Question

Using your reaction roadmap as a guide, show how to convert 1-bromopentane and sodium cyanide into \(N\)-hexylhexanamide.You must use 1-bromopentane and sodium cyanide as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way.

Short answer

Question: Describe the process to synthesize N-hexylhexanamide from 1-bromopentane and sodium cyanide. Answer: The synthesis of N-hexylhexanamide from 1-bromopentane and sodium cyanide involves three main steps: 1. Convert 1-bromopentane to 1-pentanenitrile via a nucleophilic substitution reaction with sodium cyanide (NaCN). 2. Reduce 1-pentanenitrile to hexanamide using lithium aluminum hydride (LiAlH4) as a reducing agent, followed by hydrolysis with water (H2O). 3. Convert hexanamide to N-hexylhexanamide using the Schotten-Baumann reaction, which involves forming an N-acyl imidoyl halide, and then amination using 1-bromohexane and a base like sodium hydroxide (NaOH).

Step-by-step solution

Convert 1-bromopentane to 1-pentanenitrile

To convert 1-bromopentane into 1-pentanenitrile, we will use a nucleophilic substitution reaction with sodium cyanide (NaCN). The nucleophile (CN-) will replace the bromine atom in 1-bromopentane, resulting in the formation of 1-pentanenitrile and sodium bromide (NaBr) as a side product. 1-bromopentane + NaCN → 1-pentanenitrile + NaBr

Reduce 1-pentanenitrile to hexanamide

In this step, we need to perform the reduction of the nitrile functional group (C≡N) in 1-pentanenitrile to an amide functional group (C=O-NH2). To do this, we will use a strong reducing agent, lithium aluminum hydride (LiAlH4), followed by the addition of water (H2O) to carry out the hydrolysis. 1. 1-pentanenitrile + LiAlH4 → Hexanamide-LiAlH4 complex 2. Hexanamide-LiAlH4 complex + H2O → hexanamide (C5H11CONH2) + byproducts

Convert hexanamide to N-hexylhexanamide

Now that we have hexanamide (C5H11CONH2), we need to convert it into N-hexylhexanamide (C6H13C(O)NHC5H11). To achieve this, we will use the Schotten-Baumann reaction, which is a two-step process involving the formation of an N-acyl imidoyl halide and then its amination to form the final product. The required reagents for this reaction are 1-bromohexane and a base, such as sodium hydroxide (NaOH). 1. Hexanamide + 1-bromohexane + NaOH → N-hexylhexanimidoyl bromide + NaBr 2. N-hexylhexanimidoyl bromide + hexanamide + NaOH → N-hexylhexanamide (C6H13C(O)NHC5H11) + NaBr + H2O Thus, we have successfully converted 1-bromopentane and sodium cyanide into N-hexylhexanamide through a series of reactions involving nucleophilic substitution, reduction, and finally, the Schotten-Baumann reaction.

Key concepts

Nucleophilic Substitution Reaction

Understanding the mechanics of a nucleophilic substitution reaction is crucial for gaining insights into many synthesis processes in organic chemistry. In essence, this type of reaction involves a nucleophile—an electron-rich species—attacking an electrophile, which is typically a carbon atom attached to a leaving group.

The nucleophile donates a pair of electrons to form a new chemical bond while the leaving group, an atom or group of atoms that can easily take a pair of electrons, departs from the molecule. A common example of this is the reaction between 1-bromopentane and sodium cyanide, where the cyanide ion (CN-) serves as the nucleophile and the bromine atom is the leaving group.

In the exercise, the goal was to transform 1-bromopentane into 1-pentanenitrile, a key step towards synthesizing N-hexylhexanamide. During the reaction, the bromine atom of 1-bromopentane is substituted by the cyanide ion. The student should note the importance of the initial choice of reagents to ensure all carbon atoms in the final product originate from specified starting materials.

Nitrile Reduction

Nitrile reduction is a transformative process that converts the nitrile group (-C≡N) into an amide group (-C(=O)NH2). This conversion is an essential step in building complex organic molecules from simpler ones. In the reduction process, we typically use a strong reducing agent like lithium aluminum hydride (LiAlH4).

The reduction occurs in two stages. First, the LiAlH4 reacts with the nitrile to form a hexanamide-LiAlH4 complex. This complex is then subsequently hydrolyzed by water (H2O) to release the amide product along with byproducts. When students work through the process, it's vital that they understand the change in the functional group from the triple bonded nitrogen in the nitrile to the double bonded oxygen in the newly formed amide—a significant transformation in the pathway to synthesize N-hexylhexanamide.

Schotten-Baumann Reaction

The Schotten-Baumann reaction is instrumental in the field of organic chemistry for the synthesis of amides from acyl chlorides or other activated acyl derivatives. It relies on the reactivity of these derivatives in the presence of a base to facilitate the coupling with a nucleophile, such as an amine or in this case, another amide molecule.

In the context of the exercise, the Schotten-Baumann reaction involves two main steps. Initially, the reaction of hexanamide with 1-bromohexane in the presence of a base like sodium hydroxide, forms N-hexylhexanimidoyl bromide. This intermediate is then subject to a reaction with additional hexanamide, encouraging the formation of N-hexylhexanamide, the target molecule. During this sequence, it’s important that students recognize the role of the base in deprotonating the amine to enhance nucleophilicity, facilitating the second amination step necessary to complete the synthesis.