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Chapter 14
Structure and Synthesis
of Alcohols
Classification
• Primary: carbon with –OH is bonded to
one other carbon.
• Secondary: carbon with –OH is bonded
to two other carbons.
• Tertiary: carbon with –OH is bonded to
three other carbons.
• Aromatic (phenol): -OH is bonded to a
benzene ring.
Chapter 14
2
Classify these:
CH3
CH3
CH3
CH CH2OH
CH3
C OH
CH3
OH
OH
CH3
Chapter 14
CH CH2CH3
=>
3
IUPAC Nomenclature
• Find the longest carbon chain
containing the carbon with the -OH
group.
• Drop the -e from the alkane name, add ol.
• Number the chain, starting from the end
closest to the -OH group.
• Number and name all substituents. =>
Chapter 14
4
Name these:
CH3
CH3
CH CH2OH
2-methyl-1-propanol
OH
CH3
CH CH2CH3
2-butanol
CH3
CH3
OH
C OH
CH3
2-methyl-2-propanol
Br
CH3
3-bromo-3-methylcyclohexanol
=>
Chapter 14
5
Unsaturated Alcohols
• Hydroxyl group takes precedence. Assign
that carbon the lowest number.
• Use alkene or alkyne name.
OH
CH2
CHCH2CHCH3
4-penten-2-ol (old)
pent-4-ene-2-ol
(1997 revision of IUPAC rules)
=>
Chapter 14
6
Naming Diols
• Two numbers are needed to locate the two
-OH groups.
• Use -diol as suffix instead of -ol.
OH
HO
1,6-hexanediol
=>
Chapter 14
7
Practice
• TIME FOR PRACTICE.
• Try the handout review of alcohol
naming
Chapter 14
8
Physical Properties
• Unusually high boiling points due to
hydrogen bonding between molecules.
• Small alcohols are miscible in water, but
solubility decreases as the size of the
alkyl group increases.
=>
Chapter 14
9
Boiling Points
FUNCTIONAL GROUP RANKING BY BOILING POINTS
Name
Amide
B.P
Rank
Name
Brief Explanation
222o
1
ethanamide
hydrogen bonds on both the oxygen and the nitrogen
hydrogen bonding from the of 2 oxygen atoms.
hydrogen bonding from the presence one oxygen
Acid
118o
2
ethanoic acid
or
acetic acid
Alcohol
117o
3
propanol
Chapter 14
10
Solubility in Water
Solubility decreases as the size
of the alkyl group increases.
Chapter 14
=>
11
USES FOR ALCOHOLS
1)Fuels
2)Solvents
Chapter 14
12
Methanol
• “Wood alcohol”
• Common industrial solvent
• Fuel at Indianapolis 500
Chapter 14
13
Ethanol
• Fermentation of sugar and starches in grains
• 12-15% alcohol, then yeast cells die.
• MUST BE ANEROBIC or vinegar results
•
•
•
•
Distillation produces “hard” liquors for drinking
alcohol used as solvent
Gasahol: 10% ethanol in gasoline
Toxic dose: 200 mL ethanol, 30 mL methanol
Chapter 14
14
Chapter 14
15
2-Propanol
• “Rubbing alcohol”
• Catalytic hydration of propene
• Much more efficient makes 100%
alcohol
CH2 CH
CH2 + H2O
100-300 atm, 300°C
catalyst
CH3 CH
CH3
OH
=>
Chapter 14
16
Biofuels
Pros:
-Some argue it’s
Carbon neutral, but
not quite, why?
-Protects nonrenewable fuels
needed for drugs
and plastics
• Cons
• Food is scarce,
should we burn it
when people are
hungry?
• Uses a lot of land,
water, energy to
make
Chapter 14
17
https://www.youtube.com/watch?v=C-qJ-6XJKBk
Chapter 14
18
Reactions of Alcohols
1) Combustion: R-OH  CO2
2) Dehydration: R-OH  C=C (Zaitsev)
Major and Minor Product
3) Oxidation:
a) R-OH (Primary)  R-C-HO  R-COOH
b) R-OH (secondary)  R1R2-C=O
c) Tertiary alcohols do not oxidize as the
alpha carbon (containing OH) has no H’s
Chapter 14
19
Combustion
Complete
C3H7OH(g) + 4.5O2  3CO2 + 4H2O
_________________________________
Incomplete Combustion
C3H7OH(g) + 3O2  3 CO + 4 H2O
ALCOHOL OXIDATION
10
• gentle heating, a primary alcohol can
be oxidised to produce an aldehyde.
With H2SO4 or acidified [O]
BUT
• With strong heating and excess [O]
(acidified) a carboxylic acid is formed.
To heat strongly we need a Reflux
apparatus is generally used to produce
carboxylic acids.
Note: Aldehydes must be distilled as they are formed to
prevent further oxidation to carboxylic acids. Simply
take away the reflux equipment and replace with
distillation, see next slide.
.
OXIDIZING REAGENTS
OXIDAIZING AGENT
OBSERVATIONS
OXIDIZING AGENTS
REACTION
ACIDIFIED DICROMATE
ORANGE TO
GREEN
Cr2O72-(aq) to Cr3+(aq)
ORANGE
GREEN
BENEDICT’S REAGENT
BLUE TO RED
Cu2+ + e-  Cu+
BLUE
Tollen's Reagent
RED
COLOURLESS TO Ag+(aq) + e-  Ag(s)
SILVER MIRROR
COLOURLESS SILVER
MIRROR
Chapter 14
22
Ag+  Ag (s)
Tollen's Reagent
Chapter 14
23
No hard, nothing special
• This reaction goes easily and quickly,
• Nothing special is needed, other than
the oxidizing agent is required, a hot
water bath will do. The problem is when
the aldehyde is made it boils off
immediately. To form the –COOH we
need to trap the aldehyde and force it
back with reflux, or a condenser.
Chapter 14
24
OXIDATION OF PRIMARY ALCOHOLS
Controlling the products
e.g.
CH3CH2OH(l) + [O]
——>
CH3CHO(l) + H2O(l)
then
CH3CHO(l) + [O]
——>
CH3COOH(l)
OXIDATION TO ALDEHYDES
DISTILLATION
OXIDATION TO CARBOXYLIC ACIDS
REFLUX
Aldehyde has a lower boiling point so
distils off before being oxidised further
Forces Aldehyde to condenses back
into the mixture and gets oxidised to
the acid
OXIDATION OF PRIMARY ALCOHOLS
Primary alcohols are easily oxidised to aldehydes
e.g.
CH3CH2OH(l) + [O]
ethanol
——>
CH3CHO(l) + H2O(l)
ethanal
it is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O]
ethanal
——>
CH3COOH(l)
ethanoic acid
Practical details
•
•
•
•
the alcohol is dripped into a warm solution of acidified K2Cr2O7
aldehydes have low boiling points - no hydrogen bonding - they distil off immediately
if it didn’t distil off it would be oxidised to the equivalent carboxylic acid
to oxidise an alcohol straight to the acid, reflux the mixture
compound
formula
intermolecular bonding
ETHANOL
C2H5OH
HYDROGEN BONDING
ETHANAL
CH3CHO
DIPOLE-DIPOLE
ETHANOIC ACID
CH3COOH
HYDROGEN BONDING
boiling point
78°C
23°C (volatile)
118°C
Primary OH to HO
green
orange
Simplified Molecular Equation
Chapter 14
27
Stop OR Go
• We can stop at OH to HO (aldehydes),
• But the oxidizing agent is powerful and
will continue unless we stop it.
• 1) Use excess Alcohol reagent, so all
the KMNO4 / K2Cr2O7 is used up
• 2) Remove the Aldehyde as it forms via
distillation (bp)
Chapter 14
28
OH all the way to COOH
Alcohol 10
Alcohol 10
Aldehyde
C.A.
A more simplified ONE step Version
1o Alcohol
Carboxylic Acid
29
OH to HO, then COOH
Chapter 14
30
Secondary Alcohols
• Secondary alcohols are oxidised to
produce ketones, then stop. Gentle
heating as with aldehydes
2o OH to Ketones
e.g. CH3CHOHCH3(l) + [O]
propan-2-ol
——>
Chapter 14
CH3COCH3(l) + H2O(l)
propanone
32
ALCOHOL TO KETONE
e.g CH3CHOHCH3(l) + [O]
propan-2-ol
——> CH3COCH3(l) + H2O(l)
propanone
Chapter 14
33
Tertiary Alcohols
• These are resistant to oxidation, there
will be no colour change.
• This is because the carbon (of the
alcohol group) is not bonded to any
other hydrogen atoms.
• Thus, NO COLOUR CHANGE WITH [O]
3o lack a hydrogen
Remember , hydrogen removal is the Definition of
Oxidation, as is the addition of oxygen
Chapter 14
35
Test Results
Chapter 14
36
Dehydration of Alcohols
Alcohol  Alkene + water
• when heated with an acid catalyst:such
as a concentrated acid (a dehydrating
agent) or pumice, or Al2O3
• with the loss of H and OH to form water
alcohol
alken
e
37
Zaitsev’s Rule
the dehydration of a secondary alcohol favors the product
in which hydrogen is removed from the carbon atom in the
smaller number of H atoms, the
poor get poorer, or most substituted carbon
chain with the
38
Zaitsev’s Rule…
A shorter way of thinking of it is that the more substituted
product is the one that is preferred, since it is more
stable… so which is preferred here?
39
Disubstituted vs. monosubstituted…
The pink structure (or 2-pentene) is preferred over 1pentene since it is disubstituted (vs. monosubstituted)
40
Try to draw the product of the reaction
shown in your composition book
(there is a hint on the next slide)
Hint…
Answer…
Remember that the more substituted product is prefe
(1-butene also forms, but it is the minor product).
This one, then one
more…
44
The more substituted product
is preferred…
45
Pick out the product, last one
Recall that Zaitsev’s rule indicates that the more substituted
double bond will form…
Under the OH, CH has 3 H’s, the green arrow
the carbon in the hex chain has only 2 arrows
Minor
No reaction
product
Missing the double b