The connection
between the structures of alkenes and alkanes was previously
established, which noted that we can transform an alkene into
an alkane by adding an H2 molecule across the C=C
double bond.
The driving force behind this reaction is the difference
between the strengths of the bonds that must be broken and the
bonds that form in the reaction. In the course of this
hydrogenation reaction, a relatively strong HH bond (435
kJ/mol) and a moderately strong carbon-carbon bond (270
kJ/mol) are broken, but two strong CH bonds (439 kJ/mol) are formed.
The reduction of an alkene to an alkane is therefore an
exothermic reaction.
What about the addition of an H2 molecule across a C=O double bond?
Once again, a significant amount of energy has to be invested
in this reaction to break the HH bond (435 kJ/mol) and the
carbon-oxygen bond (375 kJ/mol). The overall reaction is still
exothermic, however, because of the strength of the CH bond (439
kJ/mol) and the OH bond (498 kJ/mol) that are formed.
The addition of hydrogen across a C=O double bond raises several important points. First, and perhaps foremost, it shows the connection between the chemistry of primary alcohols and aldehydes. But it also helps us understand the origin of the term aldehyde. If a reduction reaction in which H2 is added across a double bond is an example of a hydrogenation reaction, then an oxidation reaction in which an H2 molecule is removed to form a double bond might be called dehydrogenation. Thus, using the symbol [O] to represent an oxidizing agent, we see that the product of the oxidation of a primary alcohol is literally an "al-dehyd" or aldehyde. It is an alcohol that has been dehydrogenated.
This reaction also illustrates the importance of
differentiating between primary, secondary, and tertiary
alcohols. Consider the oxidation of isopropyl alcohol, or
2-propanol, for example.
The product of this reaction was originally called aketone,
although the name was eventually softened to azetone and
finally acetone. Thus, it is not surprising that any
substance that exhibited chemistry that resembled
"aketone" became known as a ketone.
Aldehydes can be formed by oxidizing a primary alcohol; oxidation of a secondary alcohol gives a ketone. What happens when we try to oxidize a tertiary alcohol? The answer is simple: Nothing happens.
There aren't any hydrogen atoms that can be removed from the
carbon atom carrying the OH group in a 3º alcohol. And any
oxidizing agent strong enough to insert an oxygen atom into a CC bond would
oxidize the alcohol all the way to CO2 and H2O.
A variety of oxidizing agents can be used to transform a secondary alcohol to a ketone. A common reagent for this reaction is some form of chromium(VI)chromium in the +6 oxidation state in acidic solution. This reagent can be prepared by adding a salt of the chromate (CrO42-) or dichromate (Cr2O72-) ions to sulfuric acid. Or it can be made by adding chromium trioxide (CrO3) to sulfuric acid. Regardless of how it is prepared, the oxidizing agent in these reactions is chromic acid, H2CrO4.
The choice of oxidizing agents to convert a primary alcohol to
an aldehyde is much more limited. Most reagents that can oxidize
the alcohol to an aldehyde carry the reaction one step further they
oxidize the aldehyde to the corresponding carboxylic acid.
A weaker oxidizing agent, which is just strong enough to
prepare the aldehyde from the primary alcohol, can be obtained by
dissolving the complex that forms between CrO3 and
pyridine, C6H5N, in a solvent such as
dichloromethane that doesn't contain any water.
The common names of aldehydes are derived from the names of
the corresponding carboxylic acids.
The systematic names for aldehydes are obtained by adding -al
to the name of the parent alkane.
The presence of substituents is indicated by numbering the
parent alkane chain from the end of the molecule that carries the
CHO
functional group. For example,
The common name for a ketone is constructed by adding ketone
to the names of the two alkyl groups on the C=O double bond,
listed in alphabetical order.
The systematic name is obtained by adding -one to the
name of the parent alkane and using numbers to indicate the
location of the C=O group.
Formaldehyde has a sharp, somewhat unpleasant odor. The
aromatic aldehydes in the figure below, on the other hand, have a
very pleasant "flavor." Benzaldehyde has the
characteristic odor of almonds, vanillin is responsible for the
flavor of vanilla, and cinnamaldehyde makes an important
contribution to the flavor of cinnamon.
Aldehydes and ketones play an important role in the chemistry
of carbohydrates. The term carbohydrate literally means
a "hydrate" of carbon, and was introduced to describe a
family of compounds with the empirical formula CH2O.
Glucose and fructose, for example, are carbohydrates with the
formula C6H12O6. These sugars
differ in the location of the C=O double bond on the six-carbon
chain, as shown in the figure below. Glucose is an aldehyde;
fructose is a ketone
What about the addition of an H2 molecule across a C=O double bond?
The addition of hydrogen across a C=O double bond raises several important points. First, and perhaps foremost, it shows the connection between the chemistry of primary alcohols and aldehydes. But it also helps us understand the origin of the term aldehyde. If a reduction reaction in which H2 is added across a double bond is an example of a hydrogenation reaction, then an oxidation reaction in which an H2 molecule is removed to form a double bond might be called dehydrogenation. Thus, using the symbol [O] to represent an oxidizing agent, we see that the product of the oxidation of a primary alcohol is literally an "al-dehyd" or aldehyde. It is an alcohol that has been dehydrogenated.
Aldehydes can be formed by oxidizing a primary alcohol; oxidation of a secondary alcohol gives a ketone. What happens when we try to oxidize a tertiary alcohol? The answer is simple: Nothing happens.
A variety of oxidizing agents can be used to transform a secondary alcohol to a ketone. A common reagent for this reaction is some form of chromium(VI)chromium in the +6 oxidation state in acidic solution. This reagent can be prepared by adding a salt of the chromate (CrO42-) or dichromate (Cr2O72-) ions to sulfuric acid. Or it can be made by adding chromium trioxide (CrO3) to sulfuric acid. Regardless of how it is prepared, the oxidizing agent in these reactions is chromic acid, H2CrO4.
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