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

Step 1: Part (a)(i): Reagent for oxidation of primary alcohol to aldehyde.
Pyridinium chlorochromate (PCC) in dichloromethane ($CH_2Cl_2$) selectively oxidises a primary alcohol to an aldehyde and stops there (does not over-oxidise to carboxylic acid): \[ R-CH_2OH \xrightarrow{PCC/CH_2Cl_2} R-CHO \]
Step 2: Part (a)(ii): Reagent for oxidation of primary alcohol to carboxylic acid.
Acidified potassium permanganate ($KMnO_4/H_2SO_4$) or acidified potassium dichromate ($K_2Cr_2O_7/H_2SO_4$) fully oxidises a primary alcohol to the carboxylic acid: \[ R-CH_2OH \xrightarrow{K_2Cr_2O_7/H_2SO_4} R-COOH \]
Step 3: Part (b): Kolbe's reaction of phenol.
Sodium phenoxide ($C_6H_5ONa$, prepared from phenol + NaOH) is treated with $CO_2$ under high pressure ($4-7$ atm) at $125-140^\circ C$, then acidified: \[ C_6H_5ONa + CO_2 \xrightarrow{125^\circ C, 5 \text{ atm}} \text{sodium salicylate} \xrightarrow{H^+} o\text{-HOC}_6H_4COOH \] Product: 2-hydroxybenzoic acid (salicylic acid). This is Kolbe-Schmitt carboxylation.
Step 4: Part (b): Cleavage of ether by HI.
Ethers are cleaved by hydroiodic acid (HI). First the ether oxygen is protonated; then iodide ion attacks the less hindered C in $S_N2$: \[ R-O-R' + HI \rightarrow R-I + R'-OH \] With excess HI, the alcohol is further converted: $R'-OH + HI \rightarrow R'-I + H_2O$. HI is preferred because $I^-$ is the best nucleophile among halides.
Step 5: Explain why PCC stops at aldehyde.
PCC is a mild, selective oxidant. It can oxidise primary alcohols to aldehydes but cannot oxidise aldehydes further to carboxylic acids under the mild anhydrous conditions (no water present). Strong oxidants like $KMnO_4$ or $K_2Cr_2O_7$ in acidic aqueous medium can proceed all the way to carboxylic acid.
Step 6: Summarise all answers.
(a)(i) PCC (pyridinium chlorochromate): oxidises primary alcohol to aldehyde. (a)(ii) $K_2Cr_2O_7/H_2SO_4$ or $KMnO_4/H^+$: oxidises primary alcohol to carboxylic acid. (b) Kolbe reaction of phenol: salicylic acid. Ether + HI: alkyl iodide + alcohol. \[ \boxed{\text{(a-i) PCC; (a-ii) }K_2Cr_2O_7/H_2SO_4;\text{ (b) Kolbe}\rightarrow\text{salicylic acid; ether}+HI\rightarrow\text{R-I}+\text{R-OH}} \]