[fayled]

In the OCR A2 textbook for chemistry A, the mechanism can be found on page 25 (if anyone wants to look it up).

Basically the reaction the mechanism shows is:
Propanal+ 2[H] -> Propan-1-ol
CH3CH2CHO + 2[H] -> CH3CH2CH2OH
Now these equations are fully balanced, with the H representing the reducing agent, which I easily understand.

Now, the mechanism shows:

The Hydride ion attacking the slightly positive carbon atom on the C=O bond, and forming a bond with this carbon, and at the same time the pi-bond breaking in the C=O bond so the oxygen has both these electrons and is negatively charged.

Now this is the part I don't understand - water decides to get involved...
So the negatively charged oxygen atom donates an electron pair to a hydrogen atom in a water molecule, forming a dative covalent bond, and a hydroxide ion. This completes the formation of the alcohol.

Now this equation no longer matches the one I stated earlier - now it is:
CH3CH2CHO + 2[H] -> CH3CH2CH2OH + OH-
i.e no longer balanced.

The only thing I could suggest to make this correct is that in the old equation they just simplified water to just a [H] so then could ignore the hydroxide ion.
So the above shoud have been:
CH3CH2CHO + [H] + H20 -> CH3CH2CH2OH + OH-

Anyway sorry about the length of this, would appreciate some clarification.

[thegodofgod]

(Original post by fayled)
In the OCR A2 textbook for chemistry A, the mechanism can be found on page 25 (if anyone wants to look it up).

Basically the reaction the mechanism shows is:
Propanal+ 2[H] -> Propan-1-ol
CH3CH2CHO + 2[H] -> CH3CH2CH2OH
Now these equations are fully balanced, with the H representing the reducing agent, which I easily understand.

Now, the mechanism shows:

The Hydride ion attacking the slightly positive carbon atom on the C=O bond, and forming a bond with this carbon, and at the same time the pi-bond breaking in the C=O bond so the oxygen has both these electrons and is negatively charged.

Now this is the part I don't understand - water decides to get involved...
So the negatively charged oxygen atom donates an electron pair to a hydrogen atom in a water molecule, forming a dative covalent bond, and a hydroxide ion. This completes the formation of the alcohol.

Now this equation no longer matches the one I stated earlier - now it is:
CH3CH2CHO + 2[H] -> CH3CH2CH2OH + OH-
i.e no longer balanced.

The only thing I could suggest to make this correct is that in the old equation they just simplified water to just a [H] so then could ignore the hydroxide ion.
So the above shoud have been:
CH3CH2CHO + [H] + H20 -> CH3CH2CH2OH + OH-

Anyway sorry about the length of this, would appreciate some clarification.

Try solving it like a redox half-equation (as after all, it is a reduction half-equation):

C₃H₆O + 2 H⁺ + 2 e^- ----> C₃H₈O

[illusionz]

(Original post by fayled)
In the OCR A2 textbook for chemistry A, the mechanism can be found on page 25 (if anyone wants to look it up).

Basically the reaction the mechanism shows is:
Propanal+ 2[H] -> Propan-1-ol
CH3CH2CHO + 2[H] -> CH3CH2CH2OH
Now these equations are fully balanced, with the H representing the reducing agent, which I easily understand.

Now, the mechanism shows:

The Hydride ion attacking the slightly positive carbon atom on the C=O bond, and forming a bond with this carbon, and at the same time the pi-bond breaking in the C=O bond so the oxygen has both these electrons and is negatively charged.

Now this is the part I don't understand - water decides to get involved...
So the negatively charged oxygen atom donates an electron pair to a hydrogen atom in a water molecule, forming a dative covalent bond, and a hydroxide ion. This completes the formation of the alcohol.

Now this equation no longer matches the one I stated earlier - now it is:
CH3CH2CHO + 2[H] -> CH3CH2CH2OH + OH-
i.e no longer balanced.

The only thing I could suggest to make this correct is that in the old equation they just simplified water to just a [H] so then could ignore the hydroxide ion.
So the above shoud have been:
CH3CH2CHO + [H] + H20 -> CH3CH2CH2OH + OH-

Anyway sorry about the length of this, would appreciate some clarification.

You are correct in noticing that the [H] notation is a simplification and does not give the full picture. It only tells you that two hydrogen atoms are added to the molecule in question. In reality one of these comes from the reducing agent and one from an acid (in this case water acts as the acid).

[Alchem]

(Original post by illusionz)
You are correct in noticing that the [H] notation is a simplification and does not give the full picture. It only tells you that two hydrogen atoms are added to the molecule in question. In reality one of these comes from the reducing agent and one from an acid (in this case water acts as the acid).

That is correct. So, to sum up, the correct balanced equation is:

CH₃ CH₂CHO + [H ^-] + H₂O ------> CH₃CH₂CH₂OH + OH^-