Exercise 7: HYDROCARBONS Flashcards

1
Q

organic compounds composed of carbon and hydrogen atoms
only

A

HYDROCARBONS

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2
Q

 most hydrocarbons are used as — because of its flammable property

A

fuel (petroleum products)

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3
Q

 have a special property to catenate forming seemingly endless
chain of carbons (potential combinations estimated up to )

A

HYDROCARBONS
10*6

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4
Q

HYDROCARBONS are — in nature, thereby most of them are — in
water

A

non-polar

insoluble

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5
Q

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

A

• saturated

• unsaturated

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6
Q

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

A

one sigma bond

sigma and pi bonds (one pi bond for those with double bonds and two
pi bonds for those with triple bonds)

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7
Q

this makes unsaturated hydrocarbons - than saturated
ones

A

more reactive

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8
Q

based on their structures, hydrocarbons can be classified as:

• open-chain –
• closed-chain–

A

acyclic; linear structure; could be straight or branched
chains

cyclic; ring structure; could aliphatic or aromatic

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9
Q

HYDROCARBONS are generally classified into
two:

A

aliphatic and aromatic hydrocarbons.

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10
Q

 are cyclic hydrocarbons with delocalized pi electrons between carbon
atoms of ring

 their natural characteristics are described as

A

AROMATIC HCs

aromaticity

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11
Q

 criteria for aromaticity:

A

• must be cyclic in structure
• must be flat or planar in configuration
• must have conjugated double bonds
• must follow Hückel’s rule of aromaticity

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12
Q

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

A

conjugated pi bonds

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13
Q

 according to, a compound is
particularly stable if all of its bonding molecular orbitals are filled with
paired electrons

• the number of pi electrons =

A

Hückel’s Molecular Orbital Theory

4n + 2, where n = 0 or any positive
whole integer

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14
Q

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 -

A

benzene

saturated, unsaturated hydrocarbons

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15
Q

IGNITION TEST results

FLAME* PRESENCE OF SOOT
benzene
cyclohexane
gasoline
kerosene
n-hexane

A

luminous present

luminous present

non-luminous absent

non-luminous absent

luminous present

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16
Q

*Luminous flame burns - flame. Non-luminous flame burns - flame.

A

brightly yellow

almost invisible blue

17
Q

– hydrocarbon reacts with oxygen to produce carbon dioxide, water, and heat

𝐶𝑥𝐻𝑦 + 𝑛 𝑂2 ↔ 𝑥 𝐶𝑂2 + 𝑦2 (𝐻2𝑂)

A

combustion reaction

18
Q

flame color
yellow

combustion
incomplete

soot
present

heat
produced do not produced
much energy; do not
produce much heat

A

LUMINOUS FLAME

19
Q

flame color
Blue

combustion
Complete

soot
Absent

heat
burns efficiently;
produces hotter flame

A

NON-LUMINOUS FLAME

20
Q

incomplete combustion reaction – hydrocarbons
react with insufficient amount of oxygen, instead of
producing CO2, produces as
products

A

CO, H2O and C

21
Q

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

A

non-luminous flame

black smoke/soot

22
Q

BAEYER’S TEST FOR UNSATURATION results

benzene
cyclohexane
gasoline
kerosene
n-hexane

A

purple-colored solution

purple-colored solution

dark-brown precipitate

reddish-brown precipitate

purple-colored solution

23
Q

*is a positive result. Most of the time, solution turns to (still indicate a positive result).

A

Brown precipitate, reddish-brown

24
Q

named after Adolf von Baeyer

 used as a qualitative test for unsaturation (presence of double or triple
bonds)

A

BAEYER’S TEST FOR UNSATURATION

25
Q

 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

A

potassium permanganate
(KMnO4)

purple to brown precipitate

26
Q

oxidation reaction – an alkene/alkyne is oxidized by KMnO4 producing products of

A

a diol (for alkene),
an alkane with four hydroxyl groups (for alkyne),
a manganese dioxide (MnO2) and
permanganate ion (MnO42-)

27
Q

BROMINE TEST FOR UNSATURATION results

benzene
cyclohexane
gasoline
kerosene
n-hexane

A

yellow-colored solution

yellow-colored solution

clear pinkish solution**

clear colorless solution

yellow-colored solution

28
Q
  • is a positive result.
    ** may yield clear pinkish color because of its innate color.
A

Clear colorless solution

Gasoline

29
Q

used as a qualitative test for unsaturation (presence of double or triple
bonds)

A

BROMINE TEST FOR UNSATURATION

30
Q

 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

A

dichloromethane,
chloroform, or
carbon tetrachloride

brown to clear colorless solution

31
Q

addition reaction – an alkene/alkyne
reacts with Br2 producing products of a

A

dibromo for alkene
a tetrabromo for alkyne

32
Q

 the - in alkenes and
alkynes are susceptible to addition reactions

A

weak pi bond

33
Q

TEST FOR AROMATICITY (NITRATION) results

benzene
cyclohexane
gasoline
kerosene
n-hexane

A

pale yellow-colored layer in solution

clear colorless solution

deep yellow-colored layer in solution

yellow-colored layer in solution

clear colorless solution

34
Q
  • colored layer in solution is a positive result.
A

Yellow

35
Q

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

A

nitrating acid (HNO3 + H2SO4)

yellow colored layer

aromatic compounds

36
Q

➢ 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

A

electrophilic aromatic substitution reaction:

nitronium ion

sulfuric acid

benzene