Mason Delacroix
Alcohols are versatile compounds that serve as important intermediates in organic synthesis. One common reaction involving alcohols is dehydration, which involves the removal of water. This process can be achieved by warming the alcohol with a strong dehydrating acid, such as sulfuric acid, or phosphoric acid. The hydroxyl group (-OH) is protonated and eliminated as water, leaving a carbocation that can be stabilized by losing a proton from an adjacent carbon atom, resulting in the formation of an alkene. This type of elimination reaction, known as E1, follows Zaitsev's rule, where the most substituted alkene is favored. Dehydration can also be achieved through catalytic dehydrogenation, which involves removing hydrogen molecules from the alcohol. Additionally, alcohols can undergo oxidation to form ketones, aldehydes, and carboxylic acids, or substitution reactions with halide ions to form haloalkanes. Understanding these reactions is crucial for various applications, including the production of industrial solvents and the conversion of fats and oils in food manufacturing.
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What You'll Learn
Using phosphorous oxychloride (POCl3) in pyridine
Phosphorus oxychloride (POCl3) is a reagent that can be used to remove a hydrogen from an alcohol. POCl3 is a derivative of phosphoric acid. When POCl3 is added to an alcohol, a new O-P bond is formed and a P-Cl bond is broken, resulting in the formation of a "chlorophosphate ester". This is a good leaving group.
The process of removing a hydrogen from an alcohol using POCl3 is called dehydration, where water is removed from an alcohol to form an alkene. This reaction is facilitated by pyridine, a base that performs an E2 elimination to give a more substituted double bond. Pyridine removes a β-proton, which provides the electrons for making the C=C π bond. This reaction is useful when working with compounds that decompose in the presence of strong acids.
The E2 elimination of 3º-alcohols under relatively non-acidic conditions can be achieved by treating them with POCl3 in pyridine. This procedure is also effective for hindered 2º-alcohols. However, for unhindered and 1º-alcohols, an SN2 chloride ion substitution of the chlorophosphate intermediate competes with elimination. The predominance of the non-Zaitsev product (less substituted double bond) is due to the steric hindrance of the methylene group hydrogen atoms, which interferes with the approach of the base.
Overall, the use of POCl3 in pyridine provides a useful method for cleanly performing elimination reactions on alcohols, converting them into alkenes.
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Base-induced E2 elimination
The E2 elimination reaction is a process where a hydrogen halide (HX) is eliminated from an alkyl halide (RX) when treated with a strong base such as hydroxide or alkoxides. This reaction produces an alkene. The E2 reaction is usually considered synchronous with no intermediates. The rate of the E2 reaction depends on the concentration of the substrate and the base.
In the context of removing hydrogen from an alcohol, the E2 mechanism is indeed applicable, but only to primary alcohols. Secondary and tertiary alcohols follow the E1 mechanism. The E2 mechanism is favoured because primary carbocations are highly unstable and cannot be formed, unlike in the case of secondary and tertiary alcohols. The E2 mechanism is more energetically favourable.
The E2 mechanism involves the protonation of the primary alcohol, turning it into a good leaving group. The base (water or bisulfate ion) then attacks the beta hydrogen, which leaves a pair of electrons, kicking out the protonated OH group and making a double bond. These processes happen simultaneously, which is why it is a bimolecular reaction.
An example of an E2 mechanism in alcohol elimination is seen with ethanol. The lone pair of electrons from the oxygen atom attacks a hydrogen ion from the acid catalyst, protonating the oxygen. The E2 mechanism is also applicable to 1º, 2º, and 3º alcohols when using POCl3 in the presence of a base, thus avoiding rearrangements.
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Using strong bases like hydroxide and alkoxides
When removing a hydrogen from an alcohol, one of the methods that can be used involves strong bases like hydroxide and alkoxides. This method involves the use of elimination reactions, which are favoured by stronger acids. For instance, sulfuric acid is commonly used to make olefins from alcohols. The use of hydrohalic acids like HCl, HBr, or HI can also result in the formation of alkyl halides.
Alcohols themselves do not undergo base-induced elimination reactions. Instead, they are often used as solvents for such reactions. This is because most alcohols are slightly weaker acids than water, and the left side is favoured in the acid-base equilibrium that occurs when an alcohol is treated with sodium hydroxide.
However, when it comes to using strong bases like hydroxide and alkoxides, it is important to consider the stability of leaving groups. Water, for example, is a much better leaving group than the hydroxide ion. Therefore, acid-catalysis is often preferred over base-catalysis in these reactions. Additionally, hydrohalic acids are typically avoided as catalysts due to the nucleophilic nature of their conjugate bases, which may lead to substitution products.
Alkoxides, which are the conjugate acids of alcohols, are stronger bases than hydroxide. They are good nucleophiles, but their use requires caution as their base strength tends to dominate their reactivity. Alkoxides are often used in reactions with alkyl halides, where they react with a hydrogen atom to form their conjugate acid. This type of reaction is known as the Williamson Ether Synthesis and is commonly used to synthesise ethers in laboratories.
An example of a reaction involving an alkoxide is the reaction between sodium and ethanol, where a small piece of sodium is dropped into ethanol, resulting in the formation of bubbles of hydrogen gas and a colourless solution of sodium ethoxide. This reaction can be represented by the equation:
\[\ce {2CH3CH2OH + 2Na -> 2CH3CH2ONa + H2}\]
In summary, while strong bases like hydroxide and alkoxides can be used to remove a hydrogen from an alcohol, it is important to carefully consider the specific reactants and conditions to ensure the desired outcome.
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Using heat and sulfuric acid
To remove a hydrogen from an alcohol, you can use a process called dehydration, which involves applying heat and an acid catalyst, such as sulfuric acid or phosphoric acid. This process eliminates water from the alcohol by removing the -OH group from one carbon atom and an H atom from a neighbouring carbon atom.
Sulfuric acid is a strong acid and a dehydrating agent, which means it can 'suck out' the OH group with the help of heat. This process results in the formation of an alkene, which is a type of hydrocarbon with a double bond. The specific alkene formed will depend on the type of alcohol used. For example, if you start with a secondary alcohol, such as propan-2-ol, and heat it with sulfuric acid, you will end up with an alkene called propanone.
It's important to note that not all alcohols will undergo dehydration at the same rate. Primary alcohols are less susceptible to dehydration because they follow a different reaction mechanism (E2) that progresses slowly in acidic conditions. On the other hand, secondary and tertiary alcohols are more prone to dehydration and will readily form alkenes when heated with sulfuric acid.
The amount of heat and acid concentration also play a role in the effectiveness of the process. Increasing the temperature and using excess concentrated sulfuric acid generally favours the elimination of water from the alcohol. However, it's worth mentioning that using sulfuric acid can produce messy results and side reactions, such as the oxidation of alcohol to carbon dioxide and the reduction of sulfuric acid to sulfur dioxide. These gases need to be removed from the final product.
In summary, removing hydrogen from an alcohol can be achieved through dehydration using heat and sulfuric acid. The specific conditions and outcomes depend on the type of alcohol used, and it's important to be mindful of potential side reactions when employing sulfuric acid as a catalyst.
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Using HI/red phosphorous
The HI/P reduction method, sometimes called the Nagai method, is a two-step reduction process that uses hydroiodic acid and red phosphorus to reduce organic compounds like alcohols to their respective alkanes. This method was first described by Heinrich Kiliani over 140 years ago.
The first step involves the nucleophilic substitution of the OH group by I- through the protonation of the alcohol group by hydroiodic acid. This step is facilitated by the addition of a small amount of red phosphorus to the reaction mass, which reacts with the iodine formed as a byproduct, producing phosphorus triiodide. This phosphorus triiodide then rapidly hydrolyzes in water, releasing phosphorous acid and hydrogen iodide.
In the second step, the hydrogen iodide produced in the previous step protonates the alkyl radical species, resulting in the formation of an alkane. This step can be summarized by the following equation:
> C2H5OH + HI → C2H6 + H2O + I2
Here, C2H5OH represents ethyl alcohol, C2H6 is ethane (the alkane product), and the presence of red phosphorus (P) enhances the reducing ability of HI.
This reduction process is powerful enough to remove oxygen from any organic compound, and it is commercially used in the fine chemicals and specialty chemicals industry. For example, it is used in the reduction of 4-hydroxymandelic acid to 4-hydroxyphenylacetic acid, an intermediate in the production of the beta blocker "atenolol."
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Frequently asked questions
The first step in removing hydrogen from an alcohol is to identify the type of alcohol you are working with. The process differs for primary, secondary, and tertiary alcohols.
To remove hydrogen from a primary alcohol, you would typically perform a dehydration reaction. This involves warming the alcohol in the presence of a strong acid, such as sulfuric acid or phosphoric acid. This will remove a hydroxyl group (-OH) and a hydrogen atom from neighbouring carbon atoms, forming an alkene.
Yes, another method is to use a substitution reaction with halide ions in the presence of an acid catalyst, such as sulfuric acid (H2SO4) or phosphoric acid (H3PO4). This will replace the hydroxyl group with a halide ion, forming a haloalkane.
Removing hydrogen from an alcohol is often associated with oxidation reactions. While dehydration and substitution reactions directly remove hydrogen, oxidation reactions can indirectly reduce the number of hydrogen bonds by increasing the number of carbon-oxygen bonds.
