Question

Compound A, C \(_{12} \mathrm{H}_{28} \mathrm{~N}_{2}\), was soluble in dilute acid, but did not react with benzenesulfony1 chloride. Upon addition of excess methyl iodide, Compound \(\mathrm{A}\) reacted. It was then treated with silver oxide, and the resulting product heated. Thus \(\mathrm{B}, \mathrm{C}_{8} \mathrm{H}_{14}\), was formed. B was treated with ozone and the ozonide decomposed with dilute acid in the presence of zinc dust. One of the two products isolates was

Short answer

The structure of Compound A is a secondary amine \(C_{8}H_{17}NHCH_{2}CH_{3}\). After a series of reactions, ozonolysis of compound B leads to the formation of either propanal or pentanal as one of the isolated products.

Step-by-step solution

Assessing the reactivity of compound A

Compound A has the molecular formula \(C_{12}H_{28}N_{2}\) and is soluble in dilute acid but does not react with benzenesulfonyl chloride. This suggests that Compound A does not contain a primary amine group, as primary amines react with benzenesulfonyl chloride.

Reactions involving compound A

Compound A reacts with excess methyl iodide, indicating that it must contain a nitrogen with a lone pair of electrons that can react with the electrophile. The subsequent treatment with silver oxide suggests that a quaternary ammonium salt was formed due to the excess of methyl iodide. The resulting product was then heated to form compound B.

Determining the structure of compound A

Since we know that Compound A does not contain a primary amine, it must contain either a secondary amine or a tertiary amine. It reacts with excess methyl iodide, forming a quaternary ammonium salt, which indicates that the nitrogen atom in A is bonded with a hydrogen. Thus, Compound A must be a secondary amine containing a \(C_{8}\) alkyl group and a \(C_{3}\) alkyl group. The structure of Compound A can be written as \(C_{8}H_{17}NHCH_{2}CH_{3}\).

Ozonolysis of compound B and identification of the product

Compound B is formed by heating the quaternary ammonium salt, resulting in a \(C_{8}H_{14}\) compound. This compound undergoes ozonolysis and subsequent reduction with zinc dust in the presence of dilute acid. This cleaves the carbon-carbon double bonds in compound B and forms two products. One possible structure of compound B is \(CH_3CH_2CH=CHCH_2CH_2CH_2CH_3\). After ozonolysis and reduction, its two products would be \(CH_3CH_2CHO\) (propanal) and \(CH_3CH_2CH_2CH_2CHO\) (pentanal). Thus, one of the isolated products after ozonolysis of compound B would be either propanal or pentanal.

Key concepts

Amine Reactivity

In organic chemistry, amines are versatile compounds that can participate in a variety of reactions due to the presence of a nitrogen atom with a lone pair of electrons. These Lone pairs are essential as they make amines basic and nucleophilic. When dealing with amine reactivity, it is crucial to understand how they behave in different chemical environments.

• **Primary amines** have one alkyl or aryl group attached to the nitrogen and are quite reactive with a variety of electrophiles.
• **Secondary amines**, like Compound A in the problem, have two such groups and respond differently. They are less reactive with benzenesulfonyl chloride compared to primary amines, which explains why Compound A did not react in this case.
• **Tertiary amines** have three groups attached and generally lack a hydrogen atom necessary for forming certain additional products.

Understanding the type of amine helps predict its reactivity and determines the right conditions under which it will react. For instance, in the problem, the reaction of compound A with methyl iodide suggests it contains a nitrogen capable of forming quaternary ammonium salts.

Quaternary Ammonium Salts

Quaternary ammonium salts are a unique class of compounds formed when tertiary amines react with alkyl halides. These reactions result from a nitrogen atom with a lone pair attacking the electrophile, generally an alkyl halide like methyl iodide.

• This process replaces any hydrogen atoms on the nitrogen with an alkyl group, leading to a tetrahedral, positively charged ammonium ion.
• These salts are stable due to their full positive charge and cannot loose a proton easily, hence they don't behave as bases.
• In the case of our exercise, after Compound A (a secondary amine) reacts with methyl iodide, additional methyl groups are added to the nitrogen, creating a quaternary ammonium salt.

Upon heating in the presence of a base such as silver oxide, the salt decomposes to liberate an alkene, illustrated by the conversion of the ammonium salt into Compound B (an alkene with the formula \(C_8H_{14}\)). This reaction highlights both the utility and the reactivity of quaternary ammonium salts in organic chemistry.

Ozonolysis

Ozonolysis is a powerful reaction used to cleave double bonds in alkenes, resulting in two smaller molecules bearing carbon-oxygen double bonds, typically aldehydes or ketones.

• This chemical process involves the treatment of an alkene, like Compound B, with ozone \(O_3\), a strong oxidizing agent that adds across the double bond.
• After ozone treatment, cleavage of double bonds leaves an intermediate called an ozonide, which can be unstable.
• To isolate useful products, the ozonide is typically reduced in the presence of a reducing agent like zinc dust and a dilute acid, as seen in our scenario.

The fascinating part about ozonolysis is its ability to reveal or confirm the presence of double bonds in unknown compounds by providing significant hints for the molecular structure through the resulting products, such as aldehydes or ketones. In the exercise, ozonolysis of Compound B provides propanal and pentanal, indicating where the original carbon-carbon double bonds were located.

Molecular Structure Determination

Determining a molecule's structure from its chemical reactions and characteristics is known as molecular structure determination. This analytical process relies on understanding how a compound reacts with known substances and the resulting products.

• For instance, Compound A's inability to react with benzenesulfonyl chloride suggested a lack of primary amine groups, pointing towards a secondary or tertiary amine.
• Further reactions, like with methyl iodide, help clarify its structure by indicating a nitrogen center ready to form a quaternary ammonium salt, as demonstrated in the exercise.
• Finally, identifying the products resulting from processes like ozonolysis offers a way to deduce locations of functional groups or double bonds, as seen when compound B was converted to propanal and pentanal.

These observations, when combined, create a comprehensive picture of a compound's structural framework. Organic chemists frequently use these logical deductions to determine the exact make-up of complex molecules, showcasing the intricate puzzle of molecular structure determination.