Heating \(\beta\)-hydroxyethylamide (1) with thionyl chloride affords \(\beta\)-chloroethylamide (2). In turn, amide 2 yields oxazoline 3 upon treatment with \(\mathrm{NaOH}\). (a) Write arrow formalism mechanisms for the formations of \(\mathbf{2}\) and \(\mathbf{3}\). Things are not quite as simple as they first appear. For example, if \(\mathbf{1}\) is treated with thionyl chloride below \(25^{\circ} \mathrm{C}\), compound 4 is obtained instead of 2 . Compound \(\mathbf{4}\) has the same elemental composition as \(\mathbf{2}\) but, unlike \(\mathbf{2}\), is soluble in water. Upon heating, \(\mathbf{4}\) gives \(\mathbf{2}\), and compound 4 yields 3 upon treatment with aqueous sodium carbonate. (b) Propose a structure for 4 and show how 4 is formed and converted into 2 and 3 .

Arrow formalism in organic chemistry illustrates how electrons move during a reaction. These arrows show the path electrons take, usually starting from a nucleophile (electron donor) toward an electrophile (electron acceptor).

For instance, when \beta\-hydroxyethylamide reacts with thionyl chloride, the lone pairs on the oxygen atom attack the sulfur atom in \(\text{SOCl}_2\).
Arrows also depict the breaking of bonds, like the bond between chlorine and sulfur moving to chlorine. This formalism helps us visualize and understand complex mechanisms easily.

In the transformation from compound \beta\-hydroxyethylamide (1) to \(\beta\)-chloroethylamide (2) and then to oxazoline (3), each step can be illustrated with careful arrow formalism to show how bonds form and break, making the mechanisms clear and logical.

Thionyl chloride (\(SOCl_2\)) is a reagent commonly used to substitute hydroxyl groups with chlorine atoms.
In this reaction, \(\beta\)-hydroxyethylamide (1) interacts with \(SOCl_2\) to replace its hydroxyl group with a chlorine atom, forming \(\beta\)-chloroethylamide (2).

This reaction proceeds via a nucleophilic substitution mechanism where the lone pairs on the oxygen atom attack the sulfur atom in \(SOCl_2\), resulting in the release of \(SO_2\) and \(HCl\) as by-products. The detailed steps include:

• Lone pairs on the oxygen attacking the sulfur in \(SOCl_2\)
• Formation of a transition state
• Displacement of chloride ions and formation of a sulfite ester intermediate
• Release of \(HCl\) and \(SO_2\)

Oxazolines are five-membered heterocyclic compounds containing an oxygen and a nitrogen atom within the ring.

The formation of oxazoline (3) from \(\beta\)-chloroethylamide (2) involves the deprotonation of the amide group by \(NaOH\), forming an amide anion.

This anion then performs an intramolecular nucleophilic attack on the carbon bonded to the chlorine atom, leading to the cyclization process:

• The amide nitrogen attacks the carbon bearing the chlorine.
• A five-membered oxazoline ring forms.
• Sodium chloride (\(NaCl\)) is released as a by-product.

This intramolecular reaction effectively closes the ring structure, stabilizing into the oxazoline 3.

The transformation from a hydroxyl group (OH) to a chlorine atom (Cl) is a common substitution reaction in organic chemistry.
When \(\beta\)-hydroxyethylamide reacts with thionyl chloride (\(SOCl_2\)), this substitution occurs.

More specifically, the mechanism involves:

• Nucleophilic attack by the hydroxyl's lone pairs on the sulfur in \(SOCl_2\)
• Formation of a chlorosulfite intermediate
• Elimination of \(SO_2\) and \(HCl\) to yield \(\beta\)-chloroethylamide (2)

This step is crucial for further transformations, as it sets up the molecule for subsequent cyclization to form the oxazoline ring.

Intramolecular cyclization is a reaction where a molecule forms a ring by bonding to itself internally.
For instance, the compound \(\beta\)-chloroethylamide (2) undergoes intramolecular cyclization to form oxazoline (3) in the presence of a base like \(NaOH\).

The process follows these steps:

• Base deprotonates the amide, forming an amide anion
• The amide anion attacks the carbon atom bonded to chlorine
• A five-membered ring forms, with the nitrogen and oxygen included
• Sodium chloride (\(NaCl\)) is expelled as a by-product

This cyclization not only creates a stable ring structure but also dramatically changes the molecule's properties and reactivity.