Video Transcript
In this video, we will learn how to
describe various reactions of alcohols and predict what products are formed. We’ll focus on dehydration,
substitution, and oxidation reactions.
Alcohols are compounds that contain
a hydroxy group, an oxygen atom bonded to a hydrogen atom. The generic formula of an alcohol
is ROH, where R takes the place of the rest of the molecule. The reactivity and behavior of an
alcohol depends largely on the position of the hydroxy group. Alcohols may be considered primary,
where the hydroxy group is bonded to a carbon atom that has one alkyl substituent;
secondary, where the hydroxy group is bonded to a carbon atom that has two alkyl
substituents; or tertiary, where the hydroxy group is bonded to a carbon atom that
has three alkyl substituents.
Now that we can recognize primary,
secondary, and tertiary alcohols, let’s take a look at some reactions involving
these alcohols. Alcohols can be produced by
reacting an alkene to double-bonded carbon atoms with water in the presence of
sulfuric acid. This is an example of a hydration
reaction, a chemical reaction in which a substance combines with water. The alcohol produced via the
hydration of an alkene can be predicted by using Markownikoff’s rule. This rule states that an acidic
hydrogen atom will add to the carbon of the double bond that has the greatest number
of hydrogen substituents.
Over the course of this reaction,
an acidic hydrogen atom is added to the carbon of the double bond with the greatest
number of hydrogen substituents. The hydroxy group from a water
molecule is added to the carbon of the double bond with the least number of hydrogen
substituents. And the pi bond between the carbon
atoms is broken. So the hydration of propene will
produce propan-2-ol. The hydration of an alkene is
actually an equilibrium reaction. In the reverse reaction, the
hydroxy group and a hydrogen atom bonded two positions away from the hydroxy group
are eliminated and a new carbon-carbon double bond is formed. This reaction is a dehydration
reaction, a chemical reaction in which a water molecule is eliminated from the
reactant.
The equilibrium between hydration
and dehydration can be shifted depending on the reaction conditions. Hydration is preferred when the
reaction is run with excess water and a low concentration of acid. Dehydration is preferred when the
water is distilled as it is produced. The equilibrium can also be shifted
by closely monitoring the temperature of the reaction. The hydration of an alkene to
produce a secondary alcohol is preferred at temperatures less than 100 degrees
Celsius. The dehydration of a secondary
alcohol to produce an alkene is preferred at temperatures greater than 140 degrees
Celsius. For the equilibrium reaction
between an alkene and a tertiary alcohol, hydration is preferred at temperatures
less than 25 degrees Celsius. And dehydration is preferred at
temperatures greater than 50 degrees Celsius.
We’ve looked at secondary and
tertiary alcohols. But what about primary
alcohols? As the hydration of an alkene
follows Markownikoff’s rule, primary alcohols cannot be produced via this
reaction. This means that the dehydration of
a primary alcohol is not in equilibrium with the hydration of the corresponding
alkene. However, the temperature of the
reaction should still be closely monitored. At temperatures around 180 degrees
Celsius, the dehydration reaction occurs as expected. But when the temperature is around
140 degrees Celsius, the alcohol does not tend to dehydrate to produce an
alkene. Instead, two alcohol molecules will
react with one another to produce a symmetrical ether. This reaction is a substitution
reaction.
A substitution reaction is a
chemical reaction where a part of a molecule is removed and replaced by something
else. In the substitution reaction
between two primary alcohols, the hydroxy group of one alcohol molecule is
removed. It is replaced by the other alcohol
molecule to produce the ether. This reaction is just one example
of a substitution reaction.
Primary, secondary, and tertiary
alcohols can undergo substitution in the presence of a hydrohalic acid, such as
hydrochloric acid, hydrobromic acid, or hydroiodic acid. Over the course of the reaction,
the hydroxy group is removed and is replaced by the halogen of the hydrohalic
acid. The products of this substitution
reaction are a haloalkane and water. When hydrochloric acid is used as
the reagent, zinc chloride is frequently added as a catalyst.
Another common alcohol reaction is
an oxidation reaction. In an oxidation reaction, the
hydrogen atom of the hydroxy group and a hydrogen atom bonded to the same carbon
atom as the hydroxy group will be removed. To make up for this loss of bonds,
a double bond will be formed between the oxygen atom and carbon atom that each lost
a hydrogen atom. The product of alcohol oxidation
will contain a carbonyl group, a carbon atom double bonded to an oxygen atom.
Primary and secondary alcohols can
be oxidized, but tertiary alcohols cannot. This is because the carbon atom
bonded to the hydroxy group must have a hydrogen substituent that can be removed
during oxidation. The carbon atom bonded to the
hydroxy group in a tertiary alcohol does not have a hydrogen substituent that could
be removed during oxidation. Therefore, oxidation of a tertiary
alcohol will not occur.
Oxidation of an alcohol requires an
oxidizing agent. Two common oxidizing agents include
acidified potassium dichromate and potassium permanganate. Acidified potassium dichromate is
also known as Jones reagent and is prepared by dissolving potassium dichromate in
aqueous sulfuric acid to produce chromic acid in solution. Acidified potassium dichromate is
orange in color due to the presence of chromium(VI). When added to a primary or
secondary alcohol, the alcohol is oxidized and the chromium(VI) is reduced to
chromium(III), changing the color of the solution to green. Potassium permanganate is violet in
color due to the presence of manganese(VII). When it is added to a primary or
secondary alcohol, the alcohol is oxidized and the manganese(VII) is reduced to
manganese(IV), resulting in a colorless solution. The observable color change due to
the reduction of these oxidizing agents can be useful for monitoring the reaction
progress.
Now let’s take a look at the
oxidation of secondary alcohols. In the reaction equation, a capital
O written in brackets can be used to represent the oxygen of an oxidizing agent. We know that the oxidation reaction
will involve the removal of two hydrogen atoms and the formation of a new
carbon-oxygen double bond. Regardless of whether we use
acidified potassium dichromate or potassium permanganate as the oxidizing agent, the
product of the oxidation of a secondary alcohol will be a ketone, an oxygen atom
double bonded to a carbon atom that has two alkyl substituents.
A similar, though slightly
different, process occurs when primary alcohols are oxidized. When an oxidizing agent is added to
a primary alcohol, the hydrogen atoms are removed and a new carbon-oxygen double
bond is formed. The product of this reaction is an
aldehyde, an oxygen atom double bonded to a carbon atom that has at least one
hydrogen substituent. But aldehydes are easily
oxidized. If there is still oxidizing agent
and water present in the reaction vessel, the newly formed aldehyde will be oxidized
in a series of complex steps to produce a carboxylic acid, an oxygen atom double
bounded to a carbon atom that has a hydroxy substituent. Thus, the oxidation of primary
alcohols can produce both aldehydes and carboxylic acids. The reaction conditions used will
determine which product we’re most likely to produce.
Potassium permanganate is a strong
oxidizing agent. It will oxidize the alcohol to an
aldehyde, which will rapidly oxidize to a carboxylic acid. Thus, the major product under these
reaction conditions is a carboxylic acid. Notice that this reaction scheme is
not a balanced chemical equation and only shows the starting material, reagent, and
major organic product. Another way to produce a carboxylic
acid via oxidation is by using excess acidified potassium dichromate under
reflux. These reaction conditions ensure
that any aldehydes produced via initial oxidation are further oxidized to the
carboxylic acid. If the aldehyde is the desired
product, we need to take care to prevent the aldehyde from undergoing oxidation.
Acidified potassium dichromate can
still be used, though not in excess. And we won’t heat the reaction
under reflux. In addition, we should use excess
alcohol and distill the aldehyde as it is produced. In a practical laboratory setting,
specialized reagents, such as pyridinium chlorochromate or PCC, are commonly used to
oxidize primary alcohols to aldehydes, as they do not allow the aldehyde to undergo
further oxidation.
We have now looked at dehydration,
substitution, and oxidation reactions involving alcohols. But we have one final reaction to
consider. Alcohols are both weak acids and
weak bases. When combined with a strong base,
the proton of the hydroxy group can be removed. The resulting negatively charged
species is called an alkoxide. Alkoxides are the conjugate base of
an alcohol and are important reagents for several organic syntheses. A common strong base used to
convert an alcohol into an alkoxide is sodium amide. Alkoxides can also be produced by
reacting an alcohol with sodium, potassium, or lithium metal. An oxidation–reduction reaction
occurs and hydrogen gas is liberated from the reaction.
Before we summarize what we’ve
learned, let’s take a look at a question.
The reaction scheme below shows
various reactions a molecule of propan-2-ol can undergo. Which product would be a
ketone?
Propan-2-ol is a secondary
alcohol. When a secondary alcohol is reacted
with a hydrohalic acid, such as hydrochloric acid, a substitution reaction occurs
and a haloalkane is produced. When a secondary alcohol reacts
with concentrated sulfuric acid at temperatures greater than 180 degrees Celsius, a
dehydration reaction occurs and an alkene is produced. When a secondary alcohol reacts
with acidified potassium dichromate under reflux, an oxidation reaction occurs and a
ketone, in this case propanone, is produced. Therefore, the product that would
be a ketone is B.
Part (b) what name would product C
have? (A) 2-Chloropropan-2-ol, (B)
2-chloropropane, (C) 1-chloropropane.
Product C was formed via a
substitution reaction. A substitution reaction is a
chemical reaction where a part of the molecule is removed and replaced with
something else. In this substitution reaction, the
hydroxy group is removed and is replaced by the halogen, in this case chlorine. This means that the product will
have a chlorine atom bonded to the second carbon atom, the same position as the
hydroxy group in the original molecule. To name this haloalkane, we
indicate the position of the chlorine atom by writing two dash chloro, where the
number two indicates that the chlorine atom is bonded to the second carbon atom. Then, we name the carbon base chain
propane, prop- for three carbon atoms and -ane for alkane, indicating that the
carbon atoms are single bonded. The correct name of product C is
answer choice (B) 2-chloropropane.
For the next part of the question,
I will isolate the important information.
Part (c) how many different
positional isomers would result from the reaction to form product A? (A) One isomer, (B) two isomers,
(C) four isomers, (D) three isomers.
We have already established that
this reaction is a dehydration reaction, a chemical reaction in which a water
molecule is eliminated from the reactant. Over the course of this reaction,
the hydroxy group and one hydrogen atom bonded to a carbon atom two positions away
from the hydroxy group will be removed. And a new carbon-carbon double bond
will form between the two carbon atoms that each lost a substituent. There are six possible hydrogen
atoms that can be removed in addition to the hydroxy group. If any of the leftmost hydrogen
atoms are lost during dehydration, a double bond will be formed between the leftmost
and middle carbon atoms, giving us the product propene. If any of the rightmost hydrogen
atoms are lost during dehydration, a double bond will be formed between the middle
and rightmost carbon atoms. This would also produce the product
propene. As the products are the same, the
correct answer choice is answer choice (A). Only one positional isomer would
result from this reaction.
Now let’s summarize what we’ve
learned with the key points. Alcohols can be dehydrated with
sulfuric acid to produce alkenes. The dehydration of secondary and
tertiary alcohols is an equilibrium reaction. The temperature at which the
reaction has run can shift the equilibrium. In the presence of sulfuric acid at
lower temperatures, primary alcohols may react to form symmetrical ethers. Primary, secondary, and tertiary
alcohols can undergo substitution with hydrohalic acids to produce haloalkanes. Secondary alcohols are oxidized to
ketones, while primary alcohols are oxidized to aldehydes or carboxylic acids,
depending on the reaction conditions. Reacting an alcohol with sodium
metal will produce hydrogen gas and an alkoxide, an important organic reagent.
