ADDITIONAL QUESTIONS – FIVE MARK

1. Calculate the potential of a half-cell consisting of zinc electrode in 0.01 M ZnSO4 solution 25oC [Eo = 0.763 V].
The half-cell reaction is:

Zn → Zn2+ + 2e–

The Nernst equation for the oxidation half-cell reaction is:

E = Eo – 0.0591/n. log [Zn2+] / [Zn]

The number of electrons transferred,          n = 2

Standard potential of Zn/ Zn2+ half-cell, Eo = 0.763 V

Activity / Concentration of Zn2+,      [Zn2+] = 0.01 M = 10–2 M

Activity / Concentration of solid metal [Zn] = 1 i.e., equal to unity

Potential of Zn/ Zn2+ half-cell,                  E = ?

Substituting these values in the Nernst equation we have

E = 0.763 – 0.0591 / 2 . (log 10–2)

E = 0.763 – 0.0591 / 2 . (–2 . log 10)

E = 0.763 – 0 0591 / 2 . (–2)

   = 0.763 + 0.0591

E = 0.8221 V

2. Mention three methods of preparing anisole.

1. Williamson’s synthesis: When sodium phenoxide is heated with methyl iodide anisole is obtained.

C6H5ONa +  I– CH3→ C6H5 – O – CH3 + NaI

2. Using diazomethane: Phenol reacts with diazomethane gives anisole.

C6H5 – O – H + CH2 – N2 → C6H5 – O – CH3 + N2

3. Manufacture of ether: Anisole is manufactured on large scale by reacting phenol with dimethyl sulphate in presence of sodium hydroxide.

                                NaOH

C6H5OH + (CH3)2SO4 → C6H5– O– CH3 + CH3.HSO4

3. Illustrate the reducing property of acetaldehyde with examples.

1. Acetaldehyde has   – C = O  bond which is responsible for its reducing property.

                                   |

                                  H

2. Acetaldehyde reduces Tollen’s reagent (Ammonical Silver nitrate) to metallic silver

CH3CHO + 2Ag+ + 3OH– → CH3COO– + 2Ag + 2H2O

Acetaldehyde                            Acetate ion   (silver mirror)

3. Acetaldehyde reduces Fehling’s solution (Copper sulphate, Sodium potassium tartrate) to red cuprous oxide.

CH3CHO + 2Cu2+ + 5OH– → CH3COO– + Cu2O + 3H2O

(blue)                                                               (red precipitate)

cupric ion                                                          cuprous ion

4. Acetaldehyde restores the original colour (red-pink) of the Schiff's reagent. (when SO2 is passed through intensely pink coloured Schiff’s reagent in water. It forms a colourless solution. This colourless solution is used for this test). This is characteristic test for aldehydes.

4. Explain chromophore and auxochrome theory about dyes.

According to chromophore and auxochrome theory,

(1) An organic compound appears coloured due to the presence of certain unsaturated groups (the groups with multiple bonds) in it. Such groups with multiple bonds are called chromophores.

Example:

 

2) The compounds containing the chromophore group is called chromogen. The colour intensity increases with the number of chromophores or the degree of conjugation.

Example:
ethene (CH2 = CH2) is colourless, but the compound CH3 – (CH = CH)6 – CH3 is yellow in colour.

(3) The presence of certain groups which are not chromophores themselves, but deepen the colour of the chromogen. Such supporting groups are called auxochromes. Auxochromes may be acidic (phenolic) or basic.

Example: –OH, –NH2, –NHR, NR2.

The presence of an auxochrome in the chromogen molecule is essential to make it a dye. However, if an auxochrome is present in the meta position to the chromophore, it does not affect the colour.

For example, in the compound p-hydroxyazobenzene (a bright red dye),

 

5. Discuss the factors affecting adsorption.

The magnitude of gaseous adsorption depends upon the following factors:

1. Temperature

2. Pressure

3. Nature of the gas and

4. Nature of the adsorbent.

Effect of temperature and pressure

Adsorption is invariably accompanied by evolution of heat. Therefore, in accordance with Le chatelier’s principle, the magnitude of adsorption increases with decrease in temperature. Further, since adsorption of a gas leads to decrease of pressure, the magnitude of adsorption increases with increase in pressure. Thus, decrease of temperature and increase of pressure both tend to cause increase in the magnitude of adsorption of a gas on a solid.

Nature of the gas

It is observed that the more readily soluble and easily liquefiable gases such as ammonia, chlorine and sulphur dioxide are adsorbed more than the hydrogen, nitrogen and oxygen. The reason is that Vander waal’s or intermolecular forces which are involved in adsorption are more predominant in the former than in the latter.

Nature of the adsorbent

Adsorption is a surface phenomenon. Therefore, the greater the surface area per unit mass of the adsorbent, the greater is its capacity for adsorption under the given conditions of temperature and pressure.

6. Explain the buffer action of a basic buffer with an example.

Basic buffer consists of a weak base together with a salt of the same base.

Example: Ammonium hydroxide and Ammonium chloride (NH4OH/NH4Cl).

The pH of the buffer is governed by the equilibrium

NH4OH ⇌ NH4+ + OH–       ... (1)

The buffer solution has a large excess of NH4+ ions produced by complete ionisation of Ammonium chloride,

NH4Cl → NH4+ + Cl –          ... (2)

1. Addition of HCl: When HCl is added to the buffer solution, the increase of H+ ions is counteracted by combine with the excess of NH4OH to form ionised NH4+ and H2O. Thus the added H+ ions are neutralised and the pH of the buffer solution is retained.

Fig. Mechanism of buffer action of a basic buffer

2. Addition of NaOH: When NaOH is added to the buffer solution, the additional OH– ions combine with NH4+ ions present in the buffer solution to give NH4OH and hence pH is maintained.

7. Explain : i) Hell-Volhard Zelinsky reaction (HVZ – reaction) ii) Claisen ester condensation

i) Hell-Volhard Zelinsky reaction:

Halogenation of Carboxylic acid with Halogen and Phosphorous trihalide.

Or

Conversion of Carboxylic acid to α - bromo acid with Br2 / PBr3

                    Br2/PBr3                                              H2O

RCH2COOH       →      RCH2COBr   →  RCHBrCOBr   →  RCHBrCOOH

ii) Claisen ester condensation :

In presence of strong bases like sodium ethoxide, methyl acetate undergoes condensation forming Methyl acetoacetate.

8. Write any three methods of preparing Benzylamine.

1. Reduction of benzonitriles: By the reduction of benzonitrile catalytically or with Lithium Aluminium Hydride.

             H2/Ni

C6H5CN → C6H5CH2NH2

              LiAlH4

2. Reduction benzamides: By the reduction of benzamide catalytically or with Lithium Aluminium Hydride.

                     H2/Pd

C6H5CONH2   →     C6H5CH2NH2

                    LiAlH4

3. From benzylbromide: By the action of alcoholic ammonia on benzyl bromide.

C6H5CH2 – Br + H–NH2 → C6H5CH2NH2 + HBr

9. Explain the type of hybridisation, magnetic property and geometry for [Ni(CN)4]2– and [Ni(NH3)4]2+ using VB theory

1) [Ni(CN)4]2–

Nickel atom Outer electronic configuration 3d84s2

                                                3d                                4s                      4p

Ni+2 ion

↑↓

↑↓

↑↓

↑

↑

The ligand CN– is a powerful ligand. Hence it forces the unpaired electrons to pair up in d orbitals.

[Ni(CN)4]2–

↑↓

↑↓

↑↓

↑↓

x x

x x

x x

x x

                                                                   ↑             ↑             ↑      ↑     

                                                          CN–         CN–       CN–   CN–                             

                                                                 dsp2 hybridisation

 Hence this complex ion does not contain unpaired electrons.

Paramagnetic moment, μs = 0

The molecule is diamagnetic. 
2) [Ni(NH3)4]2+

Ni+2 ion

↑↓

↑↓

↑↓

↑

↑

[Ni(NH3)4]2+

↑↓

↑↓

↑↓

↑

↑

x x

x x

x x

x x

                                                                                  ↑              ↑     ↑    ↑

                                                                                  NH3         NH3NH3NH3  

                                                                                            sp3 hybridisation

Number of unpaired electrons = 2

                                              ∴ μs = √2(2+2)  = √8

Paramagnetic moment, μs         = 2.83BM

The molecule is paramagnetic.

Since the hybridisation is sp3, the geometry of the molecule is tetrahedral.

10. Write the differences between acetaldehyde and benzaldehyde.

S.No	Reactions	Acetaldehyde, CH3CHO	Benzaldehyde, C6H5CHO
1	Heating with Fehling’s solution	Gives a red precipitate	No reaction
2	With ammonia	forms simple addition product	forms complex condensation product.
3	With caustic soda	undergoes Aldol condensation	undergoes Cannizzaro reaction
4	With primary amines	does not forms Schiff’s base	form Schiff’s base
5	With chlorine	does not form acetyl chloride	forms benzoyl chloride
6	Polymerisation	Undergoes polymerisation	does not polymerise
7	Electrophilic substitution	does not undergo	undergoes at the meta position
8	With Schiff’s reagent.	gives pink in cold.	gives pink colour

11. How are Buna–S and Nylon–66 prepared? Give their uses.

Buna–S rubber is prepared by the polymerization of butadiene and styrene in presence of sodium metal.

Uses: Manufacture of tyres, rubber tubes and other mechanical rubber goods.
Nylon–66 is prepared by condensing adipic acid with hexamethylenediamine with the elimination of water molecule. 

Uses: Usually fabricated into sheets, bristles for brushes and in textile. Crinkled nylon fibres are used for making elastic hosiery.

12. Compound (A) with molecular formula C2H4O reduces Tollen's reagent. (A) on treatment with HCN gives compound (B). Compound (B) on hydrolysis with an acid gives compound (C) with molecular formula C3H6O3 which is an optically active compound. Compound (A) on reduction with N2H4/ C2H5ONa gives a hydrocarbon (D) of molecular formula C2H6. Identify (A), (B), (C) and (D) and explain the reactions.

Solution: Compound (A) with molecular formula C2H4O reduces Tollen's reagent.

CH3 – CHO + 2Ag+ + 3OH–   →   CH3COO– + 2Ag + 2H2O

(A) Acetaldehyde                           Acetate ion   (Silver mirror)

(A) on treatment with HCN  gives compound (B).

CH3 – CHO + HCN    →    CH3 – CH – CN

                                                       |

                                                       OH

(A) Acetaldehyde                     (B) Acetaldehyde cyanohydrin

Compound (B) on hydrolysis with an acid gives compound (C) with molecular formula C3H6O3. Compound (C) is an optically active compound.

                            H+ / H2O

CH3 – CH – CN         →         CH3 – *CH – COOH

           |                                               |

          OH                                          OH

(B) Acetaldehyde cyanohydrin              (C) Lactic acid

Compound (A) on reduction with N2H4 / C2H5ONa gives hydrocarbon (D) of molecular formula C2H6.

                  N2H4 / C2H5ONa

 CH3 – CHO        →            CH3 – CH3            (Wolff-Kishner Reduction)

(A) Acetaldehyde                       (D) Ethane

ANSWER

(A) Acetaldehyde

(B) Acetaldehyde cyanohydrin

(C) Lactic acid (Optically active)

(D) Ethane

13. Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is 5.7 x 105 m / sec. (h=6.626 x 10–34 kg m2s, mass of an electron = 9.1 x 10–31 kg)

Mass of an electron,                                     m = 9.1 x 10–31 kg

Uncertainty in the velocity of an electron,     ∆v = 5.7 x 105 m/sec

Uncertainty in the position of an electron,     ∆x = ?

Substituting these values in the equation for uncertainty principle i.e.,

∆x . m. ∆v = h / 4π

we have,           ∆x = h / 4π x ∆v.m

                         ∆x = 6.626 x 10–34 / 4 x 3.14 x 9.1 x 10–31 x 5.7 x 105

                          ∆x = 1.0 × 10–10 m

i.e., Uncertainty in position = ± 10–10 m.

14. The standard electrode potentials of the half cells Ag+ / Ag and Fe3+, Fe2+ / (Pt) are 0.7991 V and 0.771 V respectively. Calculate the equilibrium constant of the reaction : Ag(s) + Fe3+ ⇌ Ag+ + Fe2+

The cell formed is Ag / Ag+ ; Fe3+, Fe2+ / Pt

At anode :                   Ag(s)               →        Ag+ + e

At cathode :                Fe3+ + e          →        Fe2+

--------------------------------------------------------------------------

Overall reaction :         Ag(s) + Fe3+   ⇌         Fe2+ + Ag+

--------------------------------------------------------------------------

emf of the cell is given by Eocell = (ER – EL)

Eocell = 0.771 – 0.7991 = – 0.0281 V

At equilibrium,   Ecell = Eocell – RT/nF . ln a_Ag . a_Fe²⁺ / a_Fe3+

Since activity of solid silver is 1.0.

∴ n = 1 electron and Keq = a_Ag . a_Fe²⁺ / a_Fe3+

∴ Ecell = Eocell – 0.0591/ n . log Keq

At equilibrium, Cell potential, Ecell = 0

∴  Eocell = 0.0591/ n . log Keq

  log Keq = Eocell x n / 0.0591

Substitute the values n = 1 & Eocell = – 0.0281 V

∴   log Keq = 0.0281 x 1 / 0.0591

                   = 0.4751

            Keq = Antilog 0.4751

∴        Keq = 2.986

15. Write a note on (i) Clemmenson reduction  and (ii) Knoevenagel reaction

(i) Clemmenson reduction: Aldehydes and ketones can be reduced to hydrocarbons by zinc amalgam and conc.HCl.

This reaction proceeds by electron addition to carbonyl carbon followed by protonation. Zinc metal is the electron source.

In the absence of mercury, hydrogen gas will be evolved and the reduction is incomplete.

This reduction is called Clemmenson reduction.

(ii) Knoevenagel reaction: Benzaldehyde condenses with Malonic acid in presence of Pyridine forming Cinnamic acid, Pyridine is the basic catalyst here.

                                              Pyridine                              CO2

C6H5–CH = O + H2C(COOH)2 → C6H5 – CH = C(COOH)2 → C6H5 – CH = CH – COOH

                                                          Malonic acid             ∆      Cinnamic acid

16. For the complex K3[Cr(C2O4)3].3H2O mention a) Name b) Central metal ion c) Ligand d) Co–ordination number e) Geometry / Shape.

a) IUPAC names                     – Potassiumtris(oxalato)chromate(III) trihydrate

b) Central metal Ion                – Cr (III) / Cr3+

c) Ligand                                – C2O42– / oxalato

d) Co–ordination number         – 6

e) Geometry / Shape                – Octahedral

17. Explain the following isomerism with examples.

(i) cis–trans isomerism in organic compounds   (ii) optical isomerism in organic compounds

(i) cis–trans isomerism

Isomerism that arises out of difference in the spatial arrangement of atoms or groups about the doubly bonded carbon atoms is called Geometrical isomerism. These isomers are not mirror images of each other. Rotation about C = C is not possible at normal conditions and hence the isomers are isolable.

If different atoms or groups are bonded to the ‘C = C’ bond in a molecule, more than one spatial arrangement is possible. For example, 2–butene exists in two isomeric forms.

The isomer in which similar groups lie on the same side is called ‘Cis isomer’ (I). The other in which similar groups lie in opposite direction is called ‘Trans isomer’ (II). This isomerism is called ‘Cis–Trans’ isomerism.

The two groups attached to the carbon atoms need not be same, it may be different also. e.g., 2 – Pentene

 

This isomerism arises out of the hindrance to rotation about the C = C bond in such molecules.

(ii) optical isomerism in organic compounds

The compounds having same molecular formula and almost having the same physical and chemical properties but differ in the rotation of plane polarized light (light vibrating in only one plane) are called optical isomers and the phenomenon is called optical isomerism.

Or The compounds having same molecular formula but different three-dimensional arrangements of atoms around one or more chiral carbon atoms.

Optical isomers having the same magnitude but which differ only in the sign of (or direction of) optical rotation are called ‘‘enantiomers’’.

The enantiomer that rotates the plane of polarized light clockwise (to the right) as seen by an observer is dextrorotatory; the enantiomer rotating to the left [Anticlockwise] is levorotatory. The symbols (+) or ‘d’ and (–) or ‘l’ designate rotation to the right and left, respectively. Because of this optical activity, enantiomers are called optical isomers. The ‘‘racemic mixture’’ (±) (equal amounts of d-isomer and l-isomer) is optically inactive since it does not rotate the plane of polarized light.

Optical isomers have configurations which are non super imposable.

Example:

Lactic acid, Tartaric acid, Glucose, Fructose, All the a-amino acids (except glycine)

18. How are the following conversions carried out

i) Lactic acid to Pyruvic acid ii) oxalic acid to oxamide iii) Methyl acetate to ethylacetate?

i) Mild oxidising agent like Fenton’s reagent, Fe2+ / H2O2 forms Pyruvic acid with Lactic acid

                                          (O)

CH3CH(OH)COOH -------------------> CH3COCOOH

                                    H2O2/Fe2+

ii) Oxalic acid reacts with ammonia to give ammonium oxalate first which then loses water molecule to give oxamide.

iii) In presence of a little acid, Methyl acetate is cleaved by Ethyl alcohol to form Ethyl acetate. This is called ‘Trans esterification’.

                                        H+

CH3COOCH3 + C2H5OH → CH3COOC2H5 + CH3OH

19. Write a note on : (i) Carbylamine reaction (ii) Mustard oil reaction (iii) Sand Meyer Reaction

i) Carbylamine reaction :

Primary amines on heating with chloroform and alcoholic potash forms a foul smelling substance called carbylamine or alkyl isocyanide.

CH3NH2 + CHCl3 + 3KOH → CH3NC + 3KCl + 3H2O

ii) Mustard oil reaction :

When primary amines are warmed with carbondisulphide and mercuric chloride, alkyl isothiocyanate, having a pungent mustard like odour is obtained.

                                 HgCl2

CH3NH2 + S = C = S     →    CH3– N = C = S + H2S

iii) Sand Meyer Reaction :

When aqueous solution of diazonium chloride is warmed with Cu2Cl2 or Cu2Br2 in halogen acid, halogenated benzene is formed

                HCl

C6H5N2Cl → C6H5 – Cl + N2

                Cu2Cl2

                 HBr

C6H5N2Cl → C6H5Br + N2

               Cu2Br2

20. How are the following conversions carried out? i) Salicylic acid → Aspirin ii) Salicyclic acid → Methyl salicylate iii) Acetamide → Methyl amine.

i) Salicylic acid → Aspirin

Salicylic acid undergoes acetylation by heating with Acetic anhydride to form Aspirin.

 

ii) Salicyclic acid → Methyl salicylate

Salicylic acid on heating with Methyl alcohol in presence of Conc. H2SO4 Methyl salicylate is formed.

iii) Acetamide → Methyl amine.

Acetamide is treated with bromine and KOH gives Methyl amine. (Hoffman’s bromamide reaction)

                    Br2 / KOH

CH3CONH2        →         CH3NH2           +     CO2

Acetamide                      Methyl amine

21. An aromatic hydrocarbon (A) with molecular formula C7H8 undergoes oxidation in the presence of V2O5 at 773 K and gives (B) C7H6O which has a smell of bitter alnonds. Compound (B) reacts with acetic anhydride and sodium acetate to gives compound (C) C9H8O2 which is an unsaturated acid. Find A, B, C and D. Identify (A), (B) and (C). Explain the reactions. O–13

ANSWER

(A) Toluene

(B) Benzaldehyde

(C) Cinnamic acid

FIVE MARK Qs : S–12 [1- 8 ] M–13 [9 – 12] J–13 [13 – 19] S – 13 [20 –22]