Post on 17-Mar-2022
10 Haloalkanes and Haloarenes
1) Name the following halides according to IUPAC system and classify them as
alkyl, allyl, benzyl (primary, secondary , tertiary), vinyl or aryl halides:
i) (CH3)2CHCH(Cl)CH3 ii) CH3CH2CH(CH3)CH(C2H5)Cl
Solution
I.U.P.A.C Name: 2-Chloro-3-methylbutane
Type: 20 – Alkyl halide
I.U.P.A.CName:3-Chloro-4-methylhexane
Type:20–Alkylhalide
iii) CH3CH2C(CH3)2CH2I iv) (CH3)3CCH2CH(Br)C6H5
Solution
iii) CH3CH2C(CH3)2CH2I
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IUPAC Name: 1-Iodo-2,2-dimethylbutane
Type: 10 – Alkyl halide
IUPAC Name: 1-Bromo-1-phenyl -3,
Type: 10 – Alkyl halide
v) CH3CH(CH3)CH(Br)CH3 vi) CH3C(C2H5)2CH2Br
Solution
v) CH3CH(CH3)CH(Br)CH3
IUPAC Name: 2-Bromo-3-methylbutane ,
Type: 10 – Alkyl halide
vi) CH3C(C2H5)2CH2Br
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IUPAC Name: 3- Bromoethyl-3-methylpentane
Type: 10 – Alkyl halide
vii) CH3C(Cl)(C2H5)CH2CH3 viii) CH3CH=C(Cl)CH2CH(CH3)2
Solution
IUPAC Name: 3- Chloro-3-methylpentane
Type: 10 – Alkyl halide
IUPAC Name: 3- Chloro-5-methyl hex-2-ene
Type: 10 –Vinyl halide
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ix) CH3CH=CHC(Br)(CH3)2 x)p-ClC6H4CH2CH(CH3)2
Solution
IUPAC Name: 4-Bromom-4-methyl pent-2-ene
Type: Allyl halide
IUPAC Name: 1-Chloro -4-(2-methylpropyl) benzene
Type: Aryl halide
xi) m-ClCH2C6H4CH2C(CH3)3 x)o-Br-C6H4CH(CH3)CH2CH3
Solution
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IUPAC Name: 1-Chloromethyl -3(,22-dimethylpropyl) benzene
Type: Alkyl halide
IUPAC Name: 1-Bromo-2-(1-methylpropyl) benzene
Type: Aryl halide
2) Give the IUPAC names of the following compounds:
i) CH3CH(Cl)CH(Br)CH3
ii) CHF2CBrClF
Solution
I.U.P.A.C Name: 2-Bromo-3-chlorobutane
ii)
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I.U.P.A.C Name: 1-Bromo-1-chloro 1,2,2-tri fluoro ethane
iii) ClCH2C ≡ CCH2Br
iv) (CCl3)3CCl
Solution
iii)
I.U.P.A.C Name: 1-Bromo4-chlorobut-2-yne
I.U.P.A.C Name: 1,1,1,2,3,3,3- hepta chloro-2-trichloromethyl propane
v) CH3C(p-ClC6H4)2CH(Br)CH3
vi) (CH3)3CCH=CClC6H4I-p
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Solution
I.U.P.A.C Name: 3,3, -(4,41-dichlorophenyl) -2- bromobutane
or
2- Bromo-3,3-bis(4-chlorophenyl)butane
I.U.P.A.C Name: 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene
3) 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
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Solution
2-Chloro-3-methylpentane p-Bromochlorobenzene
1-Chloro-4-ethylcyclohexane 2-(2-Chlorophenyl)-1-iodooctane
v) 2-Bromobutane
vi) 4-tert-butyl-3-iodoheptane
vii)1-Bromo-4-sec-butyl-2-methylbenzene
viii) 1,4-Dibromobut-2-ene
Solution
2-Bromobutane 4-Tertbutyl-3-iodoheptane
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1- Bromo 4-secbutyl-2-methylbenzene 1,4-Dibromobut-2-ene
4) Which one of the following has the highest dipole moment?
i) CH2Cl2
ii) CHCl3
iii) CCl4
Solution
ii) CHCl3
Tetrachloromethane (CCl4) is a symmetrical molecule and has zero dipole moment.
In chloroform (CHCl3), the resultant dipole moment of two C-Cl bonds is cancelled
by resultant dipole moment of C-Cl and C-H bonds. Therefore, the resultant dipole
moment (µ= 1.03 D ) is smaller.
In dichloromethane (CH2Cl2) , the resultant dipole moments of two C-Cl and two C-
H bonds reinforce one another . Therefore, dichloromethane molecule has maximum
dipole moment (µ) of 1.62 D.
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5) A hydrocarbon C5H10 does not react with chlorine in dark but gives a single
monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Solution
Cyclopentane is the hydrocarbon which gives monochloro compound C5H9Cl in
bright sunlight.
6) Write the isomers of the compound having formula C4H9Br.
Solution
The isomers of the compound having formula C4H9Br are:
7) Write the equations for the preparation of 1-iodobutane from
i) 1-butanol ii) 1-chlorobutane iii) but-1-ene
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Solution
8) What are ambident nucleophiles ? Explain with an example.
Solution
Nucleophiles which possess two nucleophilic (electron pair donor) centers are called
ambident nucleophiles .
Ex:
CN- nucleophile by linking through carbon atom results in alkyl cyanides and through
nitrogen atom forms isocyanides.
9) 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
Solution
i) CH3I
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ii) CH3Cl
10) Predict all the alkenes that would be formed by dehydrohalogenation of the
following halides with sodium ethoxide in ethanol and identify the major alkenes:
i) 1-Bromo-1-methylcylohexane.
Solution
ii) 2-Chloro-2-methylbutane
Solution
iii) 2,2,3-Trimethyl-3-bromopentane.
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Solution
2- Ethyl-3,3-dimethylbut-1-ene
( Minor Product)
In dehydrogenation reactions, the preferred product is that alkene which has the
greater number of alkyl groups attached to the doubly bonded carbon atoms.
In the above case total 3 products are possible out of which 2.2.3-trimethyl-2-pentane
is major product.
11) How will you bring about the following conversions?
i) Ethanol to but-1-yne
ii) Ethane to bromoethene
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Solution
i)
iii) Propene to 1-nitropropane
iv) Toluene to benzyl alcohol
Solution
iii)
Propene Propane 1-Nitropropane
iv)
Toulene Benzyl bromide Benzyl alcohol
v) Propene to propyne
vi) Ethanol to ethyl fluoride
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Solution
v)
Propene 1,2-Dibromopropane Propyne
vi)
Ethanol Bromoethane Ethyl fluoride
vii) Bromomethane to propanone
viii) But-1-ene to but-2-ene
Solution
vii)
Methylbromide Ethane 1,1-Dibromoethane Ethyne
Monosodium Propene Propanone
acetylide
viii)
But-1-ene 2- Bromobutane
But- 2-ene
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12) 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) In chlorobenzene C-Cl bond is less polar because of sp2 –carbon and ‘Cl’-bond
possess double bond character due to resonance, whereas in cyclohexyl chloride sp3 –
carbon and ‘Cl’ –bond is polar.
ii) Hydrogen bond is not possible in case of alkyl halides.
(Covalent character of “C-X’ bond).
iii) Grignard reagents are hydrolysed easily to give hydrocarbons(alkanes)
13) Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.
Solution
Freon12- Refrigeration gas
DDT- Fungicide and Insecticide
CCl4 – Fire extinguishes and as solvent
CHI3- as antiseptic (in preparation of Tincture of iodine)
14) Write the structure of the major organic product in each of the following
reactions:
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Solution
15) Write the mechanism of the following reaction:
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Solution
Butyl bromide Intermediate Butyl cyanide
16) 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-2methylbutane
(10) (2
0) (3
0)
ii) 1-Bromo-3-methyl butane> 2-Bromo-3-methyl-butane> 2-Bromo-2
(10) (2
0) (3
0)
methylbutane.
iii) 1-Bromobutane> 1-bromo-3-methylbutane>1-Bromo-2methyl Butane> 1-Bromo-
2,2 dimethyl propane.
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NOTE: SN2 –order of reaction 1
0 > 2
0 > 3
0.
As the bulky groups hinder the approaching nucleophiles.
17) Out of C6H5CH2Cl and C6H5CHClC6H5 , which is more easily hydrolysed by
aqueous KOH?
Solution
C6H5CHClC6H5 is more easily hydrolysed because the intermediate carbo cation
formed is more resonance stabilised.
18) p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-
isomers . Discuss.
Solution
p-Dichlorobenzene has symmetric structure close packing of molecules results in
high m.pt and solubility is more because of intermolecular hydrogen bonding which
is absent in o- and m-isomers.
19) How the following conversions can be carried out?
i) Propene to propan-1-ol
ii) Ethanol to but-1-yne
iii) 1-Bromopropane to 2-bromopropane
Solution
i)
Propene 1-Bromo propane Propan-1-ol
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ii)
Ethanol Ethene 1,2-Dibromoethane Ethyne
Monosodium But-1-yne
iii)
iv) Toluene to benzyl alcohol
v) Benzene to 4-bromon nitrobenzene
Toluene Benzyl chloride Benzyl alcohol
Benzene Bromobenzene 4-Bromonitrobenzene
ix) 2-Chlorobutane to 3,4-dimethylhexane
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x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
xi) Ethyl chloride to propanoic acid
Solution
2 – Chlorobutane 3,4 –dimethylhexane
2-Methyl-1-propene 2-chloro-2-methylpropane
Ethyl chloride Ethyl cyanide Propanoic acid
vi) Benzyl alcohol to 2-phenylethanoic acid
vii) Ethanol to propanenitrile
viii) Aniline to chlorobenzene
Solution
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Benzyl alcohol Benzyl chloride Benzyl cyanide 2-phenylethanoic
acid
Ethanol Chloro ethane Propanenitrile
Aniline Benzene diazonium chloride Chlorobenzene
xii) But-1-ene to n-butyliodide
xiii) 2-Chloropropane to 1-propanol
xiv) Isopropyl alcohol to iodoform
Solution
But-1-ene 1-Bromobutane
n-butyliodide
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2-Chloropropane Propene
1-Bromopropane 1-Propanol
Isopropyl alcohol Propanone Iodoform
xviii) Benzene to diphenyl
xix) tert-Butyl bromide to isobutyl bromide
xx) Aniline to phenylisocyanide
Solution
Benzene Chlorobenzene Diphenyl
Tert-butyl bromide Isobutylene Isobutyl bromide
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Aniline Phenylisocyanide
xv) Chlorobenzene to p-nitrophenol
xvi) 2-Bromopropane to 1-bromopropane
xvii) Chloroethane to butane
Chlorobenzene p-nitrochlorobenzene p-nitrophenol
2-Bromopropane Propene 1-bromopropane
Chloroethane Butane
20) 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.
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Solution
Reaction of alkyl chlorides with aqueous KOH proceeds through nucleophilic
substitution reaction mechanism; hence a stronger nuclephile OH- , replaces weaker
nucleophile X- of alkyl halide. While alcoholic KOH is a dehydrohalogenating agent
and forms an alkene by eliminating hydrogen halide. C2H5O- will act as strong base
rather than nucleophile hence elimination takes place.
21) Primary alkyl halide C4H9Br (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
[B] [C]
[A] [D]
A is Isobutyl bromide
B is 2-methylpropene
C is 2-Bromo-2-methylpropane
D is 2,5-Dimethylhexane
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22) What happens when
i) n-butyl chloride is treated with alcoholic KOH
ii) bromobenzene is treated with Mg in the presence of dry ether.
Solution
i) When n-butyl chloride is treated with alcoholic KOH , 1-butene is formed
(H2C=CH-CH2-CH3) by dehydrochlorination.
CH3-CH2-CH2-CH2-Cl + KOH(alc) → CH2=CH-CH2-CH3+ KCl + H2O
n-Butyl chloride 1 – Butene
ii) When bromobenzene is treated with Mg in the presence of dry ether , Phenyl
magnesium bromide is formed.
Bromobenzene Phenyl magnesium bromide
iii) When chlorobenzene is subjected to hydrolysis , phenol is formed
Chlorobenzene Phenol
iv) When ethyl chloride is treated with aqueous KOH, ethanol is formed.
Chloroethane Ethanol
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v) methyl bromide is treated with sodium in the presence of dry ether
vi) methyl chloride is treated with KCN
Solution
v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is
formed. This is called wurtz reaction
Methyl bromide Ethane
vi) When methyl chloride is treated with KCN, methyl cyanide is formed.
H3C-Cl + KCN → H3C-CN + KCl
Methyl chloride Methyl cyanide
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