Question

Answer the following: \[ (a)(i)\;\text{Write the product(s) when:} \] \[ (I)\;\text{One mol of ethanal is treated with }1\text{ mol of }CH_3OH\text{ in the presence of dry }HCl\text{ gas.} \] \[ (II)\;\text{Benzaldehyde is treated with conc. }NaOH. \] \[ (III)\;\text{Ethanoic acid is heated in the presence of }P_2O_5. \] \[ (a)(ii)\;\text{Write a simple chemical test to distinguish between ethanal and propanal.} \] \[ (a)(iii)\;\text{Write the name of the reagent to transform allyl alcohol to propenal.} \]

Hint:

One mole alcohol with aldehyde gives hemiacetal, while two moles alcohol give acetal. PCC oxidises primary alcohols to aldehydes.

Answer 1

Concept:
Aldehydes react with alcohols in dry \(HCl\) to form hemiacetals or acetals. Aldehydes without alpha hydrogen undergo Cannizzaro reaction. \(P_2O_5\) is a strong dehydrating agent.

Step 1: Reaction of ethanal with one mole methanol.
Ethanal reacts with one mole of methanol in presence of dry \(HCl\). Since only one mole of alcohol is used, hemiacetal is formed. \[ CH_3CHO+CH_3OH \xrightarrow{dry\;HCl} CH_3CH(OH)OCH_3 \] The product is \(1\)-methoxyethanol.

Step 2: Reaction of benzaldehyde with concentrated NaOH.
Benzaldehyde has no alpha hydrogen. Therefore, it undergoes Cannizzaro reaction in presence of concentrated \(NaOH\). \[ 2C_6H_5CHO+NaOH \rightarrow C_6H_5CH_2OH+C_6H_5COONa \] The products are benzyl alcohol and sodium benzoate.

Step 3: Reaction of ethanoic acid with P_2O_5.
\(P_2O_5\) is a dehydrating agent. It removes water from two molecules of ethanoic acid and forms acetic anhydride. \[ 2CH_3COOH \xrightarrow{P_2O_5} (CH_3CO)_2O+H_2O \]

Step 4: Test to distinguish ethanal and propanal.
Ethanal gives iodoform test because it contains the group: \[ CH_3CHO \] On treatment with \(I_2/NaOH\), ethanal gives yellow precipitate of iodoform. \[ CHI_3 \] Propanal does not give iodoform test. Therefore, iodoform test can distinguish ethanal and propanal.

Step 5: Reagent to convert allyl alcohol to propenal.
Allyl alcohol is: \[ CH_2=CHCH_2OH \] Propenal is: \[ CH_2=CHCHO \] This is oxidation of primary alcohol to aldehyde. A mild oxidising agent is required. The reagent is: \[ PCC \] \[ CH_2=CHCH_2OH \xrightarrow{PCC} CH_2=CHCHO \] Hence: \[ CH_3CHO+CH_3OH \xrightarrow{dry\;HCl} CH_3CH(OH)OCH_3 \] \[ 2C_6H_5CHO+NaOH \rightarrow C_6H_5CH_2OH+C_6H_5COONa \] \[ 2CH_3COOH \xrightarrow{P_2O_5} (CH_3CO)_2O+H_2O \] Iodoform test distinguishes ethanal and propanal. PCC converts allyl alcohol to propenal.