Download Study Guide Exam 2 and 3 Sp2011

CHEM 131 – Spring 2011
Study Guide and Problems – Exam 2 and 3
Nomenclature: The root is always the same no matter what ending you use for the different class of compounds
Butane
CH3CH2CH2CH3
Cyclobutane
H2C
CH2
H2C
CH2
1-Butanol
CH3CH2CH2CH2-OH
OH
1-Butanethiol
CH3CH2CH2CH2-SH
Dibutyl ether
CH3CH2CH2CH2-O-CH2CH2CH2CH3
1-Butene
CH3CH2CH=CH2
1-Butyne
H3C
SH
CH2
C
O
CH
C
Butanal
CH3 CH2 CH2
2-Butanone
C
H
C
CH3
Butanoic acid
CH2
CH2
C
OH
CH2
O
C
Methyl Butanoate
CH2
CH2
O
O
O
O
CH3
O
Carboxylic acid
that is
deprotonated
(missing a H+)
O
CH2
O
OH
Butanoate
CH3
H
Carbonyl carbon
with 1 OH and 1
alkyl group
O
CH3
O
Carbonyl carbon with
2 alkyl groups
O
CH3 CH2
CH
Carbonyl carbon
with 1 H and 1
alkyl group
O
C
O
CH3
O
1-Butamine
CH3
CH2
CH3
CH2
CH2 CH2 NH2
NH2
N-methyl-1-butamine
CH2 CH2 NH
CH3
N
H
N,N-dimethyl-1-butamine
CH3
1-Butammonium ion
CH3
CH2
CH2 CH2
N
CH3
CH2 CH2 NH3+
CH2
Butanamide
CH2
CH2
N-Methylbutanamide
C
CH2
CH2
C
NH3
NH2
NH2
O
O
CH3
Protonated
amine
O
O
CH3
N
2
NH
CH3
N
H
CHEM 131 – Spring 2011
O
N,N-Dimethylbutanamide
O
CH3
As a functional group: Butyl
CH2
CH2
C
N
CH3
N
2
R-CH2CH2CH2CH3
R
Di-, tri-, tetra-, penta-, hexa- etc. is only used if the same functional group appears several times on the same molecule.
For example: 2 methyl groups would be dimethyl or 4 methyl groups would be tetramethyl
Common Name Items:
Acetic acid
Toluene
O
CH3
OH
Acetate (acetic
acid as functional
group)
Phenol
O
OH
R
O
Benzene
Aniline
NH2
Phenyl (Benzene
as functional
group)
Benzaldehyde
O
R
H
O
Benzoic acid
OH
To determine if an alcohol is a primary, secondary or tertiary, you need to look at the carbon atom to which the hydroxyl
group is attached to. Depending on how many other carbon atoms are bound to the hydroxyl carbon determines if it is
primary, secondary or tertiary. Fill in the circles with C or H depending on your structure and then look at the table.
Type of alcohol
1 (primary)
2 (secondary)
3 (tertiary)
# of carbons
1
2
3
# of hydrogens
2
1
0
OH
C
CHEM 131 – Spring 2011
If you are trying to determine if an amine is a primary, secondary or tertiary, you now replace the C with the N and the
OH with lone pair of electrons on the picture for alcohols. Depending on how many other carbon atoms are bound to
the nitrogen determines if it is primary, secondary or tertiary. Fill in the circles with C or H depending on your structure
and then look at the table.
Type of alcohol
1 (primary)
2 (secondary)
3 (tertiary)
# of carbons
1
2
3
# of hydrogens
2
1
0
N
Some of the reactions from the past
Reduction or Hydrogenation: Addition of hydrogen across double bonds (across C=C or C=O double bonds)
H
R'
H
C
C
H
H
H2
R
Pt
R
H
H
R'
O
H2
HO
This second reaction is part of your oxidation/reduction
scheme of alcohols/aldehydes/ketones/carboxylic acids.
H
Pt
Oxidation (sometimes called elimination): Loss of hydrogen to form a C=C or C=O bond
R
H
H
C
C
H
H
R'
(ox)
H
R'
R
H
- H2
HO
O
H
This second reaction is part of your oxidation/reduction scheme
of alcohols/aldehydes/ketones/carboxylic acids.
(ox)
- H2
Strictly speaking, the formation of a disulfide bond is considered an oxidation, since hydrogen is removed.
(ox)
SH
- H2
+ HS
S
S
Hydration: Addition of water across a C=C double bond to make alcohols
H
R'
+ H2O
H
H
H2SO4
H
H
OH
C
C
H
H
R'
Remember Markovnikov’s rule: the rich get richer –
the hydrogen will add to the carbon that already has
more carbons
CHEM 131 – Spring 2011
Addition of hydrogen halide: Addition of acids such as HF, HCl, HBr or HI.
H
R'
H2SO4
+ HBr
H
H
H
H
Br
C
C
H
H
R'
Again this reaction follows Markovnikov’s rule:
the rich get richer – the hydrogen will add to the
carbon that already has more carbons
Dehydration: Loss of water from an alcohol to make a C=C double bond or an ether
H
OH
C
C
H
H
CH2
OH
H
R'
R
H
H2SO4
R
R
R'
180C
+ H O
CH2
H2SO4
R
R
140C
At higher temperature the dehydration is the
preferred reaction, but at lower temperature the
+ H2O
formation of ethers is preferred.
CH2
O
CH2
R
Oxidation/Reduction of Alcohols, Aldehydes, Ketones, Carboxylic acid
The key to remember for oxidation reactions of alcohols and aldehyde is the following table:
1° Alcohol 
Aldehyde
2° Alcohol 
Ketone
3° Alcohol 
No reaction

Carboxylic acid (the last step requires the presence of water)
The reaction schemes below show the same information on generic alcohols. In each case 2 protons and 2 electrons get
removed. If there are no H on the carbon that is involved in the oxidation, then no oxidation can occur.
O
H-OH
O
H
R
C
O
H
H
(ox)
(ox)
R
H
Aldehyde
2 H+ + 2 e-
R
2 H+ + 2 e-
1Alcohol
O
H
R
C
O
H
R'
Ketone
(ox)
R
R'
2 H+ + 2 e-
2Alcohol
R"
R
C
O
R'
3Alcohol
H
(ox)
No Reaction
OH
Carboxylic Acid
CHEM 131 – Spring 2011
For reductions, you simply march backwards on the reaction schemes. Here we are adding hydrogens back in.
O
O
H
H2, Pt
H2, Pt
R
(red)
OH
R
H-OH
Carboxylic Acid
R
(red)
H
C
O
H
H
Aldehyde
1Alcohol
H
O
H2, Pt
Ketone
R
R'
(red)
R
C
O
H
R'
2Alcohol
Review of Reactions involving amines, carboxylic acids, esters, thioesters and amides
Amines as weak base (amines accept protons to form ammonium ions):
Since amines are weak bases, they easily accept a proton form water and acids and become protonated, forming an
ammonium ion:
H
H
N
CH2CH3
+ H2O
H
H
N
CH2CH3
+
OH
H
Likewise, amines will react with strong acids to form the ammonium ion. The chlorine acts as a counter ion in the salt.
H
H
N
H
CH2CH3
+ HCl
H
N
H
CH2CH3
Cl
CHEM 131 – Spring 2011
Carboxylic acids as weak acids (they can donate a proton):
Carboxylic acids are weak acids and therefore will dissociate in water
O
O
+ H3O
+ H2O
H
R
R
O
This gives a carboxylate ion, which is the
conjugate base of the carboxylic acid.
O
And carboxylic acids will react with strong bases.
This leads to a carboxylate salt. In this case a
sodium carboxylate. Water is the side
+ H2O
product.
O
O
H
R
+
NaOH
O Na
R
O
Esters, Thioesters and Amides:
The secret of esters, thioesters and amides: they are made up of a carboxylic acid combined with either alcohol, thiol or
amine, respectively. As you can see they are virtually identical. Here a generic carboxylic acid, R-COOH, is shown along
with ethanol, ethanethiol, and ethanamine.
O
Thioester
O
Ester
R
O
O
S
CH2CH3
R
O
R
OH
Amide
CH2CH3
CH2CH3
R
O
+ HO-CH2CH3
N
H
O
R
OH
+ HS-CH2CH3
R
OH
+ H2N-CH2CH3
Ester, Thioester and Amide Formation:
The formation of these esters, thioesters and amides are very similar also. They can be formed by either using
carboxylic acid combined with acid and heat, or two carboxylic acid derivatives, which are more reactive and lead to
higher yield than carboxylic acids. These consist of a carboxylic acid chloride or a carboxylic acid anhydride. In each of
these reactions the side products are slightly different.
An example is shown for the formation of ester via carboxylic acid and alcohol coupling using acid and heat:
O
O
+
R
OH
+ HO-CH2CH3
H , heat
CH2CH3
R
O
+ H2O
CHEM 131 – Spring 2011
Below is an example of thioester formation.
O
O
+ HS-CH2CH3
R
H+, heat
CH2CH3
OH
R
+ H2O
S
As long as there is a proton on the amine, amide formation is possible.
O
O
H+, heat
+ HN-CH2CH3
R
OH
R
H
CH2CH3
+ H2O
CH2CH3
+ H2O
N
H
O
O
+
H , heat
+ HN-CH2CH3
R
OH
R
CH3
CH3
O
R
N
OH
+
CH3-N-CH2CH3
H+, heat
No Reaction
CH3
Hydrolysis of esters, thioesters and amides:
When we think about cleaving the ester, thioester or amide bond, we can think of it as getting the original components
back. This reaction is your simple hydrolysis reaction where “water” adds across the bond to be broken either directly or
indirectly, depending of the reaction conditions.
Ester

Carboxylic acid + alcohol
Thioester

Carboxylic acid + thiol
Amide

Carboxylic acid + amine
This can be done either under acidic or basic conditions. And that is where most of the confusion stems from. The only
thing to consider is, how does the reaction conditions affect the products that are being formed. We already established
that a carboxylic acid is a weak acid and that it reacts with strong bases. On the other hand, it is not affected by any acid
present. Amines are weak bases and will react with strong acids, but are unaffected by bases. Alcohols and thiols are
unaffected by both acids and bases.
How do we translate this to reality? Simple – first, write down the two products you would get if everything is neutral
(ester  carboxylic acid + alcohol). Secondly, think of the effect the reaction conditions have on the products.
Carboxylic acid will react with base; and an amine will react with acid.
So if we look at the hydrolysis of esters, we see that under acidic conditions, neither product is affected by the acid. On
the other hand, when base is used, the alcohol remains unchanged, but the carboxylic acid undergoes acid-base
chemistry and is deprotonated forming the carboxylate with sodium acting as counterion.
CHEM 131 – Spring 2011
O
Acidic hydrolysis
O
+
CH2CH3
R
Basic hydrolysis
+ H2O
H , heat
+ HO-CH2CH3
R
O
O
O
CH2CH3
R
OH
+ HO-CH2CH3
+ NaOH
R
O
O Na
Thioesters behave the same identical way as alcohols, only a thiol will result as one of the products.
Now, when we look at amides, both the carboxylic acid and the resulting amine are vulnerable to acid or base
conditions. Carboxylic acid will react with the NaOH, whereas the amine will react with HCl. See acid/base behavior of
those two compounds shown above. Again the chlorine and sodium act as counterions in the resulting salt.
O
O
Acidic hydrolysis
CH2CH3
R
+ H2O
+
R
N
H
H
N
CH2CH3
OH
H
O
O
Basic hydrolysis
CH2CH3
R
H
H+, heat
N
H
+ NaOH
R
O Na
+ HN-CH2CH3
H
Cl