Class 12 - Chemistry - Haloalkanes and Haloarenes
Name the following halides according to IUPAC system and classify them as alkyl, allyl,
and benzyl (primary, secondary, tertiary), and vinyl or aryl halides:
(i) (CH3)2CHCH (Cl) CH3 (ii) CH3CH2CH (CH3) CH (C2H5) Cl (iii) CH3CH2C (CH3)2CH2I
(iv) (CH3)3CCH2CH (Br) C6H5
(v) CH3CH(CH3)CH(Br)CH3 (vi) CH3C(C2H5)2CH2Br(vii) CH3C(Cl)(C2H5)CH2CH3
(viii) CH3CH=C (Cl) CH2CH (CH3)2
(ix) CH3CH=CHC (Br) (CH3)2 (x) p-ClC6H4CH2CH (CH3)2(xi) m-ClCH2C6H4CH2C (CH3)3
(xii) o-Br-C6H4CH (CH3) CH2CH3
(i) 2-chloro-3-methylbutane, (2◦ alkyl halide)
The Cl atom is attached with two alkyl groups therefore 2◦ alkyl halide.
(ii) 3-chloro-4-methylhexane (2◦ alkyl halide)
The Cl atom is attached with two alkyl groups therefore 2◦ alkyl halide.
(iii) 1-iodo-2,2-dimethylbutane (1◦ alkyl halide)
The atom I is attached with one alkyl group therefore 1◦ alkyl halide.
(iv) 1-bromo-3,3-dimethyl-1-phenylbutane (2◦ benzylic halide)
The atom Br is attached with two alkyl groups with the benzene group therefore 2◦ benzylic halide.
(v) 2-bromo-3-methylbutane (2◦ alkyl halide)
The atom Br is attached with two alkyl groups therefore 2◦ alkyl halide.
(vi) 1-bromo-2-ethyl-2-methylbutane (1◦ alkyl halide)
The atom Br is attached with one alkyl group therefore 1◦ alkyl halide.)
(vii) 3-chloro-3-methylpentane (3◦ alkyl halide)
The atom Cl is attached with three alkyl groups therefore 3◦ alkyl halide.
(viii) 3-chloro-5-methylhex-2-ene (vinylic halide)
The atom Cl is attached with the double bond i.e. at vinylic position therefore vinylic halide.)
(ix) 4-bromo-4-methylpent-2-ene (allylic halide)
The Br atom is attached with the carbon next to the double bond i.e. at allylic position therefore allylic halide.
(x) 1-chloro-4-(2-methylpropyl) benzene (aryl halide)
The Cl atom is attached with the benzene i.e. at aryl position therefore aryl halide.
(xi) 1-chloro-3-(2, 2-dimethylpropyl) benzene (1◦ benzylic halide)
The Cl is attached with one alkyl group at the benzene carbon therefore 1◦ benzylic halide.
(xii) 1-bromo-2-(1-methylpropyl) benzene (aryl halide)
The Br is attached with the benzene i.e. at aryl position therefore aryl halide.
Give the IUPAC names of the following compounds:
(i) CH3CH (Cl) CH (Br) CH3 (ii) CHF2CBrClF (iii) ClCH2C≡CCH2Br
(iv) (CCl3)3CCl (v) CH3C (p-ClC6H4)2CH (Br) CH3
(vi) (CH3) 3CCH=ClC6H4I-p
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
(v) Perfluorobenzene (vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene (viii) 1, 4-Dibromobut-2-ene
BrCH2-CH = CH –CH2Br
Which one of the following has the highest dipole moment?
(i) CH2Cl2 (ii) CHCl3 (iii) CCl4
CCl4 is a symmetrical molecule. Therefore, the dipole moments of all four C−Cl bonds cancel each other.
Hence, its resultant dipole moment is zero.
As shown in the above figure, in CHCl3, the resultant of dipole moments of two C−Cl bonds is opposed by the resultant of dipole moments
of one C−H bond and one C−Cl bond. Since the resultant of one C−H bond and one C−Cl bond dipole moments is smaller than two C−Cl bonds,
the opposition is to a small extent. As a result, CHCl3 has a small dipole moment of 1.08 D.
On the other hand, in case of CH2Cl2, the resultant of the dipole moments of two C−Cl bonds is strengthened
by the resultant of the dipole moments of two C−H bonds. As a result, CH2Cl2 has a higher dipole moment of 1.60 D than CHCl3 i.e., CH2Cl2
has the highest dipole moment.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as:
CCl4 < CHCl3 < CH2Cl2
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight.
Identify the hydrocarbon.
A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general molecular formula CnH2n.
Therefore, it may either be an alkene or a cycloalkane.
Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene.
Thus, it should be a cycloalkane.
Further, the hydrocarbon gives a single monochloro compound, C5H9Cl by reacting with chlorine in bright sunlight.
Since a single monochloro compound is formed, the hydrocarbon must contain H−atoms that are all equivalent.
Also, as all H−atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane., the said compound is cyclopentane.
Write the isomers of the compound having formula C4H9Br.
Double bond equivalent is used to find the level of unsaturation’s present in an organic molecule.
DBE = (C + 1-(H/2) +(X/2)-(N/2))
Where C= number of carbon atoms present
H=number of hydrogen atoms present
N=number of nitrogen atoms present
X=number of halogen atoms present
Double bond equivalent (DBE) for C4H9Br
= (4 + 1 – (9/2) + (1/2))
So none of the isomers has a ring or unsaturation, so the isomers are positions or chain isomers as shown below in the table:
The possible isomers of C4H9Br are following:-
Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.
1-butanol is treated with KI in the presence of H3PO4 where the OH is being replaced by the iodine and gives H2O and KH2PO4
as the by-product with 1-iodobutane as the final product.
What are ambident nucleophiles? Explain with an example.
Ambident nucleophiles are nucleophiles having two nucleophilic sites.
Thus, ambident nucleophiles have two sites through which they can attack.
For example, nitrite ion is an ambident nucleophile.
RX + Ag- O-N=O --> R-NO2
RX + KNO2 --> R-O-N=O
Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites.
Also, it can attack through nitrogen resulting in the formation of nitroalkanes.
Which compound in each of the following pairs will react faster in SN2 reaction with –OH?
R−F << R−Cl < R−Br < R−I
Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH−.
Hence, CH3Cl reacts faster than (CH3)3CCl in SN2 reaction with OH−.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in
ethanol and identify the major alkene:
How will you bring about the following conversions?
(i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to1-nitropropane (iv) Toluene to benzyl alcohol
(v) Propene to propyne(vi) Ethanol to ethyl fluoride (vii) Bromomethane to propanone (viii) But-1-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane (x) Benzene to biphenyl.
Since the conversion of ethanol to but-1-yne involves the addition the two extra carbons that is why the conversion is carried out in 3 steps
The Bromination of ethane (Br2 in the presence of light at 520-670K) gives bromoethane
which on further treatment with alcoholic KOH results in the alkene called ethane which again on
Bromination with Br2/CCl4 gives Dibromide(vicinal bromide) called 1,2-dibromoethane
which on treatment with alcoholic KOH gives bromoethene as the final product.
This reaction happens according to Markownikoff’s rule
Now reacting this with alcoholic KOH gives alkenes
CH3-CH2-CH (Br)-CH3 + KOH (alcohol)——-> CH3CH=CHCH3 + KBr + H2O
Firstly, the benzene is treated with Br2/FeBr3 results in the formation of bromobenzene. It is an electrophilic substitution.
Further the bromobenzene is treated with Na metal in presence of dry ether to form biphenyl as the final product and NaBr as the by-product.
If instead of Na metal, bromobenzene is treated with Cu metal then the reaction is called Ullmann’s reaction.
The final product obtained in this reaction will also be biphenyl.
(i) The dipole moment of chlorobenzene is lower than that of cyclohexylchloride?
(ii) Alkyl halides, though polar, are immiscible with water?
(iii) Grignard reagents should be prepared under anhydrous conditions?
Therefore, Grignard reagents should be prepared under anhydrous conditions.
Give the uses of Freon 12, DDT, carbon tetrachloride and Iodoform.
Uses of Freon − 12
Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners.
It is also used in aerosol spray propellants such as body sprays, hair sprays, etc.
However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994.
Uses of DDT
DDT (p, p−dichlorodiphenyltrichloroethane) is one of the best known insecticides.
It is very effective against mosquitoes and lice. But due its harmful effects, it was banned in the United States in 1973.
Uses of carbon tetrachloride (CCl4)
(i) It is used for manufacturing refrigerants and propellants for aerosol cans.
(ii) It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
(iii) It is used as a solvent in the manufacture of pharmaceutical products.
(iv) Until the mid-1960’s, carbon tetrachloride was widely used as a cleaning fluid,
a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.
Uses of Iodoform (CHI3)
Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell.
The antiseptic property of Iodoform is only due to the liberation of free iodine when it comes in contact with the skin.
Write the mechanism of the following reaction:-
The given reaction is a nucleophilic reaction:
The given reaction is an SN2 reaction. In this reaction, CN acts as the stronger nucleophile and
attacks the carbon atom to which Br is attached in nBuBr.
And, CN- ion is an ambident nucleophile and can attack through both C and N.
In this case, it attacks through the C-atom.
The mechanism is shown below:
Arrange the compounds of each set in order of reactivity towards SN2 displacement:
An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached.
When the nucleophile is sterically hindered, then the reactivity towards SN2 displacement decreases.
Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order.
1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane
Hence, the increasing order of reactivity towards SN2 displacement is:
2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane
Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity
towards SN2 displacement is 3° < 2° < 1°.
Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN2 displacement as:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane
The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents
from the atom containing the leaving group.
Further, the steric hindrance increases with an increase in the number of substituents.
Therefore, the increasing order of steric hindrances in the given compounds is as below:
1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane
< 1-Bromo-2, 2-dimethylpropane
Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is:
1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3-methylbutane < 1-Bromobutane
Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?
Hydrolysis by KOH results in the formation of the carbocation.
Those compounds which lead to the formation of stable carbocation are easily hydrolysed.
C6H5CH2Cl leads to formation of 1°-carbocation, while C6H5CHClC6H5 forms 2°-carbocation,
which is more stable than 1°-carbocation. Hence C6H5CHClC6H5 is hydrolyzed more easily than C6H5CH2Cl by aqueous KOH.
p-Dichlorobenzene has higher melting point and solubility than those of o- and m-isomers. Discuss.
p-Dichlorobenzene is more symmetrical than o-and m-isomers.
For this reason, it fits more closely than o-and m-isomers in the crystal lattice.
Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene.
As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers.
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
(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
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide.
It is an AntiMarkovnikoff reaction in which under the presence of peroxide an alkene undergoes substitution,
wherein the halo group is attached to that carbon which has least no. of alkyl groups attached to it.
(ii) Ethanol to but-1-yne
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.
In an aqueous solution, KOH almost completely ionizes to give OH− ions. OH− ion is a strong nucleophile,
which leads the alkyl chloride to undergo a substitution reaction to form alcohol.
R-Cl + KCl (aqueous) --> R-OH + KCl
Alkyl Chloride Alcohol
On the other hand, an alcoholic solution of KOH contains alkoxide (RO−) ion, which is a strong base.
Thus, it can abstract hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.
OH− ion is a much weaker base than RO− ion. Also, OH− ion is highly solvated in an aqueous solution and
as a result, the basic character of OH− ion decreases. Therefore, it cannot abstract hydrogen from the β-carbon.
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?
There are two primary alkyl halides having the formula, C4H9Br. They are n − butyl bromide and isobutyl bromide.
Therefore, compound (a) is either n−butyl bromide or isobutyl bromide.
Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C8H18,
which is different from the compound formed when n−butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.
Thus, compound (d) is 2, 5−dimethylhexane.
It is given that compound (a) reacts with alcoholic KOH to give compound (b).
Hence, compound (b) is 2−methylpropene.
Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a).
Hence, compound (c) is 2−bromo−2−methylpropane.
What happens when:-
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) Bromobenzene is treated with Mg in the presence of dry ether,
(iii) Chlorobenzene is subjected to hydrolysis,
(iv) Ethyl chloride is treated with aqueous KOH,
(v) Methyl bromide is treated with sodium in the presence of dry ether,
(vi) Methyl chloride is treated with KCN??
(i) When n−butyl chloride is treated with alcoholic KOH, the formation of but−l−ene takes place.
This reaction is a dehydrohalogenation reaction.
(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.
(iii) Chlorobenzene does not undergo hydrolysis under normal conditions.
However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K
and a pressure of 300 atm to form phenol.
(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.
CH3 – CH2 –Cl --> CH3 –CH2-OH + KCl (In presence of KOH (alcoholic) & hydrolysis)
Ethyl chloride Ethanol
(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed.
This reaction is known as the Wurtz reaction.
2CH3 –Br + KCN --> CH3-CN + KCl (In presence of Dry ether and Wurtz reaction)
Methyl bromide Ethane
(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.
CH3 –Cl + KCN --> CH3-CN + KCl (Nucleophilic substitution)
Methyl chloride Methyl cyanide