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Chapter 27: Problem 656

Pyridine methochloride is converted by aqueous alkali and a mild oxidizing agent (ferricyanide) to a solid compound of formula \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{NO}\), which shows no hydroxyl absorption in the infrared. What is a likely structure for this substance and how is it formed? Would you expect a similar reaction with pyridine?

Short Answer

Expert verified
The likely structure for the solid compound with molecular formula \(\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{NO}\) is a pyridine derivative with a carbonyl group, specifically an amide (\(\ce{C5H5N-CO-NH2}\)). This structure is formed through the hydrolysis of the methyl chloride group in pyridine methochloride by aqueous alkali and subsequent oxidation by ferricyanide to convert the hydroxyl group into a carbonyl group. A similar reaction would not be expected with pyridine, as it lacks the necessary methyl chloride group for hydrolysis and oxidation.

Step by step solution

01

Identifying the Initial Structure of Pyridine Methochloride

Pyridine methochloride is a derivative of pyridine, with a methyl chloride group attached to one of the nitrogen atoms of the pyridine ring. The structure of pyridine methochloride is: \[ \ce{ C5H5N-CH3Cl} \]
02

Analyzing the Effect of Aqueous Alkali and Ferricyanide on Pyridine Methochloride

The reaction with aqueous alkali (OH-) and ferricyanide (a mild oxidizing agent) will cause hydrolysis of the methyl chloride group and substitution of the chloride ion with a hydroxyl group (OH-). Then, the ferricyanide will oxidize the hydroxyl group to a carbonyl group (C=O). The resulting molecule has a molecular formula of \(\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{NO}\). The absence of hydroxyl absorption in the infrared spectrum indicates that the product does not have a hydroxyl group.
03

Determining the Possible Structure of the Product

Based on our analysis, we can conclude that the product is a pyridine derivative with a carbonyl group at the position where the methyl chloride group was originally present. This makes the product an amide with the molecular formula \(\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{NO}\). The structure of the product is: \[ \ce{ C5H5N-CO-NH2} \]
04

Analyzing the Possibility of a Similar Reaction with Pyridine

Pyridine itself does not have a methyl chloride group required for the hydrolysis and oxidation process discussed above. The aqueous alkali and ferricyanide would not have any significant effect on the pyridine ring without a group to hydrolyze and oxidize. Therefore, we do not expect a similar reaction to occur with pyridine.

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Most popular questions from this chapter

The alkaloid hygrine is found in the coca plant. Suggest a structure for it on the basis of the following evidence: Hygrine \(\left(\mathrm{C}_{8} \mathrm{H}_{1}{ }_{5} \mathrm{ON}\right)\) is insoluble in aqueous \(\mathrm{NaOH}\) but soluble in aqueous HC1. It does not react with benzenesulfonyl chloride. It reacts with phenylhydrazine to yield a phenylhydrazone. It reacts with NaOI to yield a yellow precipitate and a carboxylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{13} \mathrm{O}_{2} \mathrm{~N}\right)\). Vigorous oxidation by \(\mathrm{CrO}_{3}\) converts hygrine into hygrimic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{11} \mathrm{O}_{2} \mathrm{~N}\right)\)

Two mechanisms for the Skraup synthesis are: 1 2. c1ccc2ccccc2c1 Devise an experiment whereby we can tell which is the actual mechanism.

Many of the beautiful colors of butterflies' wings, autumn leaves, and flowers are due to the presence of flavone derivatives called anthocyanins. When anthocyanins are hydrolyzed in hydrochloric acid (to remove sugar residues), they produce aromatic oxygen-containing salts. One of these is delphinidin chloride. Explain why delphinidin chloride is aromatic. What is the hybridization of the heterocyclic oxygen in this compound?
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