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Rama09 [41]
2 years ago
11

Which fundamental mechanisms appear in the reaction between pyridinium chlorochromate and a secondary alcohol

Chemistry
1 answer:
oksano4ka [1.4K]2 years ago
6 0

The idea is to foretell the formation of a carbonyl compound by the reaction between alcohol and too much pyridinium chlorochromate. An oxidizing agent called pyridinium chlorochromate converts the alcohol group into the 1carbonyl group.

The carbonyl molecule that results from the reaction will depend on the reactant's OH group. Pyridinium chlorochromate [PCC] converts primary OH to aldehydes, whereas it converts secondary OH to ketones, and oxidation of tertiary OH has little effect. Alcohols and pyridinium chlorochromate [PCC] react to create a carbonyl molecule.

From primary alcohols to aldehydes and from secondary alcohols to ketones, pyridinium chlorochromate oxidizes alcohols one step up the oxidation ladder. pyridinium chlorochromate will not oxidize aldehydes to carboxylic acids, in contrast to chromic acid. Comparable to Pyridine (the Collins reagent) and CrO3 will both oxidize primary alcohols to aldehydes. Here are two instances of pyridinium chlorochromate being used.

To learn more about pyridinium chlorochromate please visit -
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Explanation:

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Explanation:

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On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed
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Answer:

On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?

A) BrCH2CH2Br

B) CH3CH2CH2Br

C) CH3CHBr2

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E) BrCH2CH2CH2CH2Br

Explanation:

The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.

This is an example of free radical substitution.

The structure of ethane and its bromination is shown below:

Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).

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