# 3-step process to make propyl ethanoate from propene

I was asked to make propyl ethanoate from propene in three steps. I have access to the reactions below.

• Nucleophilic substitution (Both SN1 and SN2)
• Hydrogenation
• Dehydration
• Hydration
• Condensation
• Oxidation of alcohols

I'm assuming that you can add whatever reactants you want, since propyl ethanoate has five carbons and you can't just chop a carbon off of a propene with the above reactions. I know that the final step has to be condensation, since that's how you make esters. So essentially, I need a two-step process to make a primary alcohol that becomes the propyl in propyl ethanoate.

If you use hydrogenation on propene first, you get propane. Once you have that, it takes two steps to make a primary alcohol: one in free radical substitution (to get 1-bromopropane) and one in nucleophilic substitution (to get propan-1-ol).

If you use hydration first, you get a secondary alcohol in one step. However, you can't make propyl ethanoate with propan-2-ol.

If you use electrophilic addition first, you get a secondary haloalkane (2-bromopropane). That can also only get you a secondary alcohol.

Does anyone know how to do this? I'm convinced that it's impossible to do.

• Indeed, with the reactions you have, I'm not sure how you get to the ester in 3 steps. Of course it becomes extremely simple if you use something outside of your syllabus, but well... – orthocresol Apr 22 '16 at 0:00
• Can you suggest something like that? That would be helpful as well. – DefinitelyNotAPlesiosaur Apr 22 '16 at 0:01
• is it ethyl ethanoate or propyl ethanoate?the heading says ethyl ethanoate and the description has matter dealing with propyl ethanoate... – Abhishek Pallippara gopakumar Apr 22 '16 at 0:53
• Sorry about that! It's propyl ethanoate. – DefinitelyNotAPlesiosaur Apr 22 '16 at 1:23

My proposal is hydroboration followed by Fischer esterification. It's 2 or 3 steps depending on how many you count hydroboration to be.

Alternatively if you are intent on going through a haloalkane pathway, you could try anti-Markovnikov addition of $\ce{HBr}$ by reacting it in the presence of $\ce{H2O2}$ to prepare 1-bromopropane. Then you could do nucleophilic substitution of $\ce{OH-}$ followed by Fischer esterification. This method would have a lower yield though due to the competing E2 reaction.

• Could you explain how one would consider the above procedure to contain either two or three steps? By the way, how do you draw such nice chemical equations? – DefinitelyNotAPlesiosaur Apr 22 '16 at 2:11
• @DefinitelyNotAPlesiosaur with ChemDraw. And it's two steps if you consider the entirety of hydroboration as one step, or three if you consider borane addition and $\ce{NaOH}$ and $\ce{H2O2}$ addition as separate steps. If you don't want to use my proposed one, the alternative one is definitely three steps. – ringo Apr 22 '16 at 2:13
• Essentially, I need to get a primary alcohol from an alkene. Do you think that I could do this with electrophilic addition in some special solution (like a protic solution) or in some special conditions (like extreme temperature)? – DefinitelyNotAPlesiosaur Apr 22 '16 at 23:21
• Yes...that is exactly what the alternative method is. Special conditions by which an acid ($\ce{HBr}$) adds to the alkene to form a primary haloalkane, which can then be used to form the primary alcohol. – ringo Apr 22 '16 at 23:33

Here's another way.

1. Oxidise propene to prop-2-en-1-ol with selenium dioxide. Here's a reference: J. Org. Chem., 1979, 44, 4683

2. Esterify. The typical Fischer esterification (with carboxylic acid and concentrated sulfuric acid) should probably work, although there are many other possibilities. One is to use the acyl chloride with a weak base. The weak base serves as a nucleophilic catalyst, and also helps to mop up the $$\ce{HCl}$$ produced in the reaction. A common choice is 4-dimethylaminopyridine, often called DMAP. Another option is to use the carboxylic acid with dicyclohexylcarbodiimide (DCC) and DMAP. The DCC activates the carboxylic acid in what is known as the Steglich esterification.

3. Reduce the double bond by catalytic hydrogenation. C=C double bonds are generally more susceptible to catalytic hydrogenation than C=O double bonds (disclaimer: this may vary depending on the catalyst and conditions chosen), so this reaction can proceed chemoselectively to give the desired product, n-propyl acetate: