The compound shown below has three different types of OH groups, all with different acidities. Show the structure produced after this compound is treated with different amounts of NaH followed by a methylating reagent. Add a brief explanation.

(a) 1 equivalent of NaH, followed by 1 equivalent of CH₃I and heat
(b) 2 equivalents of NaH, followed by 2 equivalents of CH₃I and heat
(c) 3 equivalents of NaH, followed by 3 equivalents of CH₃I and heat

Verified step by step guidance

1. Step 1: Analyze the structure of the compound. The molecule contains three different types of OH groups: (1) a phenolic OH group attached to the benzene ring, (2) an alcohol OH group attached to the side chain, and (3) a carboxylic acid group (-COOH). Each of these groups has different acidities, with the carboxylic acid being the most acidic, followed by the phenolic OH, and then the alcohol OH.
2. Step 2: Reaction with 1 equivalent of NaH followed by 1 equivalent of CH3I and heat. NaH acts as a base and selectively deprotonates the most acidic group first, which is the carboxylic acid (-COOH). This forms the carboxylate anion. The methylating reagent (CH3I) then reacts with the carboxylate anion to form a methyl ester (-COOCH3).
3. Step 3: Reaction with 2 equivalents of NaH followed by 2 equivalents of CH3I and heat. After the carboxylic acid is methylated, the second equivalent of NaH deprotonates the phenolic OH group, forming the phenoxide ion. The phenoxide ion reacts with CH3I to form a methyl ether (-OCH3) at the phenolic position.
4. Step 4: Reaction with 3 equivalents of NaH followed by 3 equivalents of CH3I and heat. After the carboxylic acid and phenolic OH groups are methylated, the third equivalent of NaH deprotonates the alcohol OH group on the side chain, forming the alkoxide ion. The alkoxide ion reacts with CH3I to form a methyl ether (-OCH3) at the alcohol position.
5. Step 5: Summarize the final structures. (a) After 1 equivalent of NaH and CH3I, the carboxylic acid is converted to a methyl ester. (b) After 2 equivalents of NaH and CH3I, both the carboxylic acid and phenolic OH are methylated. (c) After 3 equivalents of NaH and CH3I, all three OH groups are methylated, resulting in a fully methylated compound.

Key Concepts

Acidity of Alcohols

Alcohols contain hydroxyl (OH) groups, which can exhibit varying levels of acidity based on their molecular environment. The acidity is influenced by factors such as steric hindrance, the presence of electron-withdrawing or donating groups, and the overall stability of the resulting alkoxide ion. Understanding these differences is crucial for predicting how each OH group will react with bases like NaH.

Deprotonation and Alkoxide Formation

Deprotonation is the process where a hydrogen ion (H+) is removed from an alcohol, resulting in the formation of an alkoxide ion (RO-). In this context, NaH acts as a strong base that can deprotonate the OH groups in the compound. The number of equivalents of NaH used will determine how many OH groups are converted to alkoxide ions, which subsequently influences the methylation reaction with CH3I.

Methylation Reaction

Methylation is a chemical reaction where a methyl group (CH3) is added to a molecule, often through the use of methylating agents like CH3I. In the presence of alkoxide ions, the methyl group can be transferred to the oxygen atom, forming ethers. The extent of methylation depends on the number of equivalents of NaH and CH3I used, which dictates how many alkoxide groups are available for reaction.