Find the final product

Question: What will be the final product P of the following reaction?

Solution: To find out the final product let us treat the given reactant with the reagent step by step:

First the given reactant is treated with H₃O⁺.

[H₃0]+ is a hydronium cation, also known as hydroxonium. It's an acid according to Bronstend-Lowry theory, about bases and acids, because it can give of a [H]⁺ :

[H₃0] <--> H₂O + H⁺

So when given reactant reacts with H₃O⁺, it gives:

On heating the obtained product losses CO₂

Now the obtained alkene is react with O₃ and H₂O₂.

As a reagent, ozone is used to cleave alkenes. When alkenes are treated with ozone and H₂O₂, they are subjected to “Oxidative workup”, this breaks the carbon-carbon double bond to form carboxylic acid. 

Now this is final product P. 

Dizonium coupling

Diazonium Coupling (Diazo Coupling, Azo Coupling)

An azo coupling is an organic reaction between a diazonium compound and another aromatic compound that produces an azo compound. Nitrogen is used to make a bridge between two benzene rings.

Dizonium compound + Aromatic compound = Azo compound

 

This is basically an electrophilic aromatic substitution reaction. Due to the positive charge on the terminal nitrogen of the -N=N⁺ group, dizonium cation may participate as an electrophile. As a result, two aromatic compounds are coupled by a -N=N- group. This is known as the azo group (diazo group). The corresponding reaction is called diazonium coupling (diazo coupling, azo coupling).

Examples of Coupling Reactions:

1)   Reaction with Phenol

Phenol is dissolved insodium hydroxide solution to give a solution of sodium phenoxide.

When this phenoxide ion reacts with dizonium ion, an azo compound is formed in which two benzene rings are linked by a nitrogen bridge.

2)   Reaction with Phenylamine (aniline)

The reaction of aniline with dizonium salt produced a yellow solid known as aniline yellow.

 

Uses: Azo coupling is the most widely used industrial reaction in the production of dyes, lakes and pigments. 

Limitation:  The electrophilicity of diazonium ions is only relatively weak, as their positive charge is delocalized over the two nitrogen atoms.

The unsubstituted benzenediazonium cation may react only with strongly activated aromatic compounds, such as phenolates and amines, The substitution normally occurs at the para position, except when this position is already occupied, in which case ortho position is favoured. .  

Friedal Crafts Alkyation

Friedal Crafts Alkylation:

This reaction is developed by Charles Friedel and James Crafts in 1877.

In Friedal crafts alkylation, a haloalkane reacts with benzene in the presence of an aluminium halide catalyst, or any other lewis acid catalyst, and produce an alkylbenzene and a hydrogen halide.

 The reactivity and polarity of haloalkane increase as we move up the periodic table. It means:

RF > RCl > RBr > RI

This means that the Lewis acid used as catalyst in Friedel Craft alkylation reaction tend have similar halogen combination. BF₃, SbCl₅, AlCl₃ and AlBr₃ are commonly used Lewis acid in the reaction.

Mechanism of Friedel Crafts alkylation:

1)    In this step, a carbocation is formed which act as a electrophile in the reaction. This activates the haloalkane.

2)    In step two, an electrophillic attacks on benzene which results in multiple resonance forms. The halogen reacts with the intermediate and picks up the hydrogen to eliminate the positive charge.

3)    The last step shows end of the step two and give the final product.

 

Limitations:

There are few limitations of Friedel Crafts alkylation.

·         The reaction works only with benzene or ACTIVATED benzene derivatives. It will not occur if the benzene ring is deactivated.

·         The reaction works only with ALKYL halides (i.e. chlorides, bromides or iodides), but it does not work with VINYL or ARYL halides.

·         Alkylation reactions are prone to carbocation rearrangement.

·         Over alkylation can be a problem since the product is more reactive than the starting material. This can usually be controlled with an excess of the benzene.

·         The Lewis acid catalyst AlCl₃ often complexes to aryl amines making them very unreactive. 

Find the standard enthalpy of combustion

Question: The standard enthalpies of formation of CO₂ (g), H₂O (l) and glucose (s) at 25^oC are -400 kJ/mol, -300kJ/mol and -1300kJ/mol, respectively. What will be the standard enthalpy of combustion per gram glucose at 25^oC?

Solution: First let us know the meaning of standard enthalpy of combustion.

The standard enthalpy change of combustion is the enthalpy change which occurs when one mole of an element or compound is burned completely in oxygen under standard conditions.

These standard conditions are:

·       A pressure of 1 atm or 100kilopascals (10²kPa)

·       A temperature of 298K (25^oC)

·       Reactants and products in physical states, normal for the above conditions.

·       A concentration of 1.0mol dm^-3 for solutions.

The enthalpy change of combustion will always have a negative value, of couse, because burning always releases heat. 

The standard enthalpy of combustion can be denoted as ΔH^o_c  where the ^o sign indicates standard conditions

When glucose burn in the oxygen it form carbon dioxide and water. So the proper equation of combustion of glucose is:

So the standard enthalpy of combustion will be:

∆H_combustion = (6 x ∆H_f CO₂ + 6 x ∆H_f H₂O) - ∆H_f C₆H₁₂O₆

                      = (6 x -400 + 6 x -300) – (-1300)

                         = -2900 kJ /mol

 The molar mass of glucose is 180 grams/mole. So

∆H_combustion = -2900/180 kJ/g

                   = -16.11 kJ/g

Thus, the standard enthalpy of combustion of per gram glucose at 25^oC is -16.11 kJ/g.  

Finkelstein Reaction

Finkelstein Reaction:

The Finkelstein reaction is named after the German chemist Hans Finkelstein.  In this reaction, one halogen atom exchanges for another. It is basically single-step bimolecular nucleophilic substitution reaction (SN2 reaction).

According to classic Finkelstein reaction an alkyl chloride or an alkyl bromide is converted into alkyl iodide by the treatment with a solution of sodium iodide in acetone.

Sodium iodide is soluble in the acetone but sodium chloride and sodium bromide are not. So when sodium chloride or sodium bromide are formed in the reaction, it get precipitated and this help reaction to driven to completion due to mass action.

For example, bromoethane can be converted to iodoethane:

Every alkyl halide reacts differently when it undergoes the Finkelstein reaction. The reaction works well for primary (except for neopentyl) halides, and exceptionally well for allyl, benzyl, and α-carbonyl halides. Secondary halides are far less reactive. Vinyl, aryl and tertiary alkyl halides are unreactive.

Find the rate of reaction

Question: KI in acetone undergoes SN² reaction with each of P, Q, R and S. The rates of reaction vary as:

a)    P > Q > R > S

b)   S > P > R > Q

c)    P > R > Q > S

d)   R > P > S > Q

Solution: P,Q, R and S are alkyl halide. When any of them reacts with KI in the presence of acetone, alkyl iodide is formed. This is also known as Finkelstein reaction.

 In this reaction (I-) which is a nucleophile, replaces (Cl-), thus it is known as a nucleophilic substitution reaction. Since, in the reaction one bond is broken and one bond is formed in one step, it is called SN2 reaction (substitution nucleophilic bimolecular reaction).

The rate of SN2 reaction involves an interaction between nucleophile and the organic substrate. It can be described as:

rate = k [Nu][R-LG]. 

Where, Nu is nucleophile which is I- while [R-LG] is RX which are P, Q, R and S.

The relative reactivity for SN2 reaction in the given structure is:

According to this the correct option is (b). 

Find the radius of octahedral void

Question: The arrangement of X^- ions around A⁺ ion in solid AX is given in the figure (not drawn to scale). If the radius of X^- is 250pm, what is the radius of A⁺?

 

Solution:

According to given figure, A⁺ is present in the octahedral void of X-. An octahedral void is formed by six spheres which are arranged on the corner of two triangles. When these two triangles are placed with their apex faced opposite sides, octahedral void arises. 

The limiting radius in octahedral void is related to the radius of sphere as:

r_void = 0.414 r_sphere

                                        r _A+ = 0.414  r_X-

                                                   = 0.414 x 250 pm

                                                   = 103.5 pm

Thus, radius of A⁺ is 103.5 pm. 

Find the magnetic moment of the complex

Question: Consider the following complex ions, P, Q and R 

P = [FeF₆]^3-, Q = [V(H₂O)₆]²⁺, and R = [Fe(H₂O)₆]²⁺

What is the correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.)?

A) R<Q<P

B)  Q<R<P

C)  R<P<Q

D) Q<P<R

Solution:

The formula used to calculate the spin-only magnetic moment can be written as:

μ= √n(n+2) B.M.

Where n = no. of unpaired electron

 

So first let us find out number of unpaired electron in each complex.

 

1)   P = [FeF₆]^3-

Charge on Fe in the complex: [FeF₆]^3-    

                                                                                     X + (1x6) = -3

                                                        = +3

So number of unpaired electron in Fe⁺³ = 3s², 3p⁶, 3d⁵, 4s⁰            

                                                                      = 5

Magnetic moment = √5 (5+2)

                                         =√35 B.M.

 

2)   Q = [V(H₂O)₆]²⁺

Charge on Fe in the complex: X + (0x6) = +2

                                                 = +2

So number of unpaired electron in V⁺² = 3s², 3p⁶, 3d³,           

                                                                     = 3

Magnetic moment = √3 (3+2)

                                         =√15 B.M.

 

3)   R = [Fe(H₂O)₆]²⁺

Charge on Fe in the complex: X + (0x6) = +2

                                                    = +2

So number of unpaired electron in Fe⁺² = 3s², 3p⁶, 3d⁶         

                                                                     = 4

Magnetic moment = √4 (4+2)

                                         =√24 B.M.

So the correct order of the complex ions, according to their spin-only magnetic moment values is P > R > Q

Thus option B is correct. 

What is Racemic Mixture

What is Racemic Mixture?

In chemistry, a racemic mixture, or racemate, is one that has equal amounts of left- and right-handed enantiomers of a chiral molecule.

If you are not getting the point from the above definition, than let’s start with the basics.

• ·       An organic compound that contains a chiral carbon atom usually shows two non-superposable structures.
• 
  
• ·       These two structures are mirror images of each other and are, thus, commonly called enantiomers.
• 
  
• ·       These enantiomers refer identical physical and chemical properties except for their ability to rotate plane-polarized light (+/−) by equal amounts but in opposite directions.  
• 
  
• ·       The enantiomer that rotates plane polarized light to right side is known as dextro or right handed isomer. Such enantiomer is denoted by the prefix (+) or D Or d. On the other hand, the enantiomer that rotates light to left side is known as levo or left handed isomer. Such enantiomer is denoted as prefix (-) or L or l.

 

Now when equal amounts of left and right handed enatiomers of a chiral molecule are mixed together they form a racemic mixture.

Nomenclature

Since in a racemic mixture, ratio of the both right and left handed enatiomers are equal i.e. 1:1, hence it is denoted by the prefix (±)- or dl.   (for sugars the prefix dl- may be used),

Also the prefix rac- (or racem-) or the symbols RS and SR (all in italic letters) are used.

If the ratio is not 1:1 (or is not known), the prefix (+)/(−), d/l- or d/l- (with a slash) is used instead.

However the usage of d and l is strongly discouraged by IUPAC. 

Properties

• ·       Although the two enantiomers rotate plane-polarized light in opposite directions, a racemate or racemic mixture is optically inactive, meaning that there is no rotation of plane-polarized light. This is because, rotation of plane polarized light into two opposite directions and in equal amounts, cancels the net rotation. 
• 
  
• ·       In contrast to the two pure enantiomers, which have identical physical properties except for the direction of rotation of plane-polarized light, a racemate sometimes has different properties from either of the pure enantiomers.
• 
  
• ·       Different melting points are most common, but different solubilities and boiling points are also possible.
• 
  
• ·       Pharmaceuticals may be available as a racemate or as the pure enantiomer, which might have different potencies.

Crystallization

There are four ways in which a racemate can crystallize, three of which H. W. B. Roozeboom had distinguished by 1899:

• Conglomerate (sometimes racemic mixture or racemic conglomerate)

• Racemic compound (sometimes true racemate)

• Pseudoracemate (sometimes racemic solid solution)

• Quasiracemate

Resolution

The separation of a racemate into its components, the pure enantiomers, is called achiral resolution. There are various methods, including crystallization, chromatography, and the use of enzymes.

 The first successful resolution of a racemate was performed by Louis Pasteur, who manually separated the crystals of a conglomerate.

Synthesis

• Without a chiral influence i.e. without a chiral catalyst, solvent or starting material, a chemical reaction that makes a chiral product will always yield a racemate.
• That can make the synthesis of a racemate cheaper and easier than making the pure enantiomer, because it does not require special conditions.
• Thr reagents of, and the reactions that produce, racemic mixtures are said to be “not stereospecific” or “not stereoselective,” for their indecision in a particular stereoisomerism.