Even in science I'm a smartass
Going over more organic chemistry today (as I have a fairly major exam tomorrow night and have to at least pass it in order to feel comfortable about my chances for passing the course, and I came across a synth that was something analagous to the following:
From 3-pentanone, make: 2-Methyl-1-Phenyl-1-penten-3-one
In other words, from CH3CH2C(=O)CH2CH3 make the right side of the =O stay exactly the same, but severely mutilate the other side by creating a double bond between the first and second carbons, adding a methyl group (-CH3) to the second carbon, and adding a phenyl group (-C6H5, a benzene ring as a functional group, essentially) to the first carbon. This makes C6H5-CH=C(CH3)C(=O)CH2CH3
I'm sure there was a very nice and pretty way for me to make this conversion, but I was tired, and I remember the tired old adage "organic chemists are lazy" (no, seriously, it gets said... by professors of organic chemistry). So I decided to be a smartass. Did I try and find the pretty-ness? Fuck no! I threw that ketone in with a peroxy acid to make it an ester and then cleaved the gorram thing with lithium aluminum hydride in diethyl ether, followed by water. This gave me 1-propanol (CH3CH2CH2OH).
Now, for the next step of my master plan to work, I needed an aldehyde. So, being the clever organic chemist that I am, then treated my 1-propanol with PDC in dichloromethane, making it into 1-propanal (CH3CH2C(=O)H).
Here comes the tricky part. As I'm just allowed to summon my reagents from the ether (no pun intended) for the purpose of these problems, I bring forth a mighty Grignard reagent of the form C6H5-CH=C(CH3)MgBr. The portion that isn't magnesium or bromine nucleophillically attacks the carbonyl carbon when this is exposed to my 1-propanal in diethyl ether and, when treated thereafter in acid solution, forms C6H5-CH=C(CH3)CHOHCH2CH3.
Where did the Grignard reagent come from? Well, I could make it from ethylene glycol (that's right, antifreeze) if I wanted to, but it doesn't really matter. It's not part of the problem to tell them how you got your reagents!
Laughing maniacally, I realize all I have left to do is convert my secondary alcohol into a ketone. Sure, I could use K2Cr2O7 in H2SO4 (street name sulfuric acid), but being lazy I decide instead to use chromic acid, as all I need is an acid and a source of Cr (VI), and chromic acid is both!
Boom, biznatches! There's your gorram 2-Methyl-1-Phenyl-1-penten-3-one.
What's really sad is that I think my way takes less steps than the one they likely wanted me to do.
From 3-pentanone, make: 2-Methyl-1-Phenyl-1-penten-3-one
In other words, from CH3CH2C(=O)CH2CH3 make the right side of the =O stay exactly the same, but severely mutilate the other side by creating a double bond between the first and second carbons, adding a methyl group (-CH3) to the second carbon, and adding a phenyl group (-C6H5, a benzene ring as a functional group, essentially) to the first carbon. This makes C6H5-CH=C(CH3)C(=O)CH2CH3
I'm sure there was a very nice and pretty way for me to make this conversion, but I was tired, and I remember the tired old adage "organic chemists are lazy" (no, seriously, it gets said... by professors of organic chemistry). So I decided to be a smartass. Did I try and find the pretty-ness? Fuck no! I threw that ketone in with a peroxy acid to make it an ester and then cleaved the gorram thing with lithium aluminum hydride in diethyl ether, followed by water. This gave me 1-propanol (CH3CH2CH2OH).
Now, for the next step of my master plan to work, I needed an aldehyde. So, being the clever organic chemist that I am, then treated my 1-propanol with PDC in dichloromethane, making it into 1-propanal (CH3CH2C(=O)H).
Here comes the tricky part. As I'm just allowed to summon my reagents from the ether (no pun intended) for the purpose of these problems, I bring forth a mighty Grignard reagent of the form C6H5-CH=C(CH3)MgBr. The portion that isn't magnesium or bromine nucleophillically attacks the carbonyl carbon when this is exposed to my 1-propanal in diethyl ether and, when treated thereafter in acid solution, forms C6H5-CH=C(CH3)CHOHCH2CH3.
Where did the Grignard reagent come from? Well, I could make it from ethylene glycol (that's right, antifreeze) if I wanted to, but it doesn't really matter. It's not part of the problem to tell them how you got your reagents!
Laughing maniacally, I realize all I have left to do is convert my secondary alcohol into a ketone. Sure, I could use K2Cr2O7 in H2SO4 (street name sulfuric acid), but being lazy I decide instead to use chromic acid, as all I need is an acid and a source of Cr (VI), and chromic acid is both!
Boom, biznatches! There's your gorram 2-Methyl-1-Phenyl-1-penten-3-one.
What's really sad is that I think my way takes less steps than the one they likely wanted me to do.