The Student Room Group

Organic Chemistry help

When forming carboxylic acids where do the hydrogens go - for example the formula for ethanol is C2H5OH whereas the formula for ethanoic acid us CH3COOH - where did 2 of the hydrogens go?
Original post by KhaledChouman
When forming carboxylic acids where do the hydrogens go - for example the formula for ethanol is C2H5OH whereas the formula for ethanoic acid us CH3COOH - where did 2 of the hydrogens go?

I assume you are talking about reactions.
If you look at the balanced equation you will see exactly where they "go".
Reply 2
Original post by KhaledChouman
When forming carboxylic acids where do the hydrogens go - for example the formula for ethanol is C2H5OH whereas the formula for ethanoic acid us CH3COOH - where did 2 of the hydrogens go?

CH3CH2OH + [O] = CH3CHO +H2O (where your 2 Hs are found) If this were distilled only ethanal would be formed.
CH3CHO + [O] = CH3COOH under reflux
[O] = acidified potassiumdichromate VI
Original post by charco
I assume you are talking about reactions.
If you look at the balanced equation you will see exactly where they "go".

If you use dichromate(VI) ions to oxidise the ethanol then the two half equations are:

C2H5OH + H2O ==>> CH3COOH + 4H+ + 4e
Cr2O72- + 14H+ + 6e ==> 2Cr3+ + 7H2O

to add the two equations together you have to first equalise the electrons by mtb 3 in equation 1 and mtb 2 in equation 2

3C2H5OH + 3H2O ==>> 3CH3COOH + 12H+ + 12e
2Cr2O72- + 28H+ + 12e ==> 4Cr3+ + 14H2O
------------------------------------------------------------------ add
3C2H5OH + 3H2O + 2Cr2O72- + 28H+ ==>> 3CH3COOH + 12H+ + 4Cr3+ + 14H2O

now cancel out common terms from both sides

3C2H5OH + 2Cr2O72- + 16H+ ==>> 3CH3COOH + 4Cr3+ + 11H2O
Reply 4
Original post by rachellca

CH3CH2OH + [O] = CH3CHO +H2O (where your 2 Hs are found) If this were distilled only ethanal would be formed.
CH3CHO + [O] = CH3COOH under reflux
[O] = acidified potassiumdichromate VI


I think in GCSE we skip like 4 steps 😂
Reply 5
Original post by KhaledChouman
When forming carboxylic acids where do the hydrogens go - for example the formula for ethanol is C2H5OH whereas the formula for ethanoic acid us CH3COOH - where did 2 of the hydrogens go?

If you look at the displayed formulae, you'll see that, in ethanol, the carbon attached to the –OH is bonded to four atoms (C, 2 x H, and O) whereas in ethanoic acid this equivalent carbon is only bonded to three atoms (C, O, and another O). To have a stable electron configuration, a carbon atom usually needs to form 4 bonds, essentially so that it has 8 valence electrons. So in ethanoic acid, one of the oxygens is bonded to the carbon via a double bond. And because oxygens usually form 2 bonds, this double-bonded oxygen does not have a hydrogen attached as well.
In the reaction, two hydrogens are usually lost as part of a water molecule.



Chemguide has a few pages on this:
GCSE: https://www.chemguide.co.uk/14to16/organic/alcohols.html (scroll down a bit to the sections about oxidation)
A level: https://www.chemguide.co.uk/organicprops/acids/preparation.html
(edited 11 months ago)

Quick Reply