xi. Predict the products of the reaction of 1-naphthol with NaOH and bromine water. Provide the balanced chemical equations and indicate the positions of the substituents.
Answer:
Reaction with NaOH: 1-Naphthol is a phenol derivative and reacts with NaOH to form sodium 1-naphthoxide.
Equation: C₁₀H₇OH + NaOH [source symbol unavailable] C₁₀H₇ONa + H₂O
Reaction with Bromine Water: The -OH group is strongly activating and ortho/para directing. In 1-naphthol, the positions ortho and para to the -OH group are 2, 4, and 5. Reaction with bromine water will yield 2,4-dibromo-1-naphthol (or a similar polybrominated product depending on conditions).
Equation: C₁₀H₇OH + 2Br₂ [source symbol unavailable] C₁₀H₅Br₂OH + 2HBr
Part C: Long Answer Questions
1. Compare and contrast the reactivity of phenol, benzene, and ethanol in electrophilic aromatic substitution reactions. Explain the role of the hydroxyl group in phenol and its influence on these reactions.
Answer: The reactivity of these three compounds towards electrophilic substitution varies significantly due to the electronic effects of their substituents.
Benzene: Benzene is the baseline for aromatic reactivity. It undergoes electrophilic substitution reactions (e.g., nitration, halogenation) only in the presence of a strong electrophile and a catalyst (e.g., HNO₃/H₂SO₄, Br₂/FeBr₃). The reaction rate is moderate. Phenol: Phenol is much more reactive than benzene towards electrophilic aromatic substitution. The hydroxyl (-OH) group is strongly activating. It donates electron density to the ring through a resonance effect (+R), where the lone pair on the oxygen atom delocalizes into the ring. This increases the electron density at the ortho and para positions relative to the -OH group. As a result, phenol reacts with mild electrophiles without a catalyst (e.g., bromine water gives 2,4,6-tribromophenol). Ethanol: Ethanol is an aliphatic alcohol and does not undergo electrophilic aromatic substitution because it lacks an aromatic ring. It undergoes nucleophilic substitution reactions instead.
Role of the Hydroxyl Group: In phenol, the -OH group acts as a strong activating and ortho/para directing group. It stabilizes the arenium ion intermediate formed during electrophilic attack at the ortho and para positions, lowering the activation energy and increasing the reaction rate.
2. Describe the acidity of phenol in terms of its ability to donate protons. Compare its acidity to that of water and ethanol and explain the molecular basis for the differences observed.
Answer: The acidity of a compound is determined by its ability to donate a proton (H⁺) and the stability of the resulting conjugate base.
Phenol (pKa ~10): Phenol is more acidic than water and ethanol. When phenol donates a proton, it forms a phenoxide ion. The negative charge on the oxygen is delocalized into the benzene ring via resonance, creating multiple resonance structures. This delocalization significantly stabilizes the phenoxide ion, making it energetically favorable
for phenol to donate a proton. Water (pKa ~15.7): Water can donate a proton to form a hydroxide ion (OH⁻). The hydroxide ion is not resonance- stabilized. However, oxygen is highly electronegative, which helps stabilize the negative charge to some extent. Water is less acidic than phenol but more acidic than ethanol. Ethanol (pKa ~16): Ethanol is the least acidic of the three. When it donates a proton, it forms an ethoxide ion (CH₃CH₂O⁻). The ethyl group is electron-donating (+I effect), which intensifies the negative charge on the oxygen atom, destabilizing the conjugate base. This makes it harder for ethanol to donate a proton.
Conclusion: The acidity order (Phenol > Water > Ethanol) is primarily due to the resonance stabilization of the phenoxide ion, which is absent in the other two.
3. Describe how 2-naphthol will undergo a reaction with bromine water and nitric acid.
Answer: 2-Naphthol (2-hydroxynaphthalene) is a phenol derivative and is highly reactive towards electrophilic substitution due to the activating -OH group.
Reaction with Bromine Water: The -OH group at position 2 is strongly activating and directs incoming electrophiles to the ortho and para positions. The available ortho positions are 1 and 3, and the para position is 6. Reaction with bromine water will lead to polybromination. The major product is likely 1,6-dibromo-2-naphthol or 1,3-dibromo-2- naphthol, depending on the specific conditions and steric factors. Reaction with Nitric Acid: Similar to bromination, nitration will occur at the ortho and para positions relative to the - OH group. With dilute nitric acid, mono-nitration can be achieved, yielding products like 1-nitro-2-naphthol or 6- nitro-2-naphthol. With concentrated nitric acid, dinitration may occur.
4. Predict the products if methanol is used instead of ethanol in the reaction with ethanoyl chloride. Provide the balanced chemical equation.
Answer: The reaction of an alcohol with ethanoyl chloride (acetyl chloride, CH₃COCl) is an esterification reaction that produces an ester and hydrogen chloride (HCl). If methanol is used instead of ethanol, the product is methyl ethanoate (methyl acetate).
Balanced Chemical Equation: CH₃COCl + CH₃OH [source symbol unavailable] CH₃COOCH₃ + HCl
The product is methyl ethanoate, a sweet-smelling ester.
5. Write the balanced chemical equations for the reactions of phenol with NaOH and sodium metal. Describe the products formed in each reaction.
Answer:
Reaction with NaOH: Phenol reacts with sodium hydroxide to form sodium phenoxide and water. This reaction demonstrates the acidic nature of phenol.
Equation: C₆H₅OH + NaOH [source symbol unavailable] C₆H₅ONa + H₂O Product: Sodium phenoxide (a white solid) and water.
Reaction with Sodium Metal: Phenol reacts with sodium metal to form sodium phenoxide and hydrogen gas. This is a redox reaction where sodium is oxidized and hydrogen is reduced.
Equation: 2C₆H₅OH + 2Na [source symbol unavailable] 2C₆H₅ONa + H₂ Product: Sodium phenoxide and hydrogen gas (bubbles observed).
6. Describe the reaction of phenol with bromine water.
Answer: Phenol reacts vigorously with bromine water at room temperature without the need for a catalyst. The reaction is an electrophilic aromatic substitution. The -OH group is strongly activating and ortho/para directing. The reaction substitutes hydrogen atoms at the 2, 4, and 6 positions (the two ortho and one para positions) with bromine atoms.
Equation: C₆H₅OH + 3Br₂ [source symbol unavailable] C₆H₂Br₃OH + 3HBr
Observation: The reddish-brown colour of bromine water decolourises rapidly, and a white precipitate of 2,4,6-tribromophenol is formed. This reaction is used as a test for phenol.
7. Describe the reaction of phenol with dilute nitric acid at room temperature. What are the products, and why do they form in the observed positions on the aromatic ring?
Answer: Phenol reacts with dilute nitric acid at room temperature to undergo electrophilic aromatic substitution. The -OH group is strongly activating and ortho/para directing.
Products: The reaction yields a mixture of ortho-nitrophenol (2-nitrophenol) and para-nitrophenol (4-nitrophenol). With dilute acid, mono-nitration is favored.
Reason for Positions: The -OH group donates electron density to the ring through resonance, increasing the electron density at the ortho and para positions. The arenium ion intermediates for attack at these positions are resonance-stabilized, lowering the activation energy. Meta attack is less favorable because the intermediate is not stabilized by the -OH group.
8. Discuss the similarities and differences in the directing effects of the hydroxyl group in phenol and 1-naphthol.
Answer:
Similarities: Both phenol and 1-naphthol contain a hydroxyl (-OH) group directly attached to an aromatic ring. In both cases, the -OH group is strongly activating and directs incoming electrophiles to the ortho and para positions relative to itself. This is due to the electron-donating resonance effect (+R) of the oxygen atom. Differences:
Phenol: The -OH group is attached to a single benzene ring. The ortho positions are C2 and C6, and the para position is C4. 1-Naphthol: The -OH group is attached to a naphthalene ring system (two fused benzene rings). The directing effects are more complex. The ortho positions are C2 and C8 (or C9 depending on numbering), and the para position is C4. Due to the fused ring system, resonance stabilization of the intermediate can extend over both rings, influencing the reactivity and product distribution. For example, 1-naphthol is more reactive than phenol and can undergo substitution at positions that are not strictly ortho/para in the same way.
9. Phenol is more acidic than ethanol. Analyze the structure of phenol to explain this difference in acidity.
Answer: The difference in acidity between phenol and ethanol lies in the stability of their conjugate bases.
Phenol: When phenol loses a proton, it forms a phenoxide ion (C₆H₅O⁻). The negative charge on the oxygen atom is delocalized into the benzene ring via resonance. This delocalization spreads the charge over the ortho and para carbon atoms, significantly stabilizing the phenoxide ion. The more stable the conjugate base, the stronger the acid.
Ethanol: When ethanol loses a proton, it forms an ethoxide ion (CH₃CH₂O⁻). The negative charge is localized on the oxygen atom. The electron-donating ethyl group (+I effect) intensifies the negative charge, destabilizing the ethoxide ion. Therefore, ethanol is a weaker acid than phenol.
Conclusion: The resonance stabilization of the phenoxide ion makes phenol (pKa ~10) significantly more acidic than ethanol (pKa ~16).
10. Phenol reacts with bromine water to form a white precipitate without needing a catalyst. What does this suggest about the electron density of the aromatic ring in phenol?
Answer: The fact that phenol reacts with bromine water without a catalyst suggests that the aromatic ring in phenol has a much higher electron density than benzene. The hydroxyl (-OH) group is strongly electron-donating through resonance (+R effect). This increases the electron density at the ortho and para positions, making the ring highly nucleophilic and susceptible to attack by even weak electrophiles like bromine. In contrast, benzene requires a catalyst (e.g., FeBr₃) to polarize the bromine molecule and generate a stronger electrophile. The high reactivity of phenol is a direct consequence of the activating effect of the -OH group.
