5)2NH (C6H5)3N C6H11NH2

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

G-7. Arrange the following in Decreasing order of their PKa value -I. CH2=CHCH2NH2, II. CH3CH2CH2NH2 , III. CHCCH2NH2(A) I > II < III (B) II > I > III (C) III > II > I (D) II < III > > > >

(B) > > > > >

(C) > > > > >

(D) > > > > >

G-9. Which of the following is the strongest base -

(A) (B) (C) (D)

G-10. (I) (II)

(III) (IV)

The correct order of decreasing basicity of the above compound is -(A) I > II > III > IV (B) II > I > IV > III (C) III > IV > II > I (D) II > I > III > IV

G-11. The correct order of basicities of the following compounds is–

(a) (b) CH3CH2NH2 (c) (CH3)2NH (d)

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






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G-12. In the following compounds–

N

H

(I) (II) (III) (IV)The order of basicity is -(A) IV > I > III > II (B) III > I > IV > II (C) II > I > III > IV (D) I > III > II > IV

G-13. Select the order of basic strength.

(I) R4N OH¯ (II) R3N(III) R2NH (IV) RNH2 (if R = Me)

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

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

G-14. Order of basic strength in the following compounds.

(I)

NH2

. .

(II)

NH2

. .

CH3

(III)

NH2

. .

NO2

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


ICl

NH2NH2NH2NH2NH2NH2NH2NH2NH2NH2NH2NH2

II

C – CH3

O


IIICH3


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


NH2 NH NH2

(I) (II) (III)

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

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G-17. Which nitrogen in LSD is most basic ?

H—N(2)N

CH3

C—N(C H )2 5 2

O(3)

(1)

(A) 1 (B) 2 (C) 3 (D) All are same basic

PART - II : MISLLANEOUS QUESTIONS

COMPREHENSION

Comprehension # 1

But-1-ene (A) and Buta-1,3-diene (B) differ not only in the number of bonds, but (B) also has and bonds at alternate positions. This type of the system is called conjugated system. Following are some ofthe conjugate system

In such systems, electron shifting result into permanent polarity. This type of -electron shift in the conjugatesystems is called mesomeric effect.

Rules for stability :(i) Structures with more covalent bonds are more stable than other structures(ii) Structures in which all of the atoms have a complete valence shell of electrons (i.e., the noble gasstructure) are especially stable and make large contributions to the hybrid.(iii) Structure that carry negative charge on a more electronegative atom and positive charge on lesselectronegative atom are comparatively more stable.

1. In which of the following compound, delocalisation is not possible

(A) (B) (C) (D)

2. Select the correct option related to stability of following structure.

(A) > (B) <

(C) < (D) > CF – C – O3

||O






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Comprehension # 2Hyper conjugation : It is delocalisation of sigma electron with p-orbital. Also known as -conjugation orno bond resonance.May takes place in alkene, alkynes, carbocation, free radical, benzene nucleus.Necessary Condition : Presence of at least one hydrogen at saturated carbon which is with respect toalkene, alkynes, carbocation, free radical, benzene nucleus.

3. Which of the following alkenes have maximum number of hyper conjugation structure.

(A) CH3–CH=CH–CH3 (B) (C) (D)

4. In which of the following species hyperconjugation is not possible

(A) (B) (C) (D)

5. Select the incorrect statement.(A) Hyperconjugation is a permanent effect.(B) CH2=CH–CH3 shows hyperconjugation.(C) Stability of alkenes directly related to number of hyperconjugation structures.(D) C6H5–CH3 does not show hyperconjugation.

Comprehension # 3Hyperconugation describes the orbital interaction between the -system and the adjacent -bond of thesubstituent group (s) in organic compounds. Hyperconjugation is also called as Baker and Nathaneffect. The necessary and sufficient condition for the hyperconjugation are :(i) compound should have at least on sp2-hybrid carbon of either alkene, carbocation or alkyl free

radical.(ii) -carbon with respect to sp2-hybrid carbon should have at least one hydrogen.Hyperconjugation are three types :-(i) (C–H), -conjugation(ii) (C–H), positive charge conjugation(iii) (C–H), odd electron conjugation

The hyperconjugation may be represented as

H – C – CH = CH2 |

|

H

H

H – C = CH – CH2 |

H

H

H C = CH – CH2 |

|

H

H

H – C = CH – CH2 |H

H

Number of resonating structures due to hyperconjugation = (n + 1 ) where n is number of -hydrogen.Greater is the number of such forms, more is the stability of species under consideration.






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6. Hyperconjugation is possible in which of the following species ?

(A) CH –CH23(B) C6H5–CH3 (C) CH2=CH2 (D) CH – C – CH = CH3 2

|

|

CH3

CH3

7. Which of the following carbonium ions will show highest number of hyperconjugation forms ?

(A) CH –CH23

(B) CH – CH3 |CH3

(C) CH – C3

|CH3

|CH3

(D) CH – CH – C3 2 |CH3

|CH3

8. Which of the following alkenes will show maximum number of hyperconjugation forms ?(A) CH2 = CH2 (B) CH3 – CH = CH2

(C) CH3 – CH2 – CH = CH2 (D) CH – CH – CH = CH3 2

CH3|

Match the column

9. Match the column.

Structure No. of -Hydrogen C =C bond length in Å

(A) CH3–CH=CH2 (X) 0 (P) 1.34

(B) CH3–CH2–CH=CH2 (Y) 1 (Q) 1.35

(C) (Z) 2 (R) 1.37

(D) (W) 3 (S) 1.39






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10. Match the correct order for stability of following carbocation :

ASSERTION / REASONINGDIRECTIONS :

Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False.(D) Statement-1 is False, Statement-2 is True.(E) Statement-1 and Statement-2 both are False.

11. Statement-1 : m-Nitrobenzoic acid is less acidic as compared to p-nitrobenzoic acid.Statement-2 : Acetic acid is stronger acid than benzoic acid but formic acid is stronger acid than both

12. Statement-1 : Ortho substituted benzoic acids are stronger than benzoic acidStatement-2 : Ortho substituent tends to prevent coplanarity of – COOH with ring. Thus resonance isdiminished which increases acidic strength (ortho effect)

13. Statement-1 : Salicylic acid is much stronger than its m-, p-isomers and benzoic acid itself.Statement-2 : It is due to steric inhibition of resonance, as – OH group forces – COOH out of the plane of ring

14. Statement-1 : Pyrrolidine () is more basic than pyrrole ()

Statement-2 : Protonated pyrrole has delocalisation of positive charge in aromatic ring.

15. Statement-1 : The aPk value of acetic acid is lower than that of phenol.Statement-2 : Phenoxide ion is more resonance stabilized.

16. Statement-1 : Acetate ion is more basic than the methoxide ion.Statement-2 : The acetate ion is resonance stabilized.






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PART - I : MIXED OBJECTIVE

Single Correct Answer Type1. Consider the follownig carbocations -

(a) CH3 – 2HC

(b) CH2 = HC

(c) CH2 = CH – 2HC

(d) C6H5 – 2HC

Stability of these carbocations in decreasing order is -(A) d > c > a > b (B) d > c > b > a (C) c > d > b > a (D) c > d > a > b

2. Consider the following compounds -

(a) (b) (c) (d)Arrange these compounds in decreasing order of their basicity :(A) a>b> c > d (B) b > c > a > d (C) d > a > c > b (D) d > a > b > c

3.. Which free radical is the most stable -

(A) C6H5 –

CH2 (B) CH2 = CH – C

H2 (C) CH3 – C

H – CH3 (D) CH3 –

3CH|

–C

CH3

4. Which carbocation is the most stable -

(A) (B) (C)

CH2

Cl

(D)

5. Which of the following carboxylic acids is most acidic in character -(A) o-methyl benzoic acid (B) m-methyl benzoic acid(C) p-methyl benzoic acid (D) Benzoic acid

6. Which one of the following is strongest acid -(A) 2-chloropentanoic acid (B) 3-chloropentanoic acid(C) 5-chloropentanoic acid (D) 4-chloropentanoic acid

7. The most stable carbanion is-

(A) CH3CH2 2HC (B) CH3–CC (C) CH3CH= HC (D) CH3– C =CH2

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8. Correct order of nucelophilicity -

(A) –3CH < –

2NH < –OH < –F (B) –F < –OH < –3CH <

–2NH

(C) –OH < –2NH < –F < –

3CH (D) –F < –OH < –2NH <

–3CH

9. Hydrogen bonding is maximum in–(A) Ethanol (B) Diethyl ether (C) Ethyl chloride (D) Triethylamine

10. Benzoic acid is less acidic than salicylic acid because of–(A) Hydrogen bond (B) Inductive effect (C) Resonance (D) none of these

11. In a reaction of C6H5Y, the major product (>60%) is m-isomer, so the group Y is–(A) –COOH (B) –Cl (C) –OH (D) –NH2

12. Which one of the following orders is correct regarding the –I effect of the substituents ?(A) –NR2 < –OR > –F (B) –NR2 > –OR > – F(C) –NR2 < –OR < –F (D) –NR2 > –OR < –F

13. Which one of the following compounds will be most easily attacked by an electrophile ?

(A) (B) (C) (D)

14. The weakest acid among the following–(A) CH3COOH (B) CH3CO2OH (C) ClCH2COOH (D) CH3CH2COOH

15. The correct order of boiling point for primary (1º), secondary (2º) and tertiary (3º) alcohols is–(A) 2º > 1º > 3º (B) 3º > 2º > 1º (C) 1º > 2º > 3º (D) 2º > 3º > 1º

16. Which of the following is most acidic ?

(A) m-chlorophenol (B) benzyl alcohol (C) phenol (D) cyclohexanol

17. Among the following the strongest base is–(A) C6H5NH2 (B) p-NO2–C6H4NH2 (C) m–NO2–C6H4NH2 (D) C6H5CH2NH2

18. Strongest acid out of the following–(A) CH3COOH (B) ClCH2COOH (C) CH3CH2COOH (D) (CH3)2CHCOOH

19. Increasing order of acid strength among p-methoxyphenol, p-methylphenol, and p-nitrophenol is–(A)p-nitrophenol, p-methoxyphenol, p-methylphenol(B) p-methylphenol, p-methoxyphenol, p-nitrophenol(C)p-nitorphenol, p-methylphenol, p-methoxyphenol(D) p-methoxyphenol, p-methylphenol, p-nitorphenol

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20. In the following compounds–

(I) (II) (III) (IV)the order of acidity is–(A) III > IV > I > II (B) I > IV > III > II (C) II > I > III > IV (D) IV > III > I > II

21. Most powerful leaving group in following -(A) NH2 (B) OH (C) CH3 (D) F

22. (I) (II) (III) (IV)

Increasing order of stability is–(A) I < III < II < IV (B) IV < III < II < I (C) III < IV < II < I (D) II < IV < III < I

23. Increasing order of acidity in para methyl (I), chloro (II), amino (III), and nitro (IV) benzoic acids is–(A) III < II < I < IV (B) I < II < IV < III (C) III < I < II < IV (D) II < I < III < IV

24. Increasing order of basic strength of following compounds isC6H5NH2, (CH3)3N, (C6H5)2NH, C6H5NHCH3 (I) (II) (III) (IV)(A) III < IV < I < II (B) III < I < IV < II (C) I < III < II < IV (D) II < IV < III < I

25. The compound that is most reactive towards electrophilic nitration is–(A) Toluene (B) Benzene (C) Benzoic acid (D) Nitrobenzene

26. The most unlikely representation of resonance structures of p-nitrophenoxide ion is :

(A) (B) (C) (D)

27. Consider the following three halides -(1) CH3 – CH2 – Cl (2) CH2 = CH – Cl (3) C6H5 – ClArrange C – Cl bond length of these compound in decreasing order -(A) 1 > 2 > 1 (B) 1 > 3 > 2 (C) 3 > 2 > 1 (D) 2 > 3 > 1

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28. Arrange the following in increasing basic strength-

(a) CH3NH2 (b) (c) Cl – CH2 – NH2 (d) – CH2 – NH2 (e) NH3

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

29. Arrange in increasing pKa(a) CH3CH2COOH (b) CH3CH2CH2COOH (c) CH3COOHCorrect answer is-(A) (c) < (a) < (b) (B) (b) < (a) < (c) (C) (a) < (c) < (b) (D) (b) < (c) < (a)

30. Arrange in increasing basic strength(a) Cl – CH2 – COOH (b) Cl – CH2 – CH2– CH2 – COOH (c) Cl – CH2 CH2 COOHCorrect answer is-(A) (a) < (c)< (b) (B) (b) < (a) < (c) (C) (c) < (b)< (a) (D) (a) < (b) < (c)

31. Which one of the carbaions is most stable?

(A) (B) (C) (D)

32. What is the correct increasing order of bond lengths of the bonds indicated as I, II, III and IV infollowing coumpounds -

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

33. Which of the following is least stable ?

(A) (B) (C) (D)

34. Which one of the following is most stable carbocation ?

(A) 22 HC–CHCH

(B) 3HC

(C) HCCH–CH3

(D) None of these

35. Which of the following ions is most stable :

(A) CH3CH2 2HC

(B) CH3C HCH2CH3 (C) (CH3)3

C (D) (CH3)3 C . 2HC






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36. The most stable carbanion is -(A) methyl carbanion (B) primary carbanion(C) seccondary carbanion ion (D) tertiary carbanion

37. Which of the following statements is wrong ?(A) a tertiary free radical is more stable than a secondary free radical(B) a secondary free radical is more stable than a primary free radical(C) a tertiary carbonium ion is more stable than a secondary carbonium ion(D) a primary carbonium ion is more stable than a secondary carbonium ion

38. The species 33 HCHCCH

is less stable then-

(A) (CH3)3C (B) CH3CH2 2HC

(C) 23 HCCH

(D) CH3

39. Most stable carbocation is -

(A) CH3 – 2CH (B)

22 CHClCH (C) ClCHCH 22

(D) 2CH – CH2NO2

40. Strongest acid among the following is -(A) CF3COOH (B) CCl3COOH (C) CBr3COOH (D) CH3COOH

41. Among the following which one is most basic-

(A) NH3 (B) CH3NH2 (C) CH3CH2NH2 (D)

Cl|CH2 – NH2

42. Which one of the following is the weakest base -(A) (C2H5)3N (B) (C2H5)2NH (C) C2H5NH2 (D) NH3

43. Which of the following alkenes will react fastest with H2 under catalytic hydrogenation conditions -

(A) (B) (C) (D)

44. M effect of -CCl3 can be explained on the basis of(A) – effect (B) + R effect(C) negative hyper conjugative effect (D) none

45. Which of following has maximum hydrogenation energy.(A) CH3 – CH2 – CH = CH2 (B) CH3 – CH = CH – CH3

cis(C) CH2 = CH – CH2 – CH2 (D) All have the same






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46. Polarisation of electrons in acrolein may be written as -

(A) 2HC

= CH – CH = O (B) 2HC

= CH – HC

–= 0

(C) 2HC–

= HC

– CH = 0 (D) 2HC–

= CH – CH =

O

47. In which compound delocalisation is not possible -(A) 2-butene (B) 1, 3-butadiene (C) 1, 3, 5-hexatriene (D) Benzene

48. Consider the following compound :

21HC = HC

2 – HC

3 = 24

HC

carbon-carbon bond length between C2 and C3 will be -(A) 1.54 Å (B) 1.3 Å(C) Less than 1.54 and greater than 1.33 Å (D) 1.21 Å

49. + R power of the given groups -

(a) – O (b) – NH2 (c) – OH (d) – NHCOCH3

increasing order is -(A) a > b > c > d (B) d > c > b > a (C) a > c > b > d (D) a > d > c > b

50. In pyridine ; Number of conjugated electrons are -

(A) 6 (B) 8 (C) zero (D) 5

51. Which of the following is corect for stability of phennoxide ion -(A) resonating structure of benzene ring(B) localization of -electron in phenoxide ion(C) delocalization of -electrons in phenoxide ion(D) all of the above

Multiple Correct Answer Type

52. In which of the following pairs the first one is the stronger base than second.

(A) , (B) , (C) , (D) CH3NH2, CH3OH

53. Which of the following are not non-aromatic ?

(A) (B) N

(C) (D)






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54. Which of the following are aromatic ?

(A)

(+)

(B) (C)

+

(D)

55. In which of the following first carbocation is more stable than second one.

(A) , (B) ,

(C) , (D) ,

56. In which of the pair of carbanion have first more stable than second.

(A) (B)

(C) (D)

PART - II : SUBJECTIVE QUESTIONS

1. Show the direction of Inductive-effect in following compounds.

(P) CH3

COOH(Q)

COONa

OH (R) (S)

COOH

NaO

2. Arrange following compounds in decreasing order of their dipolemoment.

I II III IV

(x) CH3—CH2—NO2 CH3—CH2—NH2 CH3—CH2—F CH3—CH2—CN

(y) CH3—CH2COOH CH3—CH2—OH CH3—CH2—Cl CH3—CH2—NH2

3. Which of the following cases have incorrect direction of I effect.

(P) (Q) (R) (S) CH3 – CH2 MgBr

(T) (U) (V)






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4. Identify more stable resonating structure.

(i) and (ii) and

(iii) and (iv) and

5. Write the stability order of following resonating structures : I II I II

(P) (Q)

(R) (S)

6. Which of the following compounds have delocalized electrons ?

a. 32 CHCCHCH||O

f.

b. g.

c. h. CH3CH2NHCH2CH = CHCH3

d. i. CH3CH2NHCH = CHCH3

e. CH2=CHCH2CH = CH2 j. k.






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7. For each pair of ions, determine which ion is more stable.

(A) and (B) and

(C) and (D) and

(E) and

8. In the following sets of resonating structure, label the major and minor contributors towards resonancehybrid.

(a) (b)

(c)

(d)

9. Identify the major contribution in following resonance structures:

(a) NH2– C

NH2|

NH2 NH2– C

NH2|

(I) (II)

(b) CH3– C

O||

– – C

O||

–CH3 & CH3– C

O||

–CH= –CH3

(I) (II)

(c) &

(I) (II)

(d) CH3– –OH & CH3– = –H

(I) (II)






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10. Find the number of hydrogens involved in hyperconjugation in the following compounds ?

11. Write the stability order of following carbocations

I II III IV

12. Write stability order of C=C bond present in following compound.

13. Write the number of hyperconjugable hydrogen atoms of following species :

(a) (b) CH3—CH—CH2—CH3

(c) C=CH—CH3 (d) CH3—CH=CH—C2H5






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14. Write the stability order of following alkene :

(I) (II) CH3–CH=CH–CH3 (III) C=C (IV)

33

33

33

CHCH||

CH—C—CHCH—C—CH||CHCH

15. The central carbon-carbon bond in 1, 3-butadiene is shorter than that of n-butane. Why ? [JEE-98]

16. What is aromaticity ?

17. Explain why the following systems are not aromatic ?

(i) (ii) (iii)

18. Classify the following as aromatic, antiaromatic and nonaromatic compounds.

(1) (2) (3) (4) (5)

(6) (7) (8) (9) (10)

(11) (12)

19. Benzene and [10]– Annulene both have (4n + 2) -electron in ring and all bonds are conjugated even thenonly benzene is aromatic explain.

20. Why cyclooctatetraene is nonplanar.

21. Which of the following is aromatic

(a)

+

H H

H

(b)

+

+

(c)-

(d)

+

(e) (f) (g) (h)

(i)

H

H H

H HH






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22. Arrange the given species in decreasing order of their nucleophilicity ?

H2O

23. Write nucleophilicity order in the following compound.

24. Which of the following carbocation is more stable and why ?

25. Arrange the following carbocations in decreasing order of their stability :

(i) (A) CH3—CH2—CH2 (B)

F|

CH—CH3 —CH2 (C)

Br|

CH—CH3 —CH2 (D)

Cl|

CH—CH3 —CH2

(ii) (A) 223 HC–CH–CH

(B) 323 CH–CH–HC–CH

(C) C)CH( 33 (D) 3HC

(iii) (A) (B) (C) (D)

(iv) (A) (B) (C) (D)

(v) (A) (B) (CH3CH2)2 (C) (CH3CH2)3 (D)

26. Rank the following free radical in increasing order of their stability and give appropriate reasons.






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27. Order of stability carbocation is :

(A) CH3 – O – CH2 22

3

CHCHCH

CH

|

(B) CH3 – O – CH2 32

3

CHCCH

CH|

(C) CH3 – O – CH2 3

3

CHCHCH

CH

|

(D) CH3 – O – HC

32

3

CHCHCH|

CH

28. Write the order of Ka following compounds : I II III IV

(P)

(Q)

(R)

(S)

29. Write the order of ease of deprotonation of labelled H–atoms.

30. Although phenoxide ion has more number of resonating structures than benzoate ion, benzoic acid is astronger acid than phenol. Why ? [JEE-97]






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31. Find the order of Ka of following compounds.

(p) (A) (B) (C) (D)

(q) (A) (B) (C) (D)

(r) (A) (B) (C) (D)

(s) (A) (B) (C) (D)

32. Which nitrogen is more basic

(a) (b)

33. Arrange the following species in order of decreasing basicity.

(p) (A) (B) (C)

(q) (A) H2O (B) H2N– (C) HO–

(s) (A) (B) (C) (D)






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(t) (A) (B) (C) (D)

(u) (A) (B) (C) (D)

(w) (A) (B) (C) (D)

34. Dimethylamine is stronger base than trimethylamine. Explain. [JEE-98]

35. Write the order of Kb of following amines :I II III

(P) (CH3)3N (CH3)2NH CH3NH2 (Gaseous phase)

(Q) (CH3)2NH CH3NH2 (CH3)3N (H2O as solvent)

(R) (C2H5)2NH (CH3CH2)3N C2H5NH2 (H2O as solvent)

36. Write the increasing pkb values for following anions.

3CH ,

2NH , OH ,

F

37. What is the effect of electron releasing and electron withdrawing groups on the basic strength of aniline.

38. Write the decreasing order of basic strength of following compounds.

(I) (II) (III)






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PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions are having more than one correct option.

1. CH2 = CH¯ is more basic than HC C¯. Explain. [JEE-2K(M), 1/100]

2. Which of the following acids has the lowest value of acid dissociation constant : [JEE-02(S), 3/90](A) CH3CHFCOOH (B) FCH2CH2COOH (C) BrCH2CH2COOH (D) CH3CHBrCOOH

3. Match the Ka values : [JEE-03(M), 2/60]

4. Write resonating structure of the compound . [JEE-03(S), 2/60]

5. Which of the following represent the given mode of hybridisation sp2 – sp2 – sp – sp from left to right.[JEE-03(S), 3/84]

(A) H2C = CH – C N (B) HC C – C CH (C) H2C = C = C = CH2 (D)

6. A , The product A will be - [JEE-03(S), 3/84]

(A) (B) (C) (D)






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7. What is the acidity order of x, y & z ? [JEE-04(S), 3/84]

(A) x > y > z (B) x > z > y (C) y > z > x (D) z > y > x

8. COONaCHofExcess 3 [JEE-05(S), 3/84]

The products will be :

(A) + CH3COONa (B) + CH3COOH

(C) + CH3COOH (D) + SO3

9. Which will be the least stable resonating structure : [JEE-05(S), 3/84]

(A) CH2 = CH – – – O – CH3 (B) 2HC

– – CH = CH – OCH3

(C) 2HC

– CH = CH – CH = – CH3 (D) CH2 = CH – – CH = – CH3

10. 3NaHCO ; 3NaHCO [JEE-06, 3/184]

A mixture of benzene sulphonic acid and paranitrophenol react with NaHCO3 separately. The gases producedare respectively.(A) SO2, CO2 (B) SO2, CO (C) SO2, NO2 (D) CO2, CO2

11. Among the following, the least stable resonating structure is : [JEE-07, 3/162]

(A) (B) +

N+

O

O (C) N

+

O

O

+

(D)

12. Hyperconjugation involves overlap of the following orbitals [JEE-08, 3/163](A) – (B) –p (C) p – p (D) –

13. The correct stability order for the following species is : [JEE-08, 3/163]

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






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14. The IUPAC name of the following compound is : [JEE-2009, 3/160]

(A) 4-Bromo-3-cyanophenol (B) 2-Bromo-5-hydroxybenzonitrile(C) 2-Cyano-4-hydroxybromobenzene (D) 6-Bromo-3-hydroxybenzonitrile

15. The correct stability order of the following resonating structures is : [JEE-09, 3/160]

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

16. The correct acidity order of the following is : [JEE-09, 3/160]

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

17. The total number of basic groups in the following form of lysine is : [JEE-10, 3/163]

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

19. The total number of contributing structures showing hyperconjugation (involving C-H bonds) for the follow-ing carbocation is : [JEE-2011]

20. Which of the following molecules, in pure form, is (are) unstable at room temperature ? [JEE-12, 3/136]

(A) (B) (C)

O

(D)

O

VKP SirM.Sc. IT-BHU

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)

1. In which of the following species is the underlined carbon having sp3 hybridisation ? [AIEEE 2002](1) CH3COOH (2) CH3CH2OH (3) CH3COCH3 (4) CH2=CH–CH3.

2. Which of the following compounds has wrong IUPAC name ? [AIEEE 2002]

(1) CH3CH2CH2COOCH2CH3 Ethylbutanoate (2) 3–Methylbutanal

(3) 2–Methyl-3-butanol (4) CH –CH–C–CH –CH3 2 3

|| O

|CH3

2–Methyl-3-pentanone

3. In the following benzyl/allyl system

R – CH = CH2 and

(R is alkyl group)decreasing order of inductive effect is : [AIEEE-2002](1) (CH3)3 C— > (CH3)2CH— > CH3CH2— (2) CH3CH2— > (CH3)2CH— > (CH3)3C—(3) (CH3)2CH— > CH3CH2— > (CH3)3C— (4) (CH3)2C— > CH3CH2— > (CH3)3CH—

4. The IUPAC name of CH3COCH(CH3)2 is [AIEEE 2003](1) Isopropylmethyl ketone (2) 2-Methyl-3-butanone(3) 4-Methylisopropyl ketone (4) 3-Methyl-2-butanone.

5. The correct order of increasing basic nature for the bases NH3, CH3NH2 and (CH3)2NH is : [AIEEE-2003](1) CH3NH2 < NH3 < (CH3)2NH (2) (CH3)2NH < NH3 < CH3NH2(3) NH3 < CH3NH2 < (CH3)2NH (4) CH3NH2 < (CH3)2NH < NH3

6. The IUPAC name of the compounds is [AIEEE 2004]

(1) 3,3-Dimethyl -1-hydroxycyclohexane (2) 1,1 -Dimethyl-3-hydroxycyclohexane(3) 3,3-Dimethyl-1-cyclohexanol (4) 1,1-Dimethyl-3-cyclohexanol

7. Which one of the following does not have sp2 hybridized carbon ? [AIEEE 2004](1) Acetone (2) Acetic acid (3) Acetonitrile (4) Acetamide

8. Among the following acid which has the lowest pKa value ? [AIEEE-2005](1) CH3CH2COOH (2) (CH3)2CH – COOH (3) HCOOH (4) CH3COOH

9. Amongst the following the most basic compound is [AIEEE-2005](1) p-Nitroaniline (2) Acetanilide (3) Aniline (4) Benzylamine

10. The IUPAC name of the compound shown below is [AIEEE 2006]

(1) 2-Bromo-6-chlorocyclohex-1-ene (2) 6-Bromo-2-chlorocyclohexene(3) 3-Bromo-1-chlorocyclohexene (4) 1-Bromo-3-chlorocyclohexene

VKP SirM.Sc. IT-BHU

11. The increasing order of stability of the following free radicals is : [AIEEE-2006, 3/220]

(1) (C6H5)3•C < (C6H5)2 HC

•< (CH3)3

•C < (CH3)2 HC

•

(2) (C6H5)2 HC•

< (C6H5)3•C < (CH3)3

•C < (CH3)2 HC

•

(3) (CH3)2 HC•

< (CH3)3•C < (C6H5)3

•C < (C6H5)2 HC

•

(4) (CH3)2 HC•

< (CH3)3•C < (C6H5)2 HC

• < (C6H5)3

•C

12. The correct order of increasing acid strength of the compounds [AIEEE-2006, 3/220]

(a) CH3CO2H (b) MeOCH2CO2H (c) CF3CO2H (d)

(1) d < a < c < b (2) d < a < b < c (3) a < d < c < b (4) b < d < a < c

13. The IUPAC name of [AIEEE 2007, 3/120]

(1) 5,5-Diethyl-4, 4-dimethylpentane (2) 3-Ethyl-4, 4-dimethylheptane(3) 1,1-Diethyl-2, 2-dimethylpentane (4) 4,4-Dimethyl-5, 5-diethylpentane

14. Which one of the following is the strongest base in aqueous solution ? [AIEEE-2007, 3/120](1) Trimethylamine (2) Aniline (3) Dimethylamine (4) Methylamine

15. The IUPAC name of neopentane is : [AIEEE 2009, 4/144](1) 2, 2-dimethylpropane (2) 2-methylpropane(3) 2, 2-dimethylbutane (4) 2-methylbutane

16. Arrange the carbanions, (CH3)3 C , C Cl3, (CH3)2 C H, C6H5 C H2 in order of their decreasing stability :[AIEEE-2009, 4/144]

(1) (CH3)2 CH > CCl3 > C6H5 CH2 > (CH3)3 C (2) CCl3 > C6H5 CH2 > (CH3)2 CH > (CH3)3 C

(3) (CH3)3 C > (CH3)2 CH > C6H5 CH2 > CCCl3 (4) C6H5 CH2 > CCl3 > (CH3)3 C > (CH3)2 CH

17. The correct order of increasing basicity of the given conjugate bases (R = CH3) is : [AIEEE-2010, 4/144]

(1) 2HNRCHCORCO (2) 2HNORCOCHCR

(3) RCHC2HNORCO (4) R2HNCHCORCO






VKP SirM.Sc. IT-BHU

NCERT QUESTIONS

1. Show by arrow I effect in the following molecules

(a) CH3 – CH = CH – NO2 (b)

2. Arrange for acidic strength order CH3COOH, ClCH2COOH, Cl2CHCOOH, Cl3CCOOH

3. Draw resonating structure for following molecules , CH2 = C = O ,

4. By drawing resonating structures of following molecules, Judge whether the group attatched to ring exerts+ m or – m effect.

,

5. Which of them is more basic :

(a) (b)

6. Which of them is more basic (a) (b) (c)

7. Arrange the following free radicals in increasing order of stability

8. Write tautomer of butanone and cyclohexanone.

9. Arrange the following carbocation s in stability order-

1. H2.

2.

10. Resonance energy of benzene is 36 Kcal mol, but that of toluene is 37.5 Kcal mol. Why ?

ELECTRONIC EFFECT # 42A-479 Indra vihar, kotaPh. - 9982433693 (NV Sir) 9462729791(VKP Sir)




VKP SirM.Sc. IT-BHU

Exercise # 1PART - I

A-1. (A) A-2. (C) A-3. (B) A-4. (D) A-5. (A) B-1. (C) B-2. (B)B-3. (A) B-4. (B) B-5. (B) B-6. (D) B-7. (C) B-8. (C) B-9. (C)B-10. (A) B-11. (D) B-12. (B) C-1. (A) C-2. (D) C-3. (D) C-4. (D)C-5. (B) C-6. (B) C-7. (B) C-8. (D) C-9. (B) C-10. (B) C-11. (A)C-12. (D) C-13. (B) D-1. (A) D-2. (D) D-3. (D) D-4. (B) D-5. (D)D-6. (B) E-1. (A) E-2. (C) E-3. (D) E-4. (C) E-5. (C) E-6. (B)E-7. (C) E-8. (C) E-9. (A) E-10. (D) E-11. (C) E-12. (B) E-13. (C)E-14. (D) E-15. (C) E-16. (A) F-1. (B) F-2. (B) F-3. (B) F-4. (D)F-5. (B) F-6. (D) F-7. (B) F-8. (D) F-9. (B) F-10. (A) F-11. (C)F-12. (B) F-13. (B) F-14. (B) F-15. (A) F-16. (D) F-17. (C) F-18. (A)F-19. (A) G-1. (A) G-2. (A) G-3. (A) G-4. (B) G-5. (C) G-6. (C)G-7. (B) G-8. (C) G-9. (A) G-10. (B) G-11. (D) G-12. (D) G-13. (A)G-14. (A) G-15. (B) G-16. (A) G-17. (B)

PART - II1. (A) 2. (C) 3. (D) 4. (C) 5. (D) 6. (B) 7. (C)8. (B) 9. (A) - W, S ; (B) - Z, R ; (C) - Y, Q ; (D) - X, P10. (A) - r ; (B) - p ; (C) - q 11. (C) 12. (A) 13. (C) 14. (C) 15. (C)16. (D)

Exercise # 2PART - I

1. (A) 2. (C) 3. (A) 4. (B) 5. (A) 6. (A) 7. (B)8. (D) 9. (A) 10. (D) 11. (A) 12. (C) 13. (D) 14. (B)15. (C) 16. (A) 17. (D) 18. (B) 19. (D) 20. (D) 21. (D)22. (B) 23. (C) 24. (B) 25. (A) 26. (C) 27. (A) 28. (C)29. (A) 30. (A) 31. (A) 32. (D) 33. (A) 34. (A) 35. (C)36. (A) 37. (D) 38. (A) 39. (A) 40. (A) 41. (C) 42. (A)43. (A) 44. (C) 45. (A) 46. (A) 47. (A) 48. (C) 49. (A)

50. (A) 51. (C) 52. (ACD) 53. (ABD) 54. (ACD) 55. (BCD) 56. (BC)PART - II

1. (P) CH3

COOH (Q)

COONa

OH (R)

CH = CH2

C CH

(S)

COOH

NaO

2. Decreasing order of dipolemoment for given compounds(x) : I > IV > III > II ; (y) : I > III > II > IV

3. Cases P, S, T, V has incorrect direction of I effect.

4. (i) 2 ; (ii) 2 ; (iii) 1 ; (iv) 2

5. (P) I > II ; (Q) II > I ; (R) I > II ; (S) II > I





VKP SirM.Sc. IT-BHU

6. a, d, f, g, i, j 7. A – (II), B–(I), C–(II), D–(I), E–(I)

8. (a) I – minor (b) I – minor (c) I – minor (d) I – minor9. (A) II (B) II (C) I (D) II10. (I) 12 (II) 9 (III) 6 (IV) 4 (V) 3

(VI) 2 (VII) 1 (VIII) 6 (IX) 3 (X) 4(XI) 0 (XII) 0 (XIII) 4 (XIV) 6

11. IV > III > II > I 12. b > a 13. (a) 3, (b) 5, (c) 9, (d) 514. I > III > II > IV15. 1, 3-Butadiene shows delocalisation.

CH2 = CH – CH = CH2 – CH = CH –

So bond between second & third C-atoms show partial double bond character. Hence bond length isshorter than C – C bond in n-butane.

16. Aromaticity comes through resonance. Only those molecules can be aromatic which have very highresonance energy. Only those molecules has sufficiently high amount of resonance energy to becomesaromatic which(a) are cyclic (b) are planar (c) contains (4n +2) number of -electrons in ring.(d) Does cyclic resonance between (4n + 2) number of -electrons

Where n = 0,1,2,3,4 ..........

17. (i) It has sp3 carbon in ring. (ii) One C is sp3 hybridised. (iii) It has 8 electrons. It is Antiaromatic.

18. Aromatic : (1), (2), (3), (4), (5), (6), (8)Antiaromatic : (7), (9), (10), (12)Nonaromatic : (11)

19. Benzene has all 6– hydrogen atoms in the plane. It does not have any vander waal or steric repulsions. Soit remains planar and aromatic. On the other hand [10]-annulene has inner hydrogen atoms (1, 6) whichare very close to each other causing steric or vander waal's repulsion. To minimise this the ring is twistedthis destroys coplanarity of all 2p-orbitals. So annulene is non aromatic.

20. Cyclooctatetraene is nonplanar due to its tub-shaped structure.

21. All are aromtic 22. > > > H2O 23. III > II > I > IV 24. B > A

25. (i) (A) > (C) > (D) > (B) (ii) (C) > (B) > (A) > (D). (iii) (A) > (D) > (C) > (B)(iv) (C) > (B) > (D) > (A) (v) ((D) > (C) > (B) > (A)

26. IV > III > II > I 27. (D > C > B > A)28. (P) II > I > III > IV ; (Q) II > I > III > IV ; (R) II > III > IV > I ; (S) I > IV > III > II

29. 1 > 4 > 2 > 3Acidic strength order is – SO3H > –COOH > –CCH > –CH3

30.

VKP SirM.Sc. IT-BHU

In resonting structure of phenoxide negative charge is present on C-atom (which is less electronegative)while in benzoate ion, negative charge is present on more electronegative O atom. Hence in resonatingstructure, if negative charge is present on more electronegative atoms it is more stable in comparison tothat resonating structure which have negative charge on less electronegative atoms.

31. (p) A > B > C > D (Order of Ka) (q) A > B > C ~_ D (Order of Ka)

(r) (A) > (D) > (C) > (B) (s) (A) > (B) > (D) > (C)

32. (a) X (b) X

33. (p) C > B > A (q) B > C > A (s) (D) > (B) > (C) > (A)(t) (D) > (A) > (C) > (B) (u) (C) > (B) > (A) > (D) (w) (A) > (C) > (D) > (B)

34. Due to better solvation in polar protic solvent.

35. (P) I > II > III ; (Q) I > II > III ; (R) I > II > III

36.

3CH <

2NH < OH <

F

37. ERG increases the basic strength because they tend to stabilise protonated aniline, on the other hand,EWG decrease the basic strength they tend to destabilise the protonated aniline.

(G stabilises the cation) (G' destabilises the cation)

38. I > II > IIIExercise # 3

PART - I1. Carbon has higher percentage ‘s’ character in HCC¯, due to which higher electronegativity.

2. (C)3. (a) – (ii) (b) – (iii) (c) – (iv) (d) – (i) (e) – (v)

4. 5. (A) 6. (A) 7. (A) 8. (B) 9. (A)

10. (D) 11. (A) 12. (B) 13. (D) 14. (B) 15. (B) 16. (A)

17. 2 18. (C) 19. 6 20. (B,C)

PART - II

1. (2) 2. (3) 3. (1) 4. (4) 5. (3) 6. (3) 7. (3)

8. (3) 9. (4) 10. (3) 11. (4) 12. (2) 13. (2) 14. (3)

15. (1) 16. (2) 17. (4)

VKP SirM.Sc. IT-BHU

Exercise 4

1. (a) CH CH 3 NO2

+ –CH

+(b)

O-

- -

--

2. CH3COOH < ClCH2COOH < Cl2CHCOOH < Cl3CCOOH

3. +

+

+

CH = C = O2

. .. .

OC–HC 2, N = N = N NN–N

2–

4. (a)

O||

CH–C–NH– 3 (+M group)

NH–C–CH3

O

NH–C–CH3

O

(b)

C–CH3

O

–C–CH3

O

(–m effect group)

C–CH3

O

CO CH3 CO CH3

CO CH3

CO CH3

5. is more basic due to lone pair of N atom is not delocalized.

6. is more basic due to 3sp hybridised N atom.

7. V < III >

(2) +

>+

>

+

10. Greater resonance energy of is caused by hyperconjugation effect of CH3.

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