Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Ksharindam on April 26, 2013, 01:40:04 PM
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Which compound has higher nucleophilicity among NH2OH and NH2NH2?
I am confused that which effect is predominant among alpha effect and electron withdrawing effect of oxygen.
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ofcourse NH2NH2
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Reason : Nucleophilicity parallels basicity in a period.
So NH2NH2 is a better base than NH2OH, also a better nucleophile here in this case..!
I hope you find it reasonable enough.
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NH2NH2 is a better base than NH2OH, also a better nucleophile here in this case.
But nucleophilicity depends also on alpha effect.(such as NH3 is more basic than Hydrazine but less powerful nucleophile due to alpha effect).Here in hydroxylamine has two lone pairs on neighbouring oxygen atom(hence better alpha effect) but Hydrazine has one lone pair on neighbouring N-atom.
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But here alpha effect is not predominant. Don't stick to what you have studied.
NH2NH2 is the answer..!
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Look for the differences in the two bases: Substituent is an -NH2 group in one case and substituent is an -OH group in the other case.
I think the ease at which the substituent donates electrons is the important factor to look for here.
Nitrogen has 7 protons pulling it's electrons while Oxygen has 8 protons to pull it's electrons.
Consequently, Nitrogen pulls it's electrons less tightly than oxygen , which implying the electron-nucleus distance in case of - NH2 is greater than that of -OH. Thus electrons of NH2 are donated more easily than -OH, i.e NH2-NH2 is more basic than NH2-OH, in this case.
Thanks to @Orgopete for this wonderful approach.
Corrections and comments welcomed.
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That's what I am saying. Ease of donating electron decreases along the period..!
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That's what I am saying. Ease of donating electron decreases along the period..!
That is not a reason. That is the statement of a particular pattern - but not the reason.
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Do you have any rate constant data that bear on this question?
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I
That's what I am saying. Ease of donating electron decreases along the period..!
That is not a reason. That is the statement of a particular pattern - but not the reason.
In chemistry we don't give a reason, we see what's happening and accordingly give the theory. Any reason which explains the behavior of reaction or anything, we say that this must be happening, in chemistry there is nothing certain.
Example: VBT explained that CH4 must be tetrahedral, when it was actually determined that CH4 is tetrahedral, we said that: yes..! this must happening and this theory is correct. Reason are just manifestation. Chemistry happens at molecular level, of which many things are yet to be determined, or I can say most of it is yet to be determined.
You are giving an explanation which can be true or not. I am no one to judge it. But I go by facts, that this happens generally, so this must be happening.
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Do you have any rate constant data that bear on this question?
NH2NH2 Kb=1.7 x10-6
NH2OH Kb=1.1x10^-8
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Do you have any rate constant data that bear on this question?
NH2NH2 Kb=1.7 x10-6
NH2OH Kb=1.1x10^-8
The rate constants you mention are basicities. These are thermodynamic properties, and nucleophilicity is a kinetic property.
Bromine is more basic than Iodine but Iodine is more nucleophilic than Bromine [ in a protic solvent ].
Thus equilibrium constant [basicity] cannot and does not reflect nucleophilicity.
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I know that, here nucleophillicity and basicity have same trend, Do I need to scream about it ?This question is solved...would you mind wasting your time on some other questions.
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The rate constants you mention are basicities. These are thermodynamic properties, and nucleophilicity is a kinetic property.
Bromine is more basic than Iodine but Iodine is more nucleophilic than Bromine.
Really..! I didn't know that :P. That's a group trend, there nucleophillicity is opposite to that of basicity, and in a period both are parallel.
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I know that, here nucleophillicity and basicity have same trend, Do I need to scream about it. This question is solved...would you mind wasting your time on some other questions.
This is not the correct approach to learning.
Why does nucleophilicity parallel basicity across a period?
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^^^Don't you know why ?
Okay I will explain it : Across a row in the periodic table nucleophilicity (lone pair donation) is C- > N- > O- > F- and also increasing electronegativity decreases the lone pair availability. The same is applied for basicity..!
Understood ?
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^^^Don't you know why ?
Okay I will explain it : Across a row in the periodic table nucleophilicity (lone pair donation) is C- > N- > O- > F- and also increasing electronegativity decreases the lone pair availability. The same is applied for basicity..!
Understood ?
Why across a period the trend is C- > N-> O- > F- ?
Why is basicity not parallel to nucleophilicity down a group?
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Why across a period the trend is C- > N-> O- > F- ?
Don't you know how ?, I said lone pair donation trend is this, Can't you see why ?
Why is basicity not parallel to nucleophilicity down a group?
Within a group in the periodic table, increasing polarisation of the nucleophile as you go down a group enhances the ability to form the new C-X bond and increases the nucleophilicity, so I- > Br- > Cl- > F-. The electron density of larger atoms is more readily distorted i.e. polarised, since the electrons are further from the nucleus.
Note that is the opposite order to basicity (acidity increases down a group) where polarisability is much less important for bond formation to the very small proton.
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If you understood it good enough..! If you don't, maybe you should convince yourself. Well I am leaving this topic now..!
Hope I helped someone..!
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Sure, it is better you leave this topic.
I do not understand, though. If increasing polarization increases the ability to form new C-X bond, then why is Iodine ( more polarizable ) a better leaving group than Bromine ( less polarizable ) ?
Why does not polarizability increase across a period?
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Why does not polarizability increase across a period?
http://www.mychemistrytutor.com/questions/polarizability-and-change-across-period
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If greater polarizability means stronger C-X bond formation, and if Iodine is more polarizable than Chlorine, then according to that, Chlorine should be a better leaving group than Iodine.
But that is not the case. So there must be some factor that you are missing out.
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^^^Maybe I could offer some help..!
Tendency to form C-X bond and leaving group are two different concept, leaving group tendency is how stable the compound would be after separation.. or how easily the group bear the negative charge. :)
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and calm down big fellow @betterbesafehero
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If a negatively charged compound reacts with an electrophile, and forms a stable bond, eg. C-X bond, then the reverse reaction that is breaking of the C-X shall be unfavorable. Consequently, there is flaw in the logic mentioned previously.
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then the reverse reaction that is breaking of the C-X shall be unfavorable.
But we are not considering reverse reaction of breaking of C-X, we are dealing with breaking H-X, not C-X when we are talking about basicity or acidity..
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It was mentioned that,
Why across a period the trend is C- > N-> O- > F- ?
Don't you know how ?, I said lone pair donation trend is this, Can't you see why ?
Why is basicity not parallel to nucleophilicity down a group?
Within a group in the periodic table, increasing polarisation of the nucleophile as you go down a group enhances the ability to form the new C-X bond and increases the nucleophilicity, so I- > Br- > Cl- > F-. The electron density of larger atoms is more readily distorted i.e. polarised, since the electrons are further from the nucleus.
Note that is the opposite order to basicity (acidity increases down a group) where polarisability is much less important for bond formation to the very small proton.
Consequently this post claims better the polarisation of the nucleophile, better the nucleophilicity, stronger the C-X bond formation, as underlined.
Had things been that simple, then according to the same logic, the reverse reaction i.e C-X bond breaking should have been unfavourable for the better nucleophile. However we find, C-I bond is broken more easily than C-Br or C-Cl bond, even though Iodine is a better nucleophile. Again C-I bond is formed more easily than C-Br or C-Cl bond.
Thus things are not as simple as described.
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It's not as simple as you think, the factor favoring the formation of C-X bond can't be taken in consideration while breaking of the same bond. It's depends on how stable the leaving group is after breaking, and I- is more stable with it's charge dispersed.
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Do you know what polarisation is ?
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It's not as simple as you think, the factor favoring the formation of C-X bond can't be taken in consideration while breaking of the same bond. It's depends on how stable the leaving group is after breaking, and I- is more stable with it's charge dispersed.
Tell me a simple thing. R+ + X- ::equil:: R-X
According to the above, if Iodide is the best leaving group, it should be forming the weakest R-X bond among the halogens ?
If your answer is yes, then read on. If it's no, then give your logic at this stage.
If Iodide forms the weakest R-X bond, then how is I- a better nucleophile than other halogens?
But we know it is. If you look at thermodynamic data, you will find C-F bond is strongest and C-I bond is weakest. Then why in the above reversible reaction, Iodide reacts fastest with C+ and F- slowest ? According to thermodynamics, the more stable the product the faster it should be formed, isn't it?
Explain this apparent anomaly.
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Then why in the above reversible reaction, Iodide reacts fastest with C+ and F- slowest ?
First of all it is not a reversible reaction. I admit C-I bond is the weakest. You are saying that if C-I is weakest(rather it is), why it is not favoring the the formation C-F bond which is more stronger ? Now, nucleophilicity does not depend on how stronger the bond is gonna be, or how strong the bond it would make. It depends on polarisability.
C-F bond is stronger, does that mean F- would be a good nucleophile. It is stronger as " Bond strength (energy) can be directly related to the bond length/bond distance." C-F is the strongest single bond in organic chemistry—and relatively short—due to its partial ionic character.
Iodide is relatively good nucleophile in protic solvents. This is because its large size allows it to polarize electron density in such a way that it will still be negatively charged enough to attack an electrophile after picking up a proton from the protic solvent. In aprotic solvent, it's actually a worse nucleophile than all the other halogen anions.
According to thermodynamics, the more stable the product the faster it should be formed, isn't it?
You yourself said that 'nucleophilicity is a kinetic property'. :)
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Also refer this for your answer : http://chemistry.stackexchange.com/questions/4152/how-can-a-group-be-both-a-good-nucleophile-and-a-good-leaving-group
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According to thermodynamics, the more stable the product the faster it should be formed, isn't it?
iamback already marked this part of your post. This is definitely wrong - kinetics and thermodynamics are separate phenomena and there is no simple relationship between both.
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According to thermodynamics, the more stable the product the faster it should be formed, isn't it?
iamback already marked this part of your post. This is definitely wrong - kinetics and thermodynamics are separate phenomena and there is no simple relationship between both.
Yes, it is indeed wrong. I was arguing from the other side. :)
I was expecting a proper argument about nucleophilicity being a kinetic property.
The reasons that were being pointed out so far, weren't satisfying to me.
When someone argues nucleophilicity with pkB values, these arguments naturally arise. :P
Then why in the above reversible reaction, Iodide reacts fastest with C+ and F- slowest ?
First of all it is not a reversible reaction. I admit C-I bond is the weakest. You are saying that if C-I is weakest(rather it is), why it is not favoring the the formation C-F bond which is more stronger ? Now, nucleophilicity does not depend on how stronger the bond is gonna be, or how strong the bond it would make. It depends on polarisability.
This is the simple answer that one is looking for. Nucleophilicity does not depend on how stronger a bond is going to be, it depends on how fast the nucleophile donates it's electrons.
Now, pkB definitely depends on how stable the conjugate acid is. pKb is actually the equilibrium constant of the reversible reaction. In that case, is it plausible to provide pKb data to a question on nucleophilicity?
Hence, I argued.
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Iodide is relatively good nucleophile in protic solvents. This is because its large size allows it to polarize electron density in such a way that it will still be negatively charged enough to attack an electrophile after picking up a proton from the protic solvent. In aprotic solvent, it's actually a worse nucleophile than all the other halogen anions.
Could you post data to support this?
As I recall, fluoride does indeed jump past iodide in aprotic solvents by increasing its rate, but iodide was relatively independent of a solvent effect. That was some time ago and my memory could be incorrect.
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I just checked Table 4.9 (pp. 333-334) in Lowry and Richardson's textbook (Mechanism and Theory in Organic Chemistry). The values of nCH3I for NH2OH and NH2NH2 are 6.60 and 6.61, respectively. The solvent is methanol in these reactions. The value for NH3 is 5.50 This is a logarithmic scale, BTW.
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I love it when someone looks up actual data for their argument (snack given).
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Nucleophilicity of NH2NH2 is greater than NH2OH, however considering the fact that the conjugate acid of NH2OH is much more stable than that of NH2NH2, NH2NH2 should be a better base as well. Across a period electronegativity increase that is tendency to donate the lone pair electrons decrease and if electrons cannot be donated fast enough, nucleophilicity which is a KINETIC criterion decreases across a period. Also, for a similar reasons, the basicity also decrease across a period. SO we can say across a period, basicities parallel nucleophilicities. Hence, the result.
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The nucleophilicity order is NH2-NH2>NH2-OH>NH3
This can be explained on the basis of two factors:
1)ALPHA EFFECT:
In both NH2-NH2 that is hydrazine and NH2-OH that is hydroxylamine, nucleophilic attack by lone pair of electrons is enhanced by the lone pair of electrons on the adjacent atom. The phenomenon is called "alpha effect". This effect is more pronounced in case of hydrazine compared to hydroxylamine, because lone pair of electrons on nitrogen is more polarizable than the lone pair of electrons on oxygen atom.
2)MOLECULAR ORBITAL THEORY
in NH3, no lone pair repulsion occurs hence NUCLEOPHILICITY IS MINIMUM
in NH2-OH, oxygen with lone pair is more stable hence charge repulsion is less
in NH2-NH2, charge repulsion is more between the two nitrogen atoms, so charge repulsion being more, hence NUCLEOPHILICITY IS MAXIMUM
any suggestions are welcome and appreciated :)