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

Step 1: Product (I): Ethanal + 1 mol $CH_3OH$ in dry HCl.
Ethanal ($CH_3CHO$) undergoes acid-catalysed nucleophilic addition with methanol. With only 1 mol of methanol, the reaction stops at the hemiacetal stage: \[ CH_3CHO + CH_3OH \xrightarrow{\text{dry HCl}} CH_3CH(OH)(OCH_3) \] Product: 1-methoxy-1-ethan-1-ol (methyl hemiacetal of acetaldehyde). Two moles of methanol would give the full acetal $CH_3CH(OCH_3)_2$.
Step 2: Product (II): Benzaldehyde + conc. NaOH (Cannizzaro reaction).
Benzaldehyde ($C_6H_5CHO$) has no alpha-hydrogen. It cannot undergo aldol condensation. With conc. NaOH, it undergoes Cannizzaro reaction (one molecule oxidised, one reduced): \[ 2C_6H_5CHO \xrightarrow{\text{conc. NaOH}} C_6H_5COO^-Na^+ + C_6H_5CH_2OH \] Products: sodium benzoate ($C_6H_5COONa$) and benzyl alcohol ($C_6H_5CH_2OH$).
Step 3: Product (III): Ethanoic acid + $P_2O_5$ (dehydration).
$P_2O_5$ is a very powerful dehydrating agent. Two molecules of acetic acid lose one water to form acetic anhydride: \[ 2CH_3COOH \xrightarrow{P_2O_5, \Delta} (CH_3CO)_2O + H_2O \] Product: acetic anhydride.
Step 4: Mechanism of Cannizzaro reaction.
$OH^-$ attacks the carbonyl carbon of one $C_6H_5CHO$ to form a tetrahedral alkoxide intermediate. This intermediate transfers a hydride ion ($H^-$) to the carbonyl carbon of a second benzaldehyde molecule. One molecule is oxidised to benzoate; the other is reduced to benzyl alcohol.
Step 5: Why hemiacetal forms with 1 mol methanol (Product I).
In acid catalyst, methanol (nucleophile) adds to the carbonyl carbon of ethanal in one step: \[ CH_3CHO + CH_3OH \xrightarrow{H^+} CH_3CH(OH)(OCH_3) \] With only 1 mol methanol, the product is the hemiacetal (has both $-OH$ and $-OCH_3$ on the same carbon). A second methanol molecule would be needed to eliminate water and form the full acetal.
Step 6: Summarise all three products.
(I) Hemiacetal: $CH_3CH(OH)(OCH_3)$. (II) Cannizzaro: $C_6H_5COONa + C_6H_5CH_2OH$. (III) Acetic anhydride: $(CH_3CO)_2O$. \[ \boxed{(I)\;CH_3CH(OH)(OCH_3);\;(II)\;C_6H_5COONa+C_6H_5CH_2OH;\;(III)\;(CH_3CO)_2O} \]