Question

Piperidine is subjected to exhaustive methylation and the quaternary iodide is heated with moist \(\mathrm{Ag}_{2} \mathrm{O}\) and the number of double bonds present in the product formed is/are

Short answer

The product formed has 1 double bond.

Step-by-step solution

Identify the Starting Material and Reaction

Piperidine is a heterocyclic compound which, when subjected to exhaustive methylation, is converted into its quaternary ammonium iodide. This means that the nitrogen atom in piperidine will be fully methylated, converting it into a quaternary ammonium salt.

Analyze the Reaction with Ag2O

When the quaternary ammonium iodide is treated with moist Ag2O, it undergoes a Hofmann elimination reaction. In this reaction, the hydroxide ions generated from Ag2O cause the elimination of the least hindered alkyl group, resulting in the formation of an alkene.

Determine the Product Formed

The product formed in the Hofmann elimination will have a double bond between carbon atoms adjacent to the nitrogen. Given that piperidine has a cyclic structure, the most favorable elimination results in the formation of a double bond between the ring members.

Count the Number of Double Bonds

In the product formed, the cyclic structure of piperidine undergoes an elimination that typically results in one double bond being incorporated into the 5-membered ring. Thus, only one double bond is formed.

Key concepts

Piperidine

Piperidine is a six-membered ring compound that contains one nitrogen atom. This nitrogen atom acts as a basic center in the molecule, which can participate in various chemical reactions. Piperidine is classified as a heterocyclic compound due to the presence of a non-carbon atom (nitrogen) within its carbon-containing ring.

When piperidine undergoes exhaustive methylation, it transforms into a fully methylated quaternary ammonium compound. This process involves the addition of methyl groups to the nitrogen until it carries no hydrogen atoms anymore. Each methyl group replaces a hydrogen, resulting in a positively charged nitrogen with four attached alkyl groups.

Quaternary Ammonium Salt

A quaternary ammonium salt is a type of ionic compound. This compound forms when a nitrogen atom in a molecule gains four substituents, turning it into a positively charged ion.

In the context of piperidine, exhaustive methylation leads to the formation of a quaternary ammonium iodide. Here, the nitrogen atom now carries four methyl groups. The presence of iodide balances the positive charge of the nitrogen, leading to a neutral ionic compound overall. Quaternary salts like these are often used as intermediates in organic syntheses, as they can readily undergo elimination reactions.

Alkene Formation

Alkenes are hydrocarbons characterized by the presence of double bonds between carbon atoms. The formation of alkenes often involves elimination reactions. In the case of piperidine's quaternary ammonium salt, the reaction with moist Ag extsubscript{2}O facilitates an elimination reaction known as Hofmann elimination.

During Hofmann elimination, the hydroxide ions generated from Ag extsubscript{2}O cause the removal of a small, less hindered alkyl group from the molecule. This removal leads to the formation of a double bond. In our specific case involving piperidine, the process results in an alkene with a double bond along the existing cyclic ring, favoring the formation of a product with one double bond due to piperidine's symmetrical structure.

Cyclic Structure

The cyclic structure of piperidine plays a key role in determining the outcome of chemical reactions that it undergoes. This six-membered ring provides structural stability, but it also dictates the pattern of bond formation and breaking during reactions.

In Hofmann elimination, this cyclic nature influences the location where the double bond will form. Typically, the double bond forms between carbon atoms within the ring, adjacent to the nitrogen. Thus, the cyclic structure not only maintains the overall geometry of the molecule but also directly impacts how many and where the double bonds can be formed. In our reaction, only one double bond is incorporated into the piperidine ring as a result of these constraints.