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Alkyl Halides & Aryl Halides
Victor Grignard
François Auguste Victor Grignard ( 1871 - 1935 ) was a Nobel Prize-winning French
chemist.
He is most noted for devising a new method for creating carbon-carbon bonds (i.e. an
addition reaction) in organic synthesis (Original publication: V. Grignard, Compt. Rend.
Vol. 130, p. 1322 (1900). The synthesis occurs in two steps:
- Synthesis of the Grignard reagent: an organomagnesium compound (the Grignard
reagent) is made reacting an organohalide (R-X, where R stands for some alkyl, acyl,
or aryl radical and X is a halogen such as usually bromine or iodine) with magnesium
metal dissolved in diethyl ether. The resulting compound, named a Grignard reagent,
has the general chemical formula R-Mg-X.
- Attack on the carbonyl: A ketone or an aldehyde (both contain a carbonyl group) is
added to the solution containing the Grignard reagent. The carbon atom that is bonded
to the Mg atom bonds to the carbonyl carbon atom by nucleophilic addition, with the
formation of a new compound, which is an alcohol.
The Grignard reaction is an important means of making larger organic compounds from
smaller starting materials. By careful selection of the starting materials, a wide variety of
compounds can be made by this reaction. For this work, Grignard was awarded the Nobel
Prize in Chemistry in 1912 jointly with fellow Frenchman Paul Sabatier.
Introduction
Alkyl Halides are compounds in which a halogen atom is attached to carbon. For example,
H C Cl
H
H
Methyl chloride
H C C Br
H
H
Ethyl bromide
H
H
They have the general formula
R X or C X
Where R - alkyl group; X = Cl, Br, I or F. The halogen atom bonded to carbon is the
functional group of alkyl halides.
Alkyl halides are classified as Primary (1°), Secondary (2°), or Tertiary (3°),
depending upon whether the X atom is attached to a primary, secondary, or a tertiary
carbon.
Primary carbon
R C X
H
H
Secondary carbon
R C X
R
H
Secondary carbon
R C X
R
R
1° Alkyl Halide 2° Alkyl Halide 3° Alkyl Halide
Alkyl halides are among the most useful organic compounds. They are frequently used
to introduce alkyl groups into other molecules.
5 .1 Structure
Let us consider methyl chloride (CH 3
Cl) for illustrating the orbital make up of alkyl
halides in methyl chloride, the carbon atom is sp
hybridized. The chlorine atom has a
half-filled p orbital in valence shell. The C Cl bond is formed by the overlap of an sp
orbital of carbon and the half-filled p orbtial of chlorine atom shown in figure. Each C H
bond is formed by the overlap of an sp
orbital of carbon
C
H
Cl
H
sp -p
3
s-sp
3
H
C
H
H
H
Cl 109°
Figure: Orbital structure of Methyl chloride
and the s orbital of hydrogen. All bonds are bonds. The H C H and H C Cl bond
angles are approximately tetrahedral.
5 .2 Nomenclature
Alkyl halides are named in two ways
(1) Common system: In this system the alkyl group attached to the halogen atom is
named first. This is then followed by an appropriate word chloride, bromide, or
fluoride. Notice that the common names of alkyl halides are TWO-WORD names.
CH Br 3
CH CH Cl 3 2
CH CH 3
CH 3
Br
Methyl bromide Ethyl chloride Isopropyl bromide
(2) IUPAC system: The IUPAC names of alkyl halides are obtained by using the
following rules:
R OH + H X R X + H O 2
Alcohol
Alkyl halide
CH CH OH + HCl 3 2 CH CH Cl + H O 3 2 2
Ethyl alcohol
ZnCl 2
Ethyl chloride
CH CH OH + HBr 3 2 CH CH CH Br + H O 3 2 2 2
n-Propyl alcohol n-Propyl bromide
3 2 3 2 2
n propyl alcohol n propyl bromide
CH CH OH HBr CH CH Br H O
(4) Action of Phosphorus Halides on Alcohols. Alcohols react with phosphorus halides
(PX
or PX 3
) to form alkyl halides.
R OH + PX (or PX ) 5 3 R X
Alcohol Alkyl halide
2CH CH OH + PCl 3 2 5 2CH CH Cl + POCl + H O 3 2 3 2
Ethyl alcohol Ethyl chloride
3CH CH OH + PBr 3 2 3 3CH CH Br + H PO 3 2 3 3
Ethyl bromide
3CH OH + PI 3 3 3CH I + H PO 3 3 3
Methyl iodide
PBr 3
or PI 3
are produced in situ by the addition of Br 2
and I 2
to red phosphorus.
2 3
2 3
2P 3Br 2PBr
2P 3I 2PI
(5) Action of Thionyl chloride on alcohols. Alcohols react with thionyl chloride (SOCl 2
in the presence of pyridine to produce alkyl chlorides. Pyridine (C 5
H
N) absorbs
hydrogen chloride as it is formed.
R OH + SOCl 2
R Cl + SO + HCl 2
Alcohol
pyridine
Thionyl
chloride
CH CH OH + SOCl 3 2 2 CH CH Cl + SO + HCl 3 2 2
Ethyl Alcohol
pyridine
Ethyl chloride
Alkyl chloride
(6) Halogen Exchange reaction: This reaction is particularly suitable for preparing alkyl
iodides. The alkyl bromide or chloride is heated with a concentrated solution of
sodium iodide in acetone.
CH CH Br + NaI 3 2
CH CH I + NaBr 3 2
Ethyl bromide
acetone
Ethyl iodide
Alkyl fluorides are also prepared by treating an alkyl chloride or bromide with
inorganic fluorides.
2CH Cl + Hg F 3 2 2
2CH F + Hg Cl 3 2 2
Methyl chloride
acetone
Methyl fluoride
5 .4 Physical Properties
(1) CH
Cl, CH 3
Br, CH 3
F and CH 3
CH
Cl are gases at room temperature. Other alkyl
halides upto C 18
are colourless liquids. Those beyond C 18
are colourless solids.
(2) Alkyl halides are insoluble in water but soluble in organic solvents. The insolubility
in water is due to their inability to form hydrogen bonds with water.
(3) Alkyl bromides and iodides are denser than water. Alkyl chlorides and fluorides are
lighter than water.
(4) Alkyl halides have higher boiling points than alkanes of comparable molecular weight.
For a given halogen atom, the boiling points of alkyl halides increase with the increase
in the size of the alkyl group. For a given alkyl group, the boiling points of alkyl
halides follow the order RI>RBr>RCl>RF.
5 .5 Chemical Properties
Alkyl halides are very reactive compounds. They undergo substitution, elimination and
reduction reactions. Alkyl halides also react with metals to form organometallic
compounds.
HSAB (Hard And Soft Acid-Base) Principle
According to hard and soft acid-base principle of Pearson, hard acids are those species,
which have less tendency to accept an electron pair (like H
, Li
, Mg
2+
, Cr
3+
, Al
3+
, Al
3+
etc.) and hard bases are those species, which have less tendency to donate electron pair
(like F¯, O
2 –
etc.) A hard base prefers a hard acid whereas a soft base prefers a soft acid.
Basicity And Nucleophilicity
A negatively charged species can function as nucleophile as well as like base but its
nucleophilicity and basicity are different. Nucleophilicity of the species is the ability of the
species to attack an electrophilic carbon while basicity is the ability of the species to
remove H
from an acid. Let us have a species, B¯. Its function as a nucleophile is shown
as
C
B C
L
B
and its role as base is indicated as
B
The nucleophilicity is determined by the kinetics of the reaction, which is reflected by
its rate constant (k) while basicity is determined by the equilibrium constant, which is
reflected by its K b
The order of nucleophilicity of different species depends on the nature of solvent used.
For instance, let us take F¯, Cl¯, Br¯ and I¯ with their counter cation as Na
and see their
nucleophilicity order in different solvents. There are four categories of solvents, namely
(ii) R
or
R
RNu
+ X¯
or
R: X + Nu RNu
X¯
Rate = K [RX] (
st
order) = K[RX] [:Nu¯] (
nd
order)
TS of slow step HC
3
H C
3
X
+
H 3
C Nu C X
CH
3
HC
3
CH 3
Stereochemistry Inversion and racemization Inversion (backside
attack)
Molecularity Unimolecular Bimolecular
Reactivity
Structure of R
Determining
factor
Nature of X
Solvent effect on
rate
3° > 2° > 1° > CH
3
Stability of R
Rl > RBr > RCl > RF
Rate increases in polar
solvent
CH
3
Steric hindrance in R
group
Rl > RBr > RCl > RF
With Nu¯ there is a
large rate increase in
polar aprotic solvents.
Effect of
nucleophile
Rate depends on
nucleophilicity
I¯ > Br¯ > Cl¯ ; RS¯ >
RO¯
Catalysis Lewis acid, eg. Ag
, AlCl 3
ZnCl 2
None
Competitive
reactoin
Elimination, rearrangement Elimination
The S N
2 Reaction
Mechanism and Kinetics
The reaction between methyl bromide and hydroxide ion to yield methanol follows second
order kinetics; that is, the rate depends upon the concentration of both reactants.
3 3
CH Br OH CH OH Br
rate = K[CH 3
Br] [OH¯]
The simplest way to account for the kinetics is to assume that reaction requires a
collision between a hydroxide ion and a methyl bromide molecule. In its attack, the
hydroxide ion stays for away as possible from the bromine; i.e. it attacks the molecule from
the rear and begins to overlap with the tail of the sp
hybrid orbital holding Br. The
reaction is believed to take place as shown:
Br HO: HO
C Br
H O
(Inversion)
sp
2
(T.S.)
In the T.S. the carbon is partially bonded to both – OH and – Br; the C–OH bond is not
completely formed, the C–Br bond is not yet completely broken. Hydroxide has a
diminished – ve charge, since it has begun to share its electrons with carbon. Bromine has
developed a partial negative charge, since it has partly removed a pair of electrons from
carbon. At the same time, of course, ion dipole bonds between hydroxide ion and solvent
are being broken and ion-dipole bonds between bromide ion and solvent are being formed.
As the – OH becomes attached to C, 3 bonds are forced apart (120°) until they reach
the spike arrangement of the T.S; then as bromide is expelled, they move on to the
tetrahedral arrangement opposite to the original one.
Stereochemistry
Both 2-bromo-octane and 2-octanol are chiral
Br
H 13
C 6
H
OH
H 13
C 6
H
H 3
C H 3
C
(2S)-2-bromooctane (2S)-octan-2-ol
The (–) bromide and the (–) alcohol have similar configurations, i.e. – OH occupies the
same relative position in the (–) alcohol as – Br does in the bromide.
When (–)- 2 - bromooctane is allowed to react with sodium hydroxide under S N
2 conditions,
(+)- 2 - octanol is obtained
Br
H 13
C 6
H
NaOH
HO
C 6
H 13
CH 3
H
H 3
C
(2S)-2-bromooctane (2R)-octan-2-ol
S N
2
In Fisher projection the above reaction can be represented as follows
Br
C 6
H 13
H
CH 3
NaOH
H
C 6
H 13
H O
CH 3
S
N
2
(2R)-octan-2-ol
(2S)-2-bromooctane
We see that – OH group has not taken the position previously occupied by – Br; the
alcohol obtained has a configuration opposite to the bromide. A reaction that yields a
If the attack were purely random, we would expect euqal amounts of two isomers; i.e. we
would expect ony the racemic modification. But the product is not completely racemized,
for the inverted product exceeds its enantiomer.
We can say in contrast S N
2 reaction, which proceeds with complete inversion; an S N
reaction proceeds with racemizatoin though may not be complete.
Br
R'
R R''
R''
R
R'
(a)
attack from
top
(b)
attack from
bottom
OH
R R''
R'
(Inversion)
OH
R'
R R''
Retention
OH
OH
Both enantiomers may be
formed in equal amounts
or one may exceed the other
sp
2
r.d.s formation of carbonium ion.
Reactivity of an alkyl halide depends upon the stable carbonium ion it can form.
In S N
1 reactions the order of reactivity of alkyl halides is Allyl, benzyl > 3° > 2° > 1° >
CH
X.
Some of the important nucleophilic substitution reactions of alkyl halides are described
below:
(1) Reaction with aqueous KOH: Alkyl halides react with aqueous potassium hydroxide
to form alcohols. The halogen atom is substituted by - OH group.
CH I + KOH 3 CH OH + KI 3
Methyl iodide
H O 2
Methyl alcohol
CH CH Br + KOH 3 2 CH CH OH + KBr 3 2
Ethyl bromide
H O 2
Ethyl alcohol
MECHANISM:
In the above reaction OH¯ is the nucleophile
HO: + CH CH Br 3 2
CH CH OH 3 2
Ethyl alcohol
S 2 N
(2) Reaction with Moist Silver Oxide: Alkyl halides on treatment with a suspension of
silver oxide in moist ether produce alcohols. Halogen atom is substituted by - OH
group.
2 2
Ag O H O 2AgOH
CH CH Br + AgOH 3 2
CH CH OH + KBr 3 2
Ethyl bromide
Ethyl alcohol
(3) Reaction with sodium alkoxides: Alkyl halides react with sodium alkoxides (RONa)
to form ethers. Sodium alkoxides are prepared by dissolving metallic sodium in excess
of the appropriate alcohol. For example,
CH CH OH + Na 3 2 CH CH ONa + H 3 2 2
Sodium ethoxide
CH CH Br + NaOCH CH 3 2 2 3
CH CH OCH CH + NaBr 3 2 2 3
Diethyl ether Ethyl bromide
This method of making ethers is called Williamson Ether Synthesis.
MECHANISM:
In the above reaction CH 3
CH
O : is the nucleophile.
CH CH O: + CH CH Br 3 2 3 2
Diethyl ether
S 2 N
CH CH OCH CH + :Br 3 2 2 3
Ethers can also be produced by heating an alkyl halide with dry silver oxide.
2CH I + Ag O 3 2
CH O CH + 2AgI 3 3
Methyl iodide Diethyl ether
(4) Reaction with Ammonia: When an alkyl halide is heated with an alcoholic solution
of ammonia in a sealed tube, alkylation of ammonia takes place. A mixture of different
classes of amines results.
CH CH Br + NH 3 2 3
CH O CH + 2AgI 3 3
Ethyl bromide (in ethanol) Ethylamine(1°)
CH CH NH + CH CH Br 3 2 2 3 2 (CH CH ) NH + HBr 3 2 3
Diethylamine (2°)
(CH CH ) NH + CH CH Br 3 2 2 3 2
(CH CH ) N + HBr 3 2 2
Triethylamine (3°)
(CH CH ) N + CH CH Br 3 2 3 3 2 (CH CH ) NBr 3 2 4
Tetraethylammonium
bromide (4°)
3 2 3 3 2 3 2 4
Tetraethylammonium
bromide (4 )
(CH CH ) N CH CH Br (CH CH ) NBr
Each amine formed exists in equilibrium with its salt. For example,
3 2 2 3 2 3
Ethylamine Ethylammonium
bromide
CH CH NH HBr CH CH NH Br
(5) Reaction with Sodium Cyanide: Alkyl halides react with sodium cyanide in a suitable
solvent (generally aqueous ethanol) to form alkyl cyanides or nitriles. Halogen atom is
replaced by CN group.
CH CH Br + NaCN 3 2
CH CH CN + NaBr 3 2
aqueous
Ethyl
bromide
Ethyl cyanide
ethanol,
MECHANISM:
Cyanide ion is an excellent nucleophile. It attacks ethyl bromide by an S N
mechanism to form ethyl cyanide.
NaCN Na + :CN
solvent
:CN CH CH Br
3 2
S 2 N
CH CH CN + Br: 3 2
Ethyl cyanide
CH CH + KOH 2 2
ethanol
H Br
CH = CH + KBr + H O 2 2 2
Ethyl bromide Ethylene
CH CH CH + KOH 2 2
ethanol
H Br
CH CH = CH + KBr + H O 3 2 2
1-Bromopropane Propene
MECHANISM:
In ethanol an equilibrium occurs between the solvent and potassium hydroxide to produce
potassium ethoxide.
CH CH OH + KOH 3 2
CH CH O K + H O 3 2 2
+
Ethanol
Potassium ethoxide is a strong base. It favours elimination and substitution reactions.
There is always a competition between elimination and substitution reactions. For example,
ethyl bromide on treatment with alcoholic KOH can give either ethylene or diethyl ether.
The attacking nucleophile is CH 3
CH
O:
ELIMINATION:
H C C H
H H
H Br
H C = C H + CH CH OH + Br: 3 2
H H
CH CH O: 3 2
Ethylene
SUBSTITUTION:
CH CH Br 3 2
CH CH O 3 2
CH CH + Br: 2 3
Diethyl ether
:OCH CH 2 3
The ratio of the elimination to substitution product depends on the structure of the
alkyl halide and experimental conditions. Primary and secondary alkyl halides undergo
dehydrohalogenation by E2 mechanism. Tertiary alkyl halides do so by E1 mechanism.
Saytzeff Rule: If the dehydrohalogenation of an alkyl halide can yield more than one
alkene, then according to the Saytzeff rule, the main product is the most highly substituted
alkene. For example, two alkenes are possible when 2-bromobutane is heated with
alcoholic KOH.
H C H 3
C C C
H H H
H Br H
2-Bromobutane
CH = CH 2
CH CH 2 3
1-Butene (20%)
CH CH = 3
CH CH 3
2-Butene (80%)
Notice that the major product is 2-butene, a disubstituted alkene.
(12) Reaction with Mg-alkyl halides react with magnesium metal in dry ether to form
Grignard reagents.
Ether
3 3
Methyl iodid e Methylm agnesiu m iodid e
CH I Mg CH MgI
Ether
3 2 3 3
Ethyl bromide Ethylmagnesiu m bromide
CH CH Br Mg CH CH MgBr
(13) Reaction with Lithium-Alkyl halides react with lithium in dry ether to form
alkyllithiums.
CH CH Br + 2Li 3 2
CH CH Li + LiBr 3 2
Ethyllithium Ethyl bromide
Ether
(14) Alkyllithiums behave in the same way as Grignard reagents, but with increased
reactivity. Wurtz reaction: Alkyl halides react with metallic sodium in dry ether to give
alkanes with double the number of carbon atoms.
CH CH Br + 2Na + BrCH CH 3 2 2 3
Ether
CH CH CH CH + 2NaBr 3 2 2 3
n-Butane Ethyl bromide
(15) Halogenation: Alkyl halides react with Cl 2
or Br 2
in the presence of UV light or at
high temperature to form polyhalogenation derivatives. For example, methyl chloride
react with chlorine to yield a mixture of methylene dichloride, chloroform and carbon
tetrachloride.
Cl Cl Cl 2 2 2
3 2 2 3 4 UV light
CH Cl CH Cl CHCl CCl
(16) Friedel-Crafts Alkylation: Alkyl halides react with benzene in the presence of
anhydrous AlCl 3
to form alkylbenzenes.
Ethyl bromide
CH CH Br + C H 3 2 6 5
AlCl 3
C H CH CH + HBr 6 5 2 3
Benzene Ethylbenzene
5 .7 Nucleophilic Substitution In Neopentyl Halides
Although neopentyl halides is a 1° halides, it does not undergo nucleophilic substitution by
S
N 2 mechanism because it is highly sterically crowded to be able to form a transition state.
So, neopentyl halide has a greater tendency to undergo nucleophilic substitution by S N
mechanism. Although the initially formed carbocation is a primary carbocation, it
rearranges to give a more stable carbocation, which is then attacked by nucleophile to give
corresponding product. For example,
ILLUSTRATIONS
Illustration 1
Give the organic products of the following reactions
(a)
acetone
O
||
n Pr Br N O
(b)
acetone
i Pr Br [SC N] (thiocyanate)
(c)
acetone
EtBr [SC N] (thiocyanate)
(d)
acetone
2 2 2
ClCH CH CH I CN (one mole each)
(e)
H
2 2 2 2 2
H NCH CH CH CH Br
Solution
The nucleophiles in (a), (b) and (c) are ambident since they each have more than one
reactive site. In each case, the more nucleophilic atom reacts even through the other atom
may bear a more negative charge.
(a) n-PrNO 2
(b) i-PrSCN
(c) [EtSSO 3 ]¯
(d) ClCH 2 CH 2 CH 2 CN (I¯ is a better leaving group than Cl¯)
(e)
N
H
. When the nucleophile and leaving groups are part of the same molecule, an
intramolecular displacement occurs if a three, or five – or a six-membered ring can
form.
Illustration 2
Compare the rates of S N 1 and S N 2 reactions of cyclopropyl and cyclopentyl chloride.
Solution
Cyclopropyl chloride is much less reactive than cyclopentyl chloride in each type of
reaction because the sp
2
hybridised carbon (120° bond angle) created in each transition
state augments the ring strain.
Illustration 3
Explain the fact that a small amount of NaI catalyzes the general reaction
R – Cl + R O : ¯ Na
R – OR + NaCl
Solution
With I¯ ion, the overall reaction occurs in two steps, each of which is faster than the
uncatalyzed reaction.
Step 1. R – Cl + I¯ R – I + Cl¯. This step is faster because I¯, a soft base has
more nucleophilicity than OR¯, a hard base.
Step 2. R – I + R O : ¯ R – OR + I¯. This step is faster because I¯ is a better
leaving group than Cl¯.
PRACTICE EXERCISE
1. Give the products of the following displacement reactions.
(a) (R) – CH 3 CHBrCH 2
CH
3
(b) (S) – CH 3 CHBrCH 2
CH
3
(c) cis- 4 - iodoethylcyclohexane + OH¯
(d)
CH 3
H
CO
2
Et + CN (S) – Br
2. Optically active 2-iodobutane on treatment with NaI in acetone gives a product, which
does not show optical acitivity. Explain why?
Answers
1. (a) (S) – CH 3 CH(OMe)CH 2
CH
3 (b) (R) – CH 3 CH(OEt)CH 2
CH
3
(c) trans- 4 - ethylcyclohexanol (d)
CH
3
H
CN (S) – EtCO 2
CH
3
C
2
H
5
I H
CH
3
C 2
H 5
I H
Thus they undergo both substitution and elimination reactions.
(1) Hydrolysis with aqueous NaOH or KOH, vic-Dihalides on heating with aqueous
sodium hydroxide give glycols.
CH Cl 2
CH Cl 2
3
CH OH 2
CH OH 2
3
1,2-Dichloroethane Ethylene glycol
H O 2
gem-Dihalides on hydrolysis with aqueous KOH gives an aldehyde or a ketone.
CH C 3
H CH C H 3
OH
1,1-Dichloroethane (unstable)
Cl
Cl
2KOH, H O
-2KCl
2
OH
-H O 2 CH C H 3
O
Acetaldehyde
CH C 3
CH 3
CH C CH 3 3
OH
(unstable) 2,2-Dibromopropane
Br
Br
2KOH, H O
-2KBr
2
OH
-H O 2 CH C CH 3 3
O
Acetone
This reaction is used to distinguish vic-dihalides from gem-halides. Notice that vic-
dihalides on hydrolysis give glycols while gem-dihalides give aldehydes or ketones.
(2) Reaction with zinc: Dehalogenation vic-Dihalides and gem-dihalides on treatment with
zinc dust in methanol give alkenes.
CH Br 2
CH Br 2
CH 2
CH 2
1,2-Dibromoethane Ethylene
methanol
1,3- to 1,6-Dihalides give cycloalkanes.
CH Br 2
CH Br 2
H C 2
methanol
CH 2
CH 2
H C 2
(3) Reaction with Alcoholic KOH: Dehydrohalogenation, vic-Dihalides and gem-dihalides
on treatment with alcoholic potassium hydroxide give alkynes.
H C C H + 2KOH
1,1-Dichloroethane
( gem -dihalide)
Cl
Cl
ethanol
Acetylene
H
H
HC CH + 2KCl + H O 2
H C C H + 2KOH
1,1-Dichloroethane
( vic -dihalide)
Cl
Cl
ethanol
Acetylene
H
H
HC CH + 2KCl + H O 2
5 .1 1 Trihalogen Derivatives
Trihalogen derivatives are compounds obtained by replacing three hydrogen atoms of a
hydrocarbon by three halogen atoms. The presence of three identical halogen atoms is
indicated by the prefix tri - and the position numbers.
H C Cl
Cl
H
H C Br
Br
Br
Trichloromethane
(Chloroform)
Tribromomethane
(Bromoform)
Triiodomethane
(Iodoform)
H C I
I
I
CHLOROFORM
Trichloromethane, CHCl 3
Chloroform is an important trihalogen derivative of methane. In the past of chloroform was
extensively used as a great anesthetic for surgery but it is rarely used for this purpose now
because it causes extensive liver damage.
Preparation, Chloroform is prepared
(1) From Ethyl Alcohol (or Acetone) and Bleaching powder. By heating ethyl alcohol or
acetone with bleaching powder, Ca(OCl 2
). The bleaching powder acts as source of
chlorine and calcium hydroxide. This method is used to make chloroform in the
laboratory and on commercial scale. Reaction of ethyl alcohol with bleaching powder
takes place by the following three steps.
Step 1: Oxidation
CH CH OH + Cl 3 2 2
CH CHO + 2HCl 3
Ethyl alcohol Acetaldehyde
Step 2: Chlorination
CH CHO + 3Cl 3 2
CCl CHO + 3HCl 3
Acetaldehyde Chloral
(Trichloroacetaldehyde)
Step 3: Hydrolysis
2CCl CHO + Ca(OH) 3 2
2CHCl + (HOOC) Ca 3 2
Chloral Chloroform Calcium formate
Reaction of acetone with bleaching powder takes place by the following two steps.
Step 1: Chlorination
CH COCH + 3Cl 3 3 2
CCl COCH + 3HCl 3 3
Acetone Trichloroacetone
Step 2: Hydrolysis
CCl COCH + Ca(OH) 3 3 2
Trichloroacetone
2CHCl + (CH COO) Ca 3 3 2
Chloroform Calcium acetate
Chemical Properties: The chemical properties of chloroform are as follows:
(1) Oxidation: Chloroform undergoes oxidation the presence of light and air to form
phosgene (carbonyl chloride).
CHCl + ½O 3 2
Cl C Cl + HCl
O
Phosgene Chloroform
