Why alkyl iodides are purple in the presence of sunlight and Role of solvents in SN1 and SN2 reaction ?

Why alkyl iodides are purple in the presence of sunlight ?

The C –I bond in alkyl iodide compound is larger than the other carbon – halides bond in halo alkane. With increasing bond length,bond dissociation energy decreases . 

So C – I bond is weaker than other carbon – halides bond. Hence alkyl iodides are breaks easily in presence of sunlight and yield alkane along with I_2.

Since I₂ is purple in color, so due to presence of I₂ in product mixture alkyl iodides are seem to be purple in color.The reaction proceeds through radical mechanism.

Why  alkyl halides are insoluble in water  in spite of polar in nature ?

There are two type of force of attraction acts among the alkyl halides molecules ,  which one is Vander waals attraction force and another is dipole-dipole attraction force. 

On the other hand , water molecules  being polar, they can form inter-molecular hydrogen bond . 

Now, alkyl halides will be dissolve in water if breaks these dipole-dipole attraction force of alkyl halides and hydrogen bond of water molecules. 

Now, to break these dipole-dipole attraction force and hydrogen bond, a sufficient amount of energy is required which is not available through the formation of dipole-dipole attraction between alkyl halides and water molecules.

Consequently , alkyl halides are insoluble in water  in spite of polar in nature .

Role of solvents in SN¹ and SN² reaction .

The rate of SN¹ and SN² reaction depends on the nature of the solvents used.In case of SN¹ reaction, the nature of the solvents should be polar protic solvents, such as H₂O, CH₃OH and CH₃CH₂OH etc. 

This type of solvents are eligible for SN¹ reaction. Because, in such type of solvents, both carbo cationand halide ion gain stability through polar interaction. 

On the other hand, polar but aprotic solvents like dimethy sulphoxide ( DMSO), N,N-dimethyl formamide ( DMF), acetone ( CH₃COCH₃ ) etc are very much suitable solvents used in SN² reaction. 

Because, in such type of solvents, cation gets stability but anion is free. Consequently, the activity of anions increases as a nucleophile ( CH3O ^–, HO^– etc ).

Why backside attack occurs in SN²  reaction?

In SN²  reaction mechanism backside attackoccurs. This fact can be explained on the basis of orbital picture diagram of transition state.

The carbon atom which is attacked by the nucleophile ,are in sp2 hybridized . 

On the other hand , the p-orbital of carbon atom is perpendicular to the plane on which the three sp2hybridized  orbitals are present.

The one lobe of this p-orbital takes part in overlapping with the leaving group and another lobe overlaps with the nucleophile. For this reason, front side attack does not occur. 

If it is possible, then in the hypothetical transition, both the leaving group and the nucleophile will be overlapping in the same side of the p-orbital which is impossible in real case.Consequently, backside attack occurs in SN²  reaction. 

Summary 

• Why  alkyl halides are insoluble in water  in spite of polar in nature ?
• Why alkyl iodides are purple in the presence of sunlight ?
• Why backside attack occurs in SN²  reaction ?
• Role of solvents in SN¹ and SN² reaction ?

Calculate the no. of unpaired electron and LFSE of [ Fe ( H2O)6 ]+3 ion .Click here

Calculate the number of unpaired electrons and LFSE of [ Fe ( H2O)6 ]+3 ion .

Calculate the number of unpaired electrons and LFSE of [ Fe ( H₂O)₆ ]⁺³ ion .

The IUPAC  name of the above complex ion is hexaaquo iron(III) ion. The oxidation state of iron in this complex ion is +3 .

Coordination number( C.N ) of iron is 6 .

The valence shell electronic configuration  of Fe ⁺³ ion  : [Ar] 3d⁵ .  

The arrangement of the complex ion is octahedral . The octahedral structure of this complex are as follows ,

Ligand H₂O, which is a weak  field Ligand . Under the influence of the octahedral crystal field, the five degenerate d-orbital’s  of Fe ⁺³ ion  are splitted into two sets of energetically different orbital’s. 

These two sets are energetically lower t_2g orbital and energetically higher e_g orbital . 

As there are five  electrons  in the 3d⁵ degenerate orbital’s , so according to crystal field theory the possible arrangement  is t_2g³ e_g² .

Here, the crystal field splitting energy ( 10 Dq_o ) is lower than pairing energy (P).

That is , 10 Dq_o  <  P

So, no pairing of electrons will be occured  in any sets of orbital’s . 

The crystal field splitting of degenerate d-orbital’s of  Fe ⁺³ ion  under the influence of Ligand weak field Ligand  H₂O are as follows,

Since H₂O is a weak field Ligand , so it should be high  spin complex  and from the above splitting diagram ,it is clear that the complex ion have five unpaired electron .

Due to presence of five  unpaired electron , the [ Fe( H₂O)₆ ]⁺³ complex ion exhibit paramagnetic properties .

Calculation of Ligand field stabilization energy .

The Ligand field stabilization energy ( LFSE ) of the above complex ion is,

      3x( -4 Dq_o ) + 2 x 6 Dq_o  
      = - 12 Dq_o + 12 Dq_o 
      =  0  Dq_o .

Summary : 

Calculate the number of unpaired electrons and LFSE of [ Fe ( H₂O)₆ ]⁺³ ion .

What is ligand field stabilization energy ?

click here

Calculate the spin only magnetic momentum  µ of  K3 [ Mn(CN)6]  compound  .

Why most of the transition metals are used as catalysts ?

Why most of the transition metals are used as catalysts ?

According to the modern theory of catalysis , a catalytic substance is capable of 

forming an unstable intermediate compound which readily decomposes yielding the product and regenerating the catalyst.

The transition elements, on account of their variable valency , are able to form unstable intermediate compounds very readily.

These elements can also provide a large surface area for the reactants to be adsorbed 

and thus come closer to one another for the reaction to occur readily on the surface of the catalyst itself.

For the above reason, most of the transition metals and their compounds have good 

catalytic properties and used as catalyst. Platinum, iron, vanadium pentoxide nickel etc, are important examples.

Platinum is a general catalyst and is used particularly in the contact process involving combination of sulphur dioxide and oxygen to yield sulphur trioxide .

Vanadiumpentoxide [ V2O5 ] is also a good catalyst for the same reaction. 

Iron catalyses the combination of nitrogen and hydrogen in the Haber process for the production of ammonia. Nickel is a good catalyst in hydrogenation.

Why ZnCl₂ is fully ionized in dilute aqueous acidic solution but HgCl₂ is not ?

Zinc (Zn ) and mercury ( Hg ) are the members of same family of group IIB . ZnCl₂ is fully ionized  in dilute aqueous  acidic solution because it behave as electrolytes in such solution.

But HgCl₂ remains unionized in aqueous solution . Since the solution does not conduct electricity .

Due to lanthanide contraction and ineffective shielding of 4f¹⁴ and 5d ¹⁰ electrons, 

ionization potential and electronegativity of Hg becomes higher than that of Zn , which is reverse to the group trend .

Mercury is thus less metallic or more non metallic than Zinc . ZnCl ₂ has a considerable ionic character but Hg – Cl bonding is mostly covalent . 

This strong covalent nature of HgCl₂ prevent its ionization in aqueous solution.

Thus aqueous solution of HgCl₂ is a non electrolyte like sugar solution .

Summary :

Why most of the transition metals are used as catalysts ?

Why ZnCl₂ is fully ionized  in dilute aqueous  acidic solution but HgCl₂ is not ?

 π-acceptor strength of CO, NO + and CN-ligand .

Why the colour of trans [ Co (en)2 F2 ] + is less intense than that of cis [ Co (en)2 F2 ] + ?

Why  the colour of trans [ Co (en)₂  F₂ ] ⁺ is less intense than that of cis [ Co (en)₂  F₂ ] ⁺ ?

In the given two complexes [ cis and trans form of diflouridobis( ethylene diamine ) cobalt (III) ion ] both the ligands and the geometry of the complex are same but in the above octahedral complexes arrangements are different one is cis and other is trans.