How can prepared a named aldehyde and ketone use oxidation of alcohol

Aldehydes and ketones can be prepared by the oxidation of alcohols. The type of alcohol (primary or secondary) determines whether an aldehyde or a ketone is formed.

1. Preparation of a Named Aldehyde (e.g., Ethanal) from a Primary Alcohol

Step 1: Identify the alcohol. Aldehydes are formed by the oxidation of primary alcohols. For ethanal ($\text{CH}_3\text{CHO}$), the corresponding primary alcohol is ethanol ($\text{CH}_3CH_2\text{OH}$).

Step 2: Choose an oxidizing agent and conditions. Primary alcohols can be oxidized to aldehydes using mild oxidizing agents or by carefully controlling the reaction conditions with stronger oxidizing agents. A common method uses acidified potassium dichromate(VI) ($\text{K}_2Cr_2O_7/H^+$) or pyridinium chlorochromate (PCC). To obtain the aldehyde and prevent further oxidation to a carboxylic acid, the aldehyde is distilled off as it forms, as it has a lower boiling point than the alcohol and the carboxylic acid.

Step 3: Write the chemical equation. The oxidation of ethanol to ethanal using acidified potassium dichromate(VI) is: $CH_3CH_2OH (l) + [O] \xrightarrow{Acidified K_2Cr_2O_7, heat, distill} CH_3CHO (l) + H_2O (l)$ The named aldehyde prepared is Ethanal.

2. Preparation of a Named Ketone (e.g., Propanone) from a Secondary Alcohol

Step 1: Identify the alcohol. Ketones are formed by the oxidation of secondary alcohols. For propanone ($\text{CH}_3COCH_3$), the corresponding secondary alcohol is propan-2-ol ($\text{CH}_3CH(OH)CH_3$).

Step 2: Choose an oxidizing agent and conditions. Secondary alcohols are oxidized to ketones using strong oxidizing agents like acidified potassium dichromate(VI) ($\text{K}_2Cr_2O_7/H^+$) or acidified potassium permanganate ($\text{KMnO}_4/H^+$). Ketones are generally resistant to further oxidation under these conditions, so vigorous heating is often used.

Step 3: Write the chemical equation. The oxidation of propan-2-ol to propanone using acidified potassium dichromate(VI) is: $CH_3CH(OH)CH_3 (l) + [O] \xrightarrow{Acidified K_2Cr_2O_7, heat} CH_3COCH_3 (l) + H_2O (l)$ The named ketone prepared is Propanone.

Send me the next one 📸