chemistry complete syllabus of class xii · 2019. 7. 14. · model test paper-1 subjective test for...

Model Test Paper-1(for School / Board Exams.)

CHEMISTRY

(Complete Syllabus of Class XII)

MM : 70 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

GENERAL INSTRUCTIONS :

(i) All questions are compulsory.

(ii) Questions number 1 to 5 are very short answer type questions and carry 1 mark each.

(iii) Questions number 6 to 10 are short answer type questions and carry 2 marks each.

(iv) Questions number 11 to 22 are also short answer type questions and carry 3 marks each.

(v) Questions number 23 is a value based questions and carry 4 marks.

(vi) Questions number 24 to 26 are long answer type questions and carry 5 marks each.

(vii) There is no overall choice. However, an internal choice has been provided in one question of 2 marks,

one question of 3 marks and all three questions of 5 marks each. You have to attempt only one of the

given choices in such questions.

(viii) Use log tables if necessary, use of calculator is not allowed.

SECTION-A

1. How many atoms constitute one unit cell of a face-centered cubic crystal? [1]

2. What is the role of graphite in the electrometallurgy of aluminium? [1]

3. If the rate constant of a reaction is k = 3 × 10–4 s–1, then identify the order of the reaction. [1]

4. Account for the following. PCl5

can act as an oxidising agent but not as a reducing agent. [1]

5. State Raoult’s law. [1]

SECTION-B

6. Explain why a solution of chloroform and acetone shows negative deviation from Raoult’s law? [2]

7. State Kohlrausch’s law of independent migration of ions. Why does the conductivity of a solution decrease with

dilution? [2]

8. Which one of 4

PCl+

and 4

PCl−

is not likely to exist and why? [2]

9. Write the IUPAC name of the complex [Cr(NH3

)4

Cl2

]+. What type of isomerism does it exhibit? [2]

(2)

Model Test Paper-1 Subjective Test for Class-XII (Chemistry)

10. Outline the principles of refining of metals by the following methods :

(a) Zone refining

(b) Vapour phase refining

OR

What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic

configuration of d4 in terms of t2g

and eg

in an octahedral field when

(a) 0

> P

(b) 0

< P [2]

SECTION-C

11. The density of copper metal is 8.95 g cm–3. If the radius of copper atom is 127.8 pm, is the copper unit cell

a simple cubic, a body centred cubic or face centred cubic structure (Given atomic mass of Cu = 63.54 g

mol–1 and NA

= 6.02 × 1023 mol–1) [3]

12. On heating lead (II) nitrate gives a brown gas ‘’A’’. The gas ‘’A’’ on cooling changes to colourless solid ‘’B’’.

Solid ‘’B’’ on heating with NO changes to a blue solid ‘C’. Identify ‘A’,’B’ and ‘C’ and also write reactions involved

and draw the structures of ‘B’ and ‘C’. [3]

13. Write the Nernst equation and compute the emf of the following cell at 298 K.

2+

2 o2 Sn /Sn(0.02M)0.05M

Sn(s) Sn H H ,1atm Pt[E 0.144V] [3]

14. Define the following terms with an example in each case :

(a) Macromolecular sol

(b) Peptization

(c) Emulsion [3]

15. For the complex [NiCl4

]2–, write

(a) The IUPAC name.

(b) The hybridization type.

(c) The shape of the complex.

(atomic no. of Ni = 28) [3]

16. The following results have been obtained during the kinetic study of the reaction

2A + B C + D

Expt. No.

[A] [B] Initial rate of formation

of D.

1 0.1 M 0.1 M 6.0 × 10–3

M min–1

2 0.3 M 0.2 M 7.2 × 10–3

M min–1

3 0.3 M 0.4 M 2.88 × 10–2

M min–1

4 0.4 M 0.1 M 2.40 × 10–2

M min–1

Determine the rate law and rate constant for the reaction. [3]

(3)

Subjective Test for Class-XII (Chemistry) Model Test Paper-1

17. Rearrange the compounds of each of the following sets in order of reactivity SN

2 displacement.

(a) 2-bromo-2-methyl butane, 1-bromopentane,

2-bromopentane.

(b) 1-bromo-3-methyl butane,

2-bromo-2-methyl butane,


(c) 1-bromobutane, 1-bromo-2,

2-dimethyl-propane,


1-bromo-3-methyl butane [3]

18. Explain what is meant by

(a) Pyranose structure of glucose?

(b) Glycosidic linkage? [3]

19. Complete the following reactions

(a)2

NaCN Reduction

2 5 Ni/HC H Cl A B

(b)6 5 2 3 2 2

C H N Cl H PO H O

(c) R–C–NH2

4

2

LiAlH

H O

O

[3]

20. Write the name and structure of the monomers of the following polymers :

(a) Buna-S

(b) Neoprene

(c) Nylon-6, 6 [3]

21. Explain the following terms with example of each

(a) Food preservatives

(b) Enzymes

(c) Detergents [3]

22. An organic compound (A), molecular formula C8

H16

O2

,was hydrolysed with dilute sulphuric acid to give a

carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration

gives but-1-ene. Write chemical equations for the reactions involved.

OR

A ketone A (C4

H8

O) which undergoes a haloform reaction gives compound B on reduction. B on heating with

sulphuric acid gives a compound C which forms mono-ozonide D. D on hydrolysis with zinc dust gives

only E. Identify A, B, C, D and E. Write the reactions involved. [3]

23. Phenol associates in benzene to a certain extent to form a dimer. A solution containing 20 g of phenol in

1.0 kg of benzene has its freezing point lowered by 0.69 K. Calculate the fraction of phenol that has dimerised

(Give Kf

for benzene = 5.1 Km–1] [4]

(4)


SECTION-D

24. (a) Give reasons for the following :

(i) Bond enthalpy of F2

is lower than that of Cl2

.

(ii) PH3

has lower boiling point than NH3

.

(b) Draw the structures of the following molecules :

(i) BrF3

(ii) (HPO3

)3

(iii) XeF4

OR

(a) Account for the following :

(i) Helium is used in diving apparatus.

(ii) Fluorine does not exhibit positive oxidation state.

(iii) Oxygen shows catenation behaviour less than sulphur.

(b) Draw the structures of the following molecules :

(i) XeF2

(ii) H2

S2

O8

[5]

25. (a) Define the following terms :

(i) Limiting molar conductivity

(ii) Fuel cell

(b) Resistance of a conductivity cell filled with 0.3 mol L–1 KCl solution is 100 . If the resistance of the same

cell when filled with 0.02 mol L–1 KCl solution is 520 , calculate the conductivity and molar conductivity

of 0.02 mol L–1 KCl solution. The conductivity of 0.1 mol L–1 KCl solution is 1.29 x 10–2 –1cm–1.

OR

(a) State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction

of 1 mol of Cu2+ to Cu?

(b) Calculate emf of the following cell at 298 K :

2 2Mg(s) Mg (0.1M) Cu (0.01M) Cu(s)

[Given o 1

cellE 2.71V,1 F 96500 C mol ]

[5]

26. (a) How would you convert the following?

(i) Phenol to p-benzoquinone.

(ii) Methyl magnesium bromide to 2-methylpropan-2-ol.

(iii) Propene to propan-2-ol.

(b) A compound A (C2

H6

O) on oxidation by PCC gave B, which on treatment with aqueous alkali and

subsequent heating furnished C. B on oxidation by KMnO4

, forms a monobasic carboxylic acid with molar

mass 60 g mol–1. Deduce the structure A,B and C.

(5)


OR

(a) Give a possible explanation for each one of the following.

(i) There are two – NH2

groups in semicarbazide. However, only one such group is involved in the

formation of semicarbazones.

(ii) Cyclohexanone forms cyanohydrins in good yield but 2,2, 6-trimethyl cyclohexanone does not.

(b) Write the structures of the main products of following reactions.

(i) + C H COCl6 5

3

2

Anhy.AlCl

CS

(ii)2

2 4Hg ,H SO

3H C C C H

(iii)

CH3

NO2

2 2

3

(i)CrO Cl

(ii)H O [5]

� � �

(6)


CHEMISTRY

SECTION-A

1. In face-centered cubic crystal, there are 8 atoms at the corners and 6 at face-centres.

Total no. of atoms per unit cell = 1 1

8x 6x 48 2

2. In electrolytic reduction of Al2

O3

inner carbon lining of the cell is used as cathode and graphite rods are used

as anode. The graphite anode is useful for the reduction of Al2

O3

into Al.

2 3 2Graphite

2Al O 3C 4Al+3CO

3. Order of reaction is one.

4. Phosphorus shows highest oxidation state of +5 and in PCl5

it is already in +5 oxidation state therefore, it

cannot loose electrons and not behave like reducing agent. It can act as an oxidising agent because it can

gain electron to show lower oxidation state.

5. Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component in the

solution is directly proportional to its mole fraction.

SECTION-B

6. As a result of intermolecular H-bonding between O-atom of (CH3

)2

CO and H-atom of CHCl3

, A –– B interaction

becomes stronger than A –– A and B –– B interactions. This leads to the decrease in vapour pressure and

resulting in negative deviation.

H C3

H C3

C = O ..... H –– C

Cl

Cl

Cl

Model Test Paper-1(for School / Board Exams.)

MM : 70 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

SOLUTIONS

(7)


7. According to Kohlrausch law of independent migration of ions, the limiting molar conductivity of an electrolyte

is the sum of the limiting ionic conductivites of the cation and the anion each multiplied with the number of

ions present in one formula unit of the electrolyte.

i.e., o o o

x yfor A B = xλ λ

my

The conductivity of a solution decreases with dilution because the number of current carrying particles, i.e.,

ions per cm3 of the solution become less and less on dilution.

8.4

PCl

is not likely to exist.

9. [Cr(NH3

)4

Cl2

+

Its IUPAC name is

Tetraamminedichloridochrominum (III) ion

It shows geometrical and optical isomerism as shown below :

(i) Geometrical isomers

Cr

Cl

Cl

NH3

NH3

H N3

H N3

Trans

Cr

Cl

Cl

NH3

NH3

H N3

H N3

Cis

(ii) The cis form is optically active.

Cr

Cl

Cl

NH3

NH3

H N3

H N3

Cr

Cl

NH3

NH3

H N3

Cl NH3

Non-superimposable mirror image

10. (a) Zone refining : This method is based upon the principle that the impurities are more soluble in the melt

than in the solid state of the metal.

(b) Vapour phase refining : The metal to be refined should form volatile compound with suitable reagent and

the compound so formed should be easily decomposable for the recovery of pure metal.

OR

The d-orbitals present in metal have the same energy in the free state. This is called degenerate state of

d-orbital. But, when a complex is formed the ligands destroy the degeneracy of these orbitals. The d-orbitals

gets split into two sets one with lower and one with higher energy. The difference of energy between two sets

is called field splitting energy.

SECTION-C

11. According to question

d = 8.95 g cm–3 = 8.95 x 106 g m–3

r = 127.8 pm = 127.8 x 10–12m

M = 63.54 g mol–1

NA

= 6.02 x 1023 mol–1

(8)


By assuming that Cu crystallises in fcc unit cell

Edge length, 4 2.

42

r aa r

⎛ ⎞ ⎜ ⎟⎜ ⎟

⎝ ⎠ =

-12

4 127.8 10

2

m

3-12

3 4 127.8 10

2

a m

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

= 4.725 x 10-29m3

Rank of a unit cell

3. .

Ad a N

ZM

6 29 238.95 10 4.725 10 6.02 10

63.54

4.006 4

12.3 2 673K

Pb(NO )(A)

(Brown colour)

2

2PbO 4NO

2

On cooling

Heating(B)

(Colourless solid)

2NO2 4

N O��⇀↽��

2 4250K

2NO N O2 3

2N O

(C)

(Blue solid)

��⇀↽��

∆+

(Structure of N2

O4

)

(Structure of N2

O3

)

13. Cell reaction is

2+

2Sn 2H Sn H ;(n = 2)

o

cell R LE E E 0 ( 0.14) 0.14V

According to Nernst equation,

2

o

cell cell 2

Sn0.0591VE E log

Hn

⎡ ⎤⎣ ⎦ ⎡ ⎤⎣ ⎦

cell 2

0.050.0591VE 0.14V log

2 0.02

⎡ ⎤⎣ ⎦ ⎡ ⎤⎣ ⎦

0.0591V 2.0969

= 0.14 V2

= 0.078 V

(9)


14. (a) Macromolecular sol : These are the colloids obtained by dissolving macromolecules (big size molecules)

having large molecular masses in a suitable liquid. In this case, the dispersed phase particles have size

in the colloidal range (1 nm - 100 nm). e.g., starch, cellulose etc.

(b) Peptization : The dispersion of freshly precipitated substance into colloidal solution to the addition of some

electrolytes having one common ion is known as peptization. The electrolyte used is called peptizing agent

e.g.,A freshly precipitated Fe(OH)3

is converted into colloidal solution by adding small quantity of FeCl3

to the solution of Fe(OH)3

.

(c) Emulsion : An emulsion is a colloidal dispersion in which both the dispersed phase and the dispersion

medium are liquids which are otherwise immiscible. e.g., milk, cod liver oil etc.

15. (a) Tetrachloridonickelate (II) ion

(b) sp3

(c) Tetrahedral

16. Let the order with respect to A be and B be .

Then, rate k [A] [B]

We have for the experiments,

Rate1

= k[0.1] [0.1] ...(i)

Rate2

= k[0.3] [0.2] ...(ii)

Rate3

= k[0.3] [0.4] ...(iii)

R ate4

= k[0.4] [0.1] ...(iv)

From (i) and (iv),

1

4

Rate [0.1] [0.1]

Rate [0.4] [0.1]

l

k

3

2

6.0 10 0.1 0.1 1

0.4 0.12.40 10 4

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

1 11

4 4 ⇒

From (2) and (3),

2

3

Rate [0.3] [0.2]

Rate [0.3] [0.4]

k

k

3

2

7.2 10 0.2

2 0.22.88 10

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

1 11

4 2 ⇒

Order with respect to A = 1

Order with respect to B = 2

Thus, rate law expression is

Rate = k[A] [B]2

Now, from (1)

6 × 10–3 = k (0.1) (0.1)2

3 36 10 10

1 1 1k

Rate constant 1k 6.0 min

(10)


17. (a) In SN

2 reaction the nucleophile attacks from the back side of the halide ion, bulkier the alkyl group present

on the carbon bearing halogen lesser will be its tendency to undergo SN

2 reaction. Thus, the order of

reactivity of alkyl halides towards SN

2 mechanism is 1o > 2o > 3o.

1-bromopentane > 2-bromopentane > 2-bromo-2-methyl butane

(b) 1-bromo-3-methyl butane > 3-bromo-2-methyl butane > 2-bromo-2-methyl butane

(c) 1-bromobutane > 1-bromo-3-methyl butane > 1-bromo- 2-methyl butane > 1-bromo-2, 2 dimethyl propane

18. (a) The six membered cyclic structure of glucose is called pyranose structure (– or –), in analogy with pyran.

Pyran is a cyclic compound with an oxygen atom and five carbon atoms in the ring.

-D–Glucose

CH OH2

H

OH

HO

H

OH

OH

HOH

H

(b) The ethereal or oxygen linkage through which two monosaccharides are joined together by the loss of a

water molecule to form a molecule of disaccharide is called the glycosidic linkage. The glycosidic linkage

in maltose molecule is shown below

glycosidic linkage

CH OH2

CH OH2

H H

OH OH

H HO O

H H

OH OH

H

OH OH

H H

O

H

19. (a)2

NaCN Reduction2 5 2 2 2 5 2 2Ni/H

(A) (B)

C H Cl C H CN C H CH NH

(b) C6

H5

N2

Cl + H3

PO2

+ H2

O Reduction C

6

H6

+ N2

+ H3

PO3

+ HCl

(c) R – C – NH2

O

Alkanamide

4 2LiAlH ,H O

(reduction)

2 2

Alkanamine

R CH NH

20. (a) 1, 3-Butadiene and Styrene

CH2

= CH – CH = CH2

, C6

H5

– CH = CH2

(b) Chloroprene 2 2

Cl|

CH CH C CH

(c) Adipic acid and Hexamethylenediamine

HO – C – (CH ) – C – OH,2 4

O O

H2

N – (CH2

)6

– NH2

(11)


21. (a) Food preservatives : These are the chemical substances which are added to the food materials to prevent

their spoilage and to retain their nutritive value for long periods, e.g., vinegar, sodium benzoate, etc.

(b) Enzymes : The proteins which perform the role of biological catalysts in the body are known as enzymes.

e.g., maltase enzyme converts maltose to glucose.

Maltase12 22 11 6 12 6(hydrolysis)Maltose Glucose

C H O 2C H O

(c) Detergents : They are the sodium or potassium salts of sulphonated long chain hydrocarbons, e.g., sodium

alkylbenzene sulphonate, i.e., sodium dodecyl benzene sulphonate.

CH (CH )3 2 11

SO Na3

– +

22. Summary of the given statements.

(A) Carboxylic acid + alcoholDil. H SO

2 4

hydrolysis

C H O8 16 2

(B) (C)

CrO (Oxidation)3

Alcoholdehydration

(–H O)2

(C)But-1-ene

Conclusions :

(a) Formation of carboxylic acid (B) and an alcohol (C) by the action of dil. H2

SO4

on (A), C8

H16

O2

indicates

that (A) is an ester.

(b) Dehydration of (C) to form butene-1 indicates that the alcohol (C) is CH3

CH2

CH2

CH2

OH, but not

CH3

CH2

CH(OH)CH3

because the latter on dehydration gives but-2-ene as the major product.

2

dehydration3 2 2 2 3 2 2( H O)

Bu tanol 1 But 1 ene

CH CH CH CH OH CH CH CH CH

Butanol-2

CH CH CHCH3 2 3

|OH

2

dehydration3 3( H O)

But 2 ene(Major)

CH CH CHCH

(iii) Oxidation of (C), butanol-1 with chromic acid, a strong oxidising agent to form (B), further indicates that

(B) is CH3

CH2

CH2

COOH.

Thus, (A) an ester should be butyl butyrate which explains all the given reactions.

8 10 2

3 2 2 2 2 2 3

Butyl butyrate,A(C H O )

O||

CH CH CH C OCH CH CH CH 2 4dil.H SO

(hydrolysis) 3 2 2 2 2 2 3

(B),Carboxylic acid (C),Butanol

CH CH CH COOH HOCH CH CH CH

3 2 2 2Bu tanol,C

CH CH CH CH OH

3CrO

(oxidation) 3 2 2

Bu tanoic acid, B

CH CH CH COOH

3 2 2 2Butanol, C

CH CH CH CH OH

2

dehydration

( H O) 3 2 2

But-1-ene

CH CH CH CH

(12)


OR

As A undergoes haloform reaction it must be a methyl ketone containing –COCH3

group. Thus, its possible

formula is C2

H5

COCH3

. The equations involved are

4

[H]3 2 3 LiAlH

(A)

Bu tan 2 one

CH COCH CH

(B)

CH CH CH CH3 2 3

|OH

2 4Conc.H SO CH

3 – CH = CH – CH

3

(D)

H CCH3

CHCH3

O3

(C)CH – CH = CH – CH

3 3

2Zn/H O

3(E)

Acetaldehyde

2CH CHO

23. Given, W2

= 20 g, W1

= 1 kg = 1000 g

Tf

= 0.69 K

Kf

= 5.1 km–1

Depression in freezing point

f 2

f

2 1

K W 1000T

M W

f 2

2

f 1

K W 1000M

T W

1

2

5.1Km 20g 1000M

0.69K 1000g

M2

(observed ) = 147.82

M2

(calculated) of C6

H5

OH

= 6 × 12 + 6 × 1 + 16

= 94 g mol–1

Van’t Hoff Factor 2

2

M (Calculated)(i)

M (Observed)

940.635

147.82

6 2 6 5 22C H OH (C H OH)�

Here n = 2,

(degree of association) i 1 0.635 1

1111

2n

⎛ ⎞ ⎜ ⎟⎝ ⎠

0.3650.73

0.5

73%

(13)


SECTION-D

24. (a) (i) Bond enthalpy of F2

is lower than that of Cl2

because F – F bond is weak as F atom is very small.

This strengthen the electron-electron repulsions between the lone pairs of electrons

(ii) This is due to the fact that NH3

molecules are associated with intermolecular hydrogen bonding which

increases the bond strength. A larger amount of energy is then required to overcome attractive force

among NH3

molecules which is not so in case of PH3

.

(b) (i)

Br

F

F

F

(ii)

O

P P

OO

P

OHOH

OO

O OH

(iii)

F F

F F

Xe

OR

(a) (i) Helium is used for a number of reasons. It is light, cheap, and does not dissolve in blood the same

way that nitrogen does. Being inert it cannot be toxic to the diver or corrosive to equipment. Moreover

at extreme depths the pressure is high enough that if you had nitrogen in the tank it would become

saturated enough in your blood that you would develop nitrogen narcosis. Helium would not be

absorbed across the alveoli and will not react with your body so it is used for extreme depth’s.

(ii) Being the most electronegative element, it exhibits mostly a negative O.S. of –1. Further since it has

no vacant d-orbitals, it does not exhibit any higher O.S. of +1, +3, +5, +7.

(iii) S–S bond is much stronger than O–O bond and hence, sulphur has a much greater tendency for

catenation than oxygen.

(b) (i)

Xe

F

F

(ii)

25. (a) (i) Limiting molar conductivity : It is that value of molar conductivity which is obtained at infinite dilution

i.e., when the concentration of the solution approaches zero. It is denoted by º

m.

(ii) Fuel cell : Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen,

methane, methanol etc. directly into electrical energy are called fuel cells.

Case I Case II

(b) Given : R = 100 R = 520

C = 0.1 mol L–1 C = 0.02 mol L–1

= 1.29 × 10–2 –1 cm–1

(14)


To find : and m

of 0.02 mol L–1 KCl solution.

Case I

G*= R = 100 × 1.29 × 10–2

(Where G* = cell constant)

= 1.29 cm–1

Case II : As G* = R Conductivity

R Resistance

1

G* 1.29cm

R 52

= 2.48 × 10–3 –1 cm–1

and,

1 1

m 1

1000 1.29 1000cm

C 520 0.02mol L

= 124.038 –1 cm–1

OR

(a) According to Faraday’s first law of electrolysis, the amount of chemical reaction which occurs at any

electrode during electrolysis by a current is proportional to the quantity of electricity passed through the

electrolyte.

Reduction of 1 mol of Cu2+ to Cu is given by :

Cu2+ + 2e– Cu

The charge required for 1 mole of electrons = 1F

The charge required for 2 moles of electrons = 1F

(b) Given : Mg(s)|Mg2+ (0.1M) || Cu2+ (0.01M)| Cu(s)

º

cellE = 2.71 V, 1F = 96500 C mol–1

To find : Ecell

= ?

Using Nernst equation,

º

cell cell

0.059 [P]E E log

n [R]

The redox equation for the above cell is :

Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)

2

º

cell cell 2

0.059 [Mg ]E E log

n [Cu ]

0.059 [0.1]2.71 log

2 [0.01]

0.059 0.0592.71 log10 2.71 1

2 2

= 2.68 V ( log10 1)∵

(15)


26. (a) (i)

OH O

Phenol

p-Benzoquinone

Na Cr O2 2 7

H SO2 4

O

(ii) CH MgBr3

CH COCH3 3 H , H O+

2

CH – C – CH3 3

CH – C – CH3 3

OMgBr OH

CH3

CH3

2-Methylpropan-2-ol

(iii)

OH

CH CH = CH + H SO3 2 2 4

H O2

boilCH – CH – CH

3 3

Propan-2-olPropene

HBr

CH – CH – CH3 3

CH – CH – CH3 3

Br OH

aq. KOH

Propan-2-ol

(b) A monobasic carboxylic acid has the formula RCOOH. Given, molar mass of RCOOH = 60 g mol–1

x + 12 + 16 + 16 +1 = 60

x = 60 – 45 = 15

Thus, R = CH3

(molar mass 15) and the acid is CH3

COOH. The acid is obtained by the oxidation of

aldehyde, so B is an aldehyde, i.e., CH3

CHO and A is CH3

CH2

OH (an alcohol). The reaction is as

CH CH OH3 2

[O]CH CHO

3

PCC

(A)(B)(C H O)

2 6

OR

(a) (i) In semicarbazide the –NH2

group which is directly linked to carbonyl group is involved in resonance.

H N2

C

N

H

NH2 H N

2

+C

NHNH

2

O–O

H N2

C

NH+ NH

2

O–

Therefore, the electron density on –NH2

group involved in the resonance also decreases. As a result,

it cannot act as a nucleophile. Since, the other –NH2

group is not involved in resonance; it can act

as a nucleophile and can attack carbonyl carbon atoms of aldehydes and ketones to produce

semicarbazones.

(16)


(ii) Cyclohexanones from cyanohydrins according to the following equation:

O OH CN

HCN

Cyclohexanone Cyanohydrin

In this case, the nucleophile CN– can easily attack without any steric hindrance. However, in case of

2, 2, 6, trimethyl cyclohexanone, methyl groups at -position offer steric hindrance and as a result,

CN– cannot attack effectively.

O

H C3

CH3

CH3

2, 2, 6-trimethylcyclohexanone

(b) (i)

COC H6 5

+ C H COCl6 5

Anhydrous AlCl3

+ HCl

Benzophenone

(ii) 42

43 2

Dil.H SO

HgSOH C C C H H O

CH – C = CH3 2

CH – C – CH3 3

OH O

Prop-1-ene-2ol(unstable)

Tautomerisation

Acetone

(iii)CrO Cl

2 2H O

3

+

CHO

NO2

CH3 CH(OCrCl OH)

2 2

NO2

NO2

Para-nitrobenzaldehyde

� � �

chemistry complete syllabus of class xii · 2019. 7. 14. · model test paper-1 subjective test for...

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