3
$\begingroup$

I came across this problem very recently, but no amount of thinking could lead me to a perfect one step reagent for the conversion.

Also, I had thought of reacting ethene with:

  1. $\ce{CH3MgBr}$ leading to the formation of propane, which I now react with NaOH under vigorous conditions to get alcohol as product, then heating it somehow that we obtain propanal.

  2. $\ce{HCl}$, followed by $\ce{Mg}$ in dry ether to form Grignard reagent, then adding $\ce{HCN}$.

I'm certain that both of these propositions are wrong, so do suggest a better, newer way for the preparation.

$\endgroup$
2
  • 3
    $\begingroup$ en.wikipedia.org/wiki/Hydroformylation $\endgroup$ Commented yesterday
  • 1
    $\begingroup$ HCN is too acidic, it will just protonate your Grignard, use methyl formate instead $\endgroup$ Commented 17 hours ago

3 Answers 3

8
$\begingroup$

As first mentioned by @Mithoron, hydroformylation of ethene with CO and H2 gives propanal, for example as recently described by Bell et al using rhodium/cobalt and rhodium/zinc catalysts.

Hydroformylation reaction scheme

(picture source)

$\endgroup$
7
$\begingroup$

There’s a fundamental reason this is hard: ethene has two carbons, propanal has three, so you must form a new C–C bond. That already rules out most “simple” one-step reagents.

Your proposed routes don’t work because Grignard reagents don’t react with alkenes like ethene, and converting propane into propanal with NaOH or heat isn’t chemically realistic.

To convert it in one-step you would need to do hydroformylation (the oxo process):

Ethene + CO + H₂ → propanal (using a Rh or Co catalyst under pressure)

This works industrially, but not in a typical lab or exam setting. In normal organic chemistry, the conversion necessarily requires multiple steps.

References

  1. https://link.springer.com/article/10.1007/s40828-021-00154-x
$\endgroup$
2
$\begingroup$

According to Chemical Book

Direct formation of vinyl lithium from ethylene can also be accomplished using Potassium t-Butoxide, n-Butyllithium, and TMEDA (N,N,N′,N′-Tetramethylethylenediamine) in hexane at -40 °C, followed by treatment of the vinylpotassium with Lithium Bromide.

Reaction of this vinyl lithium with methyl formate followed by mild acidic workup will give you propanal in a single step.

$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.