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1. How do you account for formation of ethane during chlorination of methane? Ans. Chlorination of methane is a free radical reaction which occurs by the

following mechanism: Initiation: Cl Cl ClHomolytic fission

Chlorine free radical− →

2

Propagation: CH H Cl CH HCl3 3− + ⋅ → ⋅ +

⋅ + − → − + ⋅ CH Cl Cl CH Cl Cl3 3 ...(i)

Termination: ⋅ + ⋅ → − CH CH CH CHethane

3 3 3 3 ...(ii)

⋅ + ⋅ → ⋅ − + ⋅ → − CH Cl CH Cl Cl Cl Cl Cl3 3 . ...(iii)

From the above mechanism, it is evident that during propagation step, CH3 free radicals are produced which may undergo three reactions, i.e., (i)–(iii). In the chain termination step, the two CH3 free radicals combine together to form ethane (CH3–CH3) molecule.

2. Write IUPAC names of the following compounds: (a) CH3CH = C(CH3)2 (b) CH2 = CH – C ≡ C – CH3

(c) (d)

(e) (f) CH CH CH CH CH

CH CH CH

3 2 4 2 3 3

2 3 2

( ) ( )

( )|

−

(g) H C CH CH CH CH CH H C CH CH CH

C H

3 2 2 2

2 5

− = − − = − − − =|

Ans. (a) CH CH C CH

CH

Methylbut ene2 3

2 2

3

= −− − −

| (b) CH CH C C CH2 3= − ≡ −

Pent-1-en-3-yne

(c)

(d) 1

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(e)

(f)

(g)

3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated:

(a) C4H8 (one double bond) (b) C5H8 (one triple bond) Ans. (a) Isomers of C4H8 having one double bond are:

(i) (ii) C C

CH3 CH3

H H

cis-But-2-ene

(iii) C C

CH3

CH3

H

H

trans-But-2-ene

(iv) CH3

C

CH3

CH2

2-Methylprop-1-ene

(b) (i) CH CH CH C CHPent yne

3 2 2

2 1

1≡

− − (ii) CH CH C C CH

Pent yne

23

12

1

32

− ≡ −− −

(iii) CH CH C CHCH

Methylbut yne

3

3 13

− − ≡

− − −

|

4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds:(i) Pent-2-ene (ii) 3, 4-Dimethylhept-3-ene (iii) 2-Ethylbut-l-ene (iv) 1-Phenylbut-l-ene.

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Ans. (i) CH CH CH CH CHpent

i O CH Cl Kii

53

42

3 2 13

3 2 2 196− − = −-2-ene

Z

( ) / ,( ) nn/H2

Prop

O

CH CH CH

→

− −3 2aanal

= + = −O O CH CHEthanal

3

(ii)

CH2

CH2 C

C

CH3

CH3

CH2

CH3

CH3

+O

C

CH2

CH3

CH3

3, 4-Dimethylhept-3-ene

(i) O3, CH

2Cl

2, 196K

(ii) Zn / H2O

Pentan-2-one

Butan-2-one

CH3

CH2

CH2

C

CH3

O

(iii) CH CH C CH

CH CH

i O CH Cl Ki

43

32

2 3

2 12

3 2 2 196− =|

( ) . ,(

2-Ethylbut-1-ene

ii Zn H OO CH

) / 2 2 → =Methanal

+ CH CH C

O

CH CHPentan-3-one

3 2 2 3− − − −

(iv) CH CH CH CH C H i O CH Cl Kii Zn H O

43

32

2 16 5

3 2 2 1962

− = − →( ) , ,( ) /

Propanal

CH CH CH O O CH C H3 2 6 5= + = −BBenzaldehyde

5. An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and IUPAC name of ‘A’.

Ans. Step 1. Write the structure of the products side by side with their oxygen atoms pointing towards each other.

Step 2. Remove the oxygen atoms and join the two ends by a double bond,

the structure of the alkene ‘A’ is

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6. An alkene ‘A’ contains three C–C, eight C–H bonds, one C–C pi-bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 a. m. u. write the IUPAC name of ‘A’.

Ans. (i) An aldehyde with molar mass of 44 a. m. u. is ethanal, CH3CH = O (ii) Write two moles of ethanal side by side with their oxygen atoms

pointing towards each other. CH3CH = O O = CHCH3

Ethanal Ethanal (iii) Remove the oxygen atoms and join them by a double bond, the

structure of alkene ‘A’ is CH3–CH=CH–CH3

But–2–ene

As required, but-2-ene has three C–C, eight C–H and one C–C

pi–bond. 7. Propanal and pentan-3-one are the ozonolysis products of an alkene. What

is the structural formula of the alkene? Ans. (i) Write the structures of propanal and pentan-3-one with their oxygen

atoms facing each other, we have,

(ii) Remove oxygen atoms and joins the two fragments by a double

bond, the structure of the alkene is

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8. Write chemical equations for the combustion reaction of the following hydrocarbons.

(i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene

Ans. (i) C4H10 (g) + 13/2 O2 (g) ∆ → 4CO2 (g) + 5 H2O (g) Butane

(ii) C5H10 (g) + 15/2 O2 (g) ∆ → 5CO2 (g) + 5 H2O (g) Pentene

(iii) C6H10 (g) + 17/2 O2 (g) ∆ → 6CO2 (g) + 5 H2O (g) Hexyne

(iv) or C7H8 (g) + 9 O2 (g) ∆ → 7 CO2 (g) + 4 H2O (g)

9. Draw the cis-and trans-structures for hex-2-ene. Which isomer will have higher boiling point and why?

Ans. The structure of cis-and trans-isomer of hex-2-ene are:

C C

CH3

H

CH2

H

CH2

CH3

C C

CH3

H CH2CH

2CH

3

H

cis ene-Hex-2-

(Higher dipole moment,

higher boiling point)

trans ene-Hex-2-

(Lower dipole moment,

lower boiling point)

The dipole moment of a molecule depends upon dipole–dipole interactions. Since cis-isomer has higher dipole moment, therefore, it has higher boiling point.

10. Why is benzene extra-ordinarily stable though it contains three double bonds? Ans. Resonance or delocalisation of electrons usually leads to stability. Since

in benzene all the six π-electrons of the three double bonds are completely delocalised to form one lowest energy molecular orbital which surrounds all the carbon atoms of the ring, therefore, it is extra-ordinary stable.

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H

H

H H

H

H

H H

H

HH

HH

HH

H

H H

Delocalisation of 6 π–electrons in benzene. 11. What are the necessary conditions for any system to be aromatic? Ans. The necessary conditions for a molecule to be aromatic are: (i) It should have a single cyclic cloud of delocalised π-electrons above

and below the plane in the molecule. (ii) It should be planar. This is because complete delocalisation of

π-electrons is possible only if the ring is planar to allow cyclic overlap of p-orbitals.

(iii) It should contain Huckel number of electrons, i.e., (4n + 2) π-electrons where n = 0, 1, 2, 3 etc.

A molecule which does not satisfy any one or more of the above conditions is said to be non-aromatic.

12. Explain why the following systems are not aromatic?

(i) (ii)

(iii)

Ans. (i) Due to the presence of a sp3-hybridised carbon, the system is not planar. It does contain six π-electrons but the system is not fully conjugated since all the six π-electrons do not form a single cyclic electron π cloud which surrounds all the atoms of the ring. Therefore, it is not an aromatic compound.

(ii) Due to the presence of a sp3-carbon, the system is not planar.

Further, it contains only four electrons, therefore, the system is not aromatic because it does not contain planar cyclic cloud having (4n + 2) π–electrons.

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sp3

(iii)

Cyclooctatetraene is not planar rather it is tub shaped. It is, therefore, a non-planar system having 8 pi-electrons. Therefore, the molecule is not aromatic since it does not contain a planar cyclic cloud having (4n + 2)π-electrons.

13. How will you convert benzene into (i) p-Nitrobromobenzene (ii) m-Nitrochlorobenzene (ii) p-Nitrotoluene (iv) Acetophenone? Ans. (i) The two substituents in the benzene ring are present at p-positions.

Therefore,thesequenceofreactionsshouldbesuchthatfirstano–, p-directing group, i.e., Br atom should be introduced in the benzene ring and this should be followed by nitration. Thus,

Br

Benzene

Br2, Anhyd. FeBr

3

Bromination

Bromobenzene

Conc.H

NO

3

+conc.H

2 SO4 ,�

Nitration NO

2

BrBrO2

N +Separated by

Fractional distillation

BrO2

N

p-Bromonitro

benzene (major)

o-Bromonitro

benzene (minor)

(ii) Here since the two substituents are at m-position with respect to each other,therefore,thefirstsubstituentinthebenzeneringshouldbea m-directing group (i.e., NO2) and then the other group (i.e., Cl) should be introduced. Therefore, the sequence of reactions is:

(iii) Here since the two substituents are at p-position with respect to each

other,therefore,thefirstsubstituentinthebenzeneringshouldbea o-, p-directing group (i.e., CH3) and then the other group (i.e., NO2) should be introduced.

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(iv) Acetophenone can be prepared by Friedel Craft (F.C.). acylation

using either acetyl chloride or acetic anhydride

C CH3

O

Anhyd. AlCl3

FC. Alkylation+

Benzene

CH3

C Cl

O

Acetyl chloride

+ HCl

Acetophenone

OR

14. In the alkane, CH3CH2–C(CH3)2–CH2–CH(CH3)2 , identify 1°,2°,3° carbon

atoms and give H – atoms bonded to each one of these. Ans. The expanded formula of the given compound is:

15. What effect does branching of an alkane chain on its boiling point? Ans. As the branching increases, the surface area of an alkane approaches a sphere.

Since, a sphere has minimum surface area, therefore, van der waal’s force of attraction is minimum and hence the boiling point of the alkane decreases with branching.

16. Addition of HBr to propene gives 2-bromopropane, while in presence of benzoyl peroxide, the same reactions give 1-bromopropane. Explain and give mechanism.

Ans. First Case:

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Step I

Step II

First Case:

O O O

Ph C O O C Ph Ph C O Ph CO− − − − − → − − → +⋅|| || ||

∆uv 2 2 2 2

In the first step,Br radical adds to propene in such away to so as to

generate the more stable 2° free radical. In the second step, the free radical thus obtained rapidly abstracts a hydrogen atom from HBr to give 1- bromopropane.

From the above discussion, it is evident that although both reactions are

electrophilic addition reactions but it is due to different orientation of addition of H and Br atoms different products are obtained.

17. Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support KekuIe structure of benzene ?

Ans. o-Xylene may be regarded as a resonance hybrid of the following two Kekule structures. Ozonolysis of each one these gives two products as shown below:

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CH3

C2 +

O

CH 3

Glyoxal

I

CH 3

Methylglyoxal

CH3

CH3

C

1.2

+

C

O

O

CH 3

Glyoxal

-Dimethylglyoxal

II

CH 3

(i) O3 2

(ii) Zn/H O,2

(i) O3 2 2

(ii) Zn/H O,2

2

2

O

C

H

Thus, in all, three products are formed. Since all the three products cannot be obtained from any one of the two Kekule structures, this shows that -xylene is a resonance hybrid of the two Kekule structures (I and II).

18. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behavior.

Ans. The hybridisation state of carbon in these three compounds is; Benzene n-hexane ethyne Type of orbital: sp2 sp3 sp s-Character: 33.3% 25% 50% since, s-electrons are closer to the nucleus, therefore, as the s-character of

the orbital making the C–H bond increases, the electrons of C–H bond lie closer and closer to the carbon atom. In other words, the partial +ve charge on the H– atom and hence the acidic character increases as the s-character of the orbital increases. Thus, the acidic character decreases in the order: Ethyne > Benzene > Hexane

19. Why does benzene undergo electrophilic substitution reactions easily and nucleophilicsubstitutionswithdifficulty?

Ans. Due to the presence of an electron cloud containing 6π-electrons above and below the plane of the ring, benzene is a rich source of electrons. Consequently, it attracts the electrophiles (electron-deficient) reagentstowards it and repels nucleophiles (electron-rich) reagents. As a result, benzene undergoes electrophilic substitution reactions easily and nucleophilic substitutionswithdifficulty.

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20. How will you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane

Ans. (i) 3HC CHEthyne

d≡ →Re hot Fe tube873 K ∆

(ii) Etheneisfirstconvertedintoethyneandthentobenzeneasshownabove.

CH CH Br CH CH BrBr CCl KOH alc

Dibromoeth2 22 4

2 2 1 2= → − − −

−

/ ( ),,

∆aane

(Dehydro bromination)

→

HC CH Benzened≡ →Re hot Fe tube

873 K

(iii) When vapours of hexane are passed over heated catalyst consisting of Cr2O3, Mo2O3 and V2O5 at 773 K under 10–20 atm pressure, cyclisation and aromatisation occurs simultaneously to afford benzene

CH3

CH2

CH2

CH2

CH2

CH3 Cr2O

3/ V

2O

5/ Mo

2O

3

–H2

Aromatisation

–3H2

-hexanen Cyclohexane Benzene

21. Write structures of all the alkenes which on hydrogenation give 2-methylbutane?

Ans. The basic skeleton of 2-methylbutane is 4 3 2 1C

C

C C C|

− − −

Putting double bonds at various different positions and satisfying the tetracovalency of each carbon, the structures of various alkenes which give 2-methylbutane on hydrogenation are:

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22. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+.

(a) chlorobenzene, 2, 4-dinitrochlorobenzene, p–nitrochlorobenzene (b) toluene , p – H3C – C6H4 – NO2, p – O2N – C6H4 – NO2

Ans. (a) The typical reactions of benzene are electrophilic substitution reactions. Higher the electron density in the benzene ring, more reactive is the compound toward these reactions. Since NO2 is more powerful electron withdrawing group than Cl, therefore, more the number of nitro groups, less reactive is the compound. Thus, the overall reactivity decreases in order.

Chlorobenzene > p-nitrochlorobenzene > 2, 4-dinitrochlorobenzene. (b) Here, CH3 group is electron donating but NO2 group is electron–

withdrawing. Therefore, the maximum electron density will be in toluene, followed by p-nitrotoluene followed by p-dinitrobenzene.

Thus, the overall reactivity decreases in the order : Toluene > p–H3C–C6H4–NO2 > p– O2N–C6H4–NO2

23. Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why ?

Ans. CH3 group is electron-donating while –NO2 group is electron-withdrawing. Therefore, maximum electron density will be in toluene, followed by benzene and least in m-dinitrobenzene. Therefore, the ease of nitration decreases in the order: toluene > benzene > m-dinitrobenzene.

24. Suggest name and another Lewis acid instead of anhydrous aluminium chloride which can be used during ethylation of benzene.

Ans. Anhydrous FeCl3, SnCl4, BF3, etc. 25. Why is Wurtz reaction not preferred for preparation of alkanes containing

odd number of carbon atoms? Illustrate your answer by taking one example. Ans. For preparation of alkanes containing odd number of carbon atoms, a

mixture of two alkyl halides has to be used. Since two alkyl halides can react in three different ways, therefore, a mixture of three alkanes instead of the desired alkane would be formed. For example, Wurtz reaction between 1-bromopropane and 1-bromobutane gives a mixture of three alkanes, i.e., hexane, heptane and octane as shown below:

CH3CH2CH2 –Br + 2Na + Br–CH2CH2CH3

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1-Bromopropane Dry ether → CH3CH2CH2CH2CH2CH3 + 2NaBr

Hexane CH3CH2CH2– Br + 2 Na + Br–CH2CH2CH2CH3

1–Bromopropane 1–Bromobutane

Dry ether → CH3CH2CH2CH2CH2CH2CH3 + 2 NaBr

Heptane CH3CH2CH2CH2– Br + 2 Na + Br –CH2CH2CH2CH3 1–Bromobutane

Dry ether → CH3CH2CH2CH2CH2CH2CH2CH3 + 2NaBr Octane