solved ii puc paper 2020 chemistry code-34(ns) class–xii

SOLVEDPAPER

With Scheme of Valuation

II PUC2020

Class–XII

ChemistryCode-34(NS)

Time : [3 Hours 15 Minutes] [Total no. of questions : 37] Max. Marks : 70

Instructions : 1. The question paper has four parts. All parts are compulsory. 2. Part-A carries 10 marks. Each question carries 1 mark. Part-B carries 10 marks. Each question carries 2 marks. Part-C carries 15 marks. Each question carries 3 marks. Part-D carries 35 marks. Each question carries 5 marks. 3. Write balanced chemical equations and draw diagrams wherever necessary. 4. Use log tables and simple calculator if necessary. (Use of scientific calculator is not allowed.)

PART - A I. Answer all the questions. Each question carries 1 mark. (Answer each question in one word or in one sentence) :

(10 × 1 = 10)

1. What is the value of Van’t Hoff factor (i) for K2SO4 ?

2. 10 mL of liquid A is mixed with 10 mL of liquid B, the volume of the resultant solution is 19.9 mL. What type of deviation is expected from Raoult’s law ?

3. What is a secondary cell ?

4. Identify the order of the reaction from the rate constant K = 2.3 × 10–6 L mol–1 s–1.

5. Give reason, Zeolites are good shape-selective catalyst.

6. Iron scraps are advisable and advantageous than zinc scraps for reducing the low grade copper ores, why ?

7. Complete the reaction :

XeF6 + H2O → ? + 2HF

8. Give reason, in case of optically active alkyl halides SN1 reactions are accompanied by racemisation.

9. Identify ‘A’ in the reaction :

C==O

Zn Hg

HCl�� A + H2O.

10. Give an example for water soluble vitamin.

PART - B

II. Answer any five of the following. Each question carries 2 marks : (5 × 2 = 10)

11. Calculate the no. of particles present per unit cell in a B.C.C. unit cell. [2]

12. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of nickel is deposited at the cathode ? [molar mass of Ni = 58.7 gram mol–1]. [2]

13. Mention any two factors which influence the rate of the reaction. [2]

14. Give two reasons. The chemistry of actinoids is more complicated than Lanthanoids. [2]

15. How is phenol prepared from Aniline ? Write the equation. [2]

16. Explain Cannizzaro’s reaction taking benzaldehyde as an example. [2]

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17. (a) Give an example for non-narcotic analgesics. [1](b) Why the use of Aspartame is limited to cold foods and soft drinks ? [1]

18. (a) Why detergents with straight chain of hydrocarbons are prefered over branched chain hydrocarbons ? [1]

(b) Give one example for detergent with straight chain hydrocarbon. [1]

PART - C

III. Answer any five of the following. Each question carries 3 marks : (5 × 3 = 15)

19. Write the equations involved in leaching of alumina from bauxite ore. [3]

20. Mention any three anomalous properties of Nitrogen. [3]

21. In the manufacture of sulphuric acid write :

(i) The equation with condition for oxidation of SO2 to SO3. [2]

(ii) Formation of Oleum from SO3. [1]

22. (a) Complete the following reaction :

(i) NH3 + 3Cl2 → ? + 3HCl [1]

(ii) Cl2 + F2 473K� �� ? [1]

(b) Write the structure of perchloric acid (HClO4). [1]

23. (a) Transition elements show catalytic property. Give two reasons. [2]

(b) Name one 3d series element that do not show variable oxidation state. [1]

24. Write the equation for the manufacture of potassium dichromate from chromite ore. [3]

25. Using valence bond theory explain geometry, hybridisation and magnetic property of [CoF6]3–. [Given atomic no. of Co-27]. [3]

26. (a) Mention any two postulates of Werner’s theory of co-ordination compounds. [2]

(b) Indicate the type of Isomerism in the following set of complex compounds.

[Co(NH3)5SCN]Cl2 and [Co(NH3)5NCS]Cl2. [1]

PART - D

IV. Answer any three of the following. Each question carries 5 marks : (3 × 5 = 15)27. (a) Calculate the packing efficiency in F.C.C. cubic lattice. [3]

(b) Calcium metal crystallises in a face centered cubic lattice with edge length of 0.556 nm. Calculate the density of the metal.

[Atomic mass of calcium 40 g/mol. NA = 6.022 × 1023 atoms/mol]. [2]

28. (a) Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a non-volatile solute dissolved in 78 gram benzene. Benzene has vapour pressure of 195 mm of Hg. Calculate the molar mass of the solute. [molar mass of benzene is 78 gram mol–1.] [3]

(b) What are azeotropes ? Give an example for binary solutions showing minimum boiling azeotrope. [2]29. (a) Calculate the e.m.f. of the cell in which the following reaction takes place : Ni + 2Ag+

(0.002M) → Ni2+ + 2Ag(0.160M)(s)

Given E0cell

= 1.05 V. [3](b) (i) State Kohlrausch’s law of Independent Migration of ions. [1]

(ii) What is meant by limiting molar conductivity ? [1]30. (a) Derive an integrated rate equation for a first order reaction. [3]

(b) According to collision theory write two factors responsible for effective collisions. [2]31. (a) Write a note on Dialysis. [2]

(b) What is the effect on DH and DS during the process of adsorption ? [2](c) Give an example for heterogeneous catalysis. [1]


Oswaal Karnataka PUE Solved Paper - 2020, chemistry, II-PUC [ 3


V. Answer any four of the following. Each question carries 5 marks : (4 × 5 = 20)32. (a) Explain SN1 mechanism for the conversion of tertiary butyl bromide to tertiary butyl alcohol. [2]

(b) Complete the following reactions :(i) CH3—CH== CH2 + HI → [1]

(ii)

Cl

HNO

Conc SO3

2 4. H� ��

[1]

(iii) CH3CH2Br

AgCNAg Ethanol

� ��

[1]

33. (a) Explain the mechanism for acid catalysed dehydration of ethanol to ethene. [3](b) How does anisole react with methyl chloride ? [2]

34. (a) How is benzoyl chloride converted into benzaldehyde. Write the equation and name the equation. [2](b) Write a general equation for the formation of carboxylic acid from Grignard reagent. [2](c) Complete the reaction [1]

R—C—CH3

O

||

NaOX� ��

35. (a) Mention the I.U.P.A.C. name of (CH3)2—N—CH3. [1](b) How is Aniline prepared from nitrobenzene ? [2](c) Give the equation for the conversion of aniline to 4-Bromo aniline. [2]

36. (a) Write a chemical reactions to ellucidate :(i) Glucose contains five —OH groups.(ii) Glucose contains six carbon atoms in a straight chain. [2]

(b) Explain denaturation of proteins with example. [2](c) Name the sugar moiety present in DNA. [1]

37. (a) Name the monomers present in the following polymers :(i) PVC(ii) Neoprene(iii) Nylon-6. [3]

(b) Explain Vulcanisation of rubber. [2]

ANSWERS

PART - A I. 1. Any value between 2 to 3, including 3.

2. Negative deviation from Raoult’s lawOr

–ve deviation from Raoult’s law.3. A cell which can be recharged

Or A cell after use can be recharged by passing current through it in the opposite direction.

4. Order = 2, Or second order, Or 2nd order.5. Zeolites have honeycomb-like structures. [Scheme of Valuation, 2020]

Detailed answer Zeolites are good shape selective catalyst because of their honeycomb like structure.6. Zinc metal is costlier than iron Or Iron scraps are cheaper than zinc scraps.7. XeOF4 [Scheme of Valuation, 2020]

Detailed answer XeF6 + H2O → XeOF4 + 2HF.




8. Since the carbocation is planarOr

sp2 hybrid carbocationOr

Nucleophile can attack the carbocation from both the sides. [Scheme of Valuation, 2020]

Detailed answer

The carbocation has sp2-hybrid carbon atom with planar structure. Thus, nucleophile can attack from both the sides equally to form d and l-isomers in equal proportion.

9.

A = CH2

10. Vitamin-B or Vitamin-C (ascorbic acid)

Or any one of the below example

Vitamin B1 (Thiamine), Vitamin B2 (Riboflavin), Vitamin B3 (Niacin), Vitamin B5 (pantothenic acid), Vitamin B6 (pyridoxine), Vitamin B7 (biotin), Vitamin B9 (folic or folate), Vitamin B12 (Cyanocobalamins).

PART - B II. 11. No. of atoms present in a body centered cube = (No. of corner atoms) × (Share of each corner atom per unit

cell) + (No. of atoms at the center of the body) × (Share of body centered atom per unit cell)

=

8 ×

18

+ 1

= 1 + 1 = 2 atoms per unit cell. [Scheme of Valuation, 2020] Detailed answer

No. of particles present per unit cell in a B.C.C. unit cell = 8

18

× + 1 = 2.

(due to corner particle) (due to particle at body center)

12. For 58.7g 2F is required Therefore mass deposited by

5 × 20 × 60C = 58 7 60002 96500

. ××

= 1.84g

Or W = ZIt

W = 29 35 5 20 60

96500. × × ×

= 1.83 g

[Scheme of Valuation, 2020]

Detailed answer

Relation used :

Mass of nickel deposited (W) =

itE96500

g.

where, current (i) = 5 amperes

time (t) = 20 minutes

= 20 × 60 = 1200 s

Equivalent mass (E) =

Molar mass (+)ve chargetotal

=58 7

2.

= 29.35 g mol–1.

or, W =

itE96500

=

5 1200 29 3596500

× × .

Mass of Nickel deposited = 1.8248 g.




13. Rate of a reaction depends on various factors like

(i) Concentration or pressure of the reactant.

(ii) Nature of the reactants

(iii) Temperature

(iv) Catalyst

(v) Presence of light

Any two factors. Each factor carries one mark

14. 1. Actinoids are radioactive.

2. Their half-life period is very short.

3. They show more number of variable oxidation state.

Any two reasons. Each reason carries one mark

15. Aniline is treated with nitrous acid (NaNO2 + HCl) at 0-5°C benzene diazonium chloride is formed when diazonium salt is warmed with dil acid phenol is formed.

OH

��

H2O

warm+ N + HCl

2

NH2

��

HNO2

+ HCl

0-5°C

NH2Cl

2

(Explanation with equation or self explanary equation 2 marks) [Scheme of Valuation, 2020]

Detailed answer

Phenol from Aniline (equation) :NH

2

(aniline)

��

273-278 K

(diazotization)

N Cl2

(benzne

diazonium

chloride)

H O2

��

Warm

OH

(phenol)

16. Benzaldehyde when treated with strong NaOH solution, a mixture of sodium benzoate and benzyl alcohol is formed

��

�

Conc.Benzaldehyde Sodium

benzoate

Benzyl

alcohol

CHO + NaOH2

COONa + CH OH2

2

(Explanation or self explanary equation 2 marks)

17. (a) Paracetamol or aspirin

(b) It is unstable at cooking temperature. [Scheme of Valuation, 2020]

Detailed answer

(a) Non-narcotic analgesics are medications used to control pain and inflammation. These are not habit forming. The examples are :

(i) Analgin :

N

C N—CH3

CH—C

CH3




(ii) Diclofenac sodium :

NH

CH COONa2

Cl

Cl

(b) Aspartame is about 150 times sweeten than sucrose but it is unstable at cooking temperature, therefore it is limited to use in cold foods and soft-drinks.

18. (a) As they are biodegradable

(b) Sodiumlaurylaulphate

Sodium dodeceyl benzene sulphonate

Cetyltrimethyl ammonium bromide

Any one example Or formula of example [Scheme of Valuation, 2020]

Detailed answer

(a) Straight chain hydrocarbons are prefered over branched chain hydrocarbons because these are more biodegradable then branched chain hydrocarbons.

In branch chain hydrocarbons, tail is source of pollution.

(b) Example of straight chain : detergent with hydrocarbon :

CH CH CH OCH CH OHLauryl

3 2 10 2 2 2 8( ) ( ) .( alcohol ethoxylate)

PART - C

III. 19. Each balanced chemical equation carries one mark

Al2O3(S) + 2NaOH(aq) + 3H2O → 2Na[Al(OH)4](aq)

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

Al2O3 x.H2O(S) → Al2O3(S) + xH2O(g)

20. (i) Nitrogen is gas at room temperature.

(ii) Nitrogen is diatomic.

(iii) Nitrogen can form pp-pp multiple bonds

(iv) Nitrogen cannot from dp-pp bonds

(v) Nitrogen does not form penta halides

(vi) Nitrogen trihalide are not stable.

Any three properties. Each property carries one mark

21. (i) 2SO2(g) + O2 → 2SO3

Any one conditions :

Temperature = 720K Or Pressure = 2 bars Or Catalyst = V2O5

(ii) SO3 + H2SO4 → H2S2O7 (Oleum) [Scheme of Valuation, 2020]

Detailed answer

(i) In manufacturing of H2SO4, equation with condition for oxidation of SO2 to SO3 are as follows :

Equation : 2SO2 + O2 catalyst� ⇀��↽ �� 2SO3.

Condition :

(a) Oxidation is always done by passing warm SO2 bearing gas, through a catalyst like V, K, Na, Cs, SiO2 etc. Now-a-days V2O5 (Vanadium penta oxide) is preferred, as it is cheaper and not-poisoned by impurities.

(b) Low temperature (about 720 K).

(c) Excess of air.

(d) High pressure (about 2 atm).




(ii) Formation of Oleum from SO3 :

SO3 + H2SO4 → H S OOleum2 2 7

( )

SO3 cannot absorbed in water because it produces dense fog, while H2SO4 absorbs SO3 to give oleum.

22. (a) (i) NCl3

(ii) 2ClF

(b)

C

O

O

O

O

H

23. (a) (i) They show variable oxidation states

(ii) Large surface area

(iii) Presence of incompletely filled or partially filled d-orbitals or vacant d-orbitals

(iv) Formation of intermediate compounds.

(Any two reasons. Each reason carries one mark)

(b) Zinc (Zn) or Scandium (Sc) [Scheme of Valuation, 2020]

Detailed answer

(a) Two reasons, because transition elements show catalytic property :

(i) Presence of vacant d-orbitals.

(ii) Ability to exhibit variable valences.

(b) Zinc (Zn) is the element of 3d-series that does not show variable valency (oxidation state) due to its electronic configuration.

Electronic configuration of Zn : [Ar] 3d104s2.

24. Balanced equations. Each equation carries one mark.

1. 4FeOCr2O3 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2↑

2. 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O

Or

2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O

3. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl [Scheme of Valuation, 2020]

Detailed answer

Equation for manufacture of potassium dichromate from chromite ore :

(i) Conversion of chromite into sodium chromate :

4 8 72 3 2 3 2FeO Cr O Na CO OChromite

.( ore)

+ +

� �� 8 2 82 4 2 3 2Na CrO Fe O COSodium( chromate)

(ii) Conversion of sodium chromate into sodium dichromate :

2 2 4 2 4Na CrO H SO(Sodium chromate)

+

� �� Na Cr O Na SO H O2 2 7 2 4 2(Sodium dichromate)

(iii) Conversion of sodium dichromate into potassium dichromate :

Na2Cr2O7 + 2KCl

� �� K Cr O NaClPo

2 2 7 2( tassium dichromate)

25. In [Co F6]3, the cobalt ion is in +3 oxidation state and has the electronic configuration [Ar] 3d6 4 s0.

Orbitals Co3+ ion :

3d 4d4s 4p




Since F– ion aligand, one 4s-orbital, three 4p-orbitals and two outer 4d-orbitals, undergo sp3d2 hybridization to form six sp3d2 hybridized orbitals directed towards the six corners of an octahedron. The hybridization scheme is as shown below.

sp3d2 hybridised orbitals of Co3+ :

3d 4dsp d3 2-Hybridisation

Six pairs of electrons, one from each F– ion, occupy the six hybrid orbitals & form six coordinate bonds. [CoF4]3– (outer orbital or high spin complex) :

3d 4dSix pairs of electrons

from six F ligands–

×× ×× ×× ×× ×× ××

Thus, the complex has octahedral geometry and is paramagnetic due to the presence of four unpaired

electrons.

26. (a) 1. In coordination compounds, metal possess two types of valences. (i) The primary valence (ii) The secondary valence

2. The primary valence is ionisable wDhereas, the secondary valence is non-ionisable. 3. The primary valence is variable whereas, the secondary valence is fixed. 4. Primary valence are satisfied by negative ions whereas secondary valence are satisfied by

negative ions or neutral molecules (ligands) 5. The secondary valences are directed towards fixed position in space around the central ion. 6. The primary valence corresponds to oxidation state of central metal atom and secondary

valence corresponds to coordination number. (Any two postulates. Each postulate carries one mark.) (b) Linkage Isomerism [Scheme of Valuation, 2020]

Detailed answer (b) Type of isomerism in [Co(NH3)5 SCN]Cl2 and [Co(NH3)5NCS]Cl2 complex compounds.

These show linkage isomerism as central metal atom ‘cobalt’ is bonded through S-atom of SCN in one case while with N-atom of NCS in other case.

In other words ‘SCN’ and ‘NCS’ are ambidentate ligands. These have two donor atoms (S-atom and N-atom respectively).

PART - D IV. 27. (a) Let a be the unit cell edge length and b be the face diagonal In DABC, AC2 = BC2 + AB2, b2 = a2 + a2 = 2a2

b = 2a

If r is the radius of the sphere, we find

b = 4r = 2a or a =

42r

=

2 2r

(with or without diagram calculating a one mark)

f.c.c. structure has 4 spheres per unit cell and their volume is 443

3� �r and the volume of the cube

is a3 or ( )2 2 3r

Therefore, Packing efficiency =

Volume occupied by four spheres in the unit cellTotal volumme of the unit cell

×100

=

4 43

2 2100

3

3

��

�r

r( )%

=

163

16 2100

3

3

�r

r� % = 74%




(b) d = za

××

MNA

3

d = 4 40

0 555 10 6 022 107 2 23�

� ��( . ) ( . )

= 1.545 g/cm3 or 1.545 × 103 kg/m3.

28. (a) M2 =

W M P

P P W2 1 1

0

10

1 1

� �

� �( )

M2 = 2 78 200

200 195 78� �� ( )

M2 = 80 g/mol (b) Binary liquid mixtures having the same composition in liquid and vapour phase and boil at a

constant temperature.Or

Constant boiling point liquid mixture [Any suitable example for solution showing minimum boiling azeotrope.]

benzene and acetone, n-Hexane and ethanol, water and ethanol, acetone and CS2, CCl4 and CHCl3 CCl4 and Toluene, Acetone and ethanol. [Scheme of Valuation, 2020]

Detailed answer (a) Given,

• Vapourpressureofpurebenzene(P0) = 200 mm of Hg. • Massofnon-volatilesolute(WB) = 2 g • Massofbenzeneassolvent(WA) = 78 g • Molarmassofbenzene(MA) = 78 g mol–1. • Vapourpressureofsolution(P) =195mmofHg.

Q No. of moles (n) =

W massM Molar

( )( mass)

\ Moles of benzene (n) = WM

= =7878

1

and Moles of non-volatile solute (nB) = WM M

B

B B=

2

To find : Value of MB.also,

Q

Dpp0

=

nn

B

A (for dilute solutions)

p pp

0

0−

=

nn

B

A

\

Dpp0

=

200 195200

2

11

��

MB

5200

=

2MB

or, MB =

2 2005

80�

�

Molar mass of solute = 80 g mol–1.(b) Azeotropes : It is a mixture of two liquids which has a constant boiling point and composition through

out distillation. It is of two types: The minimum boiling azeotrope and the maximum boiling azeotrope. A solution that shows (+)ve deviation from Raoult’s law is called minimum boiling azeotrope while a

solution that shows (–)ve deviation from Raoult’s law is called maximum boiling azeotrope.




Mixture of ethanol and water, in which ethanol is approximately 95% by volume is an example of minimum boiling azeotrope.

29. (a) Ecell = E RT

FNiAgcell

010

2

22 303

��

�.

log[ ][ ]n

Ecell = 1 052 303 0 314 298

2 965000 16

0 002 2.

. .log

.( . )

��

Ecell = 0.91 V. (b) (i) The law states that limiting molar conductivity of a electrolyte can be represented as the sum

of the individual contribution of anion and cation of the electrolyte. (ii) Limiting molar conductivity is molar conductivity at infinite dilution.

Or When the concentration approaches Zero.

30. (a) Reaction in which the rate of the reaction is proportional to the first power of the concentration of the reactant, R is called first order reaction.

For example : R → P

Rate = −ddt[ ]R

= k[R]

d[ ][ ]

RR

= – kdt

Integrating the above equation, ln [R] = – kt + I ...(1) Where I is the constant of integration When t = 0, R = [R]0

Where [R]0 is the initial concentration of the reactant Therefore, equation (1) can be written as; ln [R]0 = k × 0 + I ln [R]0 = I Substituting the value of I in equation (1) ln [R] = kt + ln [R]0 ...(2) Rearranging this equation

ln

R]R][

[ 0 = – kt

k = 1 0t

lnR]R

[[ ]

k = 2 303 0.

log[[ ]tR]R

(b) According to collision theory (i) Reactant molecules most collide with sufficient kinetic energy known as activation energy or

threshold energy. (ii) Reactant molecules must collide with proper orientation.

Each factor carries one mark [Scheme of Valuation, 2020] Detailed answer

(a) First order integrated rate equation :

K =

2 30310

.log

( )ta

a x−

Where, K = rate constant a = initial concentration a – x = concentration at time (t). t = time x = concentration ion summed during time (t).




Derivation : For a reaction

A → Products

A = reactant with intial concentration a.

Then, according to first-order rate equation :

dxdt

= K . dt ...(1)

On integration :

dxa x( )��

=

K dt∫

– loge(a – x) = Kt + C ...(2)

where, ‘C’ is integration constant.

When t = 0

C = – logea.

Putting value of ‘C’ in eq. no. (2)

– loge(a – x) = Kt – logea

Or, logea – loge(a – x) = Kt

Or, K = 2 303

10.

log( )ta

a x−

31. (a) It is a process of removing a dissolved substance [electrolyte] from a collodial solution by means of diffusion through a suitable semipermeable membrane.

Example : Parchment Or cellophane

(Definition with diagram 2 marks or diagram with explanation 2 marks)

(b) DH = –ve Or DH decreases Or DH < 0

DS = –ve Or DS decreases Or DS < 0

(c) N2(g) + 3H2(g) Fe(s)� �� 2NH3 (g) Or Haber’s process

2SO2(g) + O2 Pt(s)� �� 2SO3(g) Or Contact process

Alkene(g) + H2(g) Ni(s)� �� Alkane Or hydrogenation

Oil + H2(g) Ni(s)� �� Vanaspathi Or hydrogenation

4NH3(g) + 5O2(g) Pt/Rh gauge catalyst

K, 9 bar500� ��

4NO(g) + 6H2O(g)

CO(g) + 3H2(g) Ni(s)� �� CH4(g)

CO(g) + H2(g) Cu(s)� �� HCHO(g)

Any one example for heterogeneous catalyst [Scheme of Valuation, 2020]

Detailed answer

(a) Dialysis is the treatment that does some of the things done by healthy kidneys.

Dialysis works on the principles of diffusion of solute and ultrafiltration of fluid across a semi-permeable membrane.

The process used to remove waste and extra fluid from the blood.

The filtered blood then returned to the body with the help of dialysis machine.

(b) Effect on DH and DS during process of adsorption are as follows :

(i) DH is the enthalphy of adsorption i.e. heat released during the adsorption, means it is an exothermic process. For physical adsorption, it has small –DH value (about 20-40 kJ/mol), while for chemical adsorption it has larger (–)ve DH value (about 80-240 kJ/mol.)

(ii) Entropy is the randomness. DS is entropy change during adsorption. As the molecule adheres to the surface, the residual force decreases and movement is restricted, so the value of DS also decreases.




(c) Example of Heterogeneous catalysis : The catalytic process, in which reactant and the catalyst are in different phases is knwon as Heterogeneous

catalysis. Oxidation of SO2 into SO3 in presence of Pt or V2O5 as catalyst in contact process for manufacture of

H2SO4 is an example of Heterogeneous Catalysis. Here reactant is in gaseous state while catalyst is in solid state.

2SO2(g) + O2(g)

Pt or V O(solid-catalyst)

SO2 5 2 3� �� (g)

V. 32. (a) In this step, polarised C Br bond undergoes cleavage to produce a carbocation and a bromide ion.

C

CH3

CH3

Br

H C3

C + Br+ –

CH3

CH3

step 1

H C3

In this step, carbocation is attacked by nucleophile to form the product (formation of C—OH bond).

CH3

CH3

+ OH

H C3

step II (CH COH� �� 3 3)

(b) (i) CH3—CHI—CH3

(ii)

Cl Cl

NO2

NO2

+

(any one product one mark to be awarded) (iii) CH3CH2NC [Scheme of Valuation, 2020]

Detailed answer (a) Conversion of tertiary butyl bromide to tertiary butyl alcohol using SN1 mechanism occurs as follows :

Step-1 : Formation of carbcation :

CH —C—Br3

CH3

|

|

CH3

(tertiary butyl bromide)

CH —C + Br3

+ –

��

slow step

(carbocation)

CH3

|

|

CH3

Step-2 : Fast step and formation of racemic mixture of product :

CH —C +3

+ –

OH

(Nucleophile)

CH3

|

|

CH3

CH —C—OH3

CH3

|

|

CH3

(tertiary butyl alcohol)

��

fast

This is an example of unimolecular nucleophilic substitution reaction i.e. rate ∝ [substrate](b) (i) CH3—CH== CH2 + HI → CH —CH—CH

3 3

|

I

(ii)

Cl Cl Cl

��

HNO3

conc. H SO2 4

NO2

NO2

+

(1-chloro 2-nitrobenzene)

(minor)

(1-chloro 4-nitrobenzene)

(major)

(iii) CH3.CH2.Br AgCNaq

ethylCH CH NC AgBr

.(

. . ethanol

isocyanide)� �� 3 2




33. (a) Mechanism involves three steps Step 1 : Formation of protonated alcohol : Alcohols react with sulphuric acid to form protonated alcohol.

H—C—C—O—H + H+

Ethanol

H

|

|

H

H

|

|

H

. .

. .

H—C—C—O —H+

Protonated alcohol(Ethyl oxonium ion)

H

|

|

H

H

|

|

H

H

|

. .

Fast��

Step 2 : Formation of Carbocation : The protonated alcohol loses a molecule of water to form ethyl carbocation.

H—C—C—O —H+

H

|

|

H

H

|

|

H

H

|

. .

Slow��

H—C—C + H O+

2

Carbocation

H

|

|

H

H

|

|

H

Step 3 : Formation of ethene by elimination of a proton : The carbocation loses a proton to form ethene.

Carbocation

H—C—C+

H

|

|

H

H

|

|

H

��

C==C + H+

H

H

H

H

Ethene

(b) Anisole undergoes Friedal crafts reaction with methyl chloride in presence of anhydrous aluminium chloride. Methyl groups are introduced at ortho and para positions.

OCH3

+ CH Cl3

2-Methoxytoluene

(Minor)

2-Methoxytoluene

(Major)

+

OCH3

CH3

OCH3

CH3

��

Anhyd. AlCl3

CS2

34. (a) Benzoyl chloride is hydrogenated over catalyst, palladium on barium sulphate, Benzaldehyde is formed. This reaction is called Rosenmund reduction.

Benzoyl chloride

C—Cl

O

||

Benzaldehyde

+ HCl

CHO

��

H2

Pd-BaSO4

[Explanation or equation two marks] (b) Grignard reagant reacts with dry ice (solid CO2) to form an addition product

RMgX + CO2 R—C—OMgX

O

||��

Dry ether

Addition product which on acid hydrolysis gives carboxylic acid.

R—C—OMgX

O

||

HH O

��

2 RCOOH + Mg(OH)X

(c) Complete the reaction

C

O

R CH3

NaOX��

CHX3 + RCOONa (Any one product one mark to be awarded) [Scheme of Valuation, 2020]




Detailed answer (a) Conversion of benzoyl chloride into benzaldehyde :

The reaction is called Rosenmund’s reduction. In this reaction H2(g) is passed over benzoyl chloride using Pd/BaSO4 in boiling xylene :

C H —C==O + H6 5 2

C H —C==O + HCl6 5

|

Cl

|

H

��

Pd/BaSO4

(boiling xylene)

(benzoyl chloride) (benzaldehyde)

(b) General equation for the formation of carboxylic acid from Grignard reagent is as follows :

(i)

R

C—OH + R'—MgX + Mg X��

+

O

R OH

R' O

(ii)

��

R

H O2

OH

R'

(unstable)

OH

R OH

R' O.MgX

(iii) ��

R R

unstable and thus

converted to ketone

OH

R' R'

(Ketone)

OH

C==O

(c)

R—C—CH3

R—C—C.X3

+ 3NaOH

(alkyl methyl ketone) (Trihalo-ketone)

||O

||O

��NaOX

35. (a) N, N-dimethylmethanamine (b)

NO2

NH2

��

H2

/ Pd

Ethanol

Or

NO2

NH2

��

Sn + HCl

or Fe + HCl

Self explanatory equation (c)

NH2

Aniline

��

(CH COO) O3 2

pyridine

NHCOCH3

Acetanlide

Or

4-bromoacetanlide

��

Br2

CH COOH3

NHCOCH3

Acetanlide

NHCOCH3

Br

Or

[Any two correct equations] [Scheme of Valuation, 2020]




Detailed answer

(a) IUPAC name of (CH3)2—N—CH3 is N, N-dimethyl methanamine.

(b) Aniline from nitrobenzene :NH

2

(nitrobenzene)

��

Zn-HCl or Sn-HCl

(acidic condition)

NO2

(aniline)

(c) Equation for conversion of aniline to 4-bromo aniline :

��

(CH CO) O3 2

Pyridine

NH2

(aniline)

��

H O(OH)2

��

Br2

CH COOH3

NHCOCH3

N-Phenylethanamide

(acetanilide)

NH2

Br

(4-bromo aniline)

NHCOCH3

Br

36. (a) (i) Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five OH groups.

Or

Glucose + Acetic anhydride → Penta-acetyl glucose

(ii) When glucose heated with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.

Or

Glucose + HI/red P → n-Hexane

(b) Denaturation is loss of biological activity of a protein.

Or

Secondary and tertiary structure of protein is destroyed due to heating and pH change.

Eg : The coagulation egg on heating, curdling of milk by bacteria.

(c) b-D-2 deoxyribose OR deoxyribose

37. (a) (i) Vinylchloride or chloroethene

(ii) Chloropene or 2-chloro 1, 3-butadiene

(iii) Caprolactam

(b) The process of heating a mixture of natural rubber with sulphur (and an appropriate additive), to improve the physical property of rubber is called Vulcanisation of rubber. [Scheme of Valuation, 2020]

Detailed answer

(a) (i) Monomers of PVC is vinyl chloride (CH2== CH.Cl).

(ii) Monomer of neoprene is chloroprene(CH ==CH—C==CH )

2 2

|

Cl

(iii) Monomer of nylon-6 is caprolactumH

|

N

C==O

||

CH2

CH2

(Caprolactum)

CH2

H C2

H C2

(b) Vulcanisation of rubber : Natural rubber is soft and sticky. Its tensile strength and elasticity is low. In order to give strength and elasticity, natural rubber is vulcanised to improve properties. The process is called vulcanisation of rubber.




In this process, rubber is treated with sulphur or any suitable sulphur compound to modify its properties. The process gives cross-linkage, which provides mechanical strength and proper elasticity to the rubber.

—CH —C==CH—CH —CH —C==CH—CH —···2 2 2 2

—CH —C—CH—CH2 2 2 2

—CH —C—CH—CH —

—CH —C==CH—CH —CH —C==CH—CH —···2 2 2 2

—CH2 2 2 2—C—CH—CH —CH —C—CH—CH —

CH3

|

|

CH3

|

CH3

|

CH3

|

CH3

CH3

|

CH3

|

CH3

|

(Natural rubber chains)

(valcanised rubber)

�

�

�

Sulphur (heat)

S S S S


solved ii puc paper 2020 chemistry code-34(ns) class–xii

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