Problem on condensation polymerisation

Question: Identify the correct functional group ‘X’ and the reagent or reaction condition ‘Y’ in the following scheme.

a)      X = COOCH₃, Y = H₂ / Ni / Heat

b)      X = CONH₂, Y = H₂ / Ni / Heat

c)       X = CONH₂, Y = Br₂ / NaOH

d)      X = CN, Y = H₂ / Ni / Heat

Solution: The end product of the following scheme is a condensation polymer. From here we can get a clue for the whole reaction. So let us recall the basics of condensation polymerisation.

1. ·         A condensation polymer is formed through a condensation reaction--where two molecules (monomers) join together and loose small molecules as by-products such as water or methanol.
2. ·         There are ester and amide linkages in the polymer.
3. ·         Monomers that are joined by condensation polymerization have two functional groups.

·         These functional groups can be a carboxylic acid and an amine which form an amide linkage

·         or a carboxylic acid and an alcohol which form an ester linkage

Now let us go back to the question and check each option whether it can fulfill the above requirement.

a)      If  X = COOCH₃, Y = H₂ / Ni / Heat

In the first step, compound will hydrogenate in the presence of catalyst Ni.

 
The obtained product is an alcohol. In the next step the alcohol will react with carboxylic acid and form an ester linkage.

 The final product, a polyester, is formed by condensation of two monomers having two different functional groups i.e. an alcohol and an acid. in this reaction a water molecule is also formed as a byproduct. The poly-ester has an ester (-COO-) linkage. Thus, option (a) is correct for the given scheme.

b)      If  X = CONH₂, Y = H₂ / Ni / Heat
In the first step, amide will hydrogenate in amine in the presence of catalyst Ni.

In the next step the amine will react with carboxylic acid and form an amide linkage.

The final product, a polyamide, is formed by condensation of two monomers having two different functional groups i.e. an amine and an acid. in this reaction a water molecule is also formed as a byproduct. The poly-amide has an amide (-HN-CO-) linkage.  Thus, option (b) is correct for the given scheme.

c)       If  X = CONH₂, Y = Br₂ / NaOH

In the first step, compound will react in the presence of sodium hydroxide and bromine. Basically it is a Hoffman Bromide Reaction.

[The Hofmann rearrangement is the organic reaction in which a primary amide (–CONH₂) group is converted into primary amino (– NH₂) group. ]

The obtained product is an amine. In the next step this amine will react with carboxylic acid and form an amide linkage.  

The final product is a polyamide which is form by condensation polymerisation. Thus, option (c) is correct for the given scheme.

d)      If  X = CN, Y = H₂ / Ni / Heat
In the first step, compound will hydrogenate in the presence of catalyst Ni.

The obtained product is an amine. In the next step this amine will react with carboxylic acid and form an amide linkage.

The final product is a polyamide which is form by condensation polymerisation. Thus, option (d) is correct for the given scheme.

Thus, all options are correct for the given scheme.

Change in a chemical reaction

Question: which of the following options are correct for the given equation?

    

a)      Phase transition

b)      Allotropic change

c)       ∆H is positive

d)       ∆S is positive 

e)      ∆S is negative   

Solution: Let us observe each option with the given chemical reaction carefully.

a)      Phase Transition:

In the given reaction, solid carbon-dioxide changes into gaseous form. This gives clear hint that option (a) is correct for the equation because there is a phase transition. If you remembered the basic than you should know that a phase transition is the transformation of a substance from one phase or state of matter to another. (here solid changes into gas)

b)      Allotropic Change

There is no allotropic change in the reaction because the term allotrope refers to one or more forms of an elementary substance. Thus, option (b) is wrong for the equation. (because Carbon dioxide is a compound)

c)       ∆H is positive

Delta H refers to change in heat in the reaction. So let us look our reaction again. 

In reaction solid carbon dioxide changes directly into gaseous form.  This type of transition is called sublimation which requires additional energy.  The system absorbs this additional energy from its surroundings in the form of heat. Such kind of reaction is known as endothermic reaction. 

The change ΔH is positive in endothermic reactions, and negative in heat-releasing exothermic processes.

Thus, option (c) is correct for the equation. 

d)      ∆S is positive 

Delta S refers to change in entropy. Entropy is a mathematically-defined thermodynamic quantity that helps to account for the flow of energy through a thermodynamic process.

Change in entropy is positive when....
-you go from a solid to a liquid
-you go from a liquid to a gas
-when you go from a solid to a gas
-when there are more mols of products than mols of reactant

And similarly change in entropy is negative when the reverse of the above happens.

In the above reaction, Solid CO₂ changes into gas, so ∆S must be positive here.

Thus, option (d) is correct for the equation. 

e)      ∆S is negative 

If option (d) s correct for the equation, then option (e) must be wrong.

Thus, options (a), (c) and (d) are correct for the equation.

Reaction of hydrogen halide with silver nitrate

Question: which of the following hydrogen halide reacts with silver nitrate (aq.) to give a precipitate that dissolves in Na₂S₂O₃ (aq.)?

a)      HCl

b)      HF

c)       HBr

d)      H I

Solution: When silver nitrate (AgNO_3(aq)) reacts with hydrogen halde (HX), a double replacement reaction takes place:

A double replacement reaction is a chemical reaction where two reactant ionic compounds exchange ions to form two new product compounds with the same ions.

where X = Cl, Br, or I.

The silver of silver nitrate reacts quickly with hydrogen halide to produce silver halides which are insoluble so they form a precipitate. Except F all other halides (Cl, Br, I) give precipitate with Ag.

AgF is soluble; no precipitate formed

AgCl forms a white precipitate

AgBr forms a cream precipitate

AgI forms a yellow precipitate

This precipitate dissolves on reaction with Na₂S₂O₃ (aq.)

Thus options a, c, d are correct and give precipitate with silver nitrate which dissolve in Na₂S₂O₃ (aq.).

Oxidation number of Nitrogen

Question: Arrange the following compounds to the decreasing order of the oxidation state of Nitrogen.

NO, HNO₃, NH₄Cl, N₂

Solution: Let us recall some basic rule of oxidation number.

• The sum of the oxidation number of all the atoms in a neutral compound is zero.
• The oxidation number of a free element is always zero. Like the atoms in He and N₂ have oxidation numbers of 0.           
• The usual oxidation number of hydrogen is +1.
• The oxidation number of Chlorine is -1.
• The oxidation number of oxygen is usually -2.

Now find out the oxidation state of nitrogen in the given compounds.

Oxidation state of N in NO:  

Oxidation state of N in HNO₃ :

Oxidation state of N in NH₄Cl : 

Oxidation state of N in N₂ : As we know the oxidation number of a free element is always zero. Thus

According to above calculation, compounds in decreasing order of oxidation state are: