Q72E - Organic Chemistry (Mcmurry)

Question

When 4-hydroxybutanal is treated with methanol in the presence of an acid catalyst, 2-methoxytetrahydrofuran is formed. Explain

Step-by-Step Solution

Answer

4-Hydroxybutanal forms a cyclic hemiacetal when the hydroxyl oxygen add to the aldehyde group. Methanol reacts with this cyclic hemiacetal to form 2-methoxytetrahydrofuran which is our desired product.2-Methoxytetrahydrofuran is a cyclic acetal. The hydroxyl oxygen of 4-Hydroxybutanal reacts with the aldehyde to form cyclic ether linkage.

Step-by-step Solution Step 1: Protonation

Protonation is defined as the addition of a proton to an atom, molecule or ion. It is different from hydrogenation because in protonation a change in charge of protonated species occurs, while the charge remain unaffected during hydrogenation. For example formation of ammonium group ${{\mathrm{NH}}_4}^{+}$ from ammonia ${\mathrm{NH}}_3$.Protonation is mainly occur in the reaction of an acid with a base to form salt.

Step 2: Formation of cyclic hemiacetal

4-Hydroxybutanal forms a cyclic hemiacetal when the hydroxyl oxygen add to the aldehyde group. The formation of hemiacetal is shown in the figure given below:

Synthesis of hemiacetal

Here steps involved in the reaction mechanism are:

1. Protonation

2. Addition of -OH

3. Loss of proton

Step 3: Synthesis of 2-methoxytetrahydrofuran

Methanol reacts with the cyclic hemiacetal to form 2-methoxytetrahydrofuran. The reaction is shown in the figure given below:

Synthesis of 2-methoxytetrahydrofuran

Here, steps involved in the reaction mechanism are:

1. Protonation

2. Loss of water

3. Addition of methanol

4. Loss of proton

2-Methoxytetrahydrofuran is a cyclic acetal. The hydroxyl oxygen of 4-Hydroxybutanal reacts with the aldehyde to form cyclic ether linkage.