Question

Answer the following case-based questions on alcohols, phenols and ethers: \[ (a)\;\text{Name the reagents used in the following reactions:} \] \[ (i)\;\text{Oxidation of a primary alcohol to aldehyde} \] \[ (ii)\;\text{Oxidation of a primary alcohol to carboxylic acid} \] \[ (b)\;\text{Write the reaction involved in Kolbe's reaction.} \] \[ (c)(i)\;\text{Why are tertiary alcohols resistant to oxidation?} \] OR \[ (c)(ii)\;\text{Write the products of the following reaction:} \] \[ (CH_3)_3C-O-C_2H_5 \xrightarrow{HI} \]

Hint:

PCC stops oxidation at aldehyde stage, while strong oxidising agents convert primary alcohols into carboxylic acids. Tertiary ethers with \(HI\) generally give tertiary alkyl iodide.

Answer 1

Concept:
Alcohols undergo oxidation depending on their type. Primary alcohols can be oxidised to aldehydes or carboxylic acids. Phenols undergo Kolbe's reaction through phenoxide ion. Ethers are cleaved by hydrogen halides such as \(HI\).

Step 1: Reagent for oxidation of primary alcohol to aldehyde.
Primary alcohols are oxidised to aldehydes by mild oxidising agents. A commonly used reagent is: \[ PCC \] PCC stands for pyridinium chlorochromate. General reaction: \[ RCH_2OH \xrightarrow{PCC} RCHO \] Example: \[ CH_3CH_2OH \xrightarrow{PCC} CH_3CHO \]

Step 2: Reagent for oxidation of primary alcohol to carboxylic acid.
Primary alcohols are oxidised completely to carboxylic acids by strong oxidising agents. Common reagents are: \[ KMnO_4/H^+ \] or \[ K_2Cr_2O_7/H^+ \] General reaction: \[ RCH_2OH \xrightarrow{KMnO_4/H^+} RCOOH \] Example: \[ CH_3CH_2OH \xrightarrow{KMnO_4/H^+} CH_3COOH \]

Step 3: Kolbe's reaction.
In Kolbe's reaction, sodium phenoxide reacts with carbon dioxide under pressure. First, phenol reacts with sodium hydroxide: \[ C_6H_5OH+NaOH\rightarrow C_6H_5ONa+H_2O \] Then sodium phenoxide reacts with carbon dioxide: \[ C_6H_5ONa+CO_2 \xrightarrow{\text{pressure}} o\text{-}HOC_6H_4COONa \] On acidification: \[ o\text{-}HOC_6H_4COONa+H^+ \rightarrow o\text{-}HOC_6H_4COOH \] The final product is salicylic acid.

Step 4: Why tertiary alcohols resist oxidation.
A tertiary alcohol has the general structure: \[ R_3C-OH \] The carbon atom attached to the \(-OH\) group is bonded to three alkyl groups. It does not contain hydrogen attached to the carbinol carbon. Oxidation of alcohol generally requires removal of hydrogen from the carbon bearing the \(-OH\) group. Since tertiary alcohols do not have this hydrogen, they resist oxidation under normal conditions.

Step 5: Ether cleavage by HI.
The given ether is: \[ (CH_3)_3C-O-C_2H_5 \] This is an unsymmetrical ether. One side is tertiary butyl group and the other side is ethyl group. With \(HI\), protonation of ether oxygen occurs first. Since the tertiary carbocation is stable, cleavage occurs at the tertiary side. \[ (CH_3)_3C-O-C_2H_5+HI \rightarrow (CH_3)_3C-I+C_2H_5OH \] Thus, the products are: \[ (CH_3)_3C-I \] and \[ C_2H_5OH \] Hence: PCC is used for oxidation of primary alcohol to aldehyde. Acidified \(KMnO_4\) or acidified \(K_2Cr_2O_7\) is used for oxidation of primary alcohol to carboxylic acid. Kolbe's reaction gives salicylic acid. Tertiary alcohols resist oxidation due to absence of hydrogen on carbinol carbon. The ether gives tert-butyl iodide and ethanol with \(HI\).