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 / /
1-Butanol / CH3CH2CH2CH2-OH /
1-Butanethiol / CH3CH2CH2CH2-SH /
Dibutyl ether / CH3CH2CH2CH2-O-CH2CH2CH2CH3 /
1-Butene / CH3CH2CH=CH2 /
1-Butyne / /
Butanal / / Carbonyl carbon with 1 H and 1 alkyl group
2-Butanone / / Carbonyl carbon with 2 alkyl groups
Butanoic acid / / Carbonyl carbon with 1 OH and 1 alkyl group
Butanoate / / Carboxylic acid that is deprotonated (missing a H+)
Methyl Butanoate / /
1-Butamine / /
N-methyl-1-butamine / /
N,N-dimethyl-1-butamine / /
1-Butammonium ion / / Protonated amine
Butanamide / /
N-Methylbutanamide / /
N,N-Dimethylbutanamide / /
As a functional group: Butyl / R-CH2CH2CH2CH3 /
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 /Acetate (acetic acid as functional group) / / Phenol /
Benzene / / Aniline /
Phenyl (Benzene as functional group) / / Benzaldehyde /
Benzoic acid /
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 / # of carbons / # of hydrogens1 (primary) / 1 / 2
2 (secondary) / 2 / 1
3 (tertiary) / 3 / 0
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.
1 (primary) / 1 / 2
2 (secondary) / 2 / 1
3 (tertiary) / 3 / 0
Some of the reactions from the past
Reduction or Hydrogenation: Addition of hydrogen across double bonds (across C=C or C=O double bonds)
This second reaction is part of your oxidation/reduction scheme of alcohols/aldehydes/ketones/carboxylic acids.
Oxidation (sometimes called elimination): Loss of hydrogen to form a C=C or C=O bond
This second reaction is part of your oxidation/reduction scheme of alcohols/aldehydes/ketones/carboxylic acids.
Strictly speaking, the formation of a disulfide bond is considered an oxidation, since hydrogen is removed.
Hydration: Addition of water across a C=C double bond to make alcohols
Remember Markovnikov’s rule: the rich get richer – the hydrogen will add to the carbon that already has more carbons
Addition of hydrogen halide: Addition of acids such as HF, HCl, HBr or HI.
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
At higher temperature the dehydration is the preferred reaction, but at lower temperature the formation of ethers is preferred.
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 à Aldehyde à Carboxylic acid (the last step requires the presence of water)
2° Alcohol à Ketone
3° Alcohol à No reaction
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.
For reductions, you simply march backwards on the reaction schemes. Here we are adding hydrogens back in.
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:
Likewise, amines will react with strong acids to form the ammonium ion. The chlorine acts as a counter ion in the salt.
Carboxylic acids as weak acids (they can donate a proton):
Carboxylic acids are weak acids and therefore will dissociate in water
This gives a carboxylate ion, which is the conjugate base of the carboxylic acid.
And carboxylic acids will react with strong bases.
This leads to a carboxylate salt. In this case a sodium carboxylate. Water is the side product.
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.
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:
Below is an example of thioester formation.
As long as there is a proton on the amine, amide formation is possible.
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 à Carboxylic acid + alcohol
Thioester à Carboxylic acid + thiol
Amide à 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 à 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.
Acidic hydrolysis
Basic hydrolysis
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.
Acidic hydrolysis
Basic hydrolysis