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Chapter 22: Problem 543

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

Expert verified
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

01

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.
02

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.
03

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 \(C_{8}H_{17}NHCH_{2}CH_{3}\).
04

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.

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Most popular questions from this chapter

While azanonatetraene is comparatively stable at room temperature, \(\mathrm{N}\) -carbethoxyazanonatetraene rapidly isomerizes under the same conditions. Account for this difference.

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