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Why is it that in order to make an carboxylic acid from an alcohol you oxidise it twice...

ie alcohol + 2[O] ---> carboxylic acid

...but to get back to the alcohol you reduce it four times?

carboxylic acid + 4[H] ---> alcohol

My understanding was that oxidation is the opposite of reduction, and in organic chemistry [O] means addition of an oxygen atom or removal of a hydrogen atom, with [H] meaning the opposite.

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    $\begingroup$ One easy way to rationalize the relationship is to consider the $\ce{H2O}$ molecule where an oxygen atom can oxidize two hydrogen atoms. $\endgroup$ – Zhe Jun 6 '17 at 19:00
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Addition of oxygen and hydrogen can be useful to predict oxidation and reduction in most of the cases but this criteria fails to classify a reaction as oxidation or reduction when there are no oxygen or hydrogen atoms involved. This method is very primitive.

A better perspective :

Oxidation : Increase in oxidation state (by loss of electrons) of central atom.

Reduction : Decrease in oxidation state (by gain of electrons) of central atom.

If you already know how to find the oxidation state of the central atom (in your case carbon), then I hope you will be able to get your answer now.

The answer to the question why oxidation of alcohol to carboxylic acid takes 2 oxidation reactions, lies in the reaction mechanism. These mechanisms cannot be challenged, they are what they are. We can only observe and provide a acceptable explanation to why they happen like the way they do.

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in reaction dont see how much O or H you are using.Just see how your derired substance is oxidized.During first oxidation to carbonyl alcohol loses hydrogen in case of methanol 1 hydroge.And nest on to convert to acid adds one oxygen. Similar is the case whil reducing acid first adds remove one Oygen atom and then in carvonyl to alcohol adds one H atom.

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    $\begingroup$ This is kind of a mess and not very coherent. Can you clean it up a bit? $\endgroup$ – jonsca Jun 7 '17 at 1:38

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