SECTION - A1. Out of NH3 and N2, which gas will be adsorbed more readily on the surface of charcoal and why?2. Why are haloalkanes insoluble in water but soluble in benzene?3. Why is acetyl chloride a better acetylating agent than acetic acid?4. Why is cellulose in our diet not nourishing?5. In the preparation of aldehydes from primary alcohols, aldehydes formed must be distilled as soon as they are formed. Why ?

SECTION - B6. A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids.

What is the formula of the compound?7. Why does zinc oxide which is white in colour become yellow upon heating?8. In the reaction A + 2B - 3C + 2D, the rate of disappearance of B is 1 102 mol L1 s1. What will be the rate of the reaction and

rate of change in concentration of A and C?9. What are fuel cells? How do they resemble and differ from galvanic cells?10. (a) SF6 is used as a gaseous electrical insulator. Why?

(b) H3PO2 and H3PO3 act as good reducing agents while H3PO4 does not. DiscussOR

(a) How is XeO3 prepared? Write chemical equation.(b) Molecular nitrogen N2 is not particularly reactive. Explain.

SECTION - C11. Consider a cell composed of two half-cells:

(i) Cu(s) |Cu2+ (aq), and (ii) Ag(s) |Ag+ (aq).Calculate

Time : 3 Hrs. Max. Marks : 70

GENERAL INSTRUCTIONSI. All questions are compulsory.II. Question numbers 1 to 5 are very short answer questions and carry 1 mark each.III. Question numbers 6 to 10 are short answer questions and carry 2 marks each.IV. Question numbers 11 to 22 are also short answer questions and carry 3 marks each.V. Question number 23 is value based question and carry 4 marks.VI. Question numbers 24 to 26 are long answer questions and carry 5 marks each.VII. Use log tables, if necessary. Use of calculators is not allowed.

CHEMISTRY SAMPLE PAPER

One of the leading publishers in India, Disha Publication provides books and study materials for

schools and various competitive exams being continuously held across the country. Disha's sole

purpose is to encourage a student to get the best out of preparation. Disha Publication offers an

online bookstore to help students buy exam books online with ease. We, at Disha provide a wide

array of Bank Exam books to help all those aspirants who wish to crack their respective different

levels of bank exams. At Disha Publication, we strive to bring out the best guidebooks that

students would find to be the most useful for Bank Probationary exams.

ABOUT DISHA PUBLICATION

2

(a) the standard cell potential, and(b) the cell potential when concentration of Cu2+ is 2M and concentration of Ag+ is 005 M, at 298 K.

Given: 2+Cu /CuE = + 0.34V,

+Ag /AgE = 080 V, R = 8314 JK1 mol1,

F = 96500 C mol1

12. (a) In some reactions, the energy possessed by colliding molecules is more than the threshold energy, yet the reaction is slow. Why?(b) State one condition in which a bimolecular reaction may be kinetically of the first order.

13. Explain the following observation:(a) Lyophilic colloid is more stable than lyophobic colloid.(b) Coagulation takes place when sodium chloride solution is added to a colloidal solution of ferric hydroxide.(c) Sky appears blue in colour.

ORAccount for the following :(a) Ferric hydroxide sol is positively charged.(b) The extent of physical adsorption decreases with rise in temperature.(c) A delta is formed at the point where river enters the sea.

14. Using the Ellingham diagram given below, predict the following:(a) At what temperature can silicon reduce MgO.(b) At what temperature can aluminium reduce MgO.(c) At what temperature can carbon reduce MgO.

0

100

200

300

400

500

600

700

800

900

1000

1100

1200

0 C 400 C 800 C 1200 C 1600 C 2000 C273K 673K 1073K 1473K 1873K 2273K

692 K1180 K 1623 K

1363 K

1966 K

A

2Mg + O 2M

gO2

4/3Al+ O 2/3A

l O2

2 3

Si + O SiO2 2

2Zn+ O 2ZnO

2

2CO+ O

2CO2

2

2Fe+ O 2FeO2

4Cu+ O 2Cu O

22

C + O CO2 2

2C + O CO

2 2

DG/k

J mol1 of O

2

Temperature

15. Explain the following observations:(a) ClF3 exists but FCl3 does not.(b) Among the hydrides of elements of Group 16, water shows unusual physical properties.(c) Justifying the order of your choice, arrange the following in decreasing order of property indicated:

HClO4, HClO3, HClO2, HClO oxidising power.

16. How are xenon fluorides XeF2, XeF4 and XeF6 obtained?17. (i) Among Ag(NH3)2Cl, [Ni(CN)4]

2 and [CuCl4]2, which

(a) has square planar geometry?(b) remains colourless in aqueous solution and why?

[Ag (Z = 47), Ni (Z = 28), Cu (Z = 29)].

(ii) Write the IUPAC name of [Co(en)2Cl2]Cl.18. (a) What is the decreasing order of reactivity of the following in SN2 reaction?

3

1-Bromo-2-methylbutane, 1-Bromo-2, 2-dimethylpropane, 1-Bromopentane.

(b) Arrange the compounds CH3F, CH3I, CH3Br, CH3Cl in order of increasing reactivity in bimolecular nucleophilic substitution(SN2) reactions.

19. Write the equation for the reaction of HI with the following:(a) 1-Propoxypropane(b) Methoxybenzene(c) Benzyl ethyl ether

20. D (+) glucose gives most of the characteristics of the aldehydic (CHO) group but it does not react with sodium bisulphite,ammonia and also does not restore the pink colour of Schiff's reagent. Explain.

21. (a) Distinguish between thermosetting and thermoplastic polymers?

(b) What is phenol-formaldehyde polymer popularly known as?

22. (a) What is the difference between bathing soap and washing soaps?

(b) Aspartame is unstable at cooking temperature, where would you suggest aspartame to be used for sweetening?

SECTION - D

23. In chemistry laboratory, the teacher asked two of her students, Rohan and Kunal to do a confirmatory test for aromatic amines.Rohan decided to perform carbyl amine test whereas Kunal decided to do it with Hinsberg's reagent. After knowing this theteacher advised Rohan not to perform this method.

(a) Why do you think teacher advised him not to do so ?

(b) What values are associated with teacher and Rohan.

SECTION - E

24. (a) The degree of dissociation of Ca(NO3)2 in dilute aqueous solution containing 7.0 g of the solute per 100g of water at 100Cis 70 percent. If the vapour pressure of water at 100C is 760 mm, calculate the vapour pressure of the solution.

(b) What is the vant Hoff factor (i) of the compound Na2SO4 if it is 100% dissociated.

(c) If glycerine [C3H5 (OH)3] and ethylene glycol (C2H6O2) are sold at same price per kg, which would be cheaper for preparingan antifreeze solution for the radiator of an automobile?

OR

(a) A solution containing 0.5126 g of naphthalene (molar mass = 128g mol1) in 50.0 g of carbon tetrachloride gave a boilingpoint elevation of 0.402 K while a solution of 0.6216 g unknown solute in the same mass of carbon tetrachloride gave aboiling point of elevation of 0.647 K. Find the molar mass of the unknown solute.

(b) Explain why is measurement of osmotic pressure more widely used for determining the molar mass of macro-molecules thanthe rise in boiling point or fall in freezing point temperature of their solutions.

(c) What will happen to the elevation in boiling point of a solution if the weight of the solute dissolved is doubled but theweight of solvent taken is halved ?

25. (a) (i) Which oxidation state of Mn is most stable and why?

(ii) Which transition element show highest oxidation state and in which compound?

(iii) Which is the densest transition element?

(b) Write balanced ionic equation for what happens when:

(i) Acidified potassium permanganate solution is treated with an oxalate ion in solution.

4

(ii) An iodide ion is treated with an acidified dichromate ion in solution.

OR

(a) Give reasons for each of the following :

(i) Size of trivalent lanthanoid cations decreases with increase in the atomic number.

(ii) Transition metal fluorides are ionic is nature, whereas bromides and chlorides are usually covalent in nature.

(b) Gas (A) and gas (B) both turn K2Cr2O7/H+ green. Gas (A) also turns lead acetate paper black. When gas (A) is passed into

gas (B) in an aqueous solution, yellowish white turbidity appears. Identify gas (A) and (B) and explain the reactions.

26. (a) An organic compound 'A' (molecular formula C3H6O) is resistant to oxidation but forms a compound 'B' (C3H8O) onreduction. 'B' reacts with HBr to form a bromide which on treatment with alcoholic KOH forms an alkene 'C' (C3H6). Deducethe structure of A, B and C.

(b) State chemical tests to distinguish between the following pairs of compounds:(i) Propanal and propanone.(ii) Phenol and benzoic acid.

OR(a) An organic compound 'A' which has characteristic odour, on treatment with NaOH forms two compounds 'B' and 'C'. Compound

'B' has the molecular formula C7H8O which on oxidation gives back compound 'A'. Compound 'C' is the sodium salt of an acidwhich when heated with soda lime yields an aromatic hydrocarbon 'D'. Deduce the structures of A, B, C and D.

(b) (i) Write the IUPAC name of the following compound :

H CCHCH 3 2 CCH3

||

||||

Cl O

(ii) Write IUPAC name of CH CHCH3 CH3||

||

Br CHO

5

1. NH3 is adsorbed more readily as it is more easily liquefiable compared to N2. Moreover, NH3 molecule has has greater molecularsize. (1 mark)

2. Haloalkanes donot dissolve in water due to the lack of hydrogen bonding in the molecules of solute and solvent. However, theydissolve in benzene (organic solvent) as the solubility is based on the principle that 'like dissolve like'. (1 mark)

3. The acetylation is carried by acetyl carbocation (CH3CO+) The cation is formed by acetylchloride and not by acetic acid

CH C Cl3 O

CH C + Cl3

O

Acetyl chloride Acetyl carbocation

CH C OH3 O

CH C O + H3 +

O

Acetate ion

(1 mark)

Therefore, acetyl chloride is a better acetylating agent than acetic acid.4. Our body does not produce an enzyme to help in the metabolism of cellulose. (1 mark)5. Aldehydes formed as a result of the oxidation of primary alcohols are prone to further oxidation and are expected to form acids.

Therefore, it is advisable to remove them from the reaction mixture by distillation in case these are low boiling. Otherwise, othermethods must be used to convert primary alcohols into aldehydes. (1 mark)

6. Let the number of atoms of element N present in ccp = x\ The number of tetrahedral voids = 2x ( mark)As 1/3rd of the tetrahedral voids are occupied by atoms of element M, therefore, number

of atoms of element M present =1 223 3

xx = ( mark)

\ N : M = x : 23x

= 3x : 2x = 3 : 2\ The formula of the compound is N3M2 or M2N3. (1 mark)

7. Zinc oxide (ZnO) crystals upon heating decompose as follows:2+

2ZnO(s) Zn (s) +1/2O (g) + 2e ( mark)

The excess of Zn2+ ions formed are trapped in the interstitial spaces and the electrons are also trapped in the neighbourhood.These electrons absorb radiations corresponding to certain specific colour from the light and emit yellow light. (1 marks)

8. Given reaction is A + 2B - 3C + 2D, thus rate of this reaction

Rate d[A] 1 d[B] 1 d[C] 1 d[D]= = = =dt 2 dt 3 dt 2 dt

(1 mark)

d[A]dt

2 21 d[B] 1= (1 10 ) 0.5 102 dt 2

= = mol L1s1 ( mark)

d[C]dt

23 d[B] 3= (1 10 )2 dt 2

= = 1.5 102 mol L1s1 ( mark)

9. Fuel cells are those cells in which chemical energy of fuel is converted into electrical energy. (1 mark)In both cells, chemical energy is converted into electrical energy. The efficiency of fuel cell is higher than that of galvanic cell.Fuel cells are non-polluting. ( + = 1 mark)

10. (a) SF6 is colourless, odourless and non-toxic gas at room temperature. It is thermally stable and chemically inert. Because ofits inertness and high tendency to suppress internal discharges, it is used as a gaseous electrical insulator in high voltagegenerators and switch gears. (1 mark)

HINTS & SOLUTIONS

6

(b) Both 1

23 OH P+

and 3

33 OH P+

can increase the oxidation state of phosphorus upto + 5. They therefore, act as good reducingagents. However, in H3PO4, oxidation state of phosphorus is already + 5 (maximum possible oxidation state). It can act asan oxidising agent and not as a reducing agent. (1 mark)

OR(a) XeO3 is prepared by hydrolysis of XeF6 or XeF4. ( mark)

XeF6 + 3H2OHydrolysis XeO3 + 6HF ( mark)

6XeF4 + 12H2O 4Xe + 2XeO3 + 24HF + 3O2 ( mark)(b) N2 is not reactive, particularly at ordinary conditions of temperature because there is triple covalent bond (N N) betweentwo N-atoms. To break this bond to form N-atoms, a large amount of energy is required which is hardly available from enthalpiesof reactions under ordinary conditions. (1 mark)

11. In this cell, Cu/Cu2+ electrode acts as anode, and Ag+/Ag electrode acts as cathode. ( + = mark)(a) Standard electrode potential,

cell cathode AnodeE E E = - = 080 V 034 V = 046 V. ( mark)

(b) The net cell reaction isCu + 2Ag+ Cu2+ + 2Ag ( mark)Here n = 2.By Nernst equation, cell potential

Ecell = cellE

2+

+ 22 303RT [Cu ]log

nF [Ag ]

( mark)

\Ecell = 046 V 22 3038 314298 2log

2 96500 (0 05)

= 046 40 0591 2 10log

2 25

= 046 0086 = 037 V. (1 mark)12. (a) Though the energy possessed by molecules is more than the threshold energy and the reaction should proceed at a

reasonable rate yet in some cases the reaction is slow. It is due to the fact that in such cases the reacting molecules are notproperly oriented. Due to this the number of effective collision decreases and so the reaction is slower than expected rate.

(1 + 1 = 2 marks)(b) By taking one of the reactants in large excess so that it may not contribute towards the order. (1 mark)

13. (a) The stability of the lyophobic colloid is only on account of charge while the stability of lyophilic colloid is on account ofcharge as well as solvation of colloidal particles. (1 mark)

(b) Fe(OH)3 sol is positively charged which is coagulated by negatively charged Cl ion present in sodium chloride solution.

(1 mark)(c) Sky appears blue in colour due to scattering of light by colloidal particles like dust, mist etc. (1 mark)

OR(a) Ferric hydroxide sol which is obtained by hydrolysis of FeCl3 is positively charged because of preferential adsorption of

ferric ions (Fe3+) by colloidal particles from the solution. (1 mark)(b) Adsorption is an exothermic process. According to Le Chateliers principle, the magnitude of adsorption should decrease

with rise in temperature and this is actually so. (1 mark)(c) River water contains colloidal particles of clay, mud, etc., which are charged. Sea water contains large quantity of number of

dissolved salts and acts as a sort of electrolyte. When river water comes in contact with sea water, coagulation of colloidalparticles occurs. These coagulated clay, sand particles, etc., settle down at the point of contact resulting in the rise of riverbed. This leads river water of adopt different course and a delta is formed in due course of time. (1 mark)

14. (a) Silicon can reduce MgO above 1966 K. (1 mark)(b) Aluminium can reduce MgO above 1623 K. (1 mark)

7

(c) Carbon cannot reduce MgO at any temperature. (1 mark)15. (a) It is because Cl has vacant d-orbitals. F does not have d-orbitals so, it cannot show higher oxidation state. (1 mark)

(b) (i) Water is liquid, others are gases (ii) Water is thermally most stable among hydrides of group 16 elements. This is due tointermolecular hydrogen bonding. ( + = 1 mark)

(c) HClO > HClO2 > HClO3 > HClO4.HClO is unstable, breaks down to HCl and [O] due to which it is strong oxidising agent whereas others give oxygen lesseasily. Greater the oxidation state of Cl in the acid, lower the oxidising power. (1 mark)

16. XeF2 , XeF4 and XeF6 are obtained by direct reaction between Xe and F2 as follows:

673 K,1bar2 2Ni tube

excessXe( ) + F ( ) XeF ( )g g s (1 mark)

873 K, 7 bar2 4

(1n 1:5 ratio)Xe ( ) 2F ( ) XeF ( )g g s+ (1 mark)

573 K, 60 70bar2 6

(1n 1:20 ratio)Xe ( ) 3F ( ) XeF ( )+ g g s (1 mark)

17. (i) (a) [Ni (CN)4]2 has square planar geometry on account of dsp2 hybridisation. (1 mark)

(b) 3 2[Ag (NH ) ] Cl+ - remains colourless in aqueous solution. The complex does not have unpaired electron in central

cation, Ag+ and thus d-d transition is not possible. Ag in + 1 oxidation state which has electronic configuration : [Kr]364d10 5s0. (1 mark)

(ii) The IUPAC name is Dichlorobis (ethylene-diammine) cobalt (III) chloride. (1 mark)18. (a) All are primary alkyl halides and their structural formulae are:

2 5 2

CH3|C H CCH Br

|H

3 2

3

CH3|CH CCH Br

|CH

H|

C H CBr4 9 |H

(I) (II) (III)The order of reactivity is : (III) > (I) > (II) ( mark)Explanation : The SN2 reactions are sensitive to steric hindrance. Greater the steric hindrance to the attacking nucleophile,lesser will be the reactivity. In the light of this, the reactivity order is justified. (1 mark)

(b) CH3F < CH3Cl < CH3Br < CH3I ( mark)In these nucleophilic substitution reactions, the nucleophile is to displace the halide ion (X). Greater the bond dissociationenthalpy of the CX bond, difficult is its cleavage and lesser will be the reactivity. The order of bond dissociation enthalpyof different CX bond isC F > C Cl > C Br > C IThe order of reactivity towards SN2 reactions is the reverse. (1 mark)

19. (a) CH3CH2CH2 O CH2CH2CH3HI

373KCH3CH2CH2 OH + CH3CH2CH2I (1 mark)

(b)

OCH3 OH

+ CH I3HI, 373 K (1 mark)

(c)

CH O C H2 2 5 CH I2

+ C H OH2 5HI, 373 K (1 mark)

20. To explain the above, it has been suggested that glucose does not have an open chain structure rather it has a cyclic structurein which the aldehydic (CHO) group is a part of the six membered oxide ring(d-oxide ring). (1 marks)

8

Sodium bisulphite, ammonia etc. fail to cleave the ring but Tollen's reagent and Fehlings solutions etc. are strong enough tocleave the ring and then CHO group is free to give its characteristic reactions. (1 marks)

21. (a) Thermosetting polymer Thermoplastic polymer(i) Cannot be reshaped on heating. (i) Can be heated and than moulded in a desired shape.(ii) Has extensive cross-linking. (ii) No cross linking between the chains.(iii) Examples are bakelite and melamine. (iii) Examples are polythene and polystyrene. (2 marks)

(b) Bakelite. (1 mark)22. (a) Bathing soaps are potassium salts of long chain fatty acids while washing soaps are sodium salts of long chain fatty acids.

(2 marks)(b) In cold foods and soft drinks. (1 mark)

23. (a) Teacher advised him not to do this test because during carbyl amine reaction Phenyl isocynaide will be produced as aproduct which is harmful. (2 marks)

(b) Value associated with teacher are concerned towards the safety of students and dedicated towards his dutyValue of Rohan are alert and conscious presence of mind (1 + 1 = 2 marks)

24. (a) Step I.Calculation of the Van't Hoff factor (i)Calcium nitrate dissociates in aqueous solution as :

23 2 3Ca(NO ) Ca ( ) 2NO ( )

+ -+ aq aq ( mark)

i 1n 1-a =- ( mark)

i 10.73 1-=-

( mark)

or i = 0.7 2 + 1 = 2.4 ( mark)Step IICalculation of vapour pressure of the solutionAccording to Raoult's Law,

A S A SB B AB

S A S B A

P P P Pn i W Mix i ;

P n P M W

- - = = =

( mark)

AP = 760 mm, i = 2.4, WB = 7.0 g, MB = 164 g mol

1 (for calcium nitrate)

WA = 100 g, MA = 18 g mol1

\ 1)

S1S

760 P 2.4 (7.0g) (18g mol 0.0184P (164 g mol ) (100 g)

-

-- = =

( mark)

or SS

760 P0.0184

P-

= or S

760 1 0.0184;P

- = S760

P 746.3 mm1.0184

= = (1 mark)

(b) Three, Na2SO4+ 2

42Na + SO (1 mark)(c) Ethylene glycol (C2H6O2). This is because, it has lower molecular weight and hence contains more moles for the same mass

of glycerine. (1 mark)OR

(a) In case of naphthaleneMass of naphthalene (WB) = 0.5126 g

9

Mass of carbon tetrachloride (WA) = 50.0 g = 0.050 kgMolar mass of naphthalene (MB) = 128 g mol

1

Elevation in boiling point (DTb) = 0.402 K

Kb = B b A

B

M T WW

D ( mark)

1(128g mol ) (0.402K) (0.050 kg)(0.5126 g)

- = ( mark)

= 5.019 K kg mol1 ( mark)In case of unknown solute,Mass of solute (WB) = 0.6216 gMass of solvent (WA) = 50.0 g = 0.05 kgElevation in b.p (DTb) = 0.647 KMolal elevation constant (Kb) = 5.019 K kg mol

1

MB = b B

b A

K WT W

D ( mark)

1(5.019 K kg mol ) (0.6216 g)

(0.647 K) (0.05 kg)

- =

( mark)

= 96.44 g mol1 ( mark)(b) Both elevation in boiling point temperature and depression in freezing point temperature are not so useful for determining

the molar mass of macro-molecules because DT is very small in both the cases. Moreover some macro-molecules, proteinsin particular break at elevated temperatures. Under the circumstances, osmotic pressure is most widely used. It can bedetermined even at room temperature. (2 marks)

(c) The elevation in boiling point will become four times because DTb = Kb2

2 1

W 1000M W

(1 mark)

25. (a) (i) + 2 oxidation state of Mn is most stable. In this oxidation sate, it has exactly half-filled d-orbitals.(ii) Osmium, (Os) shows highest oxidation state in osmium tetraoxide (OsO4).(iii) Osmium (Os) is the densest transition element. (1 + 1 + 1 = 3 marks)

(b) Balanced ionic equations for the reactions are given below:

(i) + 2+4 2[MnO + 8H + 5e Mn + 4H O] 2

| 2

+ 2+4 2 2

COO2CO +2e 5

COOCOO

|2MnO 5 16H 2Mn 10CO 8H OCOO

+ + + +(1 mark)

(ii) + 2 3+2 7 2

+ 2 3+2 7 2 2

6e 14H Cr O 2Cr +7H O 2I I + 2e ] 32

14H + Cr O + 6I 2Cr + 7H O + 3I

+ +

(1 mark)OR

(a) (i) It is due to poor shielding effect of 4f-electrons. As effective nuclear charge increases, ionic size decreases.(1 mark)(ii) Ionic character of metal halide depends upon the electronegativity difference between the metal and halogen. F is more

electronegative than Cl, Br, therefore fluorides are ionic whereas chlorides and bromides are covalent in nature.

10

(1 mark)(b) The information suggests that the gas (A) is H2S while the gas (B) is SO2. Both turn acidified K2Cr2O7 paper green.

(1 mark)

2 2 7 2 4 2 2 4 2 4 3 2(Gas A) (green)

K Cr O 4H SO 3H S K SO Cr (SO ) 7H O 3(S)+ + + + + ( mark)

2 2 7 2 4 2 2 4 2 4 3 2(Gas B) (green)

K Cr O H SO 3SO K SO Cr (SO ) H O+ + + +( mark)

2 3 2 3(Black ppt.)(Gas A)

H S (CH COO) Pb PbS 2CH COOH+ + ( mark)

2 2 2(Yellowish turbidity)(Gas A) (Gas B)

2H S SO 2H O 3S+ + ( mark)

26. (a) Structures of A, B and C are decided in the following manner:

LiAlH43 3 3 3Reduction

Propan-2-olAcetone'B''A'

CH CCH CH CHCH|| |O OH

Propene'C'

2-Bromopropane(Bromide)

CH CHCH + KOH( .) CH CH==CH + KBr + H O 3 3 3 2 2|Br

alc (1+1+1 = 3 marks)

(b) (i) Propanal and propanone : The two compounds can be distinguished by iodoform test. Propanone contains CH3COgroup and gives iodoform test while propanal (CH3CH2CHO) does not give iodoform test. (1 mark)

3 3 3 3Iodoform

CH COCH + 3NaOI CH COONa+2NaOH + CHI

(ii) Phenol and benzoic acid : The two compound can be distinguished as under:Ferric chloride test : Phenol gives a violet colour with aqueous FeCl3 while benzoic acid gives buff coloured ppt. offerric benzoate. (1 mark)

6 5 3 6 5 3Phenol Violet colour

3C H OH + FeCl (C H O) Fe +3HCl

6 5 3 6 5 3Benzoic acid Ferric benzoate

(Buff coloured ppt.)

3C H COOH + FeCl (C H COO) Fe+ 3HCl

OR

(a)

CH OH2

'B'

oxidation

CHO

'A'

50%NaOH

CH OH2

'B'Benzyl alcohol

+

COONa

'C'Sodium benzoate

Heat

NaOH(CaO)

'D'Benzene

+ Na2CO3

(1 + + + 1 = 3 marks)(b) (i) 4-Chloropentan-2-one.

(ii) 3-Bromo-2-methylbutanal. (1 + 1 = 2 marks)

11_1