# What is the best way to remove/break carbon carbon bonds?

I am working on synthesis homework and I can not figure out how to remove a carbon substituent from a cyclohexane.

What is the best way to remove the three carbons and exchange them for a ketone?

A possible pathway is as follows:

At high temperatures, formation of the more substituted alkene will be favored. Catalytic amounts of base provide a kinetic pathway for this rearrangement. $\ce{OsO4}$ creates a vicinal diol, while $\ce{NaIO4}$ cleaves the $\ce{C-C}$ bond and oxidizes the alcohols to carbonyls. Additionally, $\ce{NaIO4}$ oxidizes $\ce{(HO)2OsO2}$ back to $\ce{OsO4}$, allowing the reaction cycle to continue.

• Ok thanks, I didnt know you could use base+water to rearragne a double bond like that. Very helpful – unrnick Oct 22 '16 at 17:46
• Quick question, what program are you using to draw out the compounds? – unrnick Oct 22 '16 at 17:47
• @unrnick ChemDraw. If you are at an uni, you should be able to get it for free – ringo Oct 22 '16 at 17:54
• Worth noting that OsO4 is a nasty (and expensive) chemical, which leads to the main benefit of using NaIO4 as a co-oxidant: you get to use OsO4 in catalytic quantities. – orthocresol Oct 23 '16 at 1:38
• @orthocresol I wouldn’t be sure periodate is a co-catalyst here. From what I learnt, $\ce{OsO4}$ generates the diol which is periodate-cleaved to two carbonyl groups. – Jan Oct 23 '16 at 20:59

Here's a variation of ringo's method:

1. Treat the alkene with $\ce{OsO4}$/$\ce{NaIO4}$ to obtain a cyclohexyl methyl ketone
2. Convert the ketone to cyclohexyl acetate by using a peracid (MCPBA).
3. Cleave the ester
4. Oxidize the cyclohexanol to cyclohexanone with a mild oxidant

I can offer another two-step synthesis:

1. isomerise the double bond as ringo mentioned.

2. Criegee reaction ($\ce{O3}$) followed by workup with $\ce{PPh3}$.

Ozonolysis followed by reduction, elimination of water next ozonolysis (repeat till you have cyclohexanone) looks like the best solution for me.