Exercise 7: HYDROCARBONS Flashcards
organic compounds composed of carbon and hydrogen atoms
only
HYDROCARBONS
most hydrocarbons are used as — because of its flammable property
fuel (petroleum products)
have a special property to catenate forming seemingly endless
chain of carbons (potential combinations estimated up to )
HYDROCARBONS
10*6
HYDROCARBONS are — in nature, thereby most of them are — in
water
non-polar
insoluble
HYDROCARBONS
can be classified based on the presence of multiple bonds:
– carbon atoms are bonded with single bond
– carbon atoms are bonded with double or triple bonds
• saturated
• unsaturated
single bond possesses — which is relatively a strong type
of bond compared to double and triple bonds which possesses a —
; are relatively weak type of bond
one sigma bond
sigma and pi bonds (one pi bond for those with double bonds and two
pi bonds for those with triple bonds)
this makes unsaturated hydrocarbons - than saturated
ones
more reactive
based on their structures, hydrocarbons can be classified as:
• open-chain –
• closed-chain–
acyclic; linear structure; could be straight or branched
chains
cyclic; ring structure; could aliphatic or aromatic
HYDROCARBONS are generally classified into
two:
aliphatic and aromatic hydrocarbons.
are cyclic hydrocarbons with delocalized pi electrons between carbon
atoms of ring
their natural characteristics are described as
AROMATIC HCs
aromaticity
criteria for aromaticity:
• must be cyclic in structure
• must be flat or planar in configuration
• must have conjugated double bonds
• must follow Hückel’s rule of aromaticity
AROMATIC HCs
contain - bonds, that is alternating double bonds
this special arrangement of pi bonds exhibits resonance which causes
the conjugative stability of the aromatic compounds
conjugated pi bonds
according to, a compound is
particularly stable if all of its bonding molecular orbitals are filled with
paired electrons
• the number of pi electrons =
Hückel’s Molecular Orbital Theory
4n + 2, where n = 0 or any positive
whole integer
being the most common example of an aromatic compound, generally undergo electrophilic aromatic substitution reactions
it undergoes substitution reaction wherein the ring
system is not destroyed and therefore resist addition of
substituent groups within the pi bonds
this explains the reason why aromatic compounds
resembles reactions of - rather than -
benzene
saturated, unsaturated hydrocarbons
IGNITION TEST results
FLAME* PRESENCE OF SOOT
benzene
cyclohexane
gasoline
kerosene
n-hexane
luminous present
luminous present
non-luminous absent
non-luminous absent
luminous present
*Luminous flame burns - flame. Non-luminous flame burns - flame.
brightly yellow
almost invisible blue
– hydrocarbon reacts with oxygen to produce carbon dioxide, water, and heat
𝐶𝑥𝐻𝑦 + 𝑛 𝑂2 ↔ 𝑥 𝐶𝑂2 + 𝑦2 (𝐻2𝑂)
combustion reaction
flame color
yellow
combustion
incomplete
soot
present
heat
produced do not produced
much energy; do not
produce much heat
LUMINOUS FLAME
flame color
Blue
combustion
Complete
soot
Absent
heat
burns efficiently;
produces hotter flame
NON-LUMINOUS FLAME
incomplete combustion reaction – hydrocarbons
react with insufficient amount of oxygen, instead of
producing CO2, produces as
products
CO, H2O and C
when a fuel (e.g. gasoline, kerosene) burns in plenty
of air, it receives enough oxygen for complete
combustion, thereby producing
when a fuel burns in a limited space (e.g. car
engine), there is no enough oxygen to completely
oxidize the fuel, thereby producing
non-luminous flame
black smoke/soot
BAEYER’S TEST FOR UNSATURATION results
benzene
cyclohexane
gasoline
kerosene
n-hexane
purple-colored solution
purple-colored solution
dark-brown precipitate
reddish-brown precipitate
purple-colored solution
*is a positive result. Most of the time, solution turns to (still indicate a positive result).
Brown precipitate, reddish-brown
named after Adolf von Baeyer
used as a qualitative test for unsaturation (presence of double or triple
bonds)
BAEYER’S TEST FOR UNSATURATION
Baeyer’s reagent is an alkaline solution of (strong oxidizer)
reaction with double or triple bonds (-C=C- or -C≡C-) in an organic
material causes the color to fade from
potassium permanganate
(KMnO4)
purple to brown precipitate
oxidation reaction – an alkene/alkyne is oxidized by KMnO4 producing products of
a diol (for alkene),
an alkane with four hydroxyl groups (for alkyne),
a manganese dioxide (MnO2) and
permanganate ion (MnO42-)
BROMINE TEST FOR UNSATURATION results
benzene
cyclohexane
gasoline
kerosene
n-hexane
yellow-colored solution
yellow-colored solution
clear pinkish solution**
clear colorless solution
yellow-colored solution
- is a positive result.
** may yield clear pinkish color because of its innate color.
Clear colorless solution
Gasoline
used as a qualitative test for unsaturation (presence of double or triple
bonds)
BROMINE TEST FOR UNSATURATION
bromine is dissolved either in and the alkene/alkyne sample is added to it
reaction with double or triple bonds (-C=C- or -C≡C-) in an organic
material causes the color to fade from
dichloromethane,
chloroform, or
carbon tetrachloride
brown to clear colorless solution
addition reaction – an alkene/alkyne
reacts with Br2 producing products of a
dibromo for alkene
a tetrabromo for alkyne
the - in alkenes and
alkynes are susceptible to addition reactions
weak pi bond
TEST FOR AROMATICITY (NITRATION) results
benzene
cyclohexane
gasoline
kerosene
n-hexane
pale yellow-colored layer in solution
clear colorless solution
deep yellow-colored layer in solution
yellow-colored layer in solution
clear colorless solution
- colored layer in solution is a positive result.
Yellow
benzene is nitrated using a - to produce nitrobenzene
this nitration test predicts the presence of aromatic
ring through the formation of - in
solution after warming the test sample
only - will proceed through
electrophilic aromatic substitution
nitrating acid (HNO3 + H2SO4)
yellow colored layer
aromatic compounds
➢ nitration of benzene firstly involves the formation of a very powerful
electrophile, the - , which is linear
➢ this occurs following the interaction of sulfuric and nitric acid
➢ - is stronger and it protonates the nitric acid on the OH group so that a molecule of water can leave
➢ - attacks the positively charged nitrogen atom of the electrophile, where one of the N=O bonds is broken at the same time
➢ this is followed by rapid loss of a proton to regenerate the aromaticity
electrophilic aromatic substitution reaction:
nitronium ion
sulfuric acid
benzene
