Solving H2/Pd/C Reduction for #5 & #6 Questions

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    Pd Reduction
In summary: Can you please write a step by step guide on how to do this?In summary, a chemist would want to use a Dean Stark to deprotect the ketal, and then do an internal transesterification in the absence of water.
  • #1
duchuy
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Homework Statement
Find the mechanism
Relevant Equations
x
Hi,
I'm trying to solve the number 5 and 6. Now I'm wondering what happens when I add H2/Pd/C on the 4th (after the question 4) molecule. I know that it will reduce pi bonds to sigma bonds, but will it affect the cetal group? Because if it does't I'm not quite sure that I'll be able to do the exercice with the steps given. My solution :
After adding Hd/Pd/C :
1) H+/H20, I will form 2 alcohol molecules (with the ester group) and acetone
2)Saponification : NaOH : Here I'm turning ester to carboxylate
3Intramolecular esterification : H+/H20 : I have to reacidify my environement in order to perform the esterification.

This would take 3 steps, while I only have step to do this (if H2/pd/C doesn't interact with the ketal group.
And if H2/Pd/C does give me my alcohol, it would also take me 2 extra steps to form my lactone.
Please help me find the solution for this.

By the way, what are the groups that H2 reduce?
I know that they reduce alcyne, alcene, cetone, aldehyde, ... the rest google is giving me mixed answers.
So what about : Nitrile, imine, carboxylic acid, ester, RCOCl, and acid anhydride?

Thank you so so much for your help!
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  • #2
Unless you have an activated ester, or use high pressure and temperature, consider esters non-reactive to Pd/C hydrogenation. Acyclic aliphatic ketals are also non-reactive to Pd/C hydrogenation.

There is more than one way to deprotect ketals/acetonides. Remember if we have a carboxylic acid in 5 the cyclization reaction will produce a molecule of water…. and we don’t want water in this step. Think in terms of trans-esterification.
 
Last edited:
  • #3
chemisttree said:
Unless you have an activated ester, or use high pressure and temperature, consider esters non-reactive to Pd/C hydrogenation. Acyclic aliphatic ketals are also non-reactive to Pd/C hydrogenation.

There is more than one way to deprotect ketals/acetonides. Remember if we have a carboxylic acid in 5 the cyclization reaction will produce a molecule of water…. and we don’t want water in this step. Think in terms of trans-esterification.
chemisttree said:
Unless you have an activated ester, or use high pressure and temperature, consider esters non-reactive to Pd/C hydrogenation. Acyclic aliphatic ketals are also non-reactive to Pd/C hydrogenation.

There is more than one way to deprotect ketals/acetonides. Remember if we have a carboxylic acid in 5 the cyclization reaction will produce a molecule of water…. and we don’t want water in this step. Think in terms of trans-esterification.

Ok I see, but from what I see from writing the reaction, is that if I were to do transesterification to from 5 to 6 in acidic conditions, that would be a totally reversible reaction and how could I ensure the formation of the cycle? Since I can't use an excess of reactants right? Unless I remove the alcohol at the end?
But if I were to do saponification in order to do Fischer's esterification, I could easily remove water at the end and ensure the formation of the cycle by using a Deanstark or may be ZnCl2 no?
 
  • #4
Dean-Stark in a single step?
 
  • #5
chemisttree said:
Dean-Stark in a single step?
I would make my reaction in a Deanstark for the final fischer esterification? I honestly don't know, I'm just doing chemistry on paper I really don't have much lab experience. But what do you think about my proposition?
 
  • #6
Why I wrote, “Dean stark in a single step?” …
You want aqueous acid to deprotect the ketal and then turn around and do an internal transesterification in the absence of water… in one step.
 
  • #7
chemisttree said:
Why I wrote, “Dean stark in a single step?” …
You want aqueous acid to deprotect the ketal and then turn around and do an internal transesterification in the absence of water… in one step.
Oooh ok I see thank you so so much you're absolutely amazing
 

FAQ: Solving H2/Pd/C Reduction for #5 & #6 Questions

1. What is H2/Pd/C reduction?

H2/Pd/C reduction is a chemical reaction that involves the use of hydrogen gas (H2), palladium (Pd), and carbon (C) to reduce a compound to its corresponding alcohol. This reaction is commonly used in organic synthesis to create alcohols from aldehydes, ketones, and carboxylic acids.

2. How does H2/Pd/C reduction work?

H2/Pd/C reduction works by utilizing the strong reducing properties of hydrogen gas, which is activated by palladium metal. The carbon support (C) helps to stabilize the palladium and increase its surface area, allowing for more efficient reduction of the compound. The hydrogen gas is adsorbed onto the palladium surface and then transferred to the compound, breaking the carbon-carbon or carbon-oxygen bonds and forming the alcohol product.

3. What are the benefits of using H2/Pd/C reduction?

H2/Pd/C reduction is a mild and selective method for reducing compounds to alcohols. It does not require harsh conditions or toxic reagents, making it a more environmentally friendly option for chemical synthesis. Additionally, the use of palladium as a catalyst allows for faster reaction rates and higher yields compared to other reduction methods.

4. What are the limitations of H2/Pd/C reduction?

One limitation of H2/Pd/C reduction is that it only works on compounds that contain reducible functional groups, such as aldehydes, ketones, and carboxylic acids. It is also sensitive to air and moisture, so the reaction must be carried out under inert conditions. Additionally, the cost of palladium can make this method less economical for large-scale synthesis.

5. How can I optimize H2/Pd/C reduction for my specific compound?

The conditions for H2/Pd/C reduction can be optimized by adjusting the amount of catalyst, hydrogen gas pressure, and reaction time. The choice of solvent and temperature can also affect the efficiency of the reaction. It is important to consult the literature and conduct small-scale experiments to determine the best conditions for your specific compound.

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