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What is the oxidation product of a primary alcohol?

(A) aldehyde (B) alkene (C) ester (D) ketone

My try: So because an alcohol is $\ce{R-OH}$, and oxidation is loss of electrons, I thought that the product would be something like $\ce{R-O}$, which is an ether. However that is not an answer choice. The correct answer in this case is A. Can someone explain how?

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  • $\begingroup$ 1) An alkene has no oxygen. 2) An ester has an oxygen between two carbons. 3) A ketone does not have any oxygen on a primary carbon. $\endgroup$ – Joseph Hirsch Dec 20 '16 at 18:43
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I remember this from school. Don't mind the experts telling you other things, you don't need to know that. You need to oxidize a primary alcohol. Start with this structure, not with the one you suggested:

$$\ce{R-CH2-OH}$$

Look at the carbon atom, that's the one you oxidize! Right now, it has an oxidation number of $-1$ ($+0$ (doesn't matter) from $\ce{R}$, $2x-1$ from the $\ce{H}$ and $x+1$ from $\ce{O}$). Oxidation means increasing that number. So you need to substitute one hydrogen. Let's check your options:

Aldehyde:

$$\ce{R-CH=O}$$

Perfect. You've got $+1$ now.

Alkene: Doesn't make any sense here.

Ester:

$$\ce{R-C(=O)-O-R'}$$

Although this is an oxidation, it usually needs a second molecule (which could be the same species) of alcohol.

Ketone:

$$\ce{R-CO-R'}$$

That's an oxidation. Well, let's come back to the fact that you're in school. It's in the question really. You need to oxidize your primary alcohol. Just remember that a ketone is the product of the oxidation of a secondary alcohol, which I hope you can see makes more sense:

$$\ce{R-COH-R'}$$

Btw, don't count on my oxidation numbers. The concept is somewhat useless in actual chemistry, it's been a while since I did that.

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  • $\begingroup$ Thanks for whoever edited. In my first post I made an ether instead of an ester. Man this was last relevant to me such a long time ago ... $\endgroup$ – AMT Dec 20 '16 at 15:23
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With a weak oxidizing agent, the product would be an aldehyde. Synthetically useful strong oxidants will oxidize the primary alcohol to a carboxylic acid.

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    $\begingroup$ Well it depends on how you define "strong oxidant," but the reaction could go so far as to convert all the organic matter to water and carbon dioxide. $\endgroup$ – MaxW Dec 20 '16 at 2:56
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Alcohols on oxidation always give aldehyde first and if we are using strong oxidizing agent aldehyde will further get oxidized into carboxylic acids and the final product we get from oxidation of alcohol is carboxylic acid in that case. Thus alcohol oxidizes to aldehydes and aldehyde oxidizes to carboxylic acids. On using weak oxidizing agent product will be aldehyde.

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A primary alcohol either oxidizes to an aldehyde or carboxylic acid, and it depends on two factors: strength of the oxideser and quantity

If the alcohol is in excess, then it is oxidised once to an aldehyde, especially if this aldehyde is distilled as it is formed. Full oxidation can then occur once all of the initial oxidation occurs; this time to a carboxylic acid if there is left over oxidiser and it is heated

Sometimes by using a strong oxidiser such as $\ce{KMnO_4}$ it may directly oxidise to carboxylic acid, or by using a weak oxidiser prevent formation of the carboxylic acid

The way oxidation to aldehydes works is the formation of a double bond between the oxygen on the alcohol and central carbon (carbonyl group), with the oxidser oxygen combining with the hydrogen on the central carbon and the alcohol hydrogen to produce water.

There is a stack exchange post describing the mechanisms of oxidation using $\ce{KMnO_4}$ at Mechanism of the oxidation of alcohols with KMnO4

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  • $\begingroup$ I'll bet you I could use half as much KMnO4 as PCC and still yield more carboxylic acid. $\endgroup$ – electronpusher Dec 20 '16 at 3:53
  • $\begingroup$ The standard procedure is to put more oxidiser to oxidise to carboxylic acid $\endgroup$ – Copper Dec 20 '16 at 5:06
  • $\begingroup$ But the extent of oxidation also depend on the nature of the oxidizing agent itself, not just quantity. $\endgroup$ – electronpusher Dec 20 '16 at 5:31
  • $\begingroup$ Ok, fair enough, I will edit $\endgroup$ – Copper Dec 20 '16 at 5:45
  • $\begingroup$ Thank you. I think the second to last paragraph gives a misleading sense of the oxidation mechanism. The O that picks up the hydroxyl H will likely not be part of the transition metal oxidant complex itself, but just some added base. It is also not likely the same O that will deprotonate the C. It should also be noted that the whole idea is: deprotonate the hydroxyl, add a transition metal (oxidant) leaving group to the O, then perform E2-like elimination by deprotonating the carbon, forming the C=O pi bond. Keyword: elimination. $\endgroup$ – electronpusher Dec 20 '16 at 16:34

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