How is the reaction of Methanal with NaBH4 a reduction reaction ?

Watch
lhh2003
Badges: 16
Rep:
?
#1
Report Thread starter 8 months ago
#1
If you check the formal charges on every atom, they remain unchanged. I read that the C atom's electrons somehow indicates that reduction has occurred which doesn't make sense how.

Can someone please explain why this reaction is a reduction reaction ?

Thanks.
0
reply
Arctic Kitten
Badges: 21
Rep:
?
#2
Report 8 months ago
#2
the C = O bond got broken by reducing agent [H+] in BH4 and C-H and O-H bond formed. It got reduced down to Methanol.
The mechanism is similar to all reduction reaction from an aldehyde to alcohol.
0
reply
lhh2003
Badges: 16
Rep:
?
#3
Report Thread starter 8 months ago
#3
(Original post by Arctic Kitten)
the C = O bond got broken by reducing agent [H+] in BH4 and C-H and O-H bond formed. It got reduced down to Methanol.
The mechanism is similar to all reduction reaction from an aldehyde to alcohol.
So it is a reducing agent because it gained a H atom ?
0
reply
lhh2003
Badges: 16
Rep:
?
#4
Report Thread starter 8 months ago
#4
Also, why does naBH4 have to be dissolved in water and methanol ? I get that the water provides H atom that bonds with there O- ion, but I don't get the purpose of Methanol.

Thanks.
0
reply
Arctic Kitten
Badges: 21
Rep:
?
#5
Report 8 months ago
#5
(Original post by lhh2003)
So it is a reducing agent because it gained a H atom ?
No, the H+ atom is from NaBH4, the reducing agent. The C = O bond got reduced and become a C - O bond because it gains an electron.
(Original post by lhh2003)
Also, why does naBH4 have to be dissolved in water and methanol ? I get that the water provides H atom that bonds with there O- ion, but I don't get the purpose of Methanol.

Thanks.
NaBH4 just there to increase the number of H+ in the water from the (H - BH3)- ions. Methanal is there to be reduced, and the result of the reduction is Methanol.
0
reply
lhh2003
Badges: 16
Rep:
?
#6
Report Thread starter 8 months ago
#6
(Original post by Arctic Kitten)
No, the H+ atom is from NaBH4, the reducing agent. The C = O bond got reduced and become a C - O bond because it gains an electron.

NaBH4 just there to increase the number of H+ in the water from the (H - BH3)- ions. Methanal is there to be reduced, and the result of the reduction is Methanol.
According to CGP and Chemguide, the methanal is reduced in a solution of NaBH4 and Methanol. Water molecules provide H+ ions , NABH4 provides H- ions but neither elaborate on the purpose of the methanol. Pigster , any idea ?
0
reply
Arctic Kitten
Badges: 21
Rep:
?
#7
Report 8 months ago
#7
(Original post by lhh2003)
According to CGP and Chemguide, the methanal is reduced in a solution of NaBH4 and Methanol. Water molecules provide H+ ions , NABH4 provides H- ions but neither elaborate on the purpose of the methanol. Pigster , any idea ?
Uh it's there probably to prevent the methanal from precipitate.
1
reply
Pigster
Badges: 19
Rep:
?
#8
Report 7 months ago
#8
(Original post by lhh2003)
If you check the formal charges on every atom, they remain unchanged. I read that the C atom's electrons somehow indicates that reduction has occurred which doesn't make sense how.

Can someone please explain why this reaction is a reduction reaction ?

Thanks.
Formal charge assumes that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. Pretending for a second that NaCl had a covalent bond (which to a small degree it does) then using formal charge, they would both be 0 and hence not ionic. Which of course is daft, especially since it suggests that Na + Cl isn't redox. Formal charge isn't designed for describing redox. It also isn't expected knowledge for OCR A.

Using oxidation numbers we assume electronegativity runs wild and electrons are shared totally unequally, then you get +1 and -1 which looks far more comforting. But again, there is covalent character to the bond, so isn't fully correct. BUT at least we have shown redox.

Applying o.n. to the reduction of methanal to methanol you get C = 0 -> C = -2. Bingo.

With regards to NaBH4. First: we pretend that it is a source of H- ions NOT H+ ions (as has been suggested). Second: it is soluble in protic solvents (water, short alcohols etc.), but NaBH4(aq) won't mix with many organic chemicals which are mostly immiscible with water. So we use methanol as a common solvent - the NaBH4 and the organic can both dissolve in it. Third: the protonation step (of the O- part of the intermediate, producing the final OH group) cannot involve free H+ ions (just imagine what would happen if there were H- ions and H+ ions in the same mixture), so d+ H on water are the source of these protons.
3
reply
lhh2003
Badges: 16
Rep:
?
#9
Report Thread starter 7 months ago
#9
(Original post by Pigster)
Formal charge assumes that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. Pretending for a second that NaCl had a covalent bond (which to a small degree it does) then using formal charge, they would both be 0 and hence not ionic. Which of course is daft, especially since it suggests that Na + Cl isn't redox. Formal charge isn't designed for describing redox. It also isn't expected knowledge for OCR A.

Using oxidation numbers we assume electronegativity runs wild and electrons are shared totally unequally, then you get +1 and -1 which looks far more comforting. But again, there is covalent character to the bond, so isn't fully correct. BUT at least we have shown redox.

Applying o.n. to the reduction of methanal to methanol you get C = 0 -> C = -2. Bingo.

With regards to NaBH4. First: we pretend that it is a source of H- ions NOT H+ ions (as has been suggested). Second: it is soluble in protic solvents (water, short alcohols etc.), but NaBH4(aq) won't mix with many organic chemicals which are mostly immiscible with water. So we use methanol as a common solvent - the NaBH4 and the organic can both dissolve in it. Third: the protonation step (of the O- part of the intermediate, producing the final OH group) cannot involve free H+ ions (just imagine what would happen if there were H- ions and H+ ions in the same mixture), so d+ H on water are the source of these protons.
Hi Sir,

I have just came back to this post for revision.

I don't get where you are coming from in step 2. You say that the organic reactant, i.e the ketone or aldehyde will be immiscible with water. But carbonyls are water soluble as they have a polar Oxygen that can form H bonds, so the carbonyl and organic compound will both dissolve in water therefore mix.

Moreover, why do we say that the reaction is as follows ? :

R-CH=O + 2 [H] ---> R-CH2-OH

Surely we should write the reaction with only one [H] as only one H- (reducing agent) reacts ; H+ is not a reducing agent.

Thanks as always !!
0
reply
Pigster
Badges: 19
Rep:
?
#10
Report 7 months ago
#10
Short chain aldehydes and ketones are soluble as water can form hydrogen bonds to the carbonyl group (the O has lone pairs which can attract the d+ H on water), but they cannot form H-bonds to water as they don't have a d+ H of their own. Longer chain carbonyls might not be so soluble.

[H] is just shorthand for reducing agent, in much the same way that [O] is just shorthand for oxidising agent.
0
reply
lhh2003
Badges: 16
Rep:
?
#11
Report Thread starter 7 months ago
#11
(Original post by Pigster)
Short chain aldehydes and ketones are soluble as water can form hydrogen bonds to the carbonyl group (the O has lone pairs which can attract the d+ H on water), but they cannot form H-bonds to water as they don't have a d+ H of their own. Longer chain carbonyls might not be so soluble.

[H] is just shorthand for reducing agent, in much the same way that [O] is just shorthand for oxidising agent.
Why does that make a difference ? A bond is a bond, why does it matter what molecule the oxygen / hydrogen belongs to ?

And also [H] ; the reducing agent, only reacts once , so its molar ration value should be 1 right ?
0
reply
Pigster
Badges: 19
Rep:
?
#12
Report 7 months ago
#12
(Original post by lhh2003)
Why does that make a difference ? A bond is a bond, why does it matter what molecule the oxygen / hydrogen belongs to ?

And also [H] ; the reducing agent, only reacts once , so its molar ration value should be 1 right ?
The main thing is: if carbonyls can H-bonds to water, then why can't they form H-bonds to each other?

Converting methanal to methanol does involve the addition of two H atoms to the structure of methanal.
0
reply
lhh2003
Badges: 16
Rep:
?
#13
Report Thread starter 7 months ago
#13
(Original post by Pigster)
The main thing is: if carbonyls can H-bonds to water, then why can't they form H-bonds to each other?

Converting methanal to methanol does involve the addition of two H atoms to the structure of methanal.
But H+ is not a reducing agent ? Is it something that I just have to accept as notation because it would make sense if the reduction was in terms of adding a h atom
0
reply
Pigster
Badges: 19
Rep:
?
#14
Report 7 months ago
#14
(Original post by lhh2003)
But H+ is not a reducing agent ? Is it something that I just have to accept as notation because it would make sense if the reduction was in terms of adding a h atom
Reading back through the thread, it is possible that Arctic Kitten's notation may have confused you.

[H+] means the concentration of H+ ions.
[H] is shorthand for the H atoms added by a reducing agent.

H+ is not a reducing agent. It would have a very hard job giving away an electron.
0
reply
lhh2003
Badges: 16
Rep:
?
#15
Report Thread starter 7 months ago
#15
(Original post by Pigster)
Reading back through the thread, it is possible that Arctic Kitten's notation may have confused you.

[H+] means the concentration of H+ ions.
[H] is shorthand for the H atoms added by a reducing agent.

H+ is not a reducing agent. It would have a very hard job giving away an electron.
If [H] is a reducing agent, i.e something that adds an electron to the molecule, then H- is a reducing agent but H+ is not. A H- is added to the molecule but a H+ ion is not, so why is H+ considered to be a reducing agent by us putting 2 [H] instead of just [H]. Moreover, the H+ doesn't even come from the reducing agent so you can't argue that the [H] represents the Hydrogens added to the molecule by the reducing agent ; the H+ comes from water molecules in the reaction with NaBH4 .
0
reply
Pigster
Badges: 19
Rep:
?
#16
Report 7 months ago
#16
(Original post by lhh2003)
If [H] is a reducing agent, i.e something that adds an electron to the molecule, then H- is a reducing agent but H+ is not. A H- is added to the molecule but a H+ ion is not, so why is H+ considered to be a reducing agent by us putting 2 [H] instead of just [H]. Moreover, the H+ doesn't even come from the reducing agent so you can't argue that the [H] represents the Hydrogens added to the molecule by the reducing agent ; the H+ comes from water molecules in the reaction with NaBH4 .
Reduction is the gain of e-.

The reduction of a carbonyl to an alcohol involves two steps: the addition of H- (from NaBH4) and the addition of H+ (from water).

If you were to count the total number of e-, using methanal as the simplest example:
CH2O = 6 + 2x1 + 8 = 16
on addition of H- ion, an intermediate forms by a dative bond forming from H-, the C=O bond breaks and a pair of e- moves to the O
CH3O- = 6 + 3x1 + 8 + 1 (the charge) = 18
on addition of H+ (from water) to form alcohol
CH3OH = 6 + 3x1 + 8 + 1 = 18

The addition of H- is reduction. The addition of H+ is not redox.

I say again, H+ is not AND CANNOT (since it has no e-) be a reducing agent.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Have you experienced financial difficulties as a student due to Covid-19?

Yes, I have really struggled financially (14)
11.76%
I have experienced some financial difficulties (29)
24.37%
I haven't experienced any financial difficulties and things have stayed the same (52)
43.7%
I have had better financial opportunities as a result of the pandemic (20)
16.81%
I've had another experience (let us know in the thread!) (4)
3.36%

Watched Threads

View All