[AdiDex]

I'm Confused why we can't reduce Carboxylic Acid Directly to Alcohol ??
We Have to first convert it into ester than Ester is Catalytic Hydrogenated at high pressure in presence of H₂ gas to form Alcohol .!!!

My second Doubt
"Which is better to use to reduce  anything LiAlH₄ or Pd-C / H₂ ??" Eg. I have to reduce -NO₂ to form -NH₂ , Can we use LiAlH₄ ??

And What is Difference Between reactivity of  Pd , Pt , Rh , Ni , Ru , in catalytical Hydrogenation  ??

[kriggy]

Im not sure about your 1st question but about the second..
It depends. LAH is probably easier to work with since it is a powder while H₂ is gas and requires special setup. There is no easy answer on this question, it depends on many factors but generaly speaking, by using H₂ you can reduce pretty much anything you want into anything you want, but you need to have the right conditions

[Babcock_Hall]

This is something of a tangent, but when one uses a less active reducing agent, ones has to use an ester with a decent leaving group, such as phenoxide ion.

[AdiDex]

This is something of a tangent, but when one uses a less active reducing agent, ones has to use an ester with a decent leaving group, such as phenoxide ion.

Ohk thats ok if ester having leaving group such as phenoxide (PhO^-) or methoxide (OCH₃^-) they are pretty much good leaving group than Hydroxide (OH^-) .

But If you have ester having Higher Substituents  like pentoxide (OC₅H₁₁^-) they are bad Leaving group due to large +I effect .

So I  can conclude that We can't replace OH by anything it should be either OMe or OPh ..!! Am i right ?? But as far i know catalytical hydrogenation occurs through radical Mechanism ..!! So I think leaving group ability nothing to do with Hydrogenation .!!!

Can anybody How It get Reduce to Alcohol ??
RCOOR' + H₂ RCH₂OH + R'OH

First A Pi bond is breaks Radical forms and 2 Hydrogen get inserted but  what will happen next ?? Can It Break
Sigma bond ??

[AdiDex]

Im not sure about your 1st question but about the second..
It depends. LAH is probably easier to work with since it is a powder while H₂ is gas and requires special setup. There is no easy answer on this question, it depends on many factors but generally speaking, by using H₂ you can reduce pretty much anything you want into anything you want, but you need to have the right conditions

Ok That means We can use both method to reduce .!! Thanks
By the way we can reduce Carboxylic acid to Alcohol by Using LAH but i didn't find anywhere where reduction of Carboxylic acid by Catalytical Hydrogenation can take place..!!

Can Anybody Help me on my another Doubt No one is replying
http://www.chemicalforums.com/index.php?topic=78907.0

[Babcock_Hall]

This is something of a tangent, but when one uses a less active reducing agent, ones has to use an ester with a decent leaving group, such as phenoxide ion.

Ohk thats ok if ester having leaving group such as phenoxide (PhO^-) or methoxide (OCH₃^-) they are pretty much good leaving group than Hydroxide (OH^-) .

For groups whose structures are similar, one can use basicity to estimate leaving group ability.  Which is more basic, methoxide or phenoxide?

[Altered State]

You can chemoselectively reduce carboxylic acids to alcohols (and amides to amines) with BH3, like this:





Your second question is absolutely impossible to get answered in a short way...
What reducing agent is better depends much on the situation: What functional group/s are you trying to reduce, which other groups are present on the molecule and you don't want them to get reduced...

[AdiDex]

This is something of a tangent, but when one uses a less active reducing agent, ones has to use an ester with a decent leaving group, such as phenoxide ion.

Ohk thats ok if ester having leaving group such as phenoxide (PhO^-) or methoxide (OCH₃^-) they are pretty much good leaving group than Hydroxide (OH^-) .

For groups whose structures are similar, one can use basicity to estimate leaving group ability.  Which is more basic, methoxide or phenoxide?

Methoxide is more basic , As in case of phenoxide the lone pair are in conjugation .

[discodermolide]

There is a nice new method, but from esters, which are easy to prepare.
see Iron-Catalyzed Hydrogenation of Esters to Alcohols. (2014). Iron-Catalyzed Hydrogenation of Esters to Alcohols., 53(48), 13004–13006. doi:10.1002/anie.201407613
for the method

[AdiDex]

Your second question is absolutely impossible to get answered in a short way...
What reducing agent is better depends much on the situation: What functional group/s are you trying to reduce, which other groups are present on the molecule and you don't want them to get reduced...

Ok So that means both have their own advantages . Can you prefer me any link or book where i can get the information about reducing properties of different reducing agents ??
and what about reactivity order of these metals ??
 Pd , Pt , Rh , Ni , Ru

[AdiDex]

There is a nice new method, but from esters, which are easy to prepare.
see Iron-Catalyzed Hydrogenation of Esters to Alcohols. (2014). Iron-Catalyzed Hydrogenation of Esters to Alcohols., 53(48), 13004–13006. doi:10.1002/anie.201407613
for the method

Thank you very much

But i have same question why we can't use Carboxylic acid ??
Why usually we have to convert carboxylic acid into ester first ??

[Babcock_Hall]

This is something of a tangent, but when one uses a less active reducing agent, ones has to use an ester with a decent leaving group, such as phenoxide ion.

Ohk thats ok if ester having leaving group such as phenoxide (PhO^-) or methoxide (OCH₃^-) they are pretty much good leaving group than Hydroxide (OH^-) .

For groups whose structures are similar, one can use basicity to estimate leaving group ability.  Which is more basic, methoxide or phenoxide?

Methoxide is more basic , As in case of phenoxide the lone pair are in conjugation .

Yes, so phenoxide is the better leaving group.

[AdiDex]

Babcock_Hall Brother , Listen ..!!!

As far as I know Catalytical Hydrogenation take place by radical mechanism , so Leaving group ability will not have so much influence on reaction . As in this is case Homogeneous cleavage take place , heterogeneous cleavage will not..!!!

So why we are concerning about leaving group ability

[AdiDex]

http://www.chemicalforums.com/index.php?topic=78907.0

Someone please help me on this topic .

[k1mng]

I believe catalytic hydrogenation doesn't actually take place by a radical mechanism but this is a different discussion.

In any case, the carboxyl group seems to display very different reactivity to ester groups and is pretty much in the bottom rung of reactivity against reducing agents including H₂ and even LiAlH₄, with the exception of BH₃.

The metals used in catalytic hydrogenation actually occasionally display curiously different chemoselectivities for different reducible substrates, although there is a general list of reactivity against said substrates that applies. The list has carboxylic acids smack bang at the bottom with esters being pretty challenging substrates also. This would perhaps be to do with their poor capacities to dock to the metal surface - important in the hydrog process.