01 Free Studying Resources 07 Junior College 1 Chemistry

Chapter 13: Hydroxy Compounds

July 5, 2021 by Admin

Chapter 13
Hydroxy Compounds

Before diving into the chemical reactions that hydroxy compounds undertake, it is important for students to know some of the important physical properties of such compounds. Below is a concise table which details the physical properties of hydroxy compounds:

Boiling point	Solubility
(i) Boiling point of alcohols is much higher than alkanes of similar numbers of electrons Since the compounds have a comparable number of electrons, the alcohol have a much higher boiling point because of its ability to form hydrogen bonds between its molecules (ii) Boiling point increases with increasing length of alkyl chain More energy required to break the stronger instantaneous dipole induced dipole interactions between molecules due to the larger and more polarisable electron cloud (iii) Branching of the carbon skeleton in the alcohol molecule lowers the boiling point With branching, molecules become more spherical (more compact) hence there is less surface area of contact for id-id interactions than in straight chain isomers Less energy required to overcome the less extensive id-id interactions (iv) Boiling point increases with the number of -OH groups in the molecule An increase in the number of -OH group leads to more extensive hydrogen bonding More energy required to overcome the more extensive hydrogen bonds	In water: (i) Alcohols are soluble in water Energy released by hydrogen bonding between alcohol molecules and water is sufficient to overcome the hydrogen bonding between alcohol molecules and hydrogen bonding between water molecules (ii) As hydrocarbon chain becomes longer, the respective alcohols shows a marked decrease in solubility The hydrocarbon portion of an alcohol is non-polar and as the carbon chain increases, the id-id interactions between ROH molecule becomes more predominant
In organic solvents: (i) Alcohols are soluble in organic solvents Energy released by instantaneous dipole induced dipole interactions between alcohol molecules and the organic solvent is sufficient to overcome id-d interactions between organic solvent molecules and hydrogen bonding between alcohol molecules

Boiling point

Solubility

(i) Boiling point of alcohols is much higher than alkanes of similar numbers of electrons

Since the compounds have a comparable number of electrons, the alcohol have a much higher boiling point because of its ability to form hydrogen bonds between its molecules

(ii) Boiling point increases with increasing length of alkyl chain

More energy required to break the stronger instantaneous dipole induced dipole interactions between molecules due to the larger and more polarisable electron cloud

(iii) Branching of the carbon skeleton in the alcohol molecule lowers the boiling point

With branching, molecules become more spherical (more compact) hence there is less surface area of contact for id-id interactions than in straight chain isomers
Less energy required to overcome the less extensive id-id interactions

(iv) Boiling point increases with the number of -OH groups in the molecule

An increase in the number of -OH group leads to more extensive hydrogen bonding
More energy required to overcome the more extensive hydrogen bonds

In water:

(i) Alcohols are soluble in water

Energy released by hydrogen bonding between alcohol molecules and water is sufficient to overcome the hydrogen bonding between alcohol molecules and hydrogen bonding between water molecules

(ii) As hydrocarbon chain becomes longer, the respective alcohols shows a marked decrease in solubility

The hydrocarbon portion of an alcohol is non-polar and as the carbon chain increases, the id-id interactions between ROH molecule becomes more predominant

In organic solvents:

(i) Alcohols are soluble in organic solvents

Energy released by instantaneous dipole induced dipole interactions between alcohol molecules and the organic solvent is sufficient to overcome id-d interactions between organic solvent molecules and hydrogen bonding between alcohol molecules

Boiling point
(i) Boiling point of alcohols is much higher than alkanes of similar numbers of electrons Since the compounds have a comparable number of electrons, the alcohol have a much higher boiling point because of its ability to form hydrogen bonds between its molecules (ii) Boiling point increases with increasing length of alkyl chain More energy required to break the stronger instantaneous dipole induced dipole interactions between molecules due to the larger and more polarisable electron cloud (iii) Branching of the carbon skeleton in the alcohol molecule lowers the boiling point With branching, molecules become more spherical (more compact) hence there is less surface area of contact for id-id interactions than in straight chain isomers Less energy required to overcome the less extensive id-id interactions (iv) Boiling point increases with the number of -OH groups in the molecule An increase in the number of -OH group leads to more extensive hydrogen bonding More energy required to overcome the more extensive hydrogen bonds

Boiling point

(i) Boiling point of alcohols is much higher than alkanes of similar numbers of electrons

Since the compounds have a comparable number of electrons, the alcohol have a much higher boiling point because of its ability to form hydrogen bonds between its molecules

(ii) Boiling point increases with increasing length of alkyl chain

More energy required to break the stronger instantaneous dipole induced dipole interactions between molecules due to the larger and more polarisable electron cloud

(iii) Branching of the carbon skeleton in the alcohol molecule lowers the boiling point

With branching, molecules become more spherical (more compact) hence there is less surface area of contact for id-id interactions than in straight chain isomers
Less energy required to overcome the less extensive id-id interactions

(iv) Boiling point increases with the number of -OH groups in the molecule

An increase in the number of -OH group leads to more extensive hydrogen bonding
More energy required to overcome the more extensive hydrogen bonds

Solubility
In water: (i) Alcohols are soluble in water Energy released by hydrogen bonding between alcohol molecules and water is sufficient to overcome the hydrogen bonding between alcohol molecules and hydrogen bonding between water molecules (ii) As hydrocarbon chain becomes longer, the respective alcohols shows a marked decrease in solubility The hydrocarbon portion of an alcohol is non-polar and as the carbon chain increases, the id-id interactions between ROH molecule becomes more predominant
In organic solvents: (i) Alcohols are soluble in organic solvents Energy released by instantaneous dipole induced dipole interactions between alcohol molecules and the organic solvent is sufficient to overcome id-d interactions between organic solvent molecules and hydrogen bonding between alcohol molecules

Solubility

In water:

(i) Alcohols are soluble in water

Energy released by hydrogen bonding between alcohol molecules and water is sufficient to overcome the hydrogen bonding between alcohol molecules and hydrogen bonding between water molecules

(ii) As hydrocarbon chain becomes longer, the respective alcohols shows a marked decrease in solubility

The hydrocarbon portion of an alcohol is non-polar and as the carbon chain increases, the id-id interactions between ROH molecule becomes more predominant

In organic solvents:

(i) Alcohols are soluble in organic solvents

Energy released by instantaneous dipole induced dipole interactions between alcohol molecules and the organic solvent is sufficient to overcome id-d interactions between organic solvent molecules and hydrogen bonding between alcohol molecules

Take a look at the Tri-iodomethane Test (also known as Iodoform Test):

R can be a H, alkyl or aryl group.

Tip: Equation above should be memorised! The time taken to recall and balance this equation could have been better spent on other challenging questions during the exam.

Reagents and conditions: I₂(aq), NaOH (aq), warm
Observation: Yellow precipitate CHI₃ is formed

Note: This iodoform test can help identify the presence of a CH₃CH(OH)– group

The following table below would clearly summarise the 3 golden tests to distinguish alcohols from other organic compounds:

Reagent / Conditions	Observation with alcohols
K₂Cr₂O₇ (aq), H₂SO₄(aq), heat	Dichromate solution changes colour from orange to green. (only for 1° and 2° alcohols)
Iodoform Test I₂ (aq), NaOH (aq), warm	Yellow precipitate CHI₃ is formed
PCl₅	White HCl fumes (highly not recommended to use this method since fumes cannot be seen easily)

Reagent / Conditions

Observation with alcohols

K₂Cr₂O₇ (aq), H₂SO₄(aq), heat

Dichromate solution changes colour from orange to green.

(only for 1° and 2° alcohols)

Iodoform Test

I₂ (aq), NaOH (aq), warm

Yellow precipitate CHI₃ is formed

PCl₅

White HCl fumes

(highly not recommended to use this method since fumes cannot be seen easily)

Reagent / Conditions
K₂Cr₂O₇ (aq), H₂SO₄(aq), heat
Iodoform Test I₂ (aq), NaOH (aq), warm
PCl₅

Reagent / Conditions

K₂Cr₂O₇ (aq), H₂SO₄(aq), heat

Iodoform Test

I₂ (aq), NaOH (aq), warm

PCl₅

Observation with alcohols
Dichromate solution changes colour from orange to green. (only for 1° and 2° alcohols)
Yellow precipitate CHI₃ is formed
White HCl fumes (highly not recommended to use this method since fumes cannot be seen easily)

Observation with alcohols

Dichromate solution changes colour from orange to green.

(only for 1° and 2° alcohols)

Yellow precipitate CHI₃ is formed

White HCl fumes

(highly not recommended to use this method since fumes cannot be seen easily)

Controlled Oxidation

Aldehydes, which will be introduced in the next chapter, can be formed from primary alcohols through controlled oxidation. As the primary alcohol is heated, the resultant aldehyde formed is immediately removed by distillation to prevent further oxidation to carboxylic acid.

Reagents and conditions	K₂Cr₂O₇ (aq), H₂SO₄ (aq), heat with immediate distillation
Equation	CH₃CH₂OH + [O] → CH₃CHO + H₂O
Equation

Reagents and conditions
K₂Cr₂O₇,(aq), H₂SO₄ (aq), heat with immediate distillation

Equation
CH₃CH₂OH + [O] → CH₃CHO + H₂O

Normally, organic reactions take place with heating under reflux to increase the rate of reaction. Since organic compounds commonly are volatile, they vaporise easily when heated, hindering the reaction from proceeding. Hence when heating under reflux, the attached condenser will condense any vapour, preventing reagents from leaving.

In a controlled oxidation of a primary alcohol to an aldehyde however, heat with immediate distillation is used. The reaction mixture is warmed to a temperature above the boiling point of aldehyde but below that of the alcohol. Hence the aldehyde distils once it is formed, preventing it from being further oxidised to a carboxylic acid, while alcohol is left behind.

Relative Acidity of Phenols and Alcohols

This is an extremely important concept to understand. Pure memorisation does not completely work as exam setters tend to include this concept into an application context-based question, so it requires students to answer according to the question. Hence, we would highly recommend students to grasp this concept through understanding rather than memorisation.

Firstly, you need to know that phenol is a weak acid.
Comparing the relative acidities:

Carboxylic acid > phenol > water > ethanol

However, the pKₐ is the opposite:

Carboxylic acid < phenol < water < ethanol

Linking back to the chapter on Acid-Base Equilibria, we need to recall that relative acid strength depends on the relative stability of the conjugate base. The greater the stability of the conjugate base, the more acidic is the conjugate acid.

Also, do recall the Le Chatelier’s Principle from Chemical Equilibria:

The more stable the conjugate base, Position of Equilibrium shifts further to the right. Thus, more H⁺ is formed.

Let’s take a closer look at the various conjugate bases:

Conjugate Base	Comparison of Relative Acidities
	The negative charge on oxygen is delocalised into the benzene ring, reducing the negative charge on the oxygen. Hence, the phenoxide ion is more stable. By LCP, the equilibrium position shifts more to the right, hence phenol is more acidic and accepts protons less readily.
	The existence of an electron-donating ethyl group causes the negative charge on oxygen to intensify. Hence, ethoxide is less stable. By LCP, the equilibrium position shifts less to the right, hence ethanol is less acidic and accepts protons more readily.
	The existence of an electron-withdrawing group (–Cl) on the phenoxide ion decreases the intensity of the negative charge on oxygen, thus it is more stable. By LCP, the equilibrium position shifts more to the right, hence chlorophenol is more acidic and accepts protons less readily.

Conjugate Base

Comparison of Relative Acidities
The negative charge on oxygen is delocalised into the benzene ring, reducing the negative charge on the oxygen. Hence, the phenoxide ion is more stable. By LCP, the equilibrium position shifts more to the right, hence phenol is more acidic and accepts protons less readily.
The existence of an electron-donating ethyl group causes the negative charge on oxygen to intensify. Hence, ethoxide is less stable. By LCP, the equilibrium position shifts less to the right, hence ethanol is less acidic and accepts protons more readily.
The existence of an electron-withdrawing group (–Cl) on the phenoxide ion decreases the intensity of the negative charge on oxygen, thus it is more stable. By LCP, the equilibrium position shifts more to the right, hence chlorophenol is more acidic and accepts protons less readily.

Condensation of phenol is only able to take place with an acyl chloride. Take note that phenol does not react with carboxylic acid to give an ester, unlike alcohols.

Reactions	Ester produced?
Alcohol + Carboxylic Acid	Yes
Alcohol + Acyl Chloride	Yes
Phenol + Carboxylic Acid	No
Phenol + Acyl Chloride	Yes

Reactions	Ester produced?
Alcohol + Carboxylic Acid	Yes
Alcohol + Acyl Chloride	Yes
Phenol + Carboxylic Acid	No
Phenol + Acyl Chloride	Yes

In the reaction between phenol and acyl chloride, two different solvents can be used depending on the acyl chloride. The two solvents are basic so that the HCl produced in the reaction is neutralised. For aliphatic acyl chlorides, pyridine is typically the solvent. However for aromatic acyl chloride, NaOH is used as the solvent.

The difference in solvents used can be tied back to the differences in the nature of aromatic and aliphatic compounds.

Since the base is chosen to serve as a solvent, it should not cause additional reactions such as the hydrolysis of the acyl chloride involved. Aliphatic acyl chlorides undergo basic hydrolysis with NaOH too easily, since NaOH is a strong base. Hence a weaker base such as pyridine has to be used as the basic solvent.

Reagents and conditions	Phenol, pyridine as solvent, aliphatic acyl chloride
Equation	CH₃COCl + C₆H₅OH → CH₃CO₂C₆H₅ + HCl
Equation

Reagents and conditions	Phenol, pyridine as solvent, aliphatic acyl chloride
Equation	CH₃COCl + C₆H₅OH → CH₃CO₂C₆H₅ + HCl
Equation

On the other hand, since aromatic acyl chlorides are more resistant to hydrolysis due to the aromatic ring, NaOH can be used as the solvent. The benefit of a stronger base is that a stronger nucleophile, phenoxide anion, can be produced instead of phenol.

Reagents and conditions	Phenol, NaOH as solvent, aromatic acyl chloride
Equation	C₆H₅COCl + C₆H₅OH → C₆H₅CO₂C₆H₅ + HCl
Equation

Reagents and conditions	Phenol, NaOH as solvent, aromatic acyl chloride
Equation	C₆H₅COCl + C₆H₅OH → C₆H₅CO₂C₆H₅ + HCl
Equation

Reactions of Alcohol, Phenol and Carboxylic Acid:

	Na	NaOH	Na₂CO₃
Alcohol	✓	X	X
Phenol	✓	✓	X
Carboxylic Acid	✓	✓	✓

Alcohol
Na	✓
NaOH	X
Na₂CO₃	X

Phenol
Na	✓
NaOH	✓
Na₂CO₃	X

Carboxylic Acid
Na	✓
NaOH	✓
Na₂CO₃	✓

Worked example 1
Elucidation Question

Alcohol B has esters which are responsible for the flavours of various fruits and has the molecular formula C₅H₁₂O. Reaction of B with acidified potassium dichromate(VI) produces a compound C,C₅H₁₀O₂. Heating B over Al₂O₃ produces D, C₅H₁₀. Vigorous oxidation of D forms 2-methylpropanoic acid, CH₃CH(CH₃)CO₂H, as one of the products.

Suggest the structures of B, C and D.

Draw out the structure of 2-methylpropanoic acid to visualise better.

Observation	Type of reaction	Functional group present
Reaction of B with K₂Cr₂O₇ results in compound C which has two fewer H atoms and one more O atom.	Oxidation	Carboxylic acid present in C
Heating B over Al₂O₃ produces D, C₅H₁₀.	Elimination	Carbon carbon double bond present in D
There is a decrease in the number of carbon atoms by one from D to 2-methylpropanoic acid as vigorous oxidation takes place.	Oxidative cleavage	Carbon carbon double bond present in D is present at the end of the compound.

Observation
Reaction of B with K₂Cr₂O₇ results in compound C which has two fewer H atoms and one more O atom.
Type of reaction
Oxidation
Functional group present
Carboxylic acid present in C

Observation
Heating B over Al₂O₃ produces D, C₅H₁₀.
Type of reaction
Elimination
Functional group present
Carbon carbon double bond present in D

Observation
There is a decrease in the number of carbon atoms by one from D to 2-methylpropanoic acid as vigorous oxidation takes place.
Type of reaction
Oxidative cleavage
Functional group present
Carbon carbon double bond present in D is present at the end of the compound.

Answer:

Chapter 13: Hydroxy Compounds

Chapter 13 Hydroxy Compounds

Chapter 13
Hydroxy Compounds