Find the total number of alkene

Question: How many numbers of alkenes possible by dehydrobromination of 3-bromo 3-cyclopentylhexane using alcoholic KOH?

Solution:

When any alkyl halide is treated with alcoholic KOH, it removes hydrogen and a halogen atom from it and forms alkene. This is basically a dehydrohalogenation reaction. (You can recall it’s basic from here). The halogen atom removes from the α-carbon and hydrogen atom from β-carbon of the alkyl halide.

 In 3-bromo 3-cyclopentylhexane, α-carbon atom is at position 3 (to which halogen atom attached) and three β-carbon atom which are at 2, 4 and 3 (C of cyclopentane). 

Because of three types of β-hydrogen, there are more than one alkenes are expected.

So let us find out.

11)   Hydrogen remove from cyclopentane i.e. position 3:

2

  2   2) Hydrogen remove from position 2:

Here two alkenes are obtained, one is cis from and other one is trans.

  3) Hydrogen remove from position 4:

Here also two alkenes are obtained, one is cis from and other one is trans.

So the total numbers of alkenes obtained from dehydrobromination of 3-bromo 3- cyclopentylhexane are 5. 

Dehydrohalogenation reaction

Dehydrohalogenation reaction:

·        It is also called a β-Elimination reaction and is a type of elimination reaction.

• ·        In this reaction alkyl halide when heated with alcoholic solution of potassium or sodium hydroxide, it undergoes dehydrohalogenation (i.e. removal of hydrogen and halogen) and forms alkene.

• ·        Dehydrohalogenation involves removal of the halogen atom from the α-carbon and a hydrogen atom from the adjacent β-carbon atom.

• ·        The relative reactivity of alkyl halide are I > Br > Cl > F (elimination of F is rarely used.) 

·        Chlorobenzene does not react with potassium hydroxide due to the presence of the benzene ring, which, due to stabilization as a result of aromaticity, does not give conventional elimination, as it would lead to a very unstable benzene intermediate.

• ·        If there is more than one type of β-hydrogens (β1 and β2) in alkyl halide then there is a possibility of formation of more than one alkene. In such case, one of the alkenes is formed as a major product, which can be identified by using Saytzeff’s rule.

Saytzeff’s rule: In a dehydrohalogentaion reaction, the preferred product is that alkene which is formed by removal of the hydrogen from the β-carbon having the fewest hydrogen substitutents.

Example: When 2-bromobutane is heated with alcoholic solution of KOH, it forms but-2-ene (80%) and but-1-ene (20%).

Since there are two types of β carbon (β1 and β2), therefore two alkenes are expected.  Here β2 is having fewer number of hydrogen than β1, so according to Saytzeff’s rule preferred product is formed by the removal of the hydrogen from β2. Thus but-2-ene is obtained as major product. 

Find the number of electron

Question: What is the maximum number of electron that can have principal quantum number n = 3 and spin quantum number; m = -1/2?

Solution: The principal quantum number represents the main shells to which the electron belongs. 

We can calculate the number of electron in each shell by the formula: 2n²

Where n represent number of shell.

Here principal quantum number i.e. n is 3.

Thus, number of electron in the shell is 2(3)²

                                                        = 18

So the maximum number of electron that have principal quantum number 3 is 18.

Spin quantum number accounts for spin of electron.  An electron can spin clockwise or anticlockwise, in two opposite directions. For clockwise spin value is +1/2 and for anticlockwise spin value is -1/2.  

Generally in each orbital half electron spin clockwise direction and half in anti-clockwise direction.

So the maximum number of electron that show spin quantum number -1/2 will be 9 in principal quantum number 3. 

Find the optical isomerism of coordinate complex

Question: Which of the following complex species do not show optical isomerism?

a)     [Co(en)₃]³⁺

b)    [Co(en)₂Cl₂]⁺

c)     [Co(NH₃)₃Cl₃]

 

Solution: Lets us recall the basic of Optical Isomerism

a)      [Co(en)₃]³⁺

In this complex Co is central metal while en (ethylene diammine) is bidentate chelating ligand which is coordinate to the central metal atom Co. [Co(en)₃]³⁺ shows d²sp³ hybridisation.

Hybridisation of [Co(en)₃]³⁺:

 

This means that [Co(en)₃]³⁺  is an octahedral complex.

Such complex does not possesses any element of symmetry thus, it is optically active. It has been resolved into d- and l- forms.

 

b)    [Co(en)₂Cl₂]⁺

In this complex, two symmetrical bidentate chelating ligands ‘en’ (ethylene diammine) and two monodentate ligands ‘chlorine’ are coordinate to central metal atom Co.

[Co(en)₂Cl₂]⁺ shows d²sp³ hybridisation. This means that it is an octahedral complex.

Hybridisation of [Co(en)₂Cl₂]⁺ :

 

 [Co(en)₂Cl₂]⁺ exhibit the phenomenon of optical isomerism and can be resolved into their optical isomers. Its cis form is unsymmetrical, while the trans form is symmetrical because it contain plane of symmetry.

 Hence, optical isomerism is shown by cis form only. The cis form is resolved into d- and l- forms.

 

 c)     [Co(NH₃)₃Cl₃]

In this complex, Co is attached to three monodentate ligand ammonia and three (other) monodentate ligands chlorine.

 [Co(NH₃)₃Cl₃] shows d²sp³ hybridisation. This means that it is an octahedral complex.

Hybridisation of [Co(NH₃)₃Cl₃]:

In such octahedral complex, metal is placed at the centre and six ligands occupy their position at the vertices. Due to different spatial arrangement of the groups around the central metal, these complex show geometrical isomerism. This gives rise to two types of isomers called cis and trans isomers.

The cis and trans forms of [Co(NH₃)₃Cl₃] are given below.

 

 

 

In the cis-isomer, the three Cl- are one triangular face (position 1, 2, and 3) and the three NH₃ molecules are placed on the opposite triangular face (position 4, 5 and 6). Since three identical ligands occupy one face of an octahedron, the isomer is said to be facial or fac. 

In the trans isomer, Cl- are placed on the position 1, 2 and 3 while NH2 molecules are present on the positions 3, 4 and 5. It is also called meridional isomer (in which each set of three identical ligands occupies a plane passing through the metal atom).

[Co(NH₃)₃Cl₃] show facial as well as meridional isomerism. But both contain plane of symmetry, hence it is optically inactive.

 

 

 

Optical Isomerism in coordinate complex

What is optical isomerism?

It is a type of stereoisomerism in which the isomer molecules have the tendency to rotate plan of polarized light and also show chiral behaviour. These compounds are called as Optically active compounds.

Condition required for optical isomerism:

• The molecule should be disymmetrical means they should not contain; a plane of symmetry which divides the molecule in two equal halves.

• Presence of a chiral molecule which makes a non superimposable mirror images of the compound. (A Chiral molecule is that in which a carbon is attached with four different groups.)

• The isomers should have non superimposable mirror images, and these images are called the enentiomers.

Coordinate complex that show optical isomerism:

1) Four Coordination Compound:

• Optical isomers are possible for tetrahedral complex but not for square planar complex.

• Optical isomerism is shown by tedrahedral complexes of type [Mabcd]  where central metal (M) is attached to four different monodentate ligands (a, b, c and d).

• The tetrahedral complexes of Be(II), Zn(II)and B(III) which contain unsymmetrical bidendate ligands also show optical isomerism.

  

2) Six Coordination Compound:

• The octahedral complexes of the type [Ma₂b₂c₂], [Ma₂b₂cd], [Ma₂bcde] and [Mabcdef] are optically active where a,b,c,d,e and f are monodendate ligands.

• The octahedral complexes of the type [M(A-A) ₃]^+_n show optical isomerism where (AA) is a symmetrical bidendate ligand.

• Complexes of the type [M(A-A) ₂X₂] also show optical isomerism where(AA) is symmetrical bidentate ligand whereas X is monodentate ligand. The cis form of this type of complex is optically active whereas trans form is optically inactive.

• Similarly the cis form of the complex [M(A-A) ₂BC]^+-n is optically active.

• Complexes of the type [M(A-A)B₂C₂] also show optical isomerism.

• Complexes containing optically active ligands show optical isomerism. Example:[Co(en)₂(pn) ₂ (NO₂)₂]⁺ and the complexes containing polydendate ligands also show optical isomerism. Example:[Co(EDTA)]^-

Find the magnetic property of complex

Question: Which of the following complexes are diamagnetic in nature?

a)     K₃[Fe(CN)₆]

b)    [Co(NH₃)₆]Cl₃

c)     Na₃[Co(Oxalate)₃]

d)    [NiCl₄]^-2

Solution:

Magnetic property of a complex depends upon the number of unpaired electrons present in the orbital of its metal.

·        If unpaired electron present in the complex then it is paramagnetic in nature.

·        If there is no unpaired electron then the complex is diamagnetic in nature.

Now let us start with each option and find out its magnetic property.

a)     K₃[Fe(CN)₆]

It is a anionic complex because it carries negative charge on complex ion.

                                                                      (Complex Ion)

In this complex ion Fe is central metal and CN is ligand. To calculate number of electron in Fe we have to know its oxidation state. Since the whole complex ion show (-3) charge so the oxidation number of Fe will be:

Charge number of [Fe(CN)₆]^3-  = charge on Fe + charge on CN

                                                 -3 = x + 6 (-1)     (CN has -1 charge)

                                                  X = +3

Now let us find out number of electrons in the central atom Fe³⁺

Fe (26) = 2, 8, 14, 2

           = 3s², 3p⁶, 3d⁶, 4s²

 

 Fe³⁺ = 3s², 3p⁶, 3d⁵, 4s⁰            

 

In the presence of CN as ligand, the electrons present in 3d get shifted and undergoes d²sp³ hybridisation.  But this makes one electron unpaired in 3d.

Due to presence of this unpaired electron, K₃[Fe(CN)₆] shows paramagnetic.

 

b)    [Co(NH₃)₆]Cl₃

It is a cationic complex because it carries positive charge on complex ion.

In this complex ion Co is central metal and NH₃ is ligand. Now let us know the oxidation number of Co:

Charge number of [Co(NH₃)₆]³⁺  = charge on Co + charge on NH₃

                                                  +3 = x + 6 (0)     (NH₃ is neutral)

                                                  X = +3

Now let us find out number of electrons in the central atom Co³⁺

Co (27) = 3s², 3p⁶, 3d⁷, 4s²

 

  Co³⁺ = 3s², 3p⁶, 3d⁶, 4s⁰            

 

In the presence of NH₃ as ligand, the electrons present in 3d get shifted and undergoes d²sp³ hybridisation.  But this makes all electron paired in 3d.

Hence, [Co(NH₃)₆]Cl₃ shows diamagnetic.

 

c)     Na₃[Co(Oxalate)₃]

It is an anionic complex because it carries negative charge on complex ion.

In this complex ion Co is central metal and oxalate is ligand. Now let us know the oxidation number of Co:

Charge number of [Co(oxalate)₃]^3-  = charge on Co + charge on NH₃

                                                           -3 = x + 6 (-1)     (oxalate, C₂O₄, has -1 charge)

                                                             X = +3

Now let us find out number of electrons in the central atom Co³⁺

Co (27) = 3s², 3p⁶, 3d⁷, 4s²

  Co³⁺ = 3s², 3p⁶, 3d⁶, 4s⁰            

 

In the presence of oxalate (C₂O₄) as ligand, the electrons present in 3d get shifted and undergoes d²sp³ hybridisation.  But this makes all electron paired in 3d.

Hence, Na₃[Co(oxalate)₃] shows diamagnetic.

 d)     [NiCl₄]^-2

In this complex ion Ni is central metal and Cl is ligand. Now let us know the oxidation number of Ni:

Charge number of [NiCl₄]^2-  = charge on Ni + charge on H₂O

                                                  -2 = x + 4 (-1)     (Cl has -1 oxidation number)

                                                  X = +2

Now let us find out number of electrons in the central atom Ni²⁺

Ni (28) = 3s², 3p⁶, 3d⁸, 4s²

 

  Ni²⁺ = 3s², 3p⁶, 3d⁸, 4s⁰            

In the presence of Cl as ligand, one 4s and three 4p orbitals undergo sp3 hybridisation.  Each Cl- ion donates a pair of electrons. But this makes two electrons unpaired in 3d.

Hence, [NiCl₄] ^2- shows paramagnetic nature.

Thus, options b and c are diamagnetic in nature.