Question

$$ (\mathrm{P})+\mathrm{NaOH} \longrightarrow(\mathrm{Q}) \stackrel{\mathrm{CH}_{3}-\mathrm{I}}{\longrightarrow}(\mathrm{R}) $$ (P) has molecular formula ' \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}\) ' and ( \(\mathrm{R}\) ) is aromatic in nature. Then choose the correct option(s). (A) (R) has common name anisole (B) In the conversion (P) is used as phenoxide moiety (C) (P) is known as carbolic acid (D) Order of electron density in ring \(:(\mathrm{Q})>(\mathrm{R})>\)

Short answer

The correct options are (A) and (C). Compound (P) is phenol, also known as carbolic acid. It reacts with NaOH to form sodium phenoxide (Q), which then reacts with CH3-I to form anisole (R).

Step-by-step solution

Identify the structure of compound (P)

Based on the molecular formula, C6H6O, and the aromatic nature of compound (R), compound (P) should be phenol (C6H5OH). Step 2: Determine the structure of compound (Q) after reacting with NaOH.

Determine the structure of compound (Q) after reacting with NaOH

When phenol (P) reacts with NaOH, it forms a phenoxide ion (Q) as follows: \[C_6H_5OH + NaOH \rightarrow C_6H_5O^{-}Na^{+} + H_2O\] Compound (Q) is sodium phenoxide. Step 3: Determine the structure of compound (R) after reacting with CH3-I.

Determine the structure of compound (R) after reacting with CH3-I

When sodium phenoxide (Q) reacts with CH3-I, it forms anisole (R) as follows: \[C_6H_5O^{-}Na^{+} + CH_3I \rightarrow C_6H_5OCH_3 + NaI\] Compound (R) is anisole. Now, let's analyse the given options based on the structures of compounds (P), (Q), and (R). (A) (R) has the common name anisole - Correct: We identified that compound (R) is anisole. (B) In the conversion (P) is used as phenoxide moiety - Incorrect: (P) is phenol, not phenoxide. Phenoxide is formed from phenol after reaction with NaOH to form compound (Q). (C) (P) is known as carbolic acid - Correct: Phenol, which is compound (P), is also known as carbolic acid. (D) Order of electron density in the ring: (Q)>(R)> - Requires more information: The order of electron density in the ring can be deduced by comparing the electron-donating and withdrawing groups present in each compound. Since no specific order is provided for comparison, this option cannot be evaluated as correct or incorrect. Based on our analysis, the correct options are (A) and (C).

Key concepts

Phenol Reactions

Phenol, represented by the formula ewlineewlineewlineewlineewline

ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline\( C_6H_5OH \), is both an aromatic compound and a type of alcohol. In organic chemistry, phenol's reactivity is essential knowledge for students preparing for examinations such as JEE. When phenol reacts with bases like NaOH, it converts to a phenoxide ion, characterized by a negative charge on the oxygen atom. This conversion greatly increases its reactivity towards electrophilic aromatic substitution reactions due to the increased electron density on the aromatic ring. A classic example, as shown in our textbook exercise, involves the reaction of phenol ewline

ewline (\( C_6H_5OH \) ) with NaOH that forms sodium phenoxide \( C_6H_5O^- Na^+ \). Recognizing the formation of this phenoxide ion is key in predicting phenol's subsequent reactions and understanding the mechanism by which these reactions occur.

Aromatic Compounds

Aromatic compounds, such as phenol and anisole \( (C_6H_5OCH_3) \), are a cornerstone of organic chemistry, characterized by a planar ring structure that follows Huckel's rule—having 4n+2 π electrons. They have delocalized pi electrons above and below the plane of the molecule, which contributes to their unique chemical properties, including high stability and distinctive reactivities. The aromatic characteristic of compound \( (R) \) in our JEE problem example is maintained during the chemical transformation of phenol to anisole. By adding a methyl group \( (CH_3) \) to the oxygen atom of the phenoxide ion, we preserve the aromaticity of the compound. The understanding of how substituents like methoxy \( (-OCH_3) \) groups affect the chemical properties of aromatic compounds is crucial when predicting reactivity and designing synthesis in organic chemistry.

Molecular Formula Determination

Determining the molecular formula of a compound is a fundamental skill in organic chemistry, often tested in JEE and other competitive exams. It involves deriving the number and types of atoms in a molecule based on various analysis methods. In the given problem, compound \( (P) \) is known to have the molecular formula \( C_6H_6O \). This information alone, however, does not elucidate the exact structure of the molecule. Chemists use further structural clues, like reactivity patterns and spectral analysis, to determine the arrangement of these atoms. Knowing that compound \( (R) \) retains its aromaticity after reacting with \( CH_3-I \), and deciphering that (P) is a phenol, help to pinpoint the structure of (P) and the nature of the transformation to (R). The mastery of molecular formula determination enables the prediction of not only the physical and chemical properties of a molecule but also how it might interact with other substances in a reaction.