16. Alcohols Flashcards
(17 cards)
What are the six methods of producing alcohols?
- Hydration of alkenes (electrophilic addition)
- Reaction of alkenes with cold dilute potassium permanganate to form a diol
- Substitution of a halogenoalkane with aqueous NaOH
- Reduction of an aldehyde or ketone using NaBH₄ or LiAlH₄
- Reduction of a carboxylic acid using LiAlH₄
- Hydrolysis of an ester using dilute acid or dilute alkali
What is the equation for the complete combustion of ethanol in oxygen?
C₂H₅OH + 3O₂ –> 2CO₂ + 3H₂O
What is the reaction of ethanol with sodium?
- The H atom breaks from the alcohol and is replaced by a sodium atom, which forms an ionic bond with the oxygen
- The result is the metal alkoxide sodium ethoxide
- 2C₂H₅OH + 2Na –> 2C₂H₅ONa + H₂
Other group one metals will react analogously
How are alcohols classified?
- Based on the number of carbon atoms the carbon atom bonded to the hydroxyl group is bonded to
- Primary alcohols have one carbon bonded to the hydroxyl carbon
- Secondary alcohols have two
- Tertiary alcohols have three
- In alcohols with multiple hydroxyl groups (e.g. diols), each carbon bonded to a hydroxyl group is classified separately, as each can undergo characteristic reactions of its respective class of alcohol
What are the potential products of the oxidation of alcohols and which class of alcohols are needed to form each?
- Aldehydes, formed from primary alcohols
- Carboxylic acids, formed from primary alcohols
- Ketones, formed from secondary alcohols
Do tertiary alcohols readily undergo oxidation
No
How can aldehydes be formed through the oxidation of alcohols?
- A primary alcohol is reacted with either acidified K₂Cr₂O₇ or acidified KMnO₄ (though usually the former)
- It should be heated gently and not under reflux
- The carbon with the hydroxyl group is oxidised, forming an aldehyde
- A water molecule is produced due to the hydrogens lost gaining oxygen
- To retain the aldehyde as a product, it must be distilled off as it forms before further oxidation can occur (aldehydes have lower melting points than alcohols)
How can ketones be formed through the oxidation of alcohols?
- A secondary alcohol is reacted with either acidified K₂Cr₂O₇ or acidified KMnO₄
- It should be heated, though not necessarily under reflux
- The carbon with the hydroxyl group is oxidised, forming a ketone
- A water molecule is produced due to the hydrogens lost gaining oxygen
- The ketone is then distilled off, though there is no rush as the ketone cannot be further oxidised
How can carboxylic acids be formed through the oxidation of alcohols?
- A primary alcohol is reacted with either acidified K₂Cr₂O₇ or acidified KMnO₄
- It should be heated under reflux
- The carbon with the hydroxyl group is oxidised, forming an aldehyde
- A water molecule is produced due to the two hydrogens lost gaining oxygen
- The aldehyde is further oxidised with the addition of an oxygen to the hydrogen
- For the aldehyde to oxidise into a carboxylic acid, the reaction must be continuously heated under reflux to ensure no aldehyde escapes
What is the colour change observed when acidified K₂Cr₂O₇ is used to oxidise another species?
- Orange to green
- The orange dichromate ions (Cr₂O₇²⁻) are reduced by the H⁺ ions provided by the acid to form green chromium ions (Cr³⁺)
What is the colour change observed when acidified KMnO₄ is used to oxidise another species?
- Purple to colourless
- The purple permanganate ions (MnO₄⁻) are reduced by the H⁺ ions provided by the acid to form colourless chromium ions (Mn²⁺)
Potassium permanganate is generally a stronger oxidising agent than potassium dichromate
What are the two ways of differentiating between alcohols?
The explanations are in other cards
- Attempting to oxidise with potassium dichromate and observing a colour change to differentiate between primary and secondary alcohols and tertiary alcohols
- The iodoform test
Which alcohols will the iodoform test be positive for?
- Alcohols in which the hydroxyl group is bonded to a carbon atom with an adjacent methyl group (CH₃CH(OH))
- Ethanol is the only primary alcohol where this is the case, though there are several secondary alcohols, e.g. propan-2-ol, that will give a positive result
How is the iodoform test used to identify certain alcohols?
- Heat the alcohol with I₂ in aqueous sodium hydroxide
- If the hydroxyl group is on a carbon atom next to a methyl group, it will be oxidised by the iodine and alkaline conditions to form a methyl ketone (RCOCH₃) or ethanal (the only methyl aldehyde) in the case of ethanol
- The methyl ketone (or aldehyde) will then be halogenated (iodines replace hydrogens on methyl group)
- This intermediate undergoes base hydrolysis to form a sodium carboxylate salt, which will consist of an RCO₂⁻ ion (and an Na⁺ ion, though the salt is in aqueous solution)
- A yellow precipitate will also form called tri-iodomethane, commonly known as iodoform (this is the positive test result)
Look at the image
What is the balanced equation for the iodoform test with ethanol?
Everything in one step and omitting spectator ions
CH₃CH₂OH + 4I₂ + 6OH⁻
—> CHI₃ + HCOO⁻ + 5I⁻ + 5H₂O
To what extent are alcohols acidic?
Make a comparison to the acidity of water
- They are weakly acidic
- This is because, when dissolved in water, ROH dissociates into RO⁻ (aq) + H⁺ (aq) with the equilibrium sitting very far to the left
- Water is also a weak acid, but when it dissociates into H⁺ and OH⁻, the equilibrium sits further to the right, making it a stronger acid than alcohols
Why does the equilibrium sit further to the right when water dissociates compared to when alcohols dissociate?
- In alkoxide ions, the adjacent alkyl group donates electron density to the oxygen ion
- This makes it readily accept hydrogen
- In water, the oxygen in the OH is only attached to a hydrogen, so there is no inductive effect
- This oxygen will not accept H+ as readily, so equilibrium stays further to the right, making water a stronger acid than alcohols
For water, equilbrium still lies relatively far to the left
