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Chemistry | Class 12th

CBSE Board Paper 2019

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CBSE Board Paper 2019 Set - 1

General Instructions:

1. All questions are compulsory.

2. Section A : Q. no. 1 to 5 are very short-answer questions and carry 1 mark each.

3. Section B : Q. no. 6 to 12 are short-answer questions and carry 2 marks each.

4. Section C : Q. no. 13 to 24 are also short-answer questions and carry 3 marks each.

5. Section D : Q. no. 25 to 27 are long answer questions and carry 5 marks each.

6. There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.

7. Use log tables if necessary. Use of calculators is not allowed.

Time allowed: 3 Hours Max Marks: 70

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1. Out of NaCl and AgCl, Which one shows Frenkel defect and why?

Section A

1

Section B

6. Write balanced chemical equations for the following processes:

(i) XeF2 undergoes hydrolysis.

(ii) MnO2 is heated with conc. HCl.

2

2. Why are medicines more effective in the colloidal state?

1

3. Arrange the following in increasing order of base strength in the gas phase:

(C2H5)3N, C2H5NH2, (C2H5)2NH

1

4. Define ambidient nucleophiles with an example?

1

5. What is the basic structural difference between glucose and fructose?

1

OR

Write the products obtained after hydrolysis of lactose.

OR

What is the difference between an emulsion and a gel?

4

7. State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations.

2

2

9. When MnO2 is fused with KOH in the presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionate in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidises KI to compound (C) whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C), and (D).

2

Arrange the following in order of property indicated for each set:

(i) H2O, H2S, H2Se, H2Te – increasing acidic character

(ii) HF, HCl, HBr, HI – decreasing bond enthalpy

OR

8. For a reaction

The proposed mechanism is as given below :

(1) H2O2 + I-→ H2O + IO- (slow)

(2) H2O2 + IO-→ H2O + I- + O2 (fast)

(i) Write rate law for the reaction.

(ii) Write the overall order of reaction.

(iii) Out of steps (1) and (2), which one is rate determining step?

5

10. Write IUPAC name of the complex [Cr(NH3)4Cl2]+. Draw structures of geometricalisomers for this complex.

Tetraaminedichlorochromium(iii)

2

11. Out of [CoF6]3- and [Co(C2O4)3]3-, which one complex is

(i) diamagnetic

(ii) more stable

(iii) outer orbital complex and

(iv) low spin complex?

(Atomic no. of Co = 27)

2

12. Write structures of compounds A and B in each of the following reactions:

2

OR

Using IUPAC norms write the formulae for the following :

(i) Pentaamminenitrito-O-cobalt(III) chloride structure

(ii) Potassium tetracyanidonickelate(II)

6

13. The decomposition of NH3 on platinum surface is zero order reaction. If rate constant (k) is 4 x 10-3 Ms-1, how long will it take to reduce the initial concentration of NH3 from 0.1 M to 0.064 M.

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14. (i) What is the role of activated charcoal in gas mask?

(ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

(iii) How does chemisorption vary with temperature?

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Section C

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15. An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm-3. Calculate the number of atoms in 108 g of the element.

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16. A 4% solution(w/w) of sucrose (M = 342 g mol-1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water. (Given: Freezing point of pure water = 273.15 K).

3

17. (a) Name the method of refining which is

(i) used to obtain semiconductor of high purity,

(ii) used to obtain low boiling metal.

(b) Write chemical reactions taking place in the extraction of copper from Cu2S.

3

18. Give reasons for the following:

(i) Transition elements and their compounds act as catalysts.

(ii) E° value for (Mn2+|Mn) is negative, whereas for (Cu2+|Cu) is positive.

(iii) Actinoids show irregularities in their electronic configuration.

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19. Write the structures of monomers used for getting the following polymers:

(i) Nylon-6,6 (ii) Glyptal (iii) Buna-S

3

8

20. (i) What type of drug is used in sleeping pills?

(ii) What type of detergents are used in toothpastes?

(iii) Why the use of alitame as artificial sweetener is not recommended?

3

OR

Define the following terms with a suitable example in each:

(i) Broad-spectrum antibiotics

(ii) Disinfectants

(iii) Cationic detergents

OR

(i)

(ii) Write the monomers of the following polymer :

(iii) What is the role of Sulphur in the vulcanization of rubber?

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21. (i) Out of (CH3)3C–Br and (CH3)3C–I, which one is more reactive towards SN1 and why?

(ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K, followed by acidification.

(iii) Why dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation?

3

22. An aromatic compound ‘A’ on heating with Br2 and KOH forms a compound ‘B’ of molecular formula C6H7N which on reacting with CHCl3 and alcoholic KOH produces a foul smelling compound ‘C’. Write the structures and IUPAC names of compounds A, B and C.

3

23. Complete the following reactions:

3

OR

OR

Write chemical equations for the following reactions:

(i) Propanone is treated with dilute Ba(OH)2.

(ii) Acetophenone is treated with Zn(Hg)/Conc. HCl

(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO4.

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Section D

24. Differentiate between the following:

(i) Amylose and Amylopectin

(ii) Peptide linkage and Glycosidic linkage

(iii) Fibrous proteins and Globular proteins

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25. E° cell for the given redox reaction is 2.71

Mg(s) + Cu2+ (0.01 M) ––– Mg2+ (0.001 M) + Cu(s)

Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is

(i) less than 2.71 V and (ii) greater than 2.71 V

5

OR

(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 until 2.8 g of

Fe deposited at the cathode of cell X. How long did the current flow ? Calculate the mass of Zn deposited at the cathode of cell Y.

(Molar mass : Fe = 56 g mol-1 Zn = 65.3 g mol-1, 1F = 96500 C mol-1)

(b) In the plot of molar conductivity (^m) vs square root of concentration (c1/2), following curves are obtained for two electrolytes A and B :

OR

(i) Straight chain (ii) Five alcohol groups

(iii) Aldehyde as carbonyl group

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26. (a) How do you convert the following :

(i) Phenol to Anisole

(ii) Ethanol to Propan-2-ol

(b) Write mechanism of the following reaction :

(c) Why phenol undergoes electrophilic substitution more easily than benzene ?

5

OR

(a) Account for the following :

(i) o-nitrophenol is more steam volatile than p-nitrophenol.

(ii) t-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of t-butylmethylether.

(b) Write the reaction involved in the following :

(i) Reimer-Tiemann reaction

(ii) Friedal-Crafts Alkylation of Phenol

(c) Give simple chemical test to distinguish between Ethanol and Phenol.

Answer the following :

(i) Predict the nature of electrolytes A and B.

(ii) What happens on extrapolation of ̂ m to concentration approaching zero for electrolytes A and B?

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27. (a) Give reasons for the following :

(i) Sulphur in vapour state shows paramagnetic behaviour.

(ii) N-N bond is weaker than P-P bond.

(iii) Ozone is thermodynamically less stable than oxygen.

(b) Write the name of gas released when Cu is added to

(i) dilute HNO3 and

(ii) conc. HNO3

5

OR

(a) (i) Write the disproportionation reaction of H3PO3.

(ii) Draw the structure of XeF4.

(b) Account for the following :

(i) Although Fluorine has less negative electron gain enthalpy yet F2 is strong oxidizing agent.

(ii) Acidic character decreases from N2O3 to Bi2O3 in group 15.

(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved.

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3. Why conductivity of silicon increases on doping with phosphorus?

1

5. Write IUPAC name of the given compound:

1

8. Write two differences between an ideal solution and a non-ideal solution.

1

18. Give reasons for the following:

(i) Transition metals form alloys.

(ii) Mn2O3 is basic, whereas Mn2O7 is acidic.

(iii) Eu2+ is a strong reducing agent.

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20. (i) Why bithional is added in soap ?

(ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate ?

(iii) Why soaps are biodegradable whereas detergents are non-biodegradable ?

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Set - 2

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Define the following terms with a suitable example in each:

(i) Antibiotics

(ii) Artificial sweeteners

(iii) Analgesics

OR

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2. What type of Stoichiometric defect is shown by ZnS and why?

1

3. Write one stereochemical difference between SN1 and SN2?

1

7. State Henry’s Law and write its two application?

3 24. Define the following with a suitable example in each

(i) Oligosaccharides

(ii) Denaturation of Protein

(iii) Vitamins

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Set - 3

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1. Frenkel defect is shown by those in which there is a large difference in the size of cations and anions. Here, due to the small size of Ag+ and large size of Cl-, AgCl will show Frenkel defect whereas NaCl will not show Frenkel defect.

2. Medicines are more effective when these are provided in the colloidal state because colloids have the large surface area, these get easily absorbed by our body tissues. Antibiotics such as penicillin, streptomycin etc are usually injected in our body in colloidal form.

Emulsion- The colloidal state in which both the dispersed phase and dispersion medium are in the liquid state. But in the emulsion, both the liquids aren’t miscible with each other. Eg. Milk is an emulsion of water and fat.

Gel- Gel is a colloid of liquid in solid that forms a semisolid material. Here liquid is in the dispersed phase and solid is dispersion medium. Example: Edible gelly etc.

3. (C2H5)3N > (C2H5)2NH > C2H5NH2

According to the Lewis concept, the groups that donate electron density behave as the base. Alkylamines are basic in

Solutions (Set-1)

Section A (Solutions)

OR

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nature because there is a Nitrogen atom having a pair of electrons so that it can donate their electron density. As +I effect increases, the basic property of alkylamine increases, that's why in gaseous phase order of basic property of 1°, 2° and 3° ethylamine is as follows-

(C2H5)3N > (C2H5)2NH > C2H5NH2 {Gaseous phase}

N(C2H5)3 (due to inductive effect, here 3 ethyl groups are donating electrons to Nitrogen atom)

NH(C2H5)2 (due to inductive effect, here 2 ethyl groups are donating electrons to Nitrogen atom)

C2H5NH2 (Here in this case only 1 ethyl group is donating electrons to Nitrogen Atom)

4. Ambidient nucleophiles are those which contains more than one nucleophilic centre. For Exa: CN-, NO2

- etc.

Nucleophilic Centre is the atom that is rich in electrons and attacks an electrophile.

5. Both Glucose and Fructose are monosaccharides and hexose sugars (contain six carbons atoms). But structurally they are different as follows-

Glucose

1) It is aldohexose. Its functional group is the aldehyde.

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2) It forms a pyranose ring structure and forms a six-membered ring.

Fructose

1) It is ketohexose, and its functional group is the ketone.

2) It forms a furanose ring structure and forms a five-membered ring

Lactose is a disaccharide carbohydrate and therefore on hydrolysis it breaks down in constituent monosaccharides β-D-Glucose and β-D-Galactose.

OR

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6. (i) Here is the balanced chemical equation when XeF2 undergoes hydrolysis,

Xenon difluoride reacts with water to form Xenon, Hydrogen fluoride and oxygen.

(ii) Here is the balanced chemical equation when MnO2 is heated with conc. HCl

Chlorine Gas(Cl), Manganese(II) Chloride and(MnCl2) and Water(H2O) is obtained when Manganese(IV) Oxide reacts with conc. HCl.

(i) Increasing acidic character is

In binary acids, as the size of the element increases the bond becomes weaker and hence become a better acid due to easy release of H+ ion.

(II) Decreasing bond enthalpy is

Because as the size of halogen decreases the effective nuclear charge increases and hence, the bond enthalpy increases.

7. Raoult’s law for volatile solutes states that in a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution

Section B (Solutions)

OR

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multiplied by the vapour pressure of that component in the pure state.

Mathematically:

The vapour pressure P of a solution containing two components A and B is

PA = PoA × XA

PB = P0B × XB

P = PA + PB = PoA × XA + P0

B × XB

where, is the vapour pressure of component A

is the vapour pressure of component B

is the vapour pressure of component A in its pure state

is the vapour pressure of component B in its pure state

is the mole fraction of component A

is the mole fraction of component B

Now we know that,

XA + XB = 1

XA = 1- XB

P = (1-XB)P0A + XBP0

B

P = P0A + XB(PB

0 – P0A)

For a solution to be able to obey Raoult’s law at all concentrations, it should be an ideal solution. For a solution to be called ideal, it should have the following two characteristics:

•The enthalpy of mixing of the pure components to form the solution should be zero, that is, . It means that no heat is absorbed or released during mixing.

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•The volume of mixing the pure components should be zero, that is, . This means that the volume of the solution would be equal to the sum of volumes of the two components.

8. (i) As we know, the rate of a complex reaction depends on the slowest step of the reaction mechanism. Therefore the rate law for the reaction will be as follows

Rate= k [H2O2][I-]

(ii) The overall order of the reaction is 1+1 =2

Explanation: Order of the reaction is the sum of powers of reactants in the rate law of any reaction.

(iii) Step 1 is the rate determining step as it is the slowest step and the overall rate of a reaction depends on the slowest step.

9. Potassium permanganate is prepared by reacting MnO2 with an alkali metal hydroxide in the presence of an oxidising agent. The reaction takes place via the following steps-

K2MnO4 (A) is dark green precipitate which undergoes disproportionation reaction to form permanganate and manganese oxide

In acidic solution-

In a neutral medium-

(A) MnO2- (B) MnO4- (C) MnO2 (D) Mn2+

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10. IUPAC name of the complex [Cr(NH3)4Cl2]+ is Tetraaminedichlorochromium(iii)

Two geometrical isomers exist for this complex, these are trans-isomer and cis-isomer

(i) [Co(NH3)5ONO]Cl2,

Here, Cobalt will act as central atom whereas other ions NH3, ONO will act as ligand.

(ii) K2[Ni(CN)4],

Here, Ni will act as central atom whereas CN- ion will act as ligand.

11. Oxidation state of Co in both the complexes is +3

[CoF6]3-

F- is a weak field ligand, the cation Co3+ does not allow to pair electron in its excited state hence there is no pairing takes place, so it is paramagnetic , high spin complex, outer orbital complex with sp3d2hybridisation.

[Co(C2O4)3]3-As the oxalate is have -2 charge and is a strong field ligan, it allows to pair electron of Co in its exited state hence there is pairing of electrons, so it is diamagnetic , low spin complex, inner orbital complex with d2sp3 hybridisation.

OR

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Therefore, from above information, we can conclude

(i) Diamagnetic complex= [Co(C2O4)3]3-

(ii) More stable complex= [Co(C2O4)3]3-

(iii) Outer orbital complex= [CoF6]3-

(iv) Low spin complex= [Co(C2O4)3]3-

12. (i)The steps involve oxidation of ethyl benzene in presence of oxidising agents

So, Product A is formed by the oxidation of Ethyl Benzene and after hydrolysis of Product A, Benzoic Acid is formed, which is Product B.

(ii) The steps involve oxidation of ethyl benzene in presence of oxidising agents and

Product A is formed by the ozonolysis of Phenol and Product B is the process of addition of NNHCONH2 and removal of OOxygen atom.

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13. For a zero order reaction,

Kt = A0 - A

(4 × 10-3 × t) = 0.1 – (0.0644)

t = 0.009 x 103 sec

= 9 sec.

14. (i) What is the role of activated charcoal in gas mask?

(ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

(iii) How does chemisorption vary with temperature?

Section C (Solutions)

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(i) Activated charcoal is basically used in gas masks to prevent coal miners from suffocation from ashes and poisonous gases. The toxic gases get absorbed over the surface of activated charcoal and bind with weak forces like van-der-waal forces. The larger and heavy particles are unable to enter through the mask and the person doesn’t inhale poisonous gases and ash.

(ii) Colloidal particles adsorb charges or ions which are in excess and common to the colloidal particle. Development of charge on sol particle is due to the formulation of an electrical double layer. When FeCl3 is added in NaOH, then due to excess of OH- ions n the dispersion medium, the sol attains overall negative charge. Sol representation will be Fe2O3xH2O/OH-.

(iii) Chemisorption process involves formation of chemical bonds within adsorbate and adsorbent and therefore depends on temperature conditions. With the increase in temperature, chemisorption increases up to a certain limit and then starts decreasing.

15. Given- fcc lattice, density = 10.8 g cm-3 = β, edge length = 300pm = 300 × 10-12 m = 300 x 10-10cm = a

Find- no. Of atoms in 108g of element

Using the formula for density,

Being fcc lattice Z = 4

Putting all the values on equation we get, 10.8 =

M= 175.6 g

In 175.6 g, 6.022 X 1023 atoms are present

Therefore, in 108g, no. Of atoms will be = 0.61 x 1023 atoms.

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16. 4% solution (w/w) of sucrose means 4g of sucrose in 96g water

Wsolvent = 96g, wsolute= 4g, Msolute= 342 g mol-1

Using the formula, ∆Tf = kf x m

(273.15 – 271.15) =

Kf = 16.39 K

For 5% solution (w/w) of sucrose means 5g of sucrose in 95g water

Wsolvent = 95g, wsolute= 5g, Msolute= 180 g mol-1

Using the formula, ∆Tf = kf x m

∆Tf = K

= 0.48 K

Tf = Tf0 - ∆Tf = (273.15 – 0.48) K = 272.67 K

17. (a)

The process of purifying the crude metals is called refining.

The refining method chosen depends on the physical and chemical properties of a particular metal.

i. The refining method used to obtain semi-conductors such as germanium, silicon and gallium of high purity is Zone Refining. It was invented by William Pfann, and this method is based on the fact that impurities are more soluble than the pure metal in the melt.

ii. Distillation process is used to obtain metals which have low boiling point like zinc and mercury. In this method, the impure metal is heated to its boiling point in a vessel. The vapours of metal thus formed are collected and cooled in a separate

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vessel to get pure metal. The impurities being non-volatile remain behind.

(b) The steps involved in extraction of Copper from Cu2S are:

•Roasting of copper (I) sulphide

2Cu2S(s) + 3O2(g) → 2Cu2O(s) + 2SO2(g)

•Reduction of copper (I) oxide with copper (I) sulphide

2Cu2O(s) + Cu2S(s) → 6CuS(s) + SO2(g)

The copper obtained here is very impure. Hence it undergoes electrolytic refining. The diagram below shows the process.

•At the cathode, copper (II) ions are deposited as copper.

Cu2+(aq) + 2e-→ Cu(s)

•At the anode, copper goes into solution as copper (II) ions.

Cu(s) → Cu2+(aq) + 2e-

18. i. Transition elements belong to the d-block of the periodic table of the elements. For a d-block element to be a transition element, it must possess an incompletely filled d-orbital.

The catalytic nature of the transition elements is because of two basic facts:

a. These elements are able to exhibit variable oxidation states. For example, copper exists as .

Due to this feature, the transition elements and their compounds are able to form complexes and hence act as good catalyst.

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b. Transition elements also provide a suitable surface for the reactions to occur.

ii. is known as electrode potential and it is measured in volts. It is a measure of total energy change in a cell.

for any metal is related to the sum of enthalpy changes taking place as follows:

M(s) + ΔaH → M(g)

M(g) + ΔiH → M2+(g)

M2+(g) → M2+(aq) + ΔhydH

where,

ΔaH is enthalpy of atomisation

ΔiH is enthalpy of ionisation

ΔhydH is enthalpy of hydration

In the case of copper, it has a high enthalpy of atomisation and a low enthalpy of hydration. A high energy is required to transform and this energy is not balanced by its hydration enthalpy, that is Cu has a low hydration enthalpy. Hence, E0

Cu2+

/Cu is positive.

E0Mn

2+/Mn on the other hand is observed to be highly negative

because it difficult to reduce Mn2+ to Mn due to stability of half-filled d-subshell (3d5) in Mn2+.

iii. Actinoids are group of 15 elements with atomic number 89 to 113. These species show irregularities in their electronic configuration because actinoids involve gradual filling of 5f-subshell in their atoms. The energy differences between 5f, 6d and 7s subshells are very small, almost comparable. Therefore, an electron can go and occupy any of these subshells as they are close to each other.

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19. (i) Nylon 6,6 is a polyamide, a polymer derived from the condensation reaction of monomers containing terminal Carboxylic acid(-COOH) and amine (-NH2) groups.

(ii) Glyptal is formed using phthalic anhydride and glycerol monomer. The addition of this hydroxyl group allows for extensive branching during polymerization.

(iii) Buna-S has monomers as styrene and 1,3-butadiene. It is a synthetic rubber formed by co-polymerisation of these two monomers.

(i) This is a homopolymer formed by a propene monomer because in this polymerisation only one monomer is being used without losing any molecule.

(ii) Monomers of melamine formaldehyde is melamine and formaldehyde

OR

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(iii) Sulfur is used in vulcanisation to produce cross-links between two polymer chains to produce more tensile strength, elasticity and resistance to abrasion.

20. i. Sleeping pills are taken by people who have anxiety problems and find it difficult to get sleep. Tranquilizer is the classification of the drug used in sleeping pills. Tranquilizers help to relieve stress and decrease fatigue in the body. They act on higher centres of the nervous system. Most of these drugs are derivatives of barbituric acid. Examples: Luminal, Equanil, iproniazid, barbiturate.

ii. Anionic detergents are used in toothpaste. These are sodium salts of sulphonated long chain alcohols or hydrocarbons. In anionic detergents, the anionic part of the molecule is involved in the cleansing action. For example: alkyl benzene sulphonates are obtained by neutralising alkyl benzene sulphonic acids with alkali.

iii. Artificial sweeteners are chemicals that are used in the food to give a sweet taste. However, unlike natural sweetener, these neither add calories to our body nor cause harm to it. Some examples are saccharin, alitame and aspartame which are generally used to make sweet food for diabetic patients.

However, alitame is not recommendable as its sweetness is difficult to control. A small concentration of it can cause too

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much sweetness. It is said to be a high potent artificial sweetener.

i. Antibiotics that are effective against a wide range of gram-positive and gram-negative bacteria are known as broad-spectrum antibiotics. Chloramphenicol, below image, is an example of a broad-spectrum antibiotic.

Chloramphenicol can be used in the case of typhoid, dysentery, acute fever, urinary infections, pneumonia and meningitis.

ii. Disinfectants are antimicrobial agents that are applied to the surface of non-living objects to destroy microorganisms that are living on the objects. However, they are not safe to be applied to living tissues. They are generally used to kill microorganisms present in drain toilets, floors etc.

For example: 0.2% solution of phenol act as a disinfectant. KMnO4, Cl2 are also used as disinfectant.

iii. Cationic detergents are quaternary ammonium salts of acetates with amines chlorides or bromides as anions. They are cationic detergents because the cationic part of these detergents consists of a long hydrocarbon chain and a positive charge on the nitrogen atom.

For example, Cetyltrimethyl ammonium bromide is a popular cationic detergent used in hair-conditioners.

OR

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21. Explanation: In SN1 reactions, reactivity depends on the stability of carbocation. Carbocation is same here, therefore, we will see the reactivity of leaving group. I- is a better leaving group because of being good nucleophile. Hence Sn1 reaction is faster in (CH3)3-I

(ii) When p-nitro-chloro benzene is heated with aq. NaOH at 443 K,then NO2 will replace Cl as a substitute. Then, after acidification due OH will replace NO2 as a substitute.

(iii) dextro and laevo rotatory isomers of butan-2-ol are difficult to separate by fractional distillation because of no difference in boiling points of these two rotatory isomers.

22. •The reagents Br2 with KOH (strong base) is used in the Hoffman reaction. This reaction is used to convert amides to amines with the loss of carbonyl carbon atom.

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•The given molecular formula of ‘B’ is C6H7N, and as per Hoffman reaction, this should be an amine.

•Also, it is mentioned that the starting product, ‘A’ is aromatic, therefore ‘A’ is an aromatic amide which loses carbonyl carbon atom upon reaction with Br2 and KOH.

•Hence, for the molecular formula of ‘A’, we should add -CO to the molecular formula of ‘B’.

C7H7NO is the molecular formula of aromatic amide ‘A’.

The only aromatic amide having this molecular formula is benzamide.

Structure of ‘A’

Hoffman Reaction occurs and ‘B’ is formed.

•Aniline reacts with CHCl3 and KOH, by carbylamine reaction to give a foul smelling compound ‘C’, benzene isonitrile.

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23. (i) When Benzaldehyde reacts with NaCN/HCl in basic medium i.e. 9-10, CN- as a nucleophile will attack on aldehyde group of benzene, and the obtained product is as below

(ii) When Acetyl Chloride reacts with (C6H5-CH2)2Cd then C6H5-CH2

- as a nucleophile will attack on C which is connected with Oxygen group and will replace Cl atom.

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(iii) Below reaction is also known as Hell Volhard Zelinsky reaction, in which Br will add on tertiary carbon group because of formation of carbocation at the central atom C.

24. (i) Difference between Amylose and Amylopecting is as follows:

(ii) The difference between Peptide linkage and Glycosidic Linkage are as follows

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(iii) The difference between Fibrous Proteins and Globular Proteins are as follows

(i) Heating with HI, glucose forms n-hexane by reducing the all -OH and -CHO groups, suggesting the the straight chain of glucose molecule.

OR

37

(ii) On reaction of glucose molecule with acetyl anhydride, glucose pentaacetate is formed, which confirms the presence of five free -OH substituents on glucose molecule.

(iii) On reaction of glucose with Bromine water, which is a good oxidising agent, the -CHO group gets oxidised to carboxylic acid that is gluconic acid. This indicates the presence of aldehydic functional group in glucose molecule.

25. Mg(s) + Cu2+ (0.01 M) ––– Mg2+ (0.001 M) + Cu(s)

(i) less than 2.71V, current will flow from copper to magnesium

(ii) more than 2.71V, curret will from magnesium to copper

a) (i)

2.8g of Fe requires = (2) (96500)(2.8)/(56)

Section D (Solutions)

OR

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= 9650 Coulombs

Q= It = (9650)(2) = 4825 sec

(ii)

(2) (96500) C of electricity deposit Zn = 65.3g

Therefore, 96500 C of electricity deposit Zn = (65.3) (96500)(9650) = 3.265g

26. (a) (i) Phenol to Anisole

Through williamson synthesis phenols can be converted to anisole

(ii) Ethanol to Propan-2-ol

By the use og Grignard reagent ketones can be reduced to sec- alcohols

(b) Mechanism of ethanol to ethene

The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:

Step 1:

Protonation of ethanol to form ethyl oxonium ion:

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Step 2:

Formation of carbocation (rate determining step):

Step 3:

Elimination of a proton to form ethene:

The acid consumed in step 1 is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

(c) The increased reactivity of phenols towards electrophilic substitution reactions compared to benzene is due to the lone pair of p-orbital electrons on the oxygen atom in the phenol group. The electron pair is delocalised into the ring structure, activating the ring. This increases the electron density of the molecule and hence influence electrophiles to attack over the ring. Therefore, electrophilic substitution can then take place.

(a) (i) There is intramolecular hydrogen bonding present in o-nitrophenol which is much weaker than intermolecular hydrogen bonding in p-nitrophenol. Therefore, o-nitrophenol is steam volatile due to absence of intermolecular attractions.

and

OR

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(ii) if the alkyl halide is primary than ether is formed through SN2 reaction mechanism.

But if the alkyl haide is tertiary then alkene is formed instead of ether E1 elimination mechanism

(b) (i) The Reimer–Tiemann reaction is a chemical reaction the conversion of phenol to salicylaldehyde.

(ii) Friedel Crafts Alkylation of Phenols

(c) Chemical test to distinguish between phenol and ethanol

Phenols are unsaturated aromatic alcohols while ethanol is saturated aliphatic alcohol.

1. Phenol has benzene ring, it undergoes diazo coupling reaction, while ethanol doesn’t give this reaction.

2. Phenol decolourise bromine solution and therefore proves the unsaturation in phenol, ethanol doesn’t give this test.

3. Ethanol give iodoform test and forms the yellow precipitate, phenol doesn’t give this test.

27. (a) (i) The vapour state of Sulphur exists as S2 molecule and has two unpaired electrons in its antibonding π orbitals. Due

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to presence of these two unpaired electrons in vapour state of Sulphur, it shows paramagnetism.

(ii) N-N sigma bond is weaker than P-P sigma bond due to the small bond length between the nitrogen atoms. The non-bonding electrons (lone pair of electrons) of both the atoms repel each other making it weaker than P-P sigma bond. This is why the catenation tendency is not present in Nitrogen.

(iii) The ozone is thermodynamically less stable than oxygen because it conversion to oxygen results in ΔH= -ve and ΔS= +ve which overall results in to overall ΔG= -ve according to the equation ΔG= ΔH-TΔS and hence spontaneously ozone converts back to oxygen.

(b) (i) Nitrogen monoxide is released when Copper is added to dil HNO3

(ii) Nitrogen dioxide is released when Copper is added to conc. HNO3

(a) (i) When H3PO3 is heated then H3PO3 is heated H3PO4 and PH3 is obtained, in which P undergoes in both oxidation as well as reduction reaction.

4H3PO3→ 3H3PO4 + PH3

(ii) XeF4- structure of XeF4 is square planar having 2 lone pairs and 4 bond pairs

OR

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(b) (i) The oxidising capability of a substance depends upon three factors: bond dissociation enthalpy, electron gain enthalpy and hydration energy.

Although electron gain enthalpy of fluorine is less negative, but due the small size of F its hydration energy is high. Also, bond dissociation enthalpy of F2 molecule is low due to repulsion between lone pairs of two F-atoms which are quite closely place in the molecule. Therefore, F2 is strong oxidizing agent.

(ii) Acidic character of oxides of group 15 decreases and basicity increases down the group. Down the group, atomic size increases, electronegativity decreases and metallic character increases. Oxides of N and P are acidic, oxides of As and Sb are amphoteric and oxides of Bi are basic.

(c) Sulfur dioxide gas turns acidified potassium dichromate(VI) solution from orange to green. Chromium is reduced from (VI) to (III) K2Cr2O7(aq) + 3SO2(g) + H2SO4(aq)→Cr2(SO4)3(aq) + K2SO4(aq) + H2O(l).

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3. Silicon has four valence electrons. When a silicon atom is replaced with phosphorus with 5 valence electrons atom, it can share one electron with each of the four silicon atoms surrounding it. This leaves the fifth valence electron of phosphorus free to move. Hence doping of silicon with phosphorus atom induces conductivity in silicon. Such type of doping is called n-type doping.

5. IUPAC name of the compound is 4-chlorobenzenesulphonic acid.

Explanation: Chlorine is considered to be at the 4th position on benzene ring. With respect to SO3H, Cl is at para position. Therefore, Cl will be taken at 4 position.

8. Ideal solution-

1) The homogenous solutions of liquids which obey Raoult’s Law at entire range of concentration and temperatures are ideal Solutions.

2) The Δmix H = 0 and Δmix V = 0, which means for ideal solutions there is no release or absorption of heat and no change in volume while mixing at constant pressure.

Non- Ideal solutions-

1) The homogenous solutions of liquids which don’t obey Raoult’s Law at entire range of concentration and temperatures are called non- ideal Solutions.

Solutions (Set-2)

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2) The Δmix H≠0 and Δmix V ≠0, which means for non-ideal solutions there is some release or absorption of heat and some change in volume while mixing at constant pressure.

18. (i) Explanation: Alloys are solid solutions. Solids in alloys get intermixed with each other such that individually these can’t be recognised. Since the transition metals are similar in sizes therefore get easily settled in the crystal lattice of another metal, hence transition metal form alloys. Example of some alloys are- Brass, Bronze etc.

(ii) Explanation: Mn2O3 is basic because the oxidation state of Mn is +3 and exist in d4 electronic configuration, Mn can still lose more electron to achieve +7 oxidation state of d0 state. Hence it is basic in nature. On the other side, Mn2O7 is acidic because it already exist in its highest possible oxidation state and has no more electrons left to lose. Therefore Mn2O7 is acidic.

(iii) Explanation: Europium (Eu) can exist in +2, +3 and +4 oxidation states. +3 oxidation state of Eu is very stable therefore Eu get readily oxidised from Eu+2 to Eu+3 state by losing one more electron and hence it is a strong reducing agent.

20. (i) To keep soap protected from bacteria, bithional is added to soap. Bithional is an antibacterial agent and can fight against bacteria.

(ii) The pH of sodium bicarbonate is near 8.25, which is quite alkaline. If NaHCO3 is used then being soluble, it makes the stomach alkaline and triggers even more production of acid. On the other hand, Mg(OH)2 is insoluble, and therefore, the pH

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remain neutral. Hence, Mg(OH)2 is a better antacid then NaHCO3.

(iii) Soaps are sodium or potassium salts of fatty acids. Fats and oils can easily be broken down into simple molecules by microbes, and hence soaps are biodegradable. Detergents are synthetic compounds, generally ammonium or sulfonate salts of long chain carboxylic acids. These synthetic compounds cannot be broken down into simple molecules by microbes, and hence detergents are non-biodegradable.

(i) Antibiotics are drugs to treat bacterial infections. These inhibit the growth of bacteria by blocking the biochemical pathways essential for their survival like protein synthesis. Eg. Tetracyclin, Penicillin.

(ii) Artificial sweetners are the synthetically prepared sweetners that are comparable in its sweetness with sugar. These are added to food items because these provides sweetness to food without adding much calories like sugar. Eg. Aspartame, Saccharin.

(iii) Analgesics are the medicines that has tendency to reduce pain without the loss of consciousness. Analgesics are commonly called pain-killers. Eg. Paracetamol.

OR

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,

2. Frenkel defect is shown by ZnS because there is a large difference in between the size of Zn2+ and S2- ion. Frenkel defect is a type of point defect. Here, Zn2+ ion which is very small in size compare to S2-ion.

3. Although both SN1 and SN2 are nucleophilic substitution reaction. But, the differences between these two are as follows

SN1: It is a two step mechanism. Also, the rate determining step is unimolecular.

whereas

SN2: It is one step mechanism. Also, the rate determining step is bimolecular.

7. Henry’s law states that “The Solubility of Gas is proportional to its partial pressure”. It can be written mathematically as follows

P=Kh*X

Here the importance of the Henry’s constant is as follows

(1) Increase solubility of Co2 in soft drinks.

(2) During Scuba Diving

(3) Environmental Chemistry

Solutions (Set-3)

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24. (i) An Oligosaccharide is a saccharide Polymer contain simple sugar as its monomer. For Example, Glicopilids, maltodextrins.

(ii) When a solution of protein is boiled, Protein loses their quartenary structure, tertiary structure and secondary structure from their native state. Denaturation of Protein is done by external stress(by applying heat, acid or alkali).

(iii) A vitamin is an organic molecule that is essential micronutrient that an organism needs in small quantities for the proper functioning of its metabolism. There are different types of vitamins like Vitamin A, Vitamin B, Vitamic C etc. The sources of these vitamins are quinones, Milk and other drugs etc.

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