[RainbowMoo]

I can't figure out the balanced equation for the Luche Reduction of L-Menthone to methanol with NaBH4 and CeCl3*7H2O.

So far I have

C10H18O + NaBH4 + CeCl3 + H2O = C10H20O

I don't know if that's even correct. Please help. Thanks so much.

Is this perhaps a re-dox and I should omit spectator ions(Na and Ce)?

[spirochete]

Is this question for a general chemistry class or an organic chemistry class?

As a starting point you need to finish drawing the other side products of the reaction. To understand the side products you need to draw the mechanism of the reaction. It's a named reaction so you can easily google it. Some of those things are catalysts and won't be included in the balanced equation.

[RainbowMoo]

Is this question for a general chemistry class or an organic chemistry class?

As a starting point you need to finish drawing the other side products of the reaction. To understand the side products you need to draw the mechanism of the reaction. It's a named reaction so you can easily google it. Some of those things are catalysts and won't be included in the balanced equation.

Its organic which is why I posted it in this section.

I have Googled the reaction and already have the mechanism. What I can't find out is how to balance the overall/general reaction.

[spirochete]

Before you decide what  you're going to leave out you should show everything on the product side rather than just the alcohol. Then you can start to sort through what parts are extraneous.

[spirochete]

I read a little bit more about it and the mechanism is more complicated than I thought: http://www.organic-chemistry.org/namedreactions/luche-reduction.shtm .

As you can see from that link the reaction is usually performed in methanol as a solvent, but the methanol also acts as a reactant.

[RainbowMoo]

Before you decide what  you're going to leave out you should show everything on the product side rather than just the alcohol. Then you can start to sort through what parts are extraneous.

NaBH4 --> -BH4
CeCl3*7H2O --> -CeCl3

These make up the redox reactions?

[RainbowMoo]

Before you decide what  you're going to leave out you should show everything on the product side rather than just the alcohol. Then you can start to sort through what parts are extraneous.

NaBH4 --> -BH4
CeCl3*7H2O --> -CeCl3

These make up the redox reactions?

For sodium borohydride I have BH4 + 1e- --> -BH4

Also,
CeCl3 + 7H2O --> -CeCl3 + 1e- + 14H+..I think this is wrong. I forgot how to balance the H2O. Is this acidic or basic? Acidic right?

[spirochete]

Before you decide what  you're going to leave out you should show everything on the product side rather than just the alcohol. Then you can start to sort through what parts are extraneous.

NaBH4 --> -BH4
CeCl3*7H2O --> -CeCl3

These make up the redox reactions?

For sodium borohydride I have BH4 + 1e- --> -BH4

Also,
CeCl3 + 7H2O --> -CeCl3 + 1e- + 14H+..I think this is wrong. I forgot how to balance the H2O. Is this acidic or basic? Acidic right?

You don't need to use that gen chem half reaction method to balance the equation, although it is redox reaction. The easiest way to think about it is that BH4- acts as a source of H minus and reduces the  ketone to an alcohol. No need to think about free electrons at all.

If you look at the mechanism you'll see that CeCl3 is not part of the balanced equation. It simply acts as a catalyst.  Specifically a lewis acid catalyst.

Again, you need to think about the mechanism and draw all of the side products. Once you've figured out the side products you can give them coefficients to balance the equation. As far as I know using the half reaction method will only complicate things. But if somebody else knows how to balance it that way they can chime in.

[RainbowMoo]

Before you decide what  you're going to leave out you should show everything on the product side rather than just the alcohol. Then you can start to sort through what parts are extraneous.

NaBH4 --> -BH4
CeCl3*7H2O --> -CeCl3

These make up the redox reactions?

For sodium borohydride I have BH4 + 1e- --> -BH4

Also,
CeCl3 + 7H2O --> -CeCl3 + 1e- + 14H+..I think this is wrong. I forgot how to balance the H2O. Is this acidic or basic? Acidic right?

You don't need to use that gen chem half reaction method to balance the equation, although it is redox reaction. The easiest way to think about it is that BH4- acts as a source of H minus and reduces the  ketone to an alcohol. No need to think about free electrons at all.

If you look at the mechanism you'll see that CeCl3 is not part of the balanced equation. It simply acts as a catalyst.  Specifically a lewis acid catalyst.

Again, you need to think about the mechanism and draw all of the side products. Once you've figured out the side products you can give them coefficients to balance the equation. As far as I know using the half reaction method will only complicate things. But if somebody else knows how to balance it that way they can chime in.

But, what are these side products? I'm not that great at chemistry, but I'm trying. The problem is we're given this reaction for lab, yet the prof hasn't gone over it in class, its not in the textbook, its not in the posted material for our class website, so all I have is Google. This is very frustrating, I'm not getting it even though it looks simple enough(for everyone else). This has taken up too much of my time and I'll probably bs it and get it wrong. I'm learning so much, college is great(sarcasm).

[spirochete]

Assuming Each Mole of NaBH4 provide 4 moles of hydride, and methanol reacts with various intermediates to produce B(OCH3)4:

4(Ketone) + NaBH₄ + 4(CH3OH)  4(Alcohol) + 4(NaB(OCH3)₄