Samacheer Class 11 Chemistry - Haloalkanes and Haloarenes

i) CH 3 – CH = CH – Cl = Allylic halide
ii) C 6 H 5 CH 2 I = Benzylic halide
iii)
= Alkyl halide
iv) CH 2 = CH – Cl = Vinyl halide

• Chlorination of methane is a free radical substitution reaction.
• Before chlorine reacts with methane, the Cl-Cl single bond must break to form free radicals and this can only be done in the presence of ultraviolet light.
• In dark, chlorine-free radicals formation is not possible and so chlorination of methane is not possible in dark.
• The ultraviolet light is a source of energy and is being used to break of Cl-Cl and produce Cl free radical Free radicals which can attack methane. in dark, this is not possible.

Finkelstein reaction,
nCH 3 – CH 2 – CH 2 – Br + NaI
n – CH 3 – CH 2 – CH 2 – I + NaBr
n – propyl iodide n- propyl bromide

Haloarenes react with sodium metal in dry ether, two aryl groups combine to give biaryl products.
This reaction is called fittig reaction.
C 6 H 5 Cl + 2Na + Cl – C 6 H 5
C 6 H 5 – C 6 H 5 + 2NaCl
Chlorobenzene Biphenyl

Carbon halogen bond is a polar bond as halogens are more electronegative than carbon. The carbon atom exhibits a partial positive charge (δ + ) and halogen atom a partial negative charge (δ – )
The C -X bond is formed by overlap of sp 3 orbital of a carbon atom with half-filled p- orbital of the halogen atom. The atomic size of halogen increases from fluorine to iodine, which increases the C – X bond length. Larger the size, greater is the bond length, and the weaker is the bond formed. The bond strength of C – X decreases from C – F to C – I in CH 3 X.

Grignard reagents are mostly reactive and react with the source of the product to give hydrocarbons. Even alcohols, amines, H 2 O are sufficiently acidic to convert them to corresponding hydrocarbons.
R Mg X + H 2 O → RH +
Due to its high reactivity, it is necessary to avoid even traces of moisture from the Grignard reagent.

C 5 H 11 Br – Possible isomers
1. CH 3 – CH 2 – CH 2 – CH 2 – CH 2 – Br → 1 – bromo pentane
2.
→ 2 – bromo pentane
3.
→ 3 – bromo pentane
4.
→ 1 – bromo 2, 2 – dimethyl propane
5.
→ 1 – bromo 3 – methyl butane
6.
→ 2 – bromo 3 – methyl butane
7.
→ 2 – bromo 2 – methyl butane
8.
→ 1 – bromo 2- methyl butane
9.
→ (2S) – 1 – bromo 2 – methyl butane
10.
→ (2R) – 1 – bromo 2 – methyl butane

Haloalkanes are prepared by the following methods.
From alcohols: Alcohol can be converted into halo alkenes by reacting it with any one of the following reagents.
* Hydrogen halide
* Phosphorous halides
* Thionyl chloride.
a) Reaction with hydrogen halide:
Mixture of con. HCl and anhydrous ZnCl 2 is called Lucas Reagent.
The order of reactivity of halo acids with alcohol is in the order HI > HBr > HCl.
The order of reactivity of alcohols with halo acid is tertiary > secondary > primary.
b) Reaction with phosphorous halides:
Alcohols react with PX 5 or PX 3 to form haloalkanes.
Example:
CH 3 CH 2 OH + PCl 5 → CH 3 CH 2 Cl + POCl 3 + HCl
Ethane Chloro ethane
3CH 3 CH 2 OH + PCl 3 → 3 CH 3 CH 2 Cl + H 3 PO 3
Ethanol Chloro ethane
c) Reaction with Thionyl chloride(Sulphonyl Chloride)
CH 3 CH 3 OH + SOCl 2
CH 3 CH 2 Cl + SO 2 ↑ + HCl↑
Ethanol Chloro ethane

S N 1
S N 2
Rate law
Unimolecular (Substrate only)
Biomolecular (substrate and nucleophile)
“Big Barrier”
Carbocation stability
Steric hindrance
Alkyl halide (electrophile)
3° > 2° > 1°
1° > 2° > 3°
Nucleophile
Weak (generally neutral)
Strong (generally bearing a negative charge)
Solvent
Polar protic (e.g., alcohols)
Polar aprotic (e.g., DMSO, acetone)
Stereo Chemistry
Mix of retention and inversion
inversion

The halogen of haloarenes can be substituted by OH –, NH 2 – or CN – with appropriate nucleophilic reagents at high temperature and pressure.
Example:
(i) Chlorobenzene reacts with ammonium at 250 and at 50 atm to give aniline.
C 6 H 5 Cl + 2NH 3
C 6 H 5 NH 2 + NH 4 Cl
Chlorobenzene Aniline
(ii) Chlorobenzcne reacts with CuCN in presence of pyridine at 250 to give phenyl cyanide.
C 6 H 5 Cl + CuCN
C 6 H 5 CN + CuCl
Chlorobenzene Phenyl cyanide
(iii) Dows process:
C 6 H 5 Cl + NaOH
C 6 H 5 OH + NaCl
Chlorobenzene Phenol
This reaction is known as Dow’s process.

(i) t – butyl chloride reacts with aqueous KOH by S N 1 mechanism while n – butyl chloride reacts with S N 2 mechanism.
In general, the S N 1 reaction proceeds through the formation, of carbocation, The tert-butyl chloride readily loses Cl ion to form stable 3° carbocation. Therefore, it reacts with aqueous KOH by S N 1 mechanism as:
On the other hand, n-Butyl chloride does not undergo ionization to form n-Butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo reaction by an S N 2 mechanism, which occurs is one step through a transition state involving the nucleophilic attack of OH – ion from the backside with simultaneous expulsion of Cl – ion from the front side.
S N 1 mechanism follows the reactivity order as 3° > 2°> 1° while S N 2 mechanism follows the reactivity order as 1° > 2° > 3°. Therefore, tert-butyl chloride (3°) reacts by S N 1 mechanism while n-butyl chloride (1°) reacts by an S N 2 mechanism. (ii) p – dichlorobenzene has a higher melting point than those of o – and m – dichloro benzene. The higher melting point of p – isomer is due to its symmetry which leads to more close packing of its molecules in the crystal lattice and consequently strong intermolecular attractive force which requires more energy for melting. p – Dihalo benzene > o – Dichloro benzene> m – Dichioro benzene
Melting point: 323 K 256 K 249 K

a) Name of the product and write the equation for the reaction.
CH 3 I + Mg
CH 3 MgI
b) Why all the reagents used in the reaction should be dry? Explain.
All the reagents used in the reaction should be dry because reagent reacts with H20 to produce alkane. This is the reason that everything has to be very dry during the preparation of Grignard reagents.
CH 3 – MgI + H 2 O → CH 4 +
Methane
c) How is acetone prepared from the product obtained in the experiment?

i) Freon – 12 from Carbon tetrachloride:
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride.
CCl 4 + 2HF
2HCl + CCl 2 F 2
Carbon tetrachloride Freon – 12
ii) Carbon tetrachloride from carbon disulphide.
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl3 as catalyst giving carbon tetrachloride.
CS 2 + 3 Cl 2
CCl 4 + S 2 Cl 2
Carbon disulfide Carbon tetrachloride

The chloro fluoro derivatives of methane and ethane are called freons.
Nomenclature:
Freon is represented as Freon – cba
Where a = number of carbon atoms – 1;
b = number of hydrogen atoms + 1
a = total number of fluorine atoms
Uses:
* Freons are used as a refrigerant in refrigerators and air conditioners.
* It is used as a propellant for aerosols and foams
* It is used as a propellant for foams to spray out deodorants, shaving creams, and insecticides.

i) KNO 2:
Bromo ethane reacts with an alcoholic solution of NaNO2 or KNO2 to form ethyl nitrite.
CH 3 CH 2 Br + KNO 2 → CH 3 CH 2 – O – N = O + KBr
Bromoethane Ethyl nitrite
ii)AgNO 2:
Bromo ethane reacts with an alcoholic solution of AgNO2 to form nitroethane.
CH 3 CH 2 Br + AgNO 2 → CH 3 CH 2 NO 2 + AgBr
Bromoethane Nitroethane

In S N 1 reactions, if the alkyl halide is optically active, the product obtained in a racemic mixture. The intermolecular carbocation formed in slowest step being sp2 hybridized is planar species. Therefore the attack of the nucleophile OH on it, can occur from both the faces with equal case forming a mixture of two enantiomers. Thus S N 1 reaction of optically active alkyl halides is accompanied by racemization.

i) Raschig process:
Chloro benzene is commercially prepared by passing a mixture of benzene vapour, air and HCl overheated cupric chloride, this reaction is called the Raschig process,
ii) Dows Process:
C 6 H 5 Cl + NaOH
C 6 H 5 OH + NaCl
This reaction is known as Dows process.
iii) Darzen’s process:
CH 3 CH 2 OH + SOCl
CH 3 CH 2 Cl + SO 2 ↑ + HCl↑
Ethanol Chloro ethane
This reaction is known as Darzen’s process.

i) Acetic acid:
Solid carbon dioxide reacts with methyl magnesium iodide to form additional product which on hydrolysis yields acetic acid.
ii) Acetone:
Acetyl chloride reacts with methyl magnesium iodide and followed by acid hydrolysis to give acetone.
iii) Ethyl Acetate:
Ethyl chloroformate reacts with methyl magnesium iodide to form ethyl acetate.
iv) Isopropyl alcohol:
Aldehydes (Acetaldehyde) other than formaldehyde, react with methyl magnesium iodide to give additional product which on hydrolysis yields isopropyl alcohol.
v) Methyl cyanide:
Methyl magnesium iodide reacts with cyanogen chloride to give methyl cyanide.

:
i) CH 3 – CH = CH 2 + HBr
CH 3 – CH 2 – CH 2 – Br
Propene n – propyl bromide
ii) CH 3 – CH 2 – Br + NaSH
CH 3 – CH 2 – SH + NaBr
Propyl bromide Ethanethiol
iii) C 6 H 5 Cl (Chloro benzene) + Mg
C 6 H 5 MgCl (Phenyl magnesium chloride)
iv) CHCl 3 + HNO 3
CCl 3 NO 2 + H
Chloroform Chloropicrin
v) CCl 4 (Carbon tetrachloride) + H 2 O
COCl 2 (Carbonyl chloride) + 2HCl

i) DDT:
DDT can be prepared by heating a mixture of chlorobenzene with chloral (Trichloro acetaldehyde) in the presence of con.H 2 SO 4.
ii) Chloroform:
Preparation:
Chloroform is prepared in the laboratory by the reaction between ethyl alcohol with bleaching powder followed by the distillation of the product chloroform. Bleaching powder act as a source of chlorine and calcium hydroxide. This reaction is called the haloform reaction. The reaction proceeds in three steps as shown below.
Step – 1: Oxidation
CH 3 CH 2 OH + Cl 2 → CH 3 CHO + 2HCl
Ethyl alcohol Ethanal (Acetaldehyde)
Step – 2: Chlorination
CH 3 CHO + 3Cl 2 → CCl 3 CHO + 3HCl
Acetaldehyde Trichloro acetaldehyde
Step – 3: Hydrolysis
2CCl 3 CHO + Ca(OH) 2 → 2CHCl 3 + (HCOO) 2 Ca
Chloral chloroform
iii) Biphenyl:
Chloro benzene reacts with sodium metal in dry ether, to give biphenyl. This reaction is called a fitting reaction.
C 6 H 5 Cl + 2 Na + Cl – C 6 H 5
C 6 H 5 – C 6 H 5 + 2NaCl
Chloro benzene Biphenyl
iv) Chloropicrin:
Chloroform reacts with nitric acid to form chloropicrin. (Trichloro nitromethane)
CHCl 3 + HNO 3
CCl 3 NO 2 + H 2 O
Chloroform Chloropicrin
v) Freon – 12
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of the catalytic amount of antimony pentachloride
CCl 4 + 2 HF
2 HCl + CCl 2 F 2
Carbon tetrachloride Freon – 12