1. Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 56/1/1
Candidates must write the Code on
Roll No. the title page of the answer-book.
 Code number given on the right hand side of the question paper should be written on the title page of
the answer-book by the candidate.
 Please check that this question paper contains 30 questions.
 Please write down the Serial Number of the questions before attempting it.
 15 minutes time has been allotted to read this question paper. The question paper will be distributed at
10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not
write any answer on the answer script during this period.
CHEMISTRY
[Time allowed : 3 hours] [Maximum marks : 70]
General Instructuions:
(i) All questions are compulsory.
(ii) Questions numbered 1 to 8 are very short-answer questions and carry 1 mark each.
(iii) Questions numbered 9 to 18 are short-answer questions and carry 2 marks each.
(iv) Questions numbered 19 to 27 are also short-answer questions and carry 3 marks each.
(v) Questions numbered 28 to 30 are long-answer questions and carry 5 marks each.
(vi) Use Log Tables, if necessary. Use of calculators is not allowed.
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2. STUDYmate
1. How many atoms constitute one unit cell of a face-centered cubic crystal?
1 1
Ans. 8 × +6× =4
8 2
In one unit cell, 4 atoms are present.
2. Name the method used for the refining of Nickel metal.
Ans. Mond’s Process
3. What is the covalency of nitrogen in N2O5?
Ans. 4 (Four)
O O
N–O–N
O O
4. Write the IUPAC name of CH3 – CH – CH2 – CH = CH2
Cl
Ans. 4-Chloropent-1-ene
5. What happens when CH3 – Br is treated with KCN?
Ans. Nucleophilic substitution occurs
+–
CH3Br + KC  N  CH3CN + KBr
6. Write the structure of 3-methyl butanal.
CH3 O
Ans. H3C – C – CH2 – C – H
H
3-Methyl butanal
7. Arrange the folllwing in increasing order of their basic strength in aqueous solution:
CH3NH2, (CH3)3N, (CH3)2NH
Ans. (CH3)2 N H > CH3 N H2 > (CH3)3 N
8. What are the three types of RNA molecles which perform different functions?
Ans. m – RNA
t – RNA
r – RNA
9. 18 g of glucose, C6H12O6 (Molar mass = 180 g mol–1) is dissolved in 1 kg of water in a sauce
pan. At what temperature will this solution boil?
(Kb for water = 0.52 K kg mol–1, boiling point of pure water = 373.15 K)
Ans. Tb – Tb° = Kb × m
18 1 kg
Tb – 373.15 = 0.52 × 
180 1 kg
Tb = 0.052 + 373.15
Tb = 373.202 K
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3. STUDYmate
10. The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm–1. Calculate its molar
conductivity.
k  1000
Ans. m = ; m is molar conductivity.
M
0.025  1000
=
0.20
= 125 S cm2 mol–1.
11. Write the dispersed phase and dispersion medium of the following colloidal systems:
(a) smoke (b) milk
OR
What are lyophilic and lyophobic colloids? Which of these sols can be easily coagulated on the
addition of small amounts of electrolytes?
Ans.
Dispersed phase Dispersion medium
Smoke Solid Gas or Air
Milk Liquid Liquid
OR
Ans. Lyophilic colloids are liquid loving. Certain substances mixed with a suitable liquid (dispersion
medium) directly forms the colloid. Such colloids are lyophilic colloids. Example, gum, gelatin,
starch, etc.
Lyophobic colloids are liquid hating. When substances like metals or metal sulphides are
mixed with the dispersion medium do not form the colloidal sol.
Such type of sols can be prepared by special methods.
Example: Metal sulphides.
On addition of small amount of electrolyte lyophobic collides can be precipitated.
12. Write the differences between physisorption and chemisorption with respect to the following:
(a) specificity (b) temperature dependence
(c) reversibility and (d) enthalpy change
Ans. Physisorption Chemisorption
1. Physisorption is non- specific in nature. 1. Chemisorption is highly specific in
As physisorption involves Vander nature as chemisorptions includes
Waal’s interaction between adsorbate chemical bond formation between
and adsorbent. adsorbate and adsorbent.
2. On increasing temperature 2. Chemisorption increases with increase
physisorption decreases. It is because in temperature. After reaching a
desorption occurs rapidl y. maximum value, extent of adsorption
Physisorption occurs usually at low decreases with temperature.
temperature.
3. Physisorption is reversible in nature. 3. Chemisorption is irreversible in
nature.
4. In case of physisorption enthalpy 4. In case of chemisorptions enthalpy
change is small. change is high.
For example: 20-40 kJ mol–1 (approx). For example: 40-400 kJ mol–1 (approx).
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4. STUDYmate
13. (a) Which solution is used for the leaching of silver metal in the presence of air in the
metallurgy of silver?
(b) Out of C and CO, which is better reducing agent at the lower temperature range in the
blast furnance to extract iron from the oxide ore?
Ans. (a) Aqueous NaCN
(b) CO is a better reducing agent at lower temperature.
rG° would be more negative for reduction in blast furnace to extract Fe from the oxide
ore by using CO.
14. What happens when
(a) PCl5 is heated? (b) H3PO3 is heated?

Ans. (a) 
PCl5  g   PCl3  g   Cl2  g  (b) 
4H3PO3  3H3PO4 + PH3

15. (a) Which metal in the first transition series (3d series) exhibits +1 oxidation state most
frequently and why?
(b) Which of the following cations are coloured in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
Ans. (a) Copper shows +1 oxidation state extensively. Cu+ has 3d10 configuration.
(b) V3+ = 3d2+ 4s0
Mn2+ = 3d5 4s0
V3+ as well as Mn2+ both have unpaired electrons in their d-orbitals which can show d-
d transition.
16. Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give
two reasons for the same.
Ans. Chlorobenzene is extremely less reative towards nucleophilic substitution reaction due to
conjugation.
(a) Partial double bond character between carbon of benzene and halogen causes cleavage
of bond with difficulty.
(b) Phenyl cation is quite unstable, So SN1 mechanism is not possible too.
17. Explain the mechanism of the following reaction:

H
2CH3  CH2  OH  CH3CH2  O  CH2  CH3  H2O

413K 
Ans. Step 1: Formation of Protonated alcohol
H
CH3CH2 – OH + H CH3 – CH2 – O – H
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5. STUDYmate
Step 2: Formation of protonated ether
H H
H
H3C – CH2 – O + C – OH2 CH3 – CH2 – O – C + H2O
H CH3 H H CH3
Step 3: Formation of ether
H3C – CH2 – O – CH2 – CH3 CH3 – CH2 – O – CH2 – CH3 + H
H
18. How will you convert:
(a) Propene to Propan-2-ol (b) Phenol to 2, 4, 6–trinitrophenol
Ans. Conversion
Propene  Propan-2-ol
H O/H
CH3  CH  CH2  CH3  CH CH3
2

|
OH
Phenol  2,4,6 trinitrophenol (Picric acid)
OH OH
NO2 NO2
conc. HNO3
NO2
(Picric acid)
19. (a) What type of semiconductor is obtained when silicon is doped with boron?
(b) What type of magnetism is shown in the following alignment of magnetic moments?
     
(c) What type of point defect is produced when AgCl is doped with CdCl2?
Ans. (a) Silicon is an element of group 14, Boron belongs to group 13.
So, a p-type semiconductor is resulted.
(b) Ferromagnetism (as all magnetic moments lie in same direction)
(c) AgCl with CdCl2 causes impurity defect leading to cationic vacancies.
20. Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10–2 g of K2SO4 in
2L of water at 25°C, assuming that it is completely dissociated.
(R = 0.0821 L atm K–1 mol–1, Molar mass of K2SO4 = 174 g mol–1)
Ans. K2SO4  2K+ + SO42–
i=2+1=3
Volume of solution (v) = 2L
Mass of K2SO4 (WB) = 2.5 × 10–2
= 0.025 g
Molar mass of K2SO4 (MB) = 174 g/mol
T = 25°C = 298.15 K
 = i CRT
WBRT
=i M V
B
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6. STUDYmate
3  0.025  0.0821  298.15
=
174  2
= 5.27 × 10–3 atm.
21. Calculate the emf of the following cell at 298 K:
Fe(s) | Fe2+ (0.001 M) || H+ (1 M) | H2(g) (1 bar), Pt(s)
(Given E°cell = +0.44 V)
Ans. Cell Reaction
Fe  2H  Fe2  H2
(n 2)

According to Nernst equation
0.0591 [H ]2
Ecell  Eo 
cell log
2 [Fe2 ]
0.0591
= 0.44  (3)
2
= 0.44 + 0.0887
= 0.528 V
22. How would you account for the following?
(a) Transition metals exhibit variable oxidation states.
(b) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(c) Transition metals and their compounds act as catalyst.
OR
Complete the following chemical equations:
(a) Cr2O7  + 6Fe2+ + 14H+ 
2
(b) 2CrO2  + 2H+ 
4
(c) 2MnO4 + 5C2O2  + 16H+ 

4
Ans. (a) Transition metals exhibit variable oxidation states due to the participation of ns and (n
– 1) d electrons in bond formation. Both ns and (n – 1) d electrons have almost equal
energies.
(b) Zr (160 pm) and Hf (159 pm) have almost identical radii due to Lanthanoid contraction.
Lanthanoid contraction is steady decrease in the size of lanthanoid ions (M3+) with the
increase in atomic number. The reason for this is a gradual increase in the effective
nuclear charge experienced by the outer electron.
(c) Transition metal and their compounds act as catalysts,
(i) because of their variable valencies. Transition metals sometimes form unstable
intermediate compounds and provide a new path with lower activation energy for
the reaction.
(ii) in some cases they provide suitable surface for the reaction to take place. The
reactants are adsorbed on the surface of the catalyst where reaction occurs.
OR
Ans. (a) Cr2O7 + 14H+ + 6Fe2+  2Cr3+ + 6Fe3+ +
2– 7H2O
(b) 2CrO4 + 2H+  Cr2 O7 + H2O
2– 2–
(c) 2MnO4 + 5C2O4 + 16H+  2Mn2+ + 8H2O
– 2– + 10CO2
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7. STUDYmate
23. Write the IUPAC names of the following coordination compounds :
(i) [Cr(NH3)3Cl3] (ii) K3[Fe(CN)6]
(iii) [CoBr2(en)2]+, (en = enthylenediamine)
Ans. (i) Triamminetrichloridochromium(III)
(ii) Potassium hexacyanoferrate(III)
(iii) Dibromidobis(ethane–1,2–diamine)cobalt(III)
24. Give the structures of A, B and C in the following reactions:
H O/H NH
C6H5 N2Cl   A  B  C
 CuCN
(i)  2  3 

Sn  HCl NaNO  HCl H O/H
(ii) C6H5 NO2  A  B  C
 2  2 
273K 
O
N2 Cl CN COOH C – NH2
Ans. (i) CuCN H2O/H NH 3

(A) (B) (C)
NO2 NH2 N2 Cl OH
Sn/HCl NaNO2 + HCl H 2O/H
(ii) 273 K 
Nitrobenzene Aniline Benzene diazonium Phenol
chloride
25. Write the names and structures of the monomers of the following polymers:
(i) Buna – S (ii) Neoprene
(iii) Nylon – 6, 6
CH = CH2
Ans. (i) Buna-S CH2 = CH – CH = CH2 +
1,3–Butadiene
Styrene
(ii) Neoprene CH2 = C – CH = CH2
Cl
2–Chloro–1,3–butadiene
COOH
(iii) Nylon 6,6 H2N(CH2)6NH2 + (CH2)4
Hexamethylene diamine
COOH
Adipic acid
26. After watching a programme on TV about the adverse effects of junk food and soft drinks on
the health of school children, Sonali, a student of Class XII, discussed the issue with the
school principal. Principal immediately instructed the canteen contractor to replace the fast
food with the fibre and vitamins rich food like sprouts, salad, fruits etc. This decision was
welcomed by the parents and the students.
After reading the above passage, answer the following questions:
(a) What values are expressed by Sonali and the Principal of the school?
(b) Give two examples of water-soluble vitamins.
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8. STUDYmate
Ans. (a) They are good educated citizens who believe in the saying “healthy mind in a healthy
body.”
This shows they value education and are caring towards the school inmates.
(b) Vitamin B and Vitamin C
27. (a) Which one of the following is a food preservative?
Equanil, Morphine, Sodium benzoate
(b) Why is bithional added to soap?
(c) Which class of drugs is used in sleeping pills?
Ans. (a) Sodium Benzoate
(b) To impart antiseptic properties
(c) Narcotic Analgesics
28. (a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decoposition. Calculate t1/2 for this
reaction.
(Given log 1.428 = 0.1548)
OR
(a) For a first order reaction, show that time required for 99% completion is twice the time
required for the completion of 90% of reaction.
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ accordig to the equation:
Ea 1
log k  log A   
2.303 R  T 
1
where Ea is the activation energy. When a graph is plotted for log k Vs. , a straight
T
line with a slope of –4250 K is obtained. Calculate ‘Ea’ for the reaction.
(R = 8.314 JK–1 mol–1)
dx
Ans. (a) (i)  k[A]2 [B]
dt
(ii) The rate will become 9 times.
(iii) The rate will become 8 times.
(b) t = 40 min.
Finding rate constant by the given data.
2.303 A
k log 0
t At
2.303 1
k log
40 0.7
2.303
 log1.428
40
2.303
  0.1548
40
= 0.0089 min–1
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9. STUDYmate
0.693
Now, t1/2 
k
0.693
t1/2 
0.0089
= 77.865 min.
OR
2.303 a
Ans. (a) t 90%  log
k a  0.9a
2.303
 log10
k
2.303

k
2.303 a
t 99.9%  log
k a  0.999a
2.303 a
 log
k a  0.001a
2.303
t99.9%  3
k
 t 99.9%  t90%  3
E a
(b) The slope of the reaction = 2.303 R
Ea
Now, 2.303 R  4250 K
Ea = 4250 × 2.303 × 8.314
= 81375.35 Joules
= 81.375 kJ
29. (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) H2S2O8
Ans. (a) (i) In F2, the bond between two F-atoms is formed by the head on overlapping of 2p
orbitals, while in Cl2 it is formed by the head on overlapping of 3p orbitals.
2p being smaller in size than 3p has more interelectronic repulsions, which makes
it weaker, thus its bond enthalpy is lower than that of Cl2.
(ii) Due to presence of H-bonding in NH3 its boiling point becomes more than PH3.
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10. STUDYmate
(b) (i) BrF3
F
Br F
F
T-shaped
(ii) (HPO3)3
O
O O
P P
HO OH
O O
P
O OH
Cyclic trimer
(iii) XeF4
F F
Xe
F F
Square Planar
OR
Ans. (a) (i) Deep sea divers depend upon compressed air for their oxygen supply. Thus, a
deep sea diver has both N2 and O2 dissolve considerably in the blood and other
body fluids.
Oxygen is used up for metabolism but due to high partial pressure and greater
solubility of N2, it will remain dissolved and will form bubbles when the diver
comes to the atmospheric pressure. These bubbles affect nerve impulses and
give rise to a disease called bends or decompression sickness. To avoid bends and
also the toxic effects of high concentration of nitrogen in the blood, the cylinders
used by the divers are filled with air diluted with helium.
(ii) Fluorine does not exhibit positive oxidation state because fluorine is the most
electronegative element.
(iii) Because of stronger S–S bonds as compared to O–O bonds, sulphur has a stronger
tendency for catenation than oxygen.
F
O O
(b) (i) Xe (ii) H–O–S–O–O–S–O–H
O O
F
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11. STUDYmate
30. (a) Although phenoxide ion has more number of resonating structures than carboxylate
ion, carboxylic acid is a stronger acid than phenol. Give two reasons.
(b) How will you bring about the following conversions?
(i) Propanone to propane
(ii) Benzoyl chloride to benzaldehyde
(iii) Ethanal to but-2-enal
OR
(a) Complete the following reactions:
Conc. KOH Br /P
(i) 2H  C  H   (ii) CH3COOH 
2

|
|
O
CHO
HNO3/H2SO4
(iii)
273-283 K
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanal (ii) Benzoic acid and Phenol
Ans. (a) The two reasons are :
(i) In carboxylate anion, the resonating structures are exactly equal in energy,
equivalent resonating forms. In phenol the energies of the resonating forms differ.
O O
O O etc.
R–C–O R–C=O
(ii) In the resonating structures of phenol, the negative charge is placed on O-atom
in one of the resonating form and on C-atom in another resonating form. In
carboxylic acid, the negative charge is only on O-atom.
We know that the stability of resonating form is more if negative charge is on
more electronegative atom. So the resonating forms of carboxylate anion are more
stable than phenol.
O
||
Zn  Hg/conc. HCl
(b) (i) CH3  C  CH3  CH3  CH2  CH3

This is known as Clemmenson’s Reduction.
O
C—Cl
Pd-BaSO4, S CHO
(ii) + HCl
boiling xylene
This reaction is called Rosenmund reduction
H
|
(iii) dil NaOH/100C
CH3  CH  O  CH3  C  O  CH3  CH  CH  CHO

The above reaction is aldol condensation
OR
conc.
Ans. (a) (i) 2H – C – H
+
HCOOK + CH3OH [Canizzaro reaction]
KOH
O
O O
P/Br 2
(ii) CH3 – C – O – H CH2 – C – OH [HVZ reaction]
Br
-Bromoacetic acid
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12. STUDYmate
CHO HNO3/H 2SO 4
CHO
(iii) + H2O
273-283 K
NO2
m-Nitrobenzaldehyde
(b) (i) We can use iodoform test to distinguish between ethanal and propanal.
H
CH3 – C = O + 3I2 + NaOH  CHI3 + HCOONa + 3HI
yellow ppt
This test is not given by propanal.
(ii) We can use sodium bicarbonate to distinguish between benzoic acid and phenol.
Benzoic acid is a stronger acid than phenol, so it reacts with NaHCO3 (weak base)
liberating CO2.
C6H5–COOH + NaHCO3  C6H5COONa + H2O + CO2 (brisk effervescence)
Phenol shows no reaction with NaHCO3.
We can also used neutral FeCl3. It reacts with phenol to give violet coloured
solution.
×·×·×·×·×
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13. STUDYmate
Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 56/1/2
UNCOMMON QUESTIONS ONLY
1. What type of stoichiometric defect is shown by AgCl?
Ans. Frenkel Defect
CH3
2. Write the IUPAC name of CH3CH = CH – C – CH3
Br
Ans. 4-Bromo-4-methylpent-2-ene.
4. What type of bonding helps in stabilising the -helix structure of proteins?
O
Ans. Hydrogen bonding between – C – and –NH– groups of the peptide bond.
6. What inspired N. Bartlett for carrying out reaction between Xe and PtF6?
Ans. Neil Bartlett prepared a red compound having the formula O+2PtF6–. He realised that the first
ionization enthalpy of molecular oxygen (1175 kJ mol–1) is similar to Xe (1170 kJ mol–1). So, he
was inspired to synthesise a similar compound using Xenon.
7. What happens when ethyl chloride is treated with aqueous KOH?
Ans. CH3 – CH2 – Cl + aq. KOH  CH3 – CH2 – OH + KCl
Nucleophillic substitution reaction takes place and ethanol is formed.
8. Write the structure of 4-chloropentan-2-one.
H O H H
Ans. H – C – C – C – C – CH3
H H Cl
11. What is the difference between oil/water (O/W) type and water/oil (W/O) type emulsions?
Give an example of each type.
Ans. In oil/water type emulsion, water is the dispersion medium. Example: milk and vanishing
cream.
In water/oil type emulsion, oil is the dispersion medium. Example: butter and cream.
17. (a) Which of the following ores can be concentrated by froth floatation method and why?
Fe2O3, ZnS, Al2O3
(b) What is the role of silica in the metallurgy of copper?
Ans. (a) ZnS because sulphide ores have wettabilty with pine oil which adsorb ore particles on
them.
(b) It acts as flux so that another impurity slag may be formed.
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14. STUDYmate
18. (a) Why does p-dichlorobenzene have a higher m.p. than its o– and m–isomers?
(b) Why is () – Butan-2-ol is optically inactive?
Ans. (a) p-dichlorobenzene have a higher m.p. because its packing is more efficient due to
symmetry.
(b) Racemic mixture.
23. Write the names and structures of the monomers of the following polymers:
(a) Polystyrene
(b) Dacron
(c) Teflon
CH = CH2
Ans. (a) Styrene
(b) Ethylene glycol + Terephthalic acid/Benene-1, 4-dicarboxylic acid
HOH2C – CH2OH + HOOC COOH
(c) Tetra fluoroethene
CF2 = CF2
×·×·×·×·×
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15. STUDYmate
Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 56/1/3
UNCOMMON QUESTIONS ONLY
1. What type of substances would make better permanent magnets, ferromagnetic or
ferrimagnetic?
Ans. Ferromagnetic substances
3. What is the composition of ‘Copper matte’?
Ans. Copper matte chiefly consists of Cu2S and some uncharged FeS.
5. What is a glycosidic linkage?
Ans. Glycosidic linkage formed when two mono-saccharides are held together with the loss of
small molecule’s like H2O. It is an oxide linkage which joins two monosaccharides.
7. Which compound in the following pair undergoes faster SN1 reaction?
Cl Cl
and
Cl
Ans.
8. Write the structure of p-Methylbenzaldehyde molecule.
CHO
Ans.
CH3
9. What is the difference between multi-molecular and macromolecular colloids? Give one
example of each.
Ans. When large number of atoms/molecules of a substance on dissolution aggregate together to
form species having size in the colloidal range, then species thus formed is called
multimolecular colloids. E.g. sulphur sol consisting of S8 particles.
When macromolecules in a suitable solvents form solutions in which the size of the
macromolecules is in colloidal range, then such systems are macromolecular colloids e.g.
starch, cellulose.
17. Account for the following:
(i) The C–Cl bond length in chlorobenzene is shorter than that in CH3 – Cl.
(ii) Chloroform is stored in closed dark brown bottles.
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16. STUDYmate
Ans. As resonance imparts partial double bond character
(i) to C–Cl bond in chlorobenzene.
Cl Cl
(ii) As it get oxidised in presence of sunlight
O2  2CHCl3  2COCl2  2HCl
sunlight

Poisonous gas
23. Give the structures of products A, B and C in the followng reactions:
KCN LiAlH HNO
(i) CH3CH2Br  A  B  C
 4
 0C
2

NH NaOH Br CHCl  Alc. KOH
(ii) CH3COOH  A  B  C

3 2
 3

KCN LiAlH HNO , 0C
Ans. (i) CH3CH2Br  CH3CH2CN  CH3CH2CH2NH2  CH3CH2CH2OH
 4
 2

(A) (B) Pr op 1ol
NH NaOH/Br CHCl
(ii) CH3 COOH  CH3CONH2  CH3NH2  CH3 NC
3 2 3

 Alc. KOH
(A) (B) (C)
27. Write the names and structures of the monomers of the following polymers:
(i) Bakelite (ii) Nylon-6
(iii) Polythene
OH
Ans. (i) & HCHO
Formaldehyde
Phenol
H
O
N
C
CH2 CH2
(ii)
CH2 CH2
CH2
Caprolactam
(iii) CH2 = CH2 Ethene
×·×·×·×·×
-(16)-