Identifying the mistake in this reaction scheme to form 2-hydroxyisobutyramide

[etotheipi]

Homework Statement

Determine why this sequence of steps doesn't give the product as desired!

Relevant Equations

N/A

The steps are as follows (apologies for my awful formatting, I'm not sure which Chemistry latex package there is!), $$(CH_3)_2CO \xrightarrow{HCN} (CH_3)_2C(OH)CN \xrightarrow{H_2SO_4} (CH_3)_2C(OH)COOH \xrightarrow{PCl_5} (CH_3)_2C(OH)COCl \xrightarrow{NH_3} (CH_3)_2C(OH)CONH_2$$ As far as I can tell, all of the steps seem valid. That is,

1. Nucleophilic addition of $CN^-$, with a $H^+$ ion also joining to the $O^-$ to form the alcohol group
2. Hydrolysis of the $CN$ group to a $COOH$ group
3. $Cl$ substitutes for $OH$ to produce an acyl chloride, as well as $POCl_3$ and $HCl$
4. Addition-elimination reaction of the acyl chloride to produce an amide

I thought it might perhaps due conditions in the first steps. The pH of the first step I think has to be around 8 so that there are sufficient $CN^-$ as well as $HCN$ molecules in solution. Though I can't think of a reason as to why that would cause a problem in the above scheme.

Any hints would be appreciated! Thank you!

 

[TeethWhitener]

Look at your reagents and see if any might generate unwanted side reactions. I don’t know if I can give you a much better hint than that without giving too much away.

 

[etotheipi]

Look at your reagents and see if any might generate unwanted side reactions. I don’t know if I can give you a much better hint than that without giving too much away.

Ah okay, I can think of maybe two things which might cause a problem. Firstly, the $PCl_5$ might also react with the hydroxyl group on the second carbon. And secondly, the cyanide ions from the first step could (I'm not sure) also undergo an addition elimination reaction with the acyl chloride to produce an acetyl cyanide.

 

[TeethWhitener]

The lack of detail in the reaction schemes means we kind of have to guess at the intention of the question-askers, but yes, my first thought was that PCl₅ will undergo a side reaction with the hydroxyl.

 

[TeethWhitener]

And secondly, the cyanide ions from the first step could (I'm not sure) also undergo an addition elimination reaction with the acyl chloride to produce an acetyl cyanide.

Also, it’s important to point out that these reactions are going to be carried out sequentially, not all at once. One pot reactions are usually written using one arrow with numbered steps (sometimes no numbers if it’s truly all at once). Sequential reactions are written such that each reaction has its own arrow. What this means for your case is that there shouldn’t be any cyanide by the time you get to the PCl₅ step.

 

[etotheipi]

Also, it’s important to point out that these reactions are going to be carried out sequentially, not all at once. One pot reactions are usually written using one arrow with numbered steps. Sequential reactions are written such that each reaction has its own arrow. What this means for your case is that there shouldn’t be any cyanide by the time you get to the PCl₅ step.

Right, so we essentially assume that the species written above the arrow in a sequential reaction is limiting. Or that we have some means of removing the unreacted reactants before the next step.

Also, I didn't know "one-pot" was an actual chemical term! It sounds a bit like something that would be written in a textbook for a potions class in Hogwarts...

 

[TeethWhitener]

Right, so we essentially assume that the species written above the arrow in a sequential reaction is limiting. Or that we have some means of removing the unreacted reactants before the next step.

Usually (but not always), there’s an implied workup between arrows. So for instance, if you were carrying out this reaction in practice, you’d isolate the carboxylic acid at the end of the second step before proceeding with the chlorination.

This is where a lack of detail can be frustrating or misleading. The hydrolysis of the cyanohydrin in step two is generally carried out using dilute aqueous acid (not concentrated H₂SO₄), and if you were to try to follow this step without workup, there’d be a ton of leftover water in your system and you’d hydrolyze all the PCl₅ before you got any product whatsoever.