(B) Study Material for High achievers -:

1.Solutions:

Solutions are homogeneous mixtures of two or more than two components. Binary

solution → consists of two components

Types of solutions:

Gaseous solution → solvent is gas, e.g., mixture of oxygen and nitrogen gases Liquid

solution → solvent is liquid, e.g., oxygen dissolved in water

Solid solution → solvent is solid, e.g., solution of hydrogen in palladium

Expressing concentration of solutions:

Mass percentage (w/w) of a component

Volume percentage ( v/v) of a component =

Mass by volume percentage (w/v) =

Parts per million =

Mole fraction of a component =

Molarity (M) =

Molality (m) =

Solubility:

Solubility of a substance is the maximum amount of the substance that can be dissolved in a specified amount of a solvent at a specified temperature.

Solubility of a solid in a liquid:

-Effect of temperature (Le Chatelier’s principle) –

When the dissolution process is endothermic, the solubility will increase with increase in temperature.

When the dissolution process is exothermic, the solubility will decrease with increase in temperature.

-Effect of pressure: Pressure does not affect solubility.

Solubility of a gas in a liquid:

Henry’s law → The solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Or

The partial pressure of a gas in vapour phase (p) is proportional to the mole fraction

of the gas (x) in the solution.

Where, KH → Henry’s law constant

Some applications of Henry’s law –

-The solubility of CO2 in soft drinks and soda water is increased by sealing the bottles under high pressure.

-At high pressure underwater, scuba divers have to cope with high concentrations of dissolved gases while breathing air.

-At high altitudes, climbers become weak and are unable to think clearly, which are symptoms of a condition called anoxia

Effect of temperature –

With increase in temperature, the solubility of gases in liquids decreases.

Vapour pressure of liquid solutions:

Vapour pressure of liquid–liquid solutions:

Raoult’s law → For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.

(For an ideal solution at constant temperature)

Vapour pressure of solutions of solids in liquids:

Ideal and non-ideal solutions:

Ideal solutions –

Solutions which obey Raoult’s law over the entire range of concentrations

-In an ideal solution, the solute–solute and solvent–solvent interactions are nearly equal to the solute–solvent interactions.

Non-ideal solutions – Solutions which do not obey Raoult’s law over the entire range of concentrations

-Positive deviation from Raoult’s law –

-Vapour pressure of the solution is higher than that predicted by Raoult’s law.

-Solute–solvent interactions are weaker than solute–solute and solvent–solvent interactions.

-Negative deviation from Raoult’s law –

-Vapour pressure of the solution is lower than that predicted by Raoult’s law.

-Solute–solute and solvent–solvent interactions are weaker than solute–solvent interactions.

2

Electrochemical cells:

Two types – Galvanic cell or voltaic cell, and Electrolytic cell

Galvanic cell:

Converts the chemical energy of a spontaneous redox reaction into electrical energy

Daniell cell -

The standard potential of a cell is given by

The potential of an individual half cell cannot be measured.

Relation between the standard potential of a cell and standard Gibbs energy:

Relation between standard Gibbs energy and the equilibrium constant:

Nernst equation:

Gives the concentration dependence of the potentials of the electrodes and the cells

For the electrode reaction

Nernst equation is given by

For a general electrochemical reaction of the type

Conductance of electrolytic solutions:

Resistance, R =

Where,

l → Length

A → Area of cross-section

Ρ → Resistivity or specific resistance

Conductance, G =

Where, Conductivity or specific conductance

The SI unit of conductance is Ω–1 (siemens or mho).

The conductivity of an electrolyte depends upon

nature of the solvent

nature of the electrolyte added

concentration of the electrolyte

temperature

Molar conductivity,

Variation of conductivity

For both strong and weak electrolytes, conductivity decreases with decrease in concentration.

Variation of molar conductivity

For both strong and weak electrolytes, molar conductivity increases with decrease in concentration.

Limiting molar conductivity – molar conductivity when concentration approaches zero

Degree of dissociation,

Kohlrausch law of independent migration of ions:

According to this law, for an electrolyte, the molar conductivity at infinite dilution is the sum of the contribution of the molar conductivity of the ions in which it dissociates.

Electrolytic cells and electrolysis

1F = 96487 C mol-1

Faraday’s first law of electrolysis: The amount of chemical reaction occurring at any electrode during the process of electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

Second law of electrolysis: The amounts of different substances liberated when same quantity of electricity is passed through the electrolytic solution are proportional to their chemical equivalent weights.

Battery is a galvanic cell in which chemical energy of the redox reaction is converted into electrical energy.

Mainly two types:

Primary batteriesSecondary batteries

Primary Batteries

In primary batteries, reaction occurs only once.

After use over a period of time, these become dead and cannot be reused.

Examples: Dry cell (or Leclanche cell), Mercury cell

Secondary Batteries

Secondary batteries can be recharged again by passing current through them in the opposite direction.

Examples:Lead storage battery, Nickel-cadmium cell

3.

For a reaction R → P

Rate of reaction

Factors influencing the rate of a reaction:

Rate of a reaction depends upon the concentration of reactants (pressure in the case of gas), temperature and catalyst.

Rate expression and rate constant

aA + bB → cC + dD

Rate expression

Differential rate equation ®

Where, k is called rate constant

Order of a reaction:

x is Order of the reaction with respect to A

y is Order of the reaction with respect to B

x + y is Overall order of the reaction

1. Order of a reaction can be 0, 1, 2, 3 and even a fraction

2. Units of rate constant aA

+ bB → cC + dD

Rate = k [A]x [B]y

x + y = n = Order of the reaction

[[A] = [B] and x + y = n = Order of the reaction] 1. For

a zero-order reaction, n = 0

Unit of

1. For a first-order reaction, n = 1

Unit of

1. For a second-order reaction, n = 2

Unit of

Molecularity of a reaction:

The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction

Order versus molecularity

1. Order can be zero and even a fraction. But molecularity cannot be zero or a non-integer.

2. Order is applicable to both elementary and complex reactions whereas molecularity is applicable to elementary reactions only.

Integrated rate equations:

Zero-order reactions:

R → P

First-order reactions:

R → P

4.Occurrence of metals:

Metal Ores

Aluminium Bauxite

Kaolinite (a form ofclay)

IronHaematiteMagnetiteSideriteIron pyrites

CopperCopper pyritesMalachiteCupriteCopper glanceZinc blende or

ZincSphaleriteCalamineZincite

The major steps involved in the extraction and isolation of metals from ores are:

3. Concentration of the ore 4. Isolation of the metal from the concentrated ore 5. Purification of the metal

Concentration of ores:

Hydraulic washing: It is the washing away of lighter gangue particles from the heavier ore. It is based on the gravity difference between the ore and the gangue particles.

Magnetic separation: This separation is carried out if either the ore or the gangue is attracted by a magnetic field.

Froth floatation method: This method is used for removing gangue from sulphide ores. ‘Depressants’ are used for separating two sulphide ores. E.g., for separating ZnS and PbS, NaCN is used as the depressant.

Leaching: If the ore is soluble in some suitable solvent, then this process is used. For example, ores of aluminium (bauxite), silver and gold

1. Leaching of alumina

Al2O3(s) + 2NaOH(aq) + 3H2O(l) ® 2Na[Al(OH)4](aq)

2Na[Al(OH)4](aq) + CO2(g) ® Al2O3.xH2O(s) + 2NaHCO3(aq)

Al2O3.xH2O(s) Al2O3(s) + xH2O(

Isolation of crude metal from concentrated ore: It involves two steps –

(i) Conversion into oxide and (ii) Reduction of the oxide to metal

Conversion into oxide:

1. Calcination → Involves heating

Generally, carbonate ores are converted into oxides by this process.

ZnCO3(s)

ZnO(s) + CO

2(g)

1. Roasting → Involves heating in a regular supply of air, at a temperature below the melting point of the metal.

2ZnS + 3O2 → 2ZnO + 2SO2

Generally, sulphide ores are converted into oxides by this process.

Slag → FeSiO3

Reduction of the oxide to metal:

Involves heating with some reducing agents such as C, CO or another metal.

Thermodynamic principles of metallurgy: For any process, the change in Gibbs energy at a temperature is given by

ΔG = ΔH – TΔS and

ΔGƟ = –RT ln K

A reaction will proceed when the value of ΔG is negative.

Applications:

If ΔG(X, XO) is lower than ΔG(Y, YO), then X can reduce YO.

5.

Classification: Mono, di and polyhalogen (tri, tetra, etc.)

1. Alkyl halides or haloalkanes (R−X) → They form homologous series of general formula CnH2n+1X. They are further classified into primary, secondary, and tertiary.

2. Allylic halides → Compounds containing halogen atom bonded to an allylic carbon

(iii) Benzylic halides → Compounds containing halogen atom bonded to an sp3 hybridised carbon atom next to an aromatic ringMethods of preparation:

From alcohols –

From hydrocarbons –

By free radical halogenations- Yields a complex mixture of isomeric mono– and polyhaloalkanes

By electrophilic substitution

Sandmeyer’s reaction –

From alkenes –

Addition of hydrogen halides to unsaturated hydrocarbons (Markovnikov’s rule)

Addition of halogens

(Method used for detecting double bond in a molecule) In this

method, a reddish-brown colour is discharged.

Halogen exchange –

Finkelstein reaction

Swarts reaction (synthesis of alkyl fluoride) –

Physical properties:

Melting and boiling points –

Halides have higher boiling points than hydrocarbons of comparable molecular mass because of having stronger dipole–dipole and van der Waals’ forces of attraction.

The order of increasing boiling points of the different haloalkanes is:

RF < RCl < RBr < RI

The boiling points of isomeric haloalkanes decrease with increase in branching.

Density –

The density of halides increases with increase in the number of carbon atoms, halogen atoms and atomic mass of the halogen atoms.

Solubility –

Soluble in organic solvents, but only slightly soluble in water

Reactions of haloalkanes –

Nucleophilic substitution reaction:

Mechanism

Substitution nucleophilic bimolecular (SN2) (Inversion of configuration)

The increasing order of reactivity is

3o halide < 2o halide < 1o halide < CH3X

(Due to increase in hindrance by bulky substituent in the case of 2o and 3o halides)

Substitution nucleophilic unimolecular (SN1) (Two-step mechanism)

Step I is the slowest; it is reversible and rate-determining step.

6Classification of alcohols and phenols:

Mono, di, tri or polyhydric –

On the basis of hybridisation –

6. – OH Primary, secondary and tertiary Allylic alcohols

2.

3.

1. Benzylic alcohols

2. – OH

a. Vinylic alcohol

b. Phenol

Classification of ethers:

Simple or symmetrical –

C2H5OC2H5, C6H5OC6H5

Mixed or unsymmetrical –

C2H5OCH3, CH3OC6H5

Common names of some phenols:

Preparation of alcohols –

1. From alkenes 2. Acid catalysed hydration

[According to Markovnikov’s rule]

1. Hydroboration – oxidation

1. From carbonyl compounds 2. By reduction of aldehydes and ketones.

Aldehydes give 1° and ketones give 2°alcohols.

1. By reduction of carboxylic acids and esters

7.Nomenclature of aldehydes and ketones –

Aldehydes:

Often called by their common names instead of IUPAC names.

Ketones:

Derived by naming two alkyl or aryl groups bonded to the carbonyl group

Alkyl phenyl ketones are usually named by adding the acyl group as prefix to phenone.

IUPAC Nomenclature

7. For open-chain aliphatic aldehydes and ketones, IUPAC names are derived from the names of the corresponding alkanes by replacing the ending ‘−e’ with ‘−al’ and ‘−one’ respectively.

8. In the case of aldehydes, the longest chain is numbered starting from the carbon of the aldehydic group.

9. In the case of ketones, the numbering begins from the end nearer to the carbonyl group.

10. Substituents are prefixed in the alphabetical order along with the numerals indicating their positions in the carbon chain.

11. Same rule is applicable to cyclic ketones.

12. If the aldehydic group is attached to a ring, then the suffix carbaldehyde is added to the full name of cyclohexane.

Example:

Structure of the carbonyl group –The carbonyl carbon atom is sp2 hybridised and forms three bonds and one π bond.

C=O double bond is polarised due to higher electronegativity of oxygen relative to carbon.

Carbonyl carbon − an electrophile (Lewis acid)

Carbonyl oxygen − a nucleophile (Lewis base)

Preparation of aldehydes and ketones:

(I) By oxidation of alcohols –

2. Primary alcohols Aldehydes

3. Secondary alcohols Ketones

(II) By dehydrogenation of alcohols

(i) Primary alcohols Aldehydes

(ii) Secondary alcohols Ketones

Preparation of aldehydes:

(I) From acyl chloride –(Rosenmund reduction)

(II) From nitriles – (Stephen reaction)

(III) From esters –

(IV) From hydrocarbons –

By oxidation of methylbenzene (toluene) (Etard reaction)

By side-chain chlorination, followed by hydrolysis

Gatterman–Koch reaction

Preparation of ketones:

(I) From acyl chlorides –

8.

Amines:

13.Derivatives of ammonia

14. Obtained by the replacement of one, two or all the three H-atoms of ammonia by alkyl and/or aryl group

15. Structure of Amines

Nitrogen on amines is sp3 hybridised.

Geometry − Pyramidal

Example: Pyramidal shapes of trimethylamine

Classification: Primary (1 ), secondary (2 ) and tertiary (3 )

4. If one H-atom of NH3 is replaced by R or Ar, RNH2 or ArNH2 is obtained (primary amine, 1°).

5. If two H-atoms of NH3 or one H-atom of RNH2 are replaced by alkyl or aryl group (R′), R2NH is obtained (secondary amine, 2°).

6. On the replacement of another hydrogen atom by alkyl or aryl group, R3N is obtained (tertiary amine, 3°).

Nomenclature

Common System

Aliphatic amine: Named by prefixing alkyl group to amine, i.e., alkylamine.

3. Reduction of Amides

4. Gabriel Pthalimide Synthesis: Used for the preparation of primary amine.

(vi) Hoffmann Bromamide Degradation Reaction:

Physical properties:

Solubility –3. Lower aliphatic amines are soluble in water, but higher amines are insoluble in water. This is because lower amines can form H-bonds with water, but higher amines cannot.

4. Solubility of amines in water decreases with increase in molar mass of amines as the size of the hydrophobic alkyl part increases.

5. The order of increasing boiling points of isomeric amines is 3o amine < 2 oamine < 1oamine. This because of intermolecular association of 1amines due to H-bonding.

Practice QuestionsA

Q1) What will happen to the boiling point of a solution if the weight of the solute dissolved in it is doubled and the weight of the solvent taken is reduced by half?Answer-If the weight of the solute dissolved is doubled and the weight of the solvent is reduced by half, then an elevation takes place in the boiling point of the solution. The boiling point of the solution becomes four times the original value.

Q2) What is meant by ‘10% aqueous solution of sodium carbonate’?Answer

'10% aqueous solution of sodium carbonate' means that 10 g of sodium carbonate (solute) is present in

100 g of solution containing water and sodium carbonate.

Q3) When is the value of Van’t Hoff factor more than unity?Answer

The value of Van’t Hoff factor is more than unity when a solute undergoes dissociation in the solution.

Q4)-Under what conditions do non-ideal solutions exhibit negative deviations from Raoult’s law?Answer

Non-ideal solutions exhibit negative deviations from Raoult’s law if the vapour pressure of the solution is

lower than the value predicted by Raoult’s law. This happens when the forces of interaction between the

compounds are greater than those in the pure components.

Q5) Between 0.1 molal solutions of glucose and sodium chloride, which will have a higher boiling point?Answer

0.1 M solution of NaCl will have a higher boiling point than 0.1 M solution of glucose. This is because

sodium chloride undergoes dissociation in the solution. Thus, an elevation is observed in the boiling point.

Q6) The mole fraction of the solute of an X molal solution of a compound in benzene is 0.2. What is the value of X?Answer

Let the mole fraction of solute X be x2.

Let the mole fraction of the solvent be x1.

Hence,

And,

Taking,  (Molecular mass of C6H6 = 78)

Hence, the molalityof the solutionis 3.20. 

Q7) What happens to molarity when the temperature of a solution is increased?Answer

When the temperature of a solution is increased, the molarity decreases. This is because the volume of

the solution increases with an increase in temperature but the number of moles of the solute remains the

same.

Molarity = 

Hence, molarity is a function of temperature. 

Q8) a. The freezing point of a solution containing 0.3 g of acetic acid in 30.0 g of benzene is lowered by 0.45 K. Calculate Van’t Hoff factor. (Kffor benzene = 5.12 Kkgmol−1)

b. The osmotic pressure of a 0.0103 molar solution of an electrolyte is found to be 0.70 atm

at 27°C. Calculate Van’t Hoff factor. (R = 0.082 Latmmol−1K−1)Answer

a. Depression in freezing point can be calculated as follows:

ΔTf(calculated) 

Q9)

a. The outer shell of two eggs is removed and kept in dilute HCl. Then, one shell is placed in

distilled water, while the other is placed in a saturated solution of NaCl. What will be

observed?

b. Explain why a bottle containing liquid ammonia is kept in ice before it is opened.

Answer

a. The egg shell kept in distilled water will get swollen. On the other hand, the egg shell kept in NaCl

solution will shrink. This happens because of osmosis, in which the net flow of solvent from the less

concentrated to the more concentrated solution takes place. Here, the membrane beneath the outer shell

of the egg acts as a semi-permeable membrane.

b. The vapour pressure of liquid ammonia at room temperature is very high. On cooling, the vapour

pressure inside the bottle containing liquid ammonia decreases. As a result, on opening the bottle, liquid

ammonia does not splash out.

Q10) a. Name and explain the factor introduced in 1880 to account for the extent of association or dissociation.

b. Arrange the following solutions in the increasing order of their osmotic pressure. Give

reasons for your answer.

i. 34.2 g/L Sucrose

ii. 60 g/L Urea

iii.58.5 g/L Sodium chlorideAnswer

a. In 1880, Van’t Hoff introduced the Van’t Hoff factor i.e., ‘i’ to account for the extent of association or

dissociation. The Van’t Hoff factor i.e., 'i' is defined as the ratio of experimental value of colligative

property to its calculated value i.e.,

Q11) a. Why is molality preferred over molarity while expressing the concentration of a solution?

b. Why does the boiling point of water increase when sodium chloride is added to it?

c. Why is phenol partially soluble in water?Answer

a. While molarity decreases with an increase in temperature, molality is independent of

temperature. This happens because molality involves mass, which does not change with a

change in temperature, while molarity involves volume, which is temperature dependent.

Hence, molality is preferred over molarity while expressing the concentration of a solution.

b. When a non-volatile solute such as sodium chloride is dissolved in water, the vapour

pressure of water decreases. This happens because on addition of NaCl, some of the solvent

molecules on the surface are replaced by the non-volatile solute molecules. Hence, the

solution has to be heated at a higher temperature to make the vapour pressure equal to the

external pressure. Hence, the boiling point of the solution increases.

c. As a general rule, like dissolves like. Phenol has a polar −OH group but an aromatic

phenyl C6H5 group. Hence, it is partially soluble in water.

B.Q1) Write the structure of the compound, β-methylbutyraldehyde.Answer

The structure of β-methylbutyraldehyde is:

Q2) What type of reactions are shown by benzaldehyde towards nucleophiles?Answer

Benzaldehyde shows nucleophilic substitution reactions at ortho and para positions. This is due to the

electron withdrawing nature of carbonyl group, which decreases the electron density at ortho and para

positions.

Q3) Give a commercial method to obtain benzaldehyde from toluene.Answer

Benzaldehyde can be obtained from toluene by the following method.

 

Q4) Write the Rosenmund reduction reaction of acetyl chloride.Answer

The Rosenmund reduction reaction of acetyl chloride is:

 Q5) Write the IUPAC name of the following compound.

Answer

The IUPAC name of the given compound is

2-(2-Bromophenyl) ethanal. 

Q6)

a. What happens when acetone is treated with sodium bisulphite?

b. Where will the equilibrium lie in the reaction in part (a)?Answer

a.

Acetone, when treated with sodium bisulphite, forms an addition product i.e., acetone bisulphite.

b. The equilibrium lies to the left hand side of the reaction. Q7) Between acetaldehyde and formaldehyde, which will undergo Cannizzaro’s reaction? Also, write the chemical equation for the reaction involved?Answer

Formaldehyde (HCHO) does not contain an α-hydrogen atom. Thus, it will undergo Cannizzaro’s reaction.

On the other hand, acetaldehyde (CH3CHO) will not undergo Cannizzaro’s reaction.

 

Q8) Giving reasons, arrange the following carbonyl compounds in the decreasing order of their reactivity in nucleophilic addition reactions.

Answer

Reactivity in nucleophilic addition reactions decreases in the following order.

This is due to the following reasons:

i. Inductive effect:- The alkyl group has +I effect. The greater the number of alkyl groups

attached to carbonyl group, the greater will be electron density on carbonyl carbon. As a

result, its reactivity is lower towards nucleophilic addition reactions.

ii. Steric effect:- As the number of alkyl group increases, the attack of the nucleophile on

the carbonyl group becomes more and more difficult due to steric hindrance. Q9)

a. Give two chemical tests to distinguish between CH3CHO and CH3COCH3.b. Give a chemical test to distinguish between CH3CHO and HCHO.Answera. CH3CHO and CH3COCH3 can be distinguished by the following tests:

i. Tollen’s test:- On warming CH3CHO with Tollen’s reagent (ammonical silver nitrate), a bright silver mirror is produced. CH3COCH3, on the other hand, does not reduce Tollen’s reagent.

ii. Fehling’s test:- When CH3CHO is heated with Fehling’s reagent, a reddish brown precipitate is formed while CH3COCH3 does not give this test.

b. CH3CHO and HCHO can be distinguished by the iodoform test.CH3CHO forms a yellow precipitate of iodoform on reaction with NaOI, whereas HCHO does not give this reaction.

Q10)

An aromatic organic compound ‘A’ gives a yellow precipitate with 2, 4-DNP hydrazine. It also

gives a silver mirror on reaction with Tollen’s reagent. Now, two molecules of ‘A’ react with

concentrated NaOH forming ‘B’ and ‘C’. Out of ‘B’ and ‘C’, the less acidic product on

reaction with CrO3-H2SO4 gives ‘D’. The compound ‘D’ reacts with Br2|FeBr3 to give ‘E’.

Indentify compounds A -E. Also, give the reactions involved.Answer

Compound ‘A’ gives a yellow precipitate with 2,4-DNP hydrazine and also gives silver mirror

test with Tollen’s reagent. Thus, compound ‘A’ is an aldehyde. Also, compound ‘A’ reacts

with concentrated NaOH i.e., the compound gives Cannizzaro’s reaction. This shows that

compound ‘A’ does not have any α-hydrogen atom.

∴ Compound ‘A’ is benzaldehyde.

Reaction of benzaldehyde with Tollen’s reagent:

Cannizzaro’s reaction of benzaldehyde:

Hence,compound ‘B’ is benzyl alcohol and ‘C’ is sodium benzoate.

Between benzyl alcohol and sodium benzoate, benzyl alcohol is less acidic. It reacts with

CrO3-H2SO4 to give benzoic acid.

Hence, compound ‘D’ is benzoic acid.

Benzoic acid further reacts with Br2 | FeBr3 to give m-bromo benzoic acid.

Hence,compound ‘E’ is m-bromo benzoic acid. 

Q11)

a. Carry out the following conversions.i.Acetyl chloride to ethaneii. Acetic acid to malonic acidiii. Ethanol to 3-hydroxy butanalb. Write short notes on the following:i. Stephen reactionii. Gatterman-Koch reactionAnswera.i.

ii.

iii.

 b.i.Stephen reaction:In this reaction, nitriles are reduced to corresponding imines with stannous chloride in the presence of hydrochloric acid. Imine, on further hydrolysis, gives the corresponding aldehydes.

ii. Gatterman-Koch reaction:In this reaction, benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride forming benzaldehyde or substituted benzaldehyde.