[betterbesafehero]

^^^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 ?

[souro10]

^^^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?

[betterbesafehero]

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.

[betterbesafehero]

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..!

[souro10]

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?

[betterbesafehero]

Why does not polarizability increase across a period?

http://www.mychemistrytutor.com/questions/polarizability-and-change-across-period

[souro10]

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.

[iamback]

^^^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.

[iamback]

and calm down big fellow @betterbesafehero

[souro10]

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.

[iamback]

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..

[souro10]

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.

[iamback]

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.

[iamback]

Do you know what polarisation is ?

[souro10]

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^-  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.