Chapter 14 Flashcards

1
Q

What important properties do alcohols have?

A
  • They are polar and can form hydrogen bonds
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2
Q

List 3 differences between the properties of alkanes and alcohols.

A
  • Alcohols have higher boiling points,
  • are less volatile
  • and are more soluble
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3
Q

Why do alcohols have higher boiling points than alkanes?

A
  • They have hydrogen bonds
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4
Q

How does the boiling point of alcohols change as the chain length increases, and why?

A
  • As alcohols get longer, their boiling points become more similar to those of alkanes
  • This is because the contribution of the -OH decreases
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5
Q

Why are alcohols less volatile than alkanes of the same chain length?

A
  • The higher the boiling point of a substance, the lower its volatility
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6
Q

Why are alcohols more soluble than alkanes?

A
  • Hydrogen bonds form between the alcohol and water molecules, so alcohols are very soluble
  • Alkanes are non-polar so they are insoluble
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7
Q

How can alcohols be classified?

A
  • According to the number of other carbon atoms the carbon atom they are bonded to is bonded to
  • In a primary (1°) alcohol, the carbon the hydroxyl group is attached to is bonded to one other carbon (e.g. ethanol)
  • In a secondary (2°) alcohol, it is bonded to 2
  • In a tertiary (3°) alcohol, it is bonded to 3
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8
Q

What happens when alcohols undergo combustion in a plentiful supply of oxygen?

A
  • They produce carbon dioxide and water
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9
Q

How does the chain length of an alcohol affect its combustion?

A
  • The longer the chain, the more energy released per mole
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10
Q

What oxidising agent is used to oxidate alcohols?

A
  • Acidified potassium dichromate (VI)
  • Written as: K2Cr2O7/ H2SO4
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11
Q

How can you tell if an alcohol is oxidised in the presence of potassium dichromate(VI), and why?

A
  • The orange solution containing the dichromate ions goes from orange to green
  • This is because the solution containing dichromate (VI) ions is reduced to a solution containing chromium (III) ions
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12
Q

What can primary alcohols be oxidised into, and what causes the difference?

A
  • Aldehydes, when heated gently with acidified potassium dichromate and distilled as it’s formed
  • Carboxylic acids, when heated under reflux with an excess of potassium dichromate
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13
Q

Why must alcohols be heated gently and distilled when forming aldehydes?

A
  • To prevent further oxidation into a carboxylic acid
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14
Q

Why are alcohols heated strongly under reflux to make carboxylic acids?

A
  • This ensures any aldehyde that is initially made is oxidised
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15
Q

Why are alcohols heated with excess potassium dichromate to form carboxylic acids?

A
  • To ensure all of the alcohol is oxidised
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16
Q

How would the oxidation of propan-1-ol to form propanal be shown?

A
  • Propan-1-ol + [O]
  • Arrow: K2Cr2O7/ H2SO4, distill immediately
  • Propanal + H2O
17
Q

What does [O] stand for?

A
  • The oxidising agent
18
Q

How would the oxidation of propan-1-ol to form propanoic acid be shown?

A
  • Propan-1-ol + 2[O]
  • Arrow: K2Cr2O7/ H2SO4, heat under reflux
  • Propanoic acid + H2O
19
Q

What can secondary alcohols be oxidised into?

A
  • Ketones
20
Q

How can the oxidation of butan-2-ol be shown?

A
  • Butan-2-ol + [O]
  • Arrow: K2Cr2O7/ H2SO4, heat under reflux
  • Butanone + H2O
21
Q

Why are secondary alcohols heated under reflux during their oxidation?

A
  • To ensure the reaction goes to completion
22
Q

What happens when tertiary alcohols are mixed with potassium dichromate?

A
  • Nothing; they can’t be oxidised
  • The potassium dichromate remains orange
23
Q

How do alcohols react to form alkenes?

A
  • The alcohol is heated under reflux with either phosphoric acid or concentrated sulfuric acid catalyst
  • An alkene and water are produced
  • It is a dehydration reaction, which is a type of elimination reaction
24
Q

What do alcohols react with to form haloalkanes, and what type of reaction is this?

A
  • Alcohols react with hydrogen halides to form haloalkanes
  • Substitution
25
Q

How do alcohols form haloalkanes?

A
  • The hydrogen halide is formed ‘in situ’; within the reaction
  • Sulfuric acid and a sodium halide are used to do this
  • Step 1: NaX + H2SO4 -> NaHSO4 + HX
  • Step 2: HX + C3H7OH -> C3H7X + H2O
  • Overall equation: C3H7OH + NaX + H2SO4 -> C3H7X + NaHSO4 + H2O