What will be the product of this organic reaction?

In summary, the question asks for the major product of a reaction involving a benzene ring with a methyl group attached and a ketone. After researching, it is determined that the reaction involves oxidation with a strong oxidizing agent, KMnO4, resulting in the formation of benzyldehyde. The oxidizing agent used can affect the product, with KMnO4 producing benzyldehyde and chromyl chloride producing a different product. The reaction also involves the use of a base, such as KOH, and sulfuric acid. The correct answer to the question depends on the specific reagents used and their concentrations and conditions.
  • #1
Adesh
735
191
Summary:: I'm always unable to find the products of a chemical reaction. No Matter how much concept I study I can't ever get products from reasoning.

Here is the question:

What is the major product of this reaction?

Question.png
.

OPTIONS :
Options.png
MY ATTEMPT :
First of all I can see that in question I have a benzene ring to which a methyl group is attached, on the other side we have a ketone. The structure of the molecule in question can be drawn as:

Molecule.jpeg
. Now, I know that oxidation (with strong oxidizing agents) of alcohols, aldehydes and ketones results in carboxylic acid. Therefore, I treat the above molecule (a ketone group) with strong oxidizing agent [itex] KMnO_4[/itex] we would would get something like this

Reaction.jpeg
. (I really don't know what effect would [itex]KOH[/itex] have on our molecule because as per my knowledge [itex]OH^-[/itex] is a nucleophile and hence attacks reagents like alkyl chloride, benzyl chloride. In this case I have no idea what would happen).

This much being done, now we treat our carboxylic acid with sulfuric acid and according to me there will be no reaction. There are chances that benzene ring (which contains [itex]\pi[/itex] electrons) can act as a base, but seeing the options I have to eradicate this idea.

The methyl group [itex]CH_3[/itex] which is attached to our starting molecule is, according to my studies, is not reactive. Alkenes undergo oxidation to form alkynes or alcohol (I may be wrong here, I said alkynes because removal of hydrogen from alkenes ,i.e. oxidation, would result in alkynes or the addition of [itex]OH[/itex] , i.e. addition of oxygen to carbon, would result in alkyl alcohols). Alkanes can only be combusted,
Alkyl halides are almost never prepared by direct halogenation of alkanes. From the standpoint of synthesis in tha laboratory, an alkane is a dead end.
-Organic Chemistry by Morrison and Boyd
.

Thank you. Any help will be much appreciated.

P.S. :- I highly apologize for posting the images, but I have no other option. If you can suggest me any other way of drawing molecules (using latex or something) then please do suggest me.
 
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  • #2
1) You’re pretty much stuck posting pictures for molecular structures. A LaTeX chemistry extension does exist, but as far as I can tell, it’s not installed on this forum.

2) What happens to toluene when it is exposed to KMnO4?
 
  • #3
TeethWhitener said:
2) What happens to toluene when it is exposed to KMnO4?
I don’t know, how toluene can be oxidised?
 
  • #4
TeethWhitener said:
1) You’re pretty much stuck posting pictures for molecular structures. A LaTeX chemistry extension does exist, but as far as I can tell, it’s not installed on this forum.

2) What happens to toluene when it is exposed to KMnO4?
I have searched on internet and found that it is something called Etard reaction. Hence, benzyldehyde will be formed, am I right?
 
  • #5
Adesh said:
I have searched on internet and found that it is something called Etard reaction. Hence, benzyldehyde will be formed, am I right?
Is your oxidizer chromyl chloride?
 
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  • #6
TeethWhitener said:
Is your oxidizer chromyl chloride?
Yes,Wikipedia writes so. Does product depends on the oxdizing agent? I thought that it is oxidation( addiction of oxygen in this case) that matters.
 
  • #7
Adesh said:
Yes,Wikipedia writes so. Does product depends on the oxdizing agent?
Your OP stated the reagent was KMnO4, not chromyl chloride. And yes, the product depends on the oxidizing agent.
 
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  • #8
TeethWhitener said:
Your OP stated the reagent was KMnO4, not chromyl chloride. And yes, the product depends on the oxidizing agent.
Okay. I don’t know the action of [itex] KMnO_4[/itex] on toluene, all I know is that [itex] KMnO_4[/itex] will oxidise my toluene.
 
  • #10
And... if a carboxylic acid is the product and that is in a solution of base, is there a further (trivial) reaction that is possible?
 
  • #11
chemisttree said:
And... if a carboxylic acid is the product and that is in a solution of base, is there a further (trivial) reaction that is possible?
What is the base? I think in first reaction we have to use either KMnO4 or KOH and then we have to use the sulfuric acid.
 
  • #13
No, the first reaction involves a heated (that’s what the ##\Delta## means) basic (from KOH) solution of KMnO4. The slash isn’t an either/or symbol. It may be that the reason you’re having so much trouble with chemistry is that you’re not understanding the basic notation.
 
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  • #14
Adesh said:
So, it seems to me that option (2) is right, is it really right?
I have no idea what you mean by option 2.

edit: I see, you mean in the OP.
 
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  • #15
Yes, #2 is the product I would expect. The important point here is that benzylic alkyl carbons tends to be significantly more reactive than other alkyl groups.
 
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  • #16
TeethWhitener said:
I have no idea what you mean by option 2.

edit: I see, you mean in the OP.
Now, how KMnO4 and basic KMnO4 going to differ in their action? Will the carboxylic acid that is formed going to donate an H to the base and accepts K from KOH?
 
  • #17
Adesh said:
Now, how KMnO4 and basic KMnO4 going to differ in their action? Will the carboxylic acid that is formed going to donate an H to the base and accepts K from KOH?
It more has to do with the strength of the oxidizer. Permanganate is a far stronger oxidizer in acid than in base.

edit: in fact, it wouldn’t surprise me to learn that KMnO4 in acid simply destroys the molecule entirely.
 
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  • #18
TeethWhitener said:
It more has to do with the strength of the oxidizer. Permanganate is a far stronger oxidizer in acid than in base.

edit: in fact, it wouldn’t surprise me to learn that KMnO4 in acid simply destroys the molecule entirely.
And that reaction with sulfur acid? Is given only to confuse the students?
 
  • #19
Again, what trivial reaction occurs between a carboxylic acid and a base (KOH)? Write out the product! You should have learned this in your first chemistry class.
 
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  • #20
Adesh said:
And that reaction with sulfur acid? Is given only to confuse the students?
See @chemisttree 's post above.
 
  • #21
The KOH is required for permanganate oxidations. Neutral permanganate decomposes to Mn(IV) and acidic permanganate decomposes to Mn(II).
 
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  • #22
chemisttree said:
Again, what trivial reaction occurs between a carboxylic acid and a base (KOH)? Write out the product! You should have learned this in your first chemistry class.
I will have K+ attached to oxygen, I mean I will get potassium salt of carboxylic acid (and these acids are attached to the benzene ring) and the leaving H will form water with OH-. Am I right?
 
  • #23
chemisttree said:
The KOH is required for permanganate oxidations. Neutral permanganate decomposes to Mn(IV) and acidic permanganate decomposes to Mn(II).
These are the end products of the reaction. KOH isn’t required for the oxidation. The acid actually increases the oxidative power of KMnO4, because its end product is so much more reduced. The reason KOH is added is actually to make the oxidizer weaker, because acidic KMnO4 has a tendency to chew everything organic up to pieces.
 
  • #24
TeethWhitener said:
These are the end products of the reaction. KOH isn’t required for the oxidation. The acid actually increases the oxidative power of KMnO4, because its end product is so much more reduced. The reason KOH is added is actually to make the oxidizer weaker, because acidic KMnO4 has a tendency to chew everything organic up to pieces.
Thank you for helping me. I have really enjoyed the conversation with you and you have taught me so many things. If you don't mind, I want to make a request to please give some advice (your personal) on how to approach for products of a chemical reaction.
 
  • #25
Adesh said:
I will have K+ attached to oxygen, I mean I will get potassium salt of carboxylic acid (and these acids are attached to the benzene ring) and the leaving H will form water with OH-. Am I right?
Yes. After the reaction you will have a solution of the dicarboxylate salt and a bunch of brown mud. Filtering gives a clear solution of the dicarboxylate salt. Treating that with the sulfuric acid gets you back to the dicarboxylic acid.
 
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  • #26
TeethWhitener said:
...The reason KOH is added is actually to make the oxidizer weaker, because acidic KMnO4 has a tendency to chew everything organic up to pieces.
Like this?

image.jpg
 

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  • #27
chemisttree said:
Like this?

View attachment 254659
Yes. Check the yield. Basic KMnO4 will give you closer to quantitative.
 
  • #28
:wink:
image.jpg
 
  • #29
Yes. This corroborates what I said in post 27
 

FAQ: What will be the product of this organic reaction?

1. What is the purpose of this organic reaction?

The purpose of an organic reaction is to create a new chemical compound or change the structure of an existing compound by breaking and forming chemical bonds. This can be used to synthesize new materials, purify substances, or produce desired products for industrial or medicinal purposes.

2. What factors influence the product of an organic reaction?

The product of an organic reaction is influenced by a variety of factors, including reactant concentration, temperature, pressure, catalysts, and reaction time. The type and strength of the chemical bonds in the reactants also play a crucial role in determining the outcome of the reaction.

3. How can we predict the product of an organic reaction?

Predicting the product of an organic reaction can be challenging and requires a deep understanding of organic chemistry principles. However, some helpful tools for prediction include reaction mechanisms, thermodynamics, and spectroscopic analysis techniques.

4. Can the product of an organic reaction be controlled?

In some cases, the product of an organic reaction can be controlled by carefully selecting reaction conditions and using specific catalysts. However, there are also instances where the reaction is not controllable, and the product is determined by the reactant's inherent chemical properties and the reaction conditions.

5. How do we determine the yield of the product in an organic reaction?

The yield of a product in an organic reaction refers to the amount of the desired product produced compared to the theoretical maximum amount that could be produced. The yield can be determined through various analytical techniques such as mass spectrometry or chromatography.

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