Chapter 3 Alkenes

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ORGANIC CHEMISTRY 1
CHM 207
CHAPTER 3:
ALKENES
NOR AKMALAZURA JANI

ALKENES
•Also called olefins
•Contain at least one carbon-carbon double bond
(C=C)
•General formula, C
n
H
2n
(n=2,3,…)
•Classified as unsaturated hydrocarbons
(compound with double or triple carbon-carbon
bonds that enable them to add hydrogen atoms.
•sp
2
-hybridized
•For example:
C
2
H
4
- ethylene
CH
2CH
2

Naming AlkenesNaming Alkenes

IUPAC RULES
RULE 1. Select the longest continuous
carbon chain that contains a double bond.
This chain
contains 6
carbon atoms

RULE 2. Name this compound as you would an
alkane, but change –ane to –ene for an alkene.
This chain
contains 8
carbon atoms
This is the longest
continuous chain.
Select it as the parent
compound.
Name the parent
compound octene.

RULE 3. Number the carbon chain of the
parent compound starting with the end nearer to
the double bond. Use the smaller of the two
numbers on the double-bonded carbon to
indicate the position of the double bond. Place
this number in front of the alkene name.

IUPAC RULES
This end of the chain is closest to the
double bond. Begin numbering here.

The name of the parent compound is
1-octene.
IUPAC RULES
8
7
4 3 21
6
5

RULE 4. Branched chains and other groups are
treated as in naming alkanes. Name the
substituent group, and designate its position on
the parent chain with a number.

IUPAC RULES
This is an ethyl group.
8
7
4 3 21
6
5
The ethyl group is attached to carbon 4.
4
4-ethyl-1-octene

•A compound with more than one double bond.
-Two double bond: diene
-Three double bond: triene
-Four double bond: tetraene
* Numbers are used to specify the locations of the double bonds.
CH
2
CCCH
2
HH
IUPAC names: 1,3-butadiene 1,3,5-heptatriene
new IUPAC names: buta-1,3-diene hepta-1,3,5-triene
1 2 34
CH
3CCCCCCH
2
12
347 65
HHHHH
12
3
4
7
65
8
IUPAC names: 1,3, 5, 7-cyclooctatetraene
new IUPAC names: cycloocta-1,3,5,7-tetraene

ALKENES AS SUBSTITUENTS
•Alkenes names as substituents are called alkenyl groups.
•Can be named systematically as ethenyl, propenyl, etc. or
by common names such as vinyl, ally, methylene and
phenyl groups.
CH
2
methylene group
(methylidene group)
-CH=CH
2
vinyl group
(ethenyl group)
3-methylenecyclohexene
CHCHCH
2CHCH
2 CH
2
CH=CH
2
3-vinyl-1,5-hexadiene
3-vinylhexa-1,5-diene
-CH
2-CH=CH
2
allyl group
(2-propenyl group)

CYCLOALKENES
•Contains C=C in the ring
cyclopropenecyclobutene cyclohexenecyclopentene
•Nomenclature of cycloalkenes:
-Similar to that alkenes
-Carbons atoms in the double bond are designated
C1 and C2
CH
3
1
2
3
4
5
6
1
23
4
5
1-methylcyclohexene 1,5-dimethylcyclopentene

NOMENCLATURE OF cis-trans
ISOMERS
• cis – two particular atoms (or groups of atoms) are
adjacent to each other
• trans – the two atoms (or groups of atoms) are
across from each other
CC
H
3C
H
CH
2CH
3
H
CC
H
3C
H
H
CH
2CH
3
cis-2-pentene trans-2-pentene

PHYSICAL PROPERTIES OF ALKENESPHYSICAL PROPERTIES OF ALKENES
• Boiling points and densities:
- Most physical properties of alkenes are similar to
those alkanes.
- Example: the boiling points of 1-butene, cis-2-butene,
trans-2-butene and n-butane are close to 0
o
C.
- Densities of alkenes: around 0.6 or 0.7 g/cm
3
.
- Boiling points of alkenes increase smoothly with
molecular weight.
- Increased branching leads to greater volatility and
lower boiling points.

• Polarity:
- relatively nonpolar.
- insoluble in water but soluble in non-polar
solvents such as hexane, gasoline, halogenated
solvents and ethers.
- slightly more polar than alkanes because:
i) electrons in the pi bond is more polarizable
(contributing to instantaneous dipole moments).
ii) the vinylic bonds tend to be slightly polar
(contributing to a permanent dipole moment).

Alkyl groups are electron donating toward double
bond, helping to stabilize it. This donating slightly
polarizes the vinylic bond, with small partial positive
charge on the alkyl group and a small negative charge
on the double bond carbon atom.
For example, propene has a small dipole moment of
0.35 D.
propene, μ = 0.35 D
CC
H
3C
H
H
H
CC
H
3C
H
CH
3
H
CC
H
3C
H
H
CH
3
Vector sum =
propene, μ = 0.33 D
cis-2-butene, bp 4
o
C
Vector sum = 0
propene, μ = 0
trans-2-butene, bp 1
o
C
Vinylic bonds

•In a cis-disubstituted alkene, the vector sum of the two
dipole moments is directed perpendicular to the
double bond.
•In a trans-disubstituted alkene, the two dipole
moments tend to cancel out. If an alkene is
symmetrically trans-disubstituted, the dipole moment
is zero.
Vector sum =
propene, μ = 0.33 D
cis-2-butene, bp 4
o
C
Vector sum = 0
propene, μ = 0
trans-2-butene, bp 1
o
C

•Cis- and trans-2-butene have similar van der Waals
attractions, but only cis isomer has dipole-dipole
attractions.
•Because of its increased intermolecular attractions,
cis-2-butene must be heated to a slightly higher
temperature (4
o
C versus 1
o
C) before it begins to boil.
Vector sum =
propene, μ = 0.33 D
cis-2-butene, bp 4
o
C
Vector sum = 0
propene, μ = 0
trans-2-butene, bp 1
o
C

PREPARATION OF ALKENES
•Alkenes can be prepared in the following ways:
i) Dehydration of alcohols
conc. H
2
SO
4
R-CH
2
-CH
2
-OH R-CH=CH
2
+ H
2
O
ii) Dehydrohalogenation of haloalkanes
NaOH/ethanol
R-CH
2
-CH
2
-X
reflux
R-CH=CH
2
+ HX
NaOH can be replaced by KOH

•Saytzeff rule:
- A reaction that produces an alkene would favour the
formation of an alkene that has the greatest number of
substituents attached to the C=C group.
CH
3CH
2-CH-CH
3
OH
H
+
H
+
CH
3CH=CH-CH
3 + H
2O
CH
3CH
2-CH=CH
2 + H
2O
2-butanol
2-butene
major product
1-butene
CH
3CH-CH-CH
2
BrH H
KOH CH
3CH=CH-CH
3
CH
3CH
2CH=CH
2
alcohol
reflux
2-bromobutane
2-butene
(major product)
1-butene
Dehydration of alcohols
Dehydrohalogenation of haloalkanes

REACTIVITY OF REACTIVITY OF
ALKENESALKENES
More reactive than alkanes because:
•A carbon-carbon double bond consists of a σ and a π
bond. It is easy to break the π bond while the σ bond
remains intact.
•The π electrons in the double bond act as a source of
electrons (Lewis base). Alkenes are reactive towards
electrophiles which are attracted to the negative charge
of the π electrons.
•π bond will broken, each carbon atom becomes an
active site which can form a new covalent bond with
another atom. One π bond is converted into 2 σ bonds.

REACTIONS OF ALKENES
Catalytic hydrogenation:
- hydrogenation: addition of hydrogen to a double bond and
triple bond to yield saturated product.
- alkenes will combine with hydrogen in the present to
catalyst to form alkanes.
CC HH CC
HH
Pt or Pd
25-90
o
C
-Plantinum (Pt) and palladium (Pd) – Catalysts
-Pt and Pd: temperature 25-90
o
C
-Nickel can also used as a catalyst, but a higher temperature
of 140
o
C – 200
o
C is needed.

H
2CCH
2 H
2
Pt
CH
3CH
2CH
2CH
2CHCH
2 H
2
Pt
H
3CCH
3
CH
3CH
2CH
2CH
2CH
2CH
3
EXAMPLES:
ethylene
ethane
low pressure
low pressure
hexene hexane

Addition of halogens:
i) In inert solvent:
- alkenes react with halogens at room temperature and in dark.
- the halogens is usually dissolved in an inert solvent such as
dichloromethane (CH
2
Cl
2
) and tetrachloromethane (CCl
4
).
- Iodine will not react with alkenes because it is less reactive
than chlorine and bromine.
- Fluorine is very reactive. The reaction will produced
explosion.
CC XX CC
XX
inert solvent
XX= halogen such as Br
2 or Cl
2
Inert solvent = CCl
4 or CH
2Cl
2

EXAMPLES:
CC
HH
H H BrBr
Br
2
Br
Br
CCl
4
CH
3CH=CH
2Cl
2
CCl
4
CH
3CH
Cl
CH
2
Cl
CC
Br
HH
Br
H H
inert solvent (CCl4)
ethene
1,2-dibromoethane
* the red-brown colour of the bromine solution will fade and the
solution becomes colourless.
cyclohexene
1,2-dibromocyclohexane
propene
1,2-dichloropropane

Addition of halogens:
ii) In water / aqueous medium:
- chlorine dissolves in water to form HCl and chloric (l)
acid
(HOCl).
Cl
2 (aq) + H
2O(l) HCl(aq) + HOCl (aq)
- same as bromine
Br
2 (aq) + H
2O(l) HBr(aq) + HOBr(aq)
* Reaction of alkenes with halogens in water (eg. chlorine
water and bromine water) produced halohydrins (an
alcohol with a halogen on the adjacent carbon atom).

EXAMPLES:
CH
3CH=CH
2 + Br
2
H
2O
CH
3CH
OH
CH
2
Br
CH
3CH
Br
CH
2
Br
1-bromo-2-propanol
(major product)
1,2-dibromopropane
(minor product)
propene
* Br atom attached to the carbon atom of the double bond which has the greater
number of hydrogen atoms.
CH
3CH
2
1-chloro-2-butanol
1-butene
CH
3CH
2CH=CH
2
CH
OH
CH
2
Cl
Cl
2, H
2O

•Addition of hydrogen halides:
- Addition reaction with electrophilic reagents.
- Alkenes react with hydrogen halides (in gaseous state or
in aqueous solution) to form addition products.
- The hydrogen and halogen atoms add across the double
bond to form haloalkanes (alkyl halides).
- General equation:
CC CC
HX
HX
alkene haloalkane
-Reactivity of hydrogen halides : HF < HCl < HBr < HI

* Reaction with HCl needs a catalyst such as AlCl
3
H
2CCH
2
HCl
AlCl
3
CH
3CH
2Cl
H-I
CH
3CH=CHCH
3 + H-Br
I
CH
3CH
2CHCH
3
Br
EXAMPLES:
cyclopentene iodocyclopentane
2-butene
2-bromobutane

MARKOVNIKOV’S RULE
•There are 2 possible products when hydrogen halides react
with an unsymmetrical alkene.
•It is because hydrogen halide molecule can add to the C=C
bond in two different ways.
CC
H
HCH
3
H
H-I
CC
H
HCH
3
H
H-I
CC
H
HCH
3
H
HI
CC
H
HCH
3
H
IH
1-iodopropane
2-iodopropane
(major product)

Markovnikov’s rules:Markovnikov’s rules:
- the addition of HX to an unsymmetrical
alkene, the hydrogen atom attaches itself
to the carbon atom (of the double bond)
with the larger number of hydrogen atoms.

Step 2: Rapid reaction with a negative ion.
The negative ion (Y
-
) acts as nucleophile and attacks the
positively charged carbon atom to give product of the addition
reaction.
CC
E
Y
-
CC
EY
Mechanism of electrophilic addition reactions:
- C=C : electron rich part of the alkene molecule
- Electrophiles: electron-seeking
Step 1: Formation of carbocation.
Attack of the pi bond on the electrophile to form carbocation.
CC CC
E
EY
Y
-
carbocation
δ+
δ-

ADDITIONOFHYDROGENHALIDESTO

UNSYMMETRICALALKENESAND

MARKOVNIKOVSRULE
CH
3CH=CH
2
HCl
CH
3CHCH
2
HCl
CH
3CHCH
2
ClH
1-chloropropane
2-chloropropane
(major product)
according to Markovnikov's
rules
123
Propene

MECHANISM:
Step 1: Formation of carbocation
CC
HH
HCH
3
HCl CC
HH
HC
H
H
H
H
CC
HH
HCH
H
HH
or
less stable carbocation
(1
o
carbocation)
more stable carbocation
(2
o
carbocation)
Cl
-
- 2
o
carbocation is more stable than 1
o
carbocation.
- 2
o
carbocation tends to persist longer, making it more likely to combine with
Cl
-
ion to form 2-chloromethane (basis of Markovnikov's rule).
CC
HH
HCH
H
HH
Cl
-
Step 2: Rapid reaction with a negative ion
CC
HH
HCH
H
HHCl
2-chloromethane (major product)

Addition reaction with concentrated sulfuric
acid: hydration of alkenes
- the alkene is absorbed slowly when it passed
through concentrated sulfuric acid in the cold
(0-15
o
C).
- involves the addition of H atom and HSO
4

group across the carbon-carbon double bond.
- follows Markovnikov’s rule.

CCH
HH
H HOSO
3H
(H2SO4)
CH
3CH
2OSO
3H + H-OH
(H
2O)
CCH
HH
H
HOSO
3H
CH
3CH
2OH + H
2SO
4
ethyl hydrogensulphate
(CH
3CH
2HSO
4)
When the reaction mixture is added to water and warmed,
ethyl hydrogensulphate is readily hydrolysed to ethanol
*ethene reacts with concentrated H
2SO
4 to form ethanol*
or
*alkene reacts with concentrated H
2SO
4 to form alcohol*

•Addition reaction with acidified water (H
3
O
+
): hydration of
alkenes
•Hydration: The addition of H atoms and –OH groups from
water molecules to a multiple bond.
•Reverse of the dehydration reaction.
•Direct hydration of ethene:
- passing a mixture of ethene and steam over phosphoric (v)
acid (H
3
PO
4
) absorbed on silica pellets at 300
o
C and a
pressure of 60 atmospheres.
- H
3
PO
4
is a catalyst.
CH
2=CH
2 H
2O
H3PO4
CH
3CH
2OH
(g) (g)
300
o
C, 60 atm
(g)
ethene ethanol
CC H
2O CC
HOH
alkene
alcohol
H
+

•Markovnikov’s rule is apply to the addition of a water
molecule across the double bond of an unsymmetrical
alkene.
•For examples:
CH
3CCH
2
CH
3
HOH
H
+
CH
3CH=CH
2 + H
2O CH
3CHCH
3
OH
CH
3CCH
2
CH
3
OHH
25
o
C
2-methylpropene
tert-butyl alcohol
propene
2-propanol
H
+
H
+
= catalyst

CC
HH
HCH
3
CC
HH
HCH
H
HH
H
+
O
H
H
CH
3CHCH
3
OH
H
CH
3CHCH
3
OH
CC
HH
HCH
H
HH
CH
3CHCH
3
OH
H
H
+
MECHANISM OF ACID CATALYSED HYDRATION OF ALKENES
Step 1: Protonation to form carbocation
more stable carbocation
(2
o
carbocation)
Step 2: Addition of H
2O to form a protonated alcohol
Step 3: Loss of a proton (deprotonated) to form alcohol
H
+
= catalyst

•When HBr is added to an alkene in the absence of peroxides
it obey Markovnikov’s rule.
•When HBr (not HCl or HI) reacts with unsymmetrical alkene
in the presence of peroxides (compounds containing the O-
O group) or oxygen, HBr adds in the opposite direction to
that predicted by Markovnikov’s rule.
•The product between propene and HBr under these
conditions is 1-bromopropane and not 2-bromopropane.
ANTI-MARKOVNIKOV’S RULE: FREE RADICAL
ADDITION OF HYDROGEN BROMIDE
CH
3CH=CH
2
HBr CH
3CH
2CH
2Br
peroxide
1-bromopropane
(major product)
anti-Markovnikov's orientation

•Anti-Markovnikov’s addition:
- peroxide-catalysed addition of HBr occurs
through a free radical addition rather than a
polar electrophilic addition.
- also observed for the reaction between HBr
and many different alkenes.
- not observed with HF, HCl or HI.

Combustion of alkenes:
The alkenes are highly flammable and burn
readily in air, forming carbon dioxide and
water.
For example, ethene burns as follows :
C
2
H
4
+ 3O
2
→ 2CO
2
+ 2H
2
O

•Oxidation: reactions that form carbon-
oxygen bonds.
•Oxidation reaction of alkenes:
i) epoxidation
ii)hydroxylation
iii)Ozonolysis
OXIDATIONOXIDATION

EPOXIDATION OF ALKENES
•Epoxide / oxirane: a three-membered cyclic ether.
CC
RC
O
OOH
O
CC RC
O
OH
alkene
peroxyacid
epoxide (oxirane) acid
•Examples of epoxidizing reagent:
CH
3C
O
peroxyacetic acid
OOH C
O
peroxybenzoic acid
(PhCO
3H)
OOH
m-chloroperoxybenzoic acid
(MCPBA)
Cl
O
O
O
H

Examples:
MCPBA
MCPBA
O
O
CH
2CI
2, 25
o
C
cyclohexene 1,2-epoxycyclohexane
CH
2CI
2, 25
o
C
cycloheptene
1,2-epoxycycloheptane

•Hydroxylation:
- Converting an alkene to a glycol requires adding a
hydroxyl group to each end of the double bond.
•Hydroxylation reagents:
i) Osmium tetroxide (OsO
4
)
ii)Potassium permanganate (KMnO
4
)
HYDROXYLATION OF
ALKENES
CC OsO
4
H
2O
2 CC
OHOH(or KMnO
4,
-
OH)
glycol

CHCH
2CH
3
CH
2CH
2 CH
2CH
2
OHOH
CH
2
OH
CHCH
3
OH
MnO
2
MnO
2
KMnO
4 (aq), OH-
cold
ethene
1,2-ethanediol
KMnO
4 (aq), OH-
cold
propene
1,2-propanediol
* Also known as Baeyer’s test

•Ozonolysis:
- The reaction of alkenes with ozone (O
3) to form an ozonide, followed
by hydrolysis of the ozonide to produce aldehydes and /or ketone.
- Widely used to determine the position of the carbon-carbon double
bond.
- Ozonolysis is milder and both ketone and aldehydes can be
recovered without further oxidation.
OZONOLYSIS OF ALKENES
CC
R
R
R'
H
O
3 C
OO
C
OR'
H
R
R
(CH3)2S
CO
R
R
CO
R'
H
ozonide ketone aldehyde
or H
2O, Zn/H
+

EXAMPLES:
H
OCH
3CH
3O
H
H
O
O
OCH
3
H
O
CH
3O
O
H
O
H
O
i) O
3
ii) (CH
3)
2S
3-nonene
i) O
3
ii) (CH
3)
2S

REACTIONS OF ALKENES WITH HOT, ACIDIFIED REACTIONS OF ALKENES WITH HOT, ACIDIFIED
KMnOKMnO
44
CC
R
R
R'
H
CCH
R'R
OH
R
OH
KMnO
4/H
+
CO
R
R
CO
H
R'
CO
R
R
CO
OH
R'
ketone acid ketone aldehyde
Example:
KMnO
4/H
+
C
O
O
C
HO
4-methyl-4-octene 2-pentanone butanoic acid

•Polymer: A large molecule composed of many smaller
repeating units (the monomers) bonded together.
•Alkenes serves as monomers for some of the most
common polymers such as polyethylene (polyethene),
polypropylene, polystyrene, poly(vinyl chloride) and
etc.
•Undergo addition polymerization /chain-growth
polymer:
- a polymer that results from the rapid addition of one
molecule at a time to a growing polymer chain, usually
with a reactive intermediate (cation, radical or anion) at
the growing end of the chain.

POLYMERIZATIONOFALKENES

POLYMERIZATIONOFALKENES

CC
CI
H
H
H
CC
CI
H
H
H
CC
CI
H
H
H
CC
CI
H
H
H
C
H
H
C
Cl
H
C
H
H
C
Cl
H
n
poly(vinyl chloride)
vinyl chloride
repeating unit

•Reactions of alkenes with KMnO
4
- KMnO
4
is a strong oxidising agent.
- alkenes undergo oxidation reactions with
KMnO
4
solution under two conditions:
a) Mild oxidation conditions using cold,
dilute, alkaline KMnO
4
(Baeyer’s test).
b) Vigorous oxidation conditions using hot,
acidified KMnO
4.
UNSATURATION TESTS FOR ALKENESUNSATURATION TESTS FOR ALKENES

•Reaction of alkenes with cold, dilute, alkaline
KMnO
4
(Baeyer’s test)
- the purple colour of KMnO
4
solution disappears
and a cloudy brown colour appears caused by the
precipitation of manganese (IV) oxide, MnO
2
.
- test for carbon-carbon double or triple bonds.
- a diol is formed (containing two hydroxyl groups
on adjacent carbon atoms).

CC CC
OHOH
MnO
2
KMnO
4 (aq), OH
-
cold
a diol

b) Bromine
- A solution of bromine in inert solvent (CH
2
CI
2
or CCI
4
)
and dilute bromine water are yellow in colour.
- The solution is decolorised when added to alkenes or
organic compounds containing C=C bonds.

CC Br
2
CH
2CI
2
CC Br
2(aq)
H
2O
CC
BrBr
CC
OHBr
CC
BrBr

a) Ozonolysis of alkenes:
- For example, ozonolysis of an alkene produces
methanal and propanone.
DETERMINATION OF THE POSITION OF DETERMINATION OF THE POSITION OF
THE DOUBLE BONDTHE DOUBLE BOND
CO
methanal
H
H CO CH
3
CH
3
propanone
C
H
H CCH
3
CH
3
CC CH
3
CH
3H
H
remove the oxygen atoms from the carbonyl compounds and
joining the carbon atoms with a double bond.
2-methylpropene

b) Reaction of alkenes with hot, acidified KMnO
4
- by using hot, acidified KMnO
4
, the diol obtained is
oxidised further.
- cleavage of carbon-carbon bonds occurs and the
final products are ketones, carboxylic acids or
CO
2
.
KMnO
4/H
+
CCH
2
CH
3
CH
3
CO
CH
3
CH
3
CO
2 + H
2O
2-methylpropene
propanone
(ketone)

•Example:
An alkene with the molecular formula C
6
H
12
is oxidised with
hot KMnO
4
solution. The carboxylic acids, butanoic acid
(CH
3
CH
2
CH
2
COOH) and ethanoic acid (CH
3
COOH), are
produced. Identify the structural formula of the alkene.
CC
H
R
H
R'
CH
3CH
2CH
2COOH and CH
3COOH
CO
OH
R CO
OH
R'
CH
3CH
2CH
2CH=CHCH
3
KMnO
4/H
+
i) cleavage of the double bond gives a mixture of carboxylic acids
ii) location of the double bond is done by taking away the oxygen atoms from the
carboxylic acids and then joining the carbon atoms by the double bond.
RCOOH and R'COOH RCH=CHR'
butanoic acid ethanoic acid
2-hexene

•Ethylene and propylene are the largest-volume industrial
organic chemicals.
•Used to synthesis a wide variety of useful compounds.
USES OF ALKENESUSES OF ALKENES
CH
3C
O
OH
CH
2CH
2
CICI
Cl
2
CC
H
H
H
H
CH
3C
O
H
O
2
CC
CIH
H H
CH
3CH
2
OH
NaOH
CC
HH
HH
H
+
H
2O
CH
2CH
2
OHOH
O
H
2CCH
2
n
polyethylene
polymerize
acetaldehyde
oxidize
oxidize
acetic acid
ethylene
ethylene dichloride
vinyl chloride
H
2O
catalyst
Ag catalyst
ethylene oxide
ethylene glycol ethanol

•The most popular plastic.
•Uses:
i) Grocery bags
ii)Shampoo bottles
iii)Children's toy
iv)Bullet proof vests
v)Film wrapping
vi)Kitchenware
POLYETHENE (PE)

POLYVINYL CHLORIDE (PVC)
CC
HH
CIH CC
H
H
CI
H
C
H
H
C
CI
H
C
H
H
C
CI
Hnvinyl chloride
polymerize
poly(vinyl chloride)
PVC, "vinyl"
USES OF PVC:
Clothing
- PVC fabric has a sheen to it and is waterproof.
- coats, shoes, jackets, aprons and bags.
As the insulation on electric wires.
Producing pipes for various municipal and industrial
applications. For examples, for drinking water distribution
and wastewater mains.
As a composite for the production of accessories or
housings for portable electronics.
uPVC or Rigid PVC is used in the building industry as a
low-maintenance material.
Ceiling tiles.

USES OF ETHANOL
•Motor fuel and fuel additive.
•As a fuel to power Direct-ethanol fuel cells (DEFC) in order to
produce electricity.
•As fuel in bipropellant rocket vehicles.
•In alcoholic beverages.
•An important industrial ingredient and use as a base chemical
for other organic compounds include ethyl halides, ethyl
esters, diethyl ether, acetic acid, ethyl amines and to a lesser
extent butadiene.
•Antiseptic use.
•An antidote.
•Ethanol is easily miscible in water and is a good solvent.
Ethanol is less polar than water and is used in perfumes,
paints and tinctures.
•Ethanol is also used in design and sketch art markers.
•Ethanol is also found in certain kinds of deodorants.