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I am stuck in finding the final product $\ce{C}$ formed in the following series of reactions:

$$\ce{Bu-C#CH ->[NaNH2] A ->[Ph-CHO][H2O] B ->[MnO2] C}$$ where $\ce{Bu}$ denotes the butyl substituent and $\ce{Ph}$ the phenyl substituent.

I could easily predict product $\ce{A}$ and $\ce{B}$. $\ce{A}$ is simply formed by replacement of hydrogen with sodium and $\ce{B}$ by the meta substitution in benzene ring.

Proposed reaction scheme

However, I am having trouble predicting the final product. I know that $\ce{MnO2}$ is a mild oxidising agent.

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    $\begingroup$ Um, exactly how is your acetylide anion reacting with benzaldehyde? That anion is a nucleophile, not an electrophile, so electrophilic aromatic substitution (which involves the benzene ring reacting with an electrophile) isn't gonna work here. Is there any part of benzaldehyde that might react with nucleophiles? $\endgroup$ – orthocresol Jan 4 '17 at 14:00
  • $\begingroup$ @orthocresol yes I was mistaken.it should be para nucleophillic aromatic substitution.yes benzaldehyde also has a replaceable hydrogen with itself but why it should be the major pathway. $\endgroup$ – Pink Jan 4 '17 at 14:20
  • $\begingroup$ @aniline It can't be nucleophilic aromatic substitution either, since there is no leaving group on the aromatic ring... Most reactions are nucleophile / electrophile reactions, so the questions are: what's the most reactive nucleophile and what's the most reactive electrophile. I'll give you the hint that aromatic rings are generally very stable, so that's not likely to react as either nucleophile or electrophile... $\endgroup$ – jerepierre Jan 4 '17 at 17:21
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    $\begingroup$ One more hint... What type of functional group is MnO2 likely to react with? $\endgroup$ – jerepierre Jan 4 '17 at 17:24
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    $\begingroup$ And another hint: A will react with $\ce{Ph-CHO}$, but it will react in similar fashion with $\ce{R-CHO}$ (R = alkyl) to give a B-type product. This, in turn, will react with $\ce{MnO2}$ as well. $\endgroup$ – Klaus-Dieter Warzecha Jan 4 '17 at 18:03
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You seem very sure about your intermediate B; we’ll come back to that in a second.

For now, let’s concentrate on the reagents for your final step.

$$\ce{B ->[MnO2] C}$$

Manganese dioxide is the product you get from heating any manganese salts in the bunsen burner until red-hot; thus, one might call it the most stable or thermodynamic oxide. It seems very unreactive and in fact I only know of one type of reaction that it is able to perform on organic molecules: allylic and benzylic alcohols are oxidised to the corresponding carbonyl compounds while fully saturated (i.e. not allylic) alcohols are untouched.

Since you have that reagent, it might be helpful to go backwards; assuming that the final step is an oxidation of a benzylic or allylic alcohol to a corresponding carbonyl. Thus, $\ce{C}$ would either be a ketone or an aldehyde and $\ce{B}$ would have to be an alcohol. But your proposed structure $\ce{B}$ does not include an alcohol! Therefore, your proposed reaction of the second step is very likely wrong.

$$\ce{A ->[Ph-CHO][H2O] B}$$

Indeed, here you assumed that the anion $\ce{A}$ you proposed reacts with the phenyl ring of benzaldehyde in an electrophilic aromatic substitution. Take a good look at the two emboldened words and consider whether an anion can really be electrophilic. Then ask yourself what anions are in philicity terms. Once you have done that, attempt to locate the position of benzaldehyde with the corresponding electronic properties.

The solution is given in the spoiler below; please try to find it yourself, though; it is a much better exercise than just looking things up.

Anions are negatively charged as are electrons. Thus, anions are practically never drawn towards negative charges — remember that like and like repulse each other. Instead, anions are drawn to positive charges: they are nucleophilic.

Nucleophiles are drawn to the most positively charged bit of a molecule they are attacking. In the case of benzaldehyde, the carbon side of the $\ce{C=O}$ bond is most positively charged due to the resonance depiction we can draw below:

$$\ce{C=O <-> \overset{+}{C}-\overset{-}{O}}$$

Therefore, the acetylide anion will attack the carbonyl carbon of benzaldehyde nucleophilicly, creating a benzylic alcohol. It is now no secret that this is cleanly oxidised to the corresponding alkynophenone.

 

$$\ce{Bu-C#C^- (A) ->[Ph-CHO][H2O] Ph-CHOH-C#C-Bu (B) ->[MnO2] Ph-CO-C#C-Bu (C)}$$

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