1. Name the following halides according to IUPAC … 12TH/CLASS...1 1. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary,

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1. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: a. (CH3)2CHCH(Cl)CH3 b. CH3CH2CH(CH3)CH(C2H5)Cl c. CH3CH2C(CH3)2CH2I

d. (CH3)3CCH2CH(Br)C6H5.

• Solution:

a. 2-chloro-3-methyl butane (sec)

b.3-chloro-4-methyl-hexane[sec]

c. 1- ido-2, 2-dimethyl butane

d. 1-bromo –3,3-dimethyl-1-phenyl-butane[sec,,benzylic]

2. Name the following halides according to IUPAC system and classify

them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: a. CH3CH(CH3)CH(Br)CH3

b. CH3C(C2H5)2CH2Br

c. CH3C(Cl)(C2H5)CH2CH3

d. CH3CH=C(Cl)CH2CH(CH3)2.

• Solution:

a. 2-bromo-3-methyl butane (sec)

b. 1-bromo-2-ethyl-2-methyl-butane[pri]

c. 3-chloro-3-methyl-pentane[ter]

d. 3-chloro-5-methyl-hex-2-ene[vinylic]

3. Name the following halides according to IUPAC system and classify

them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

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a. CH3CH=CHC(Br)(CH3)2

b. p-ClC6H4CH2CH(CH3)2

c. m-ClCH2C6H4CH2C(CH3)3

d. o-Br-C6H4CH(CH3)CH2CH3.

• Solution: a. 4-bromo-4-methyl- pent-2-ene [allylic]

b. 1-chloro-4-(2-methyl propyl ) benzene(aryl) (sec)

c. 1-chloro methyl-3-(2,2-dimethyl propyl)benzene [benzylic])

d. 1-bromo-2(1-methyl propyl) benzene.[aryl]

4. Give the IUPAC names of the following compounds:

i. CH3CH(Cl)CH(Br)CH3

ii. CHF2CBrClF iii. ClCH2C ßCCH2Br.

• Solution:

i. 2-bromo-3-chloro-butane

ii. 1-bromo,1-chloro,1,2,2-tri fluoro ethane

iii. 1-bromo-4-chloro-but-2-yne

5. Give the IUPAC names of the following compounds:

i. (CCl3)3CCl

ii. CH3C(p-ClC6H4)2CH(Br)CH3

iii. (CH3)3CCH=ClC6H4I-p.

• Solution:

i 2-(trichloro methyl)-1,1,1,2,3,3,3-hepta chloro propane.

ii 2-bromo—3,3-bis( p- chloro phenyl) butane

iii 1-chloro-1-(4-iodo phenyl) -3,3-dimethyl-but-1-ene.



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6. Write the structures of the following organic halogen compounds.

i. 2-Chloro-3-methylpentane

ii. p-Bromochlorobenzene

iii. 1-Chloro-4-ethylcyclohexane

iv. 2-(2-Chlorophenyl)-1-iodooctane.

• Solution:

i.

ii.

iii.

7. Write the structures of the following organic halogen compounds.

i. Perfluorobenzene

ii. 4-tert-Butyl-3-iodoheptane

iii. 1-Bromo-4-sec-butyl-2-methylbenzene

vi. 1,4-Dibromobut-2-ene.

• Solution:



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iv. .

8. Which one of the following has the highest dipole moment?

i. CH2Cl2 ii. CHCl3 iii. CCl4.

• Solution: i. CH2Cl2

9. A hydrocarbon C5H10 does not react with chlorine in dark but gives a

single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

• Solution: The higher alkanes do not react in dark, but in ultraviolet light or higher temperature gives monochloro derivative as main product. Pentene or chlorination in presence of sunlight gives monohalogen derivative.

10. Write the isomers of the compound having formula C4H9Br.

• Solution:

a. - 1-bromobutane

b. - 1-bromo-2-methyl propane



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c. - 2-bromo- 2-methyl propane

d. - 2-bromobutane.

11. Write the equations for the preparation of 1-iodobutane from

(i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.

• Solution: (i)

(ii)

(iii) .

12. What are ambident nucleophiles? Explain with an example.

• Solution: The nucleophiles with two nucleophilic centers are called ambident nucleophile. They can react through either of these centers. Depending on the reagent and the reaction conditions; the reaction may take place predominantly at one of these centers. Examples are C ≡ N (reagent KCN, AgCN) etc Nitrite O = N – O- (KNO2, AgNO2).

13. Which compound in each of the following pairs will react faster in

SN2 reaction with –OH?

i. CH3Br or CH3I ii. (CH3)3CCl or CH3Cl.



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• Solution:

Methyl iodide is more reactive than methyl bromide as iodide is a better leaving group.

14. Predict all the alkenes that would be formed by dehydrohalogenation

of the following halides with sodium ethoxide in ethanol and identify the major alkene: i. 1-Bromo-1-methylcyclohexane ii. 2-Chloro-2-methylbutane

iii. 2,2,3-Trimethyl-3-bromopentane.

• Solution:

i. 1 - methyl cyclohexene

ii. 2 - methyl but-2-ene

iii. 3,4,4-trimethyl-pent-2-ene

15. How will you bring about the following conversions?

i. Ethanol to but-1-yne

ii. Ethene to bromoethane

iii. Propene to 1-nitropropane

iv. Toluene to benzyl alcohol

v. Propene to propyne.

• Solution:

ii.

iii.



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vi.

v.

16. How will you bring about the following conversions?

i. Ethanol to ethyl fluoride

ii. Bromomethane to propanone

iii. But-1-ene to but-2-ene

iv. 1-Chlorobutane to n-octane

v. Benzene to biphenyl.

• Solution:

i.

ii.

iii.

iv.



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v.

17. Explain why

i. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

ii. Alkyl halides, though polar, are immiscible with water?

iii. Grignard reagents should be prepared under anhydrous conditions?

• Solution: i. The low dipolemoment of chlorobenzene is due to resonance phenomenon. The polarity and hence dipole moment is due to resonance effect as a result the transmission of π - electron in conjugate π - bonds is very little. Where as there is no resonance in cyclohexyl chloride, but the polarity and hence dipolemoment is due to induction effect which is more.

ii. All the halogen derivatives of hydrocarbon are polar in nature but they are insoluble in water, because they are unable to form hydrogen bond with water to break the hydrogen bond already present in water. They are soluble in organic solvents.

iii. Grignard reagent should be prepared under anhydrous conditions, because it is very reactive. It reacts very quickly with any source of proton to give hydrocarbon. It reacts with water very quickly. Therefore, it is necessary to avoid moisture from the Grignard reagents.

18. Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.

• Solution: Freon12 a) It is used as a refrigerant (cooling agent) in refrigerators and air conditioners,

b) It is also used as a propellant in aerosols and foams to spray out deodorants, cleaners, hair sprays, shaving creams and

c) It is also used as insecticides.

Carbon tetrachloride a) It is used as a solvent for oils, fats and waxes.

b) It is used as a fire extinguisher under the name pyrene.



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c) It is used as dry cleaning.

Iodoform a) It is used as antiseptic and this nature is due to iodine that it liberates.

b) It is used in the manufacture of pharmaceuticals.

19. Write the structure of the major organic product in each of the

following reactions:

i.

ii.

iii.

iv.

• Solution:

i.

ii.

iii.

iv.


following reactions:

i.

ii.

iii.



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iv.

• Solution:

i.

ii.

iii.

iv.

21. Write the mechanism of the following reaction:

nBuBr + KCN €€€€nBuCN EtOH – H2O

• Solution:

€n-BuBr is a primary halide

Undergoes nucleophilic substitution by SN2 mechanism

CN- appraoaches the carbon atom from the side opposite at the halogen atom and forms a covalent bond and at the same time Br leaves as Br-

The transition state is

The nucleophiles, Nu- approaches the carbon atom on the side opposite from the halogen and forms new covalent bond with the carbon of the group –C-X as the displaced base, X-, leaves Nu- + R-X slow Nu…….R…….X fast



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Nu—R + X-

RX + KCN transition state

RC = N + KX This is because KCN is an ionic compound, K+ [CN-], Since carbon carrying a lone pair of electrons is more reactive than nitrogen carrying a lone pair, the transition state is formed by the carbon of the cyanide ion forming band with the carbon of the –C-X.

22. Arrange the compounds of each set in order of reactivity towards

SN2 displacement:

i. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

ii. 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane

iii. 1- Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

• Solution:

i. 1-Bromopentane, 2-Bromopentane, 2-Bromo-2-methylbutane,[ in decreasing order]

ii .1-bromo-3-methyl butane, 2-bromo- 3 – methyl butane,2-bromo-2-methyl-butane

iii. 1- Bromobutane, 1-Bromo-3-methyl-butane,1-bromo-2-methyl-butane,1-bromo-2,2 dimethyl propane.

23. Out of C6H5CH2Cl and C6H5 CHClC6H5 , which is more easily hydrolysed

by aqueous KOH.

• Solution:

C6H5CH2Cl is a primary halide and undergoes hydrolysis bySN2 mechanism.

C6H5CHClC6H5 undergoes hydrolysis by SN1 or SN2 mechanism. Though the carbocation in this case is more stable, there is steric hindrance and hence SN2 mechanism is followed and the first one is more readily hydrolysed.

24. p-Dichlorobenzene has higher m.p. and solubility than those of o- and

m-isomers. Discuss.

• Solution: The melting of p-isomer is quite higher than ortho and meta isomer. This is due to the fact that it has symmetrical strcture and therefore its molecules can be easily packed closely in crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore greater energy is required to break its lattice and it melts at higher temperature.



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25. The treatment of alkyl chlorides with aqueous KOH leads to the

formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

• Solution:

Aqueous KOH contains only OH- ions, which acts as nucleophiles and these bring about hydrolysis of alkyl chlorides to the corresponding alchohol.

In the case of alcoholic KOH, the ethoxide ion present is a strong base and removes HCl.

26. Primary alkyl halide C4H9 Br (a) reacted with alcoholic KOH to give

compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

• Solution:

According to the problem when alkyl bromide (A) was treated with sodium, it gave a compound which is not a straight chain hydrocarbon. Therefore (A) cannot be butyl bromide. It may be isobutyl bromide. The complete reactions are

27. Write structures of the following compounds:

i. 2-Chloro-3-methylpentane

ii. 1-Chloro-4-ethylcyclohexane

iii. 4-tert. Butyl-3-iodoheptane



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iv. 1,4-Dibromobut-2-ene

v. 1-Bromo-4-sec. butyl-2-methylbenzene.

• Solution:

i.

ii.

iii.

iv.

v.

28. Why is sulphuric acid not used during the reaction of alcohols with KI?

• Solution: H2SO4 cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I2.

29. Write structures of different dihalogen derivatives of propane.

• Solution: i. ClCH2CH2CH2Cl ii. ClCH2CHClCH3 iii. Cl2CH2CH2CH3 iv. CH3CCl2CH3.

30. Among the isomeric alkanes of molecular formula C5H12, identify the

one that on photochemical chlorination yields i. A single monochloride.

ii. Three isomeric monochlorides.



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iii. Four isomeric monochlorides.

• Solution: i. All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.

ii. The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product.

iii. Similarly the equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

31. Draw the structures of major monohalo products in each of the following reactions:

i.

ii.

iii.

• Solution:

i.



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ii.

iii.

32. Arrange each set of compounds in order of increasing boiling points.

i. Bromomethane, Bromoform, Chloromethane, Dibromomethane.

ii. 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

• Solution: i. Chloromethane, Bromomethane, Dibromomethane, Bromoform. Boiling point increases with increase in molecular mass.

ii. Isopropyl chloride, 1-Chloropropane, 1-Chlorobutane. Isopropyl chloride being branched has lower b.p. than 1-Chloropropane.

33. In the following pairs of halogen compounds, which compound

undergoes faster SN1 reaction?

• Solution:

Tertiary halide reacts faster than secondary halide because of the greater stability of tert-carbocation.


following reactions

i.



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ii.

iii.

iv.

v.

vi.

vii.

viii.

• Solution:

i.

ii.

iii.

iv.

v.

vi.



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vii.

viii.

35. How can the following conversions be carried out?

i. Propene to propan-1-ol

ii. Ethanol to but-1-yne

iii.1-Bromopropane to 2-bromopropane

iv.Toluene to benzyl alcohol

v. Benzene to 4-bromonitrobenzene

vi. Benzyl alcohol to 2-phenylethanoic acid

vii. Ethanol to propanenitrile

viii. Aniline to chlorobenzene

ix. 2-chlorobutane to 3, 4-dimethylhexane

x. 2-methyl-1-propene to 2-chloro-2-methylpropane

xi. Ethyl chloride to propanoic acid

xii. But-1-ene to n-butyliodide

xiii. 2-Chloropropane to 1-propanol

xiv. Isopropyl alcohol to iodoform

xv. Chlorobenzene to p - nitrophenol

xvi. 2-Bromopropane to 1-bromopropane

xvii. Chloroethane to butane



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xviii. Benzene to diphenyl

xix. tert - Butyl bromide to isobutyl bromide

xx. Aniline to phenylisocyanide

• Solution:

i.

ii.

iii.

iv.

v.

vi.

vii.

Viii.



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ix.

x.

xi.

xii.

xiii.

xiv.



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xv.

xvi.

xvii.

xviii.

xix.



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xx.



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Documents

1. Name the following halides according to IUPAC … 12TH/CLASS...1 1. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary,