Write the product(s) when :
(I) One mol of ethanal is treated with 1 mol of \(CH_3OH\) in the presence of dry HCl gas.
(II) Benzaldehyde is treated with conc. NaOH.
(III) Ethanoic acid is heated in the presence of \(P_2O_5\).
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Solution and Explanation
(I) Reaction of ethanal with one mole of methanol in presence of dry HCl
Step 1: Nature of reaction Alcohols add to aldehydes in the presence of dry acid catalysts. When one mole of alcohol reacts with one mole of aldehyde, the product obtained is called a
hemiacetal.
Step 2: Reaction \[ CH_3CHO + CH_3OH \xrightarrow{\text{dry HCl}} CH_3CH(OH)(OCH_3) \]
Step 3: Product formed The product contains both an \(-OH\) group and an \(-OCH_3\) group attached to the same carbon atom. Hence the product is: \[ \boxed{CH_3CH(OH)(OCH_3)} \] which is called the hemiacetal of ethanal.
(II) Benzaldehyde treated with concentrated NaOH
Step 1: Identify the reaction Benzaldehyde does not contain an \(\alpha\)-hydrogen atom. Therefore it cannot undergo aldol condensation. Instead, in concentrated sodium hydroxide solution, it undergoes the Cannizzaro reaction.
Step 2: Cannizzaro reaction In this reaction one molecule of benzaldehyde gets oxidized to benzoic acid while another molecule gets reduced to benzyl alcohol. \[ 2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \]
Step 3: Products formed The products are: \[ \boxed{C_6H_5CH_2OH} \] (Benzyl alcohol) and \[ \boxed{C_6H_5COONa} \] (Sodium benzoate)
(III) Ethanoic acid heated with \(P_2O_5\)
Step 1: Role of \(P_2O_5\) Phosphorus pentoxide is a powerful dehydrating agent. It removes a molecule of water from two molecules of carboxylic acid.
Step 2: Formation of acid anhydride \[ 2CH_3COOH \xrightarrow{P_2O_5} (CH_3CO)_2O + H_2O \]
Step 3: Product formed The product obtained is acetic anhydride. \[ \boxed{(CH_3CO)_2O} \]
Final Answers \[ \boxed{CH_3CH(OH)(OCH_3)} \] \[ \boxed{C_6H_5CH_2OH + C_6H_5COONa} \] \[ \boxed{(CH_3CO)_2O} \]
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\((v)\)
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