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Metallurgy- Extraction of elements, and purification methods:

METALLURGY-INTRODUCTION:
PYROMETALLURGY:

How to determine Basic order of different amines and it derivatives ?

  1. How is base strength related to the availabihty of the electron-pair?
  2. Amines are more basic than ammonia why?
  3. What is relative basic strength order 1° amines , 2°amines and 3° amines ? Explain:
  4. Give an explanation for the fact that Guanidine NH=C(CH3)2 is a stronger base than most of amines?
  5. Arrange in correct order of basic Character of aniline, pyrrol, pyridine and piperidine?
  6. What is correct basicity order of pyridine, pyridazine, pyrimidine and pyrazine ?
  7. Give an explanation for the fact that Guanidine NH=C(CH3)2 is a stronger base than most of amines?
  8. Arrange in correct order of basic Character of aniline, pyrrol, pyridine and piperidine?
  9. What is correct basicity order of pyridine, pyridazine, pyrimidine and pyrazine ?
  10. Why pyridine is more basic than Pyrrole?
  11. Why pyrimidine is less basic than pyridine?
  12. Imidazole is more basic than pyridine? Why?
  13. Biological Important of Imidazole and structure:
  14. Pyridine is almost 1 million times less basic than piperidine? Why?
  15. Cyclohexylamine amine is the stronger base than Aniline? Why?
  16. Tetrahydroquinoline amine is the stronger base than Tetrahydroisoquinoline? Why?
  17. Arrange the following in the order of increasing basicity : p-Toluidine, N, N-Dimethyl-p-toluidine, p-Nitroaniline, Aniline. (I.I.T.1986)
  18. Arrange the following in the increasing order of their acid strength : Methyl amine, Dimethyl amine, Aniline, N-methyl aniline (I.I.T, 1988).







What are Zeeman and Stark effects? Were they explained by the Bohr’s theory?

The splitting of spectral lines into many components in the presence of a magnetic field is called Zeeman effect.

Analogously, the splitting of spectral lines into many components in the presence of an electric field is called Stark effect. 
These effects could not be explained within the framework of Bohr's theory and required the presence of multiple non-circular orbits corresponding to each energy level. An effort, with some degree of success, was made by Sommerfeld to explain these phenomena by invoking multiple elliptical orbits corresponding to each energy level. The Bohr Sommerfeld theory was ultimately discarded along with Bohr's model for other reasons.

Ionization energy of Boron is smaller than Beryllium even though effective nuclear charge is higher?

The electronic configurations of Boron and Beryllium are (B [He]1S2,2S2,2p1) and (Be [He]1S2,2S2). In Boron the outermost electron is present in the 2p orbital and is less strongly bound than the electron present in a 2S orbital of Beryllium, which will has a higher Zeff. It is easier to ionize the Boron atom.
 

The complex formation Tendency of NH3 is wisher than PH3.


SOLUTION: In NH3 lane pair is present in one of the sp3 Hybridized orbital. While in PH3 lone pair is present in pure S orbital and hence lane pair donation capacity of NH3 is stronger than PH3

What happens if two deactivating and one activating group is present on a benzene ring?

We know that reactivity of benzene ring towards electrophilic reaction depends upon presence of group on benzene ring. Benzene gives electrophilic substitution reaction due to negative pi cloud formation , if there is EWG (electron withdrawing) group on benzene ring it's negativeness decreases and reactivity also decreases and due to resonance upcoming group goes at meta position only while in the presence of ERG (electron releasing groups) it's negative ness increases and reactivity increase and due to resonance upcoming group goes at ortho and para position .
If both EWG and ERG present on benzene ring then upcoming electrophile goes to ortho and para position  according to ERG group .

What is bromoform reaction?

Haloform reaction includes Chloroform, Bromoform and Iodoform reaction, It is feature reaction of terminal methyl ketones and alpha hydroxy ketones . In which methyl ketones oxidised in the presence of I2 and NaOH ( NaOI) into Sodium salt of carboxylic acid and corresponding haloform ( Chloroform , Bromoform and Iodoform)

Is Methanal an aldehyde?

Yes ! Methanal is a aldehyde containing -CHO group. And is molecular formula is HCHO . It is more commonly known as formaldehyde. It is also first member of aldehyde homologous series.

Which is a better nucleophile, among halides ions ( fluoride, Chloride, bromide and iodide)? and Why?

Nucleophilicity of an nucleophilic depends upon nature of solvent, for example:
(1) I⁻ is a better nucleophile than F⁻ in polar protic solvents.

(2) F⁻ is a better nucleophile than Br⁻ in polar aprotic solvents.

We know that a protic solvent has an H atom bound to more electronegative elements like F, O or N. It can use its H atom to formed H-bonding with a nucleophile. Which accumulate around nucleophilic and creates a shell around the nucleophile. So that it becomes more difficult to attack the positive carbon bearing the leaving group.

F⁻ is a small ion with a high charge density. It is tightly solvated.

I⁻ is a large ion with a low charge density. It is loosely solvated. There are only a few solvent molecules to push out of the way.

Hence over all order of nucleophilicity in polar protic solvents is given as :

        [ I⁻ > Br⁻ > Cl⁻ > F⁻ ]

We know that a polar aprotic solvent does not have a hydrogen atom that can formed hydrogen bond.

But in all of them, the negative ends of the dipoles directedaway from the molecule. So that It is easy for them to solvate cations.

The positive ends of the dipoles are closer to the middle of the molecule. It is difficult for them to get close to the anions.

As the result the nucleophile has few molecules in its solvent shell. The nucleophile can more easily attack on positive carbon (electrophilic center).

Hence F⁻ becomes a much better nucleophile than chloride, 

Hence over all order of nucleophilicity in polar aprotic solvents is just reverse of that given in polar protic solvent :

        [ F⁻ > Cl⁻ > Br⁻ > I⁻ ]


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