Question

Give for each of the following pairs of compounds a chemical test, preferably a test-tube reaction, that will distinguish between the two compounds. Write a structural formula for each compound and equations for the reactions involved. (a) phenol and cyclohexanol. (b) methy1 \(p\) -hydroxybenzoate and \(p\) -methoxybenzoic acid. (c) 9,10 -anthraquinone and 1,4 -anthraquinone.

Short answer

a) Phenol and Cyclohexanol Structural formulas: Phenol: \(\text{C}_6\text{H}_5\text{OH}\), Cyclohexanol: \(\text{C}_6\text{H}_{11}\text{OH}\) Test: Reaction with bromine water. Phenol gives a white precipitate, while cyclohexanol doesn't react. Equations: Phenol: \(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{OBr}_3 + 3\text{HBr}\). Cyclohexanol: No reaction. b) Methyl \(p\)-hydroxybenzoate and \(p\)-methoxybenzoic acid Structural formulas: Methyl \(p\)-hydroxybenzoate: \(\text{CH}_3\text{O}\text{C(O)}\text{C}_6\text{H}_4\text{OH}\), \(p\)-methoxybenzoic acid: \(\text{C(O)OH}\text{C}_6\text{H}_4\text{OCH}_3\) Test: Hydrolysis with dilute NaOH. Methyl \(p\)-hydroxybenzoate reacts, while \(p\)-methoxybenzoic acid doesn't react. Equations: Methyl \(p\)-hydroxybenzoate: \(\text{CH}_3\text{O}\text{C(O)}\text{C}_6\text{H}_4\text{OH} + \text{NaOH} \rightarrow \text{C(O)ONa}\text{C}_6\text{H}_4\text{OH} + \text{CH}_3\text{OH}\). \(p\)-methoxybenzoic acid: No reaction. c) 9,10-Anthraquinone and 1,4-Anthraquinone Structural formulas: Both: \(\text{C}_{14}\text{H}_8\text{O}_2\) Test: Reaction with sodium bisulfite. Only 1,4-Anthraquinone forms an adduct. Equations: 9,10-Anthraquinone: No reaction. 1,4-Anthraquinone: \(\text{C}_{14}\text{H}_8\text{O}_2 + 2\text{NaHSO}_3 \rightarrow \text{C}_{14}\text{H}_8\text{O}_2(\text{HSO}_3)_2 + 2\text{Na}^+\).

Step-by-step solution

a) Phenol and Cyclohexanol

1. Write structural formula: Phenol has a hydroxyl group attached to a benzene ring, whereas cyclohexanol has a hydroxyl group attached to a cyclohexane ring. Their structural formulas are: Phenol: \(\text{C}_6\text{H}_5\text{OH}\) Cyclohexanol: \(\text{C}_6\text{H}_{11}\text{OH}\) 2. Find distinguishing tests: A test that can distinguish phenol and cyclohexanol is the reaction with bromine water. Phenol undergoes electrophilic aromatic substitution, giving a white precipitate, while cyclohexanol doesn't react. 3. Write equations for the reactions: Phenol + Bromine water: \(\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{OBr}_3 + 3\text{HBr}\) (white precipitate) Cyclohexanol + Bromine water: No reaction

b) Methyl \(p\)-hydroxybenzoate and \(p\)-methoxybenzoic acid

1. Write structural formula: Methyl \(p\)-hydroxybenzoate and \(p\)-methoxybenzoic acid both have a benzene ring with substituents at para positions. Their structural formulas are: Methyl \(p\)-hydroxybenzoate: \(\text{CH}_3\text{O}\text{C(O)}\text{C}_6\text{H}_4\text{OH}\) \(p\)-methoxybenzoic acid: \(\text{C(O)OH}\text{C}_6\text{H}_4\text{OCH}_3\) 2. Find distinguishing tests: A test that can distinguish between these two compounds is the hydrolysis with a dilute base, such as sodium hydroxide. Methyl \(p\)-hydroxybenzoate will react with the base, forming sodium \(p\)-hydroxybenzoate and methanol, while \(p\)-methoxybenzoic acid will not react. 3. Write equations for the reactions: Methyl \(p\)-hydroxybenzoate + NaOH: \(\text{CH}_3\text{O}\text{C(O)}\text{C}_6\text{H}_4\text{OH} + \text{NaOH} \rightarrow \text{C(O)ONa}\text{C}_6\text{H}_4\text{OH} + \text{CH}_3\text{OH}\) \(p\)-methoxybenzoic acid + NaOH: No reaction

c) 9,10-Anthraquinone and 1,4-Anthraquinone

1. Write structural formula: 9,10 and 1,4-Anthraquinone have quinone structures with different positions of the carbonyl groups. Their structural formulas are: 9,10-Anthraquinone: \(\text{C}_{14}\text{H}_8\text{O}_2\) 1,4-Anthraquinone: \(\text{C}_{14}\text{H}_8\text{O}_2\) 2. Find distinguishing tests: A test that can differentiate between these two compounds is the reaction with sodium bisulfite. 1,4-Anthraquinone will react with sodium bisulfite to form an adduct, whereas 9,10-Anthraquinone will not react. 3. Write equations for the reactions: 9,10-Anthraquinone + Sodium bisulfite: No reaction 1,4-Anthraquinone + Sodium bisulfite: \(\text{C}_{14}\text{H}_8\text{O}_2 + 2\text{NaHSO}_3 \rightarrow \text{C}_{14}\text{H}_8\text{O}_2(\text{HSO}_3)_2 + 2\text{Na}^+\) (adduct formation)

Key concepts

Electrophilic Aromatic Substitution

Electrophilic aromatic substitution (EAS) is a common reaction mechanism in organic chemistry, especially when dealing with aromatic compounds like benzene derivatives. Aromatic compounds have a stable ring structure, often a benzene ring, characterized by alternating single and double bonds.
This stability is known as aromaticity. The process of EAS involves an electrophile replacing a hydrogen atom on the aromatic ring. The reaction typically begins with the formation of a reactive intermediate, known as a carbocation, which is then quickly stabilized by electron delocalization across the ring.

• The substituents already present on the ring can influence both the rate and regioselectivity of the reaction.
• If a substituent is electron-donating, it can activate the ring towards further substitution.
• Conversely, electron-withdrawing substituents deactivate the ring.

An example of EAS is the reaction between phenol and bromine water. Phenol's hydroxyl group is strongly activating, facilitating a rapid reaction with bromine. This results in the substitution of the hydrogen atoms on the ortho and para positions with bromine, forming a tribromophenol precipitate.

Hydrolysis Reaction

Hydrolysis reactions involve the breakdown of a compound due to its reaction with water. It's a fundamental process in both organic and inorganic chemistry. In aqueous solutions, especially those catalyzed by acids or bases, these reactions are crucial in transforming or decomposing various substances.In organic chemistry, such as with esters or amides, hydrolysis involves the cleavage of bonds through the incorporation of water molecules. Let's consider an example using esters: when an ester react with an aqueous base such as sodium hydroxide, hydrolysis can occur.

• This creates an alcohol and a carboxylate ion.
• The overall process is sometimes referred to as saponification when fats are involved.

A pertinent example is the reaction between methyl \( p \)-hydroxybenzoate and sodium hydroxide. The ester bond in methyl \( p \)-hydroxybenzoate undergoes hydrolysis resulting in the formation of sodium \( p \)-hydroxybenzoate and methanol. However, \( p \)-methoxybenzoic acid does not undergo hydrolysis under these conditions, allowing for distinctions between the two compounds.

Anthraquinone Chemistry

Anthraquinones are a class of aromatic organic compounds containing a quinone structure fused to an aromatic ring system. These compounds are pivotal in numerous industrial and biological settings.
They are known for their vivid color properties and are often utilized as dyes or pigments.The structure of anthraquinone involves two carbonyl groups (\( \text{C} = \text{O} \) groups) connected to an aromatic ring. This structure is what defines the quinone aspect of anthraquinones.

• Different isomers exist, such as 9,10-anthraquinone and 1,4-anthraquinone, and these can be differentiated based on their chemical behavior.
• Although both molecules share the same molecular formula, their structural orientation causes different chemical reactivity.

Specifically, 1,4-anthraquinone can form an adduct with sodium bisulfite, a reaction that does not occur with 9,10-anthraquinone. Such differences in reactivity provide an approach to differentiate these isomers using chemical testing.

Bromine Water Test

The bromine water test is a straightforward chemical test used for detecting the presence of unsaturated compounds, such as alkenes and phenols. The test relies on bromine molecules (\( \text{Br}_2 \)) reacting with double bonds or electron-rich aromatic systems.When bromine water, which is an aqueous solution of bromine, is added to a solution containing an unsaturated compound:

• The reaction typically results in the loss of the brownish coloration of the bromine water.
• This decolorization indicates a chemical reaction taking place.
• Often, a precipitate might form, such as in the reaction with phenol where a white precipitate of tribromophenol is produced.

In the case of phenol versus cyclohexanol, bromine water can be used to distinguish between the two since phenol undergoes electrophilic aromatic substitution to form a precipitate, whereas cyclohexanol does not react with bromine water, keeping the solution color unchanged. This test provides a visual and straightforward method for identifying certain functional groups in organic compounds.