NCTC (2009) ─ DETAILED SOLUTIONS
Q1
CORRECT ANSWER: E
This question tests students on the understanding of ideal gas behavior and the
application of the ideal gas equation of state, as well as the mass, mole and molar mass
relations. First step is using the ideal gas equation of state to calculate the number of
moles of the unknown gas as follows:
n = (PV)/(RT) = (98130/101325)(0.125)/(0.08205)(293.15) = 5 x 10-3. The molar mass
can be calculated from the relation: M = 0.1727/n = 0.1727/(5 x 10-3) = 34.3 The
unknown gas is then identified as O2 which has a molar mass of 32, nearest to the
calculated value.
Common mistakes here are: using the incorrect values for pressure, volume, the R
constant, and/or temperature.
Q2
CORRECT ANSWER: D
The stoichiometry of the reaction shows that 1 mole of CO2 will be absorbed by 1 mole
of Li2O. So the number of moles of CO2 that can be absorbed by 1 kg of Li2O is
1000/29.8 = 33.56, where 29.8 is the molar mass of Li2O. The volume of CO2 absorbed
can then be calculated using the ideal gas equation of state as follows:
(33.36)(0.08205)(298.15)/(1) = 821 L. Again, common mistakes students make here are
the use of incorrect values of the R constant, and the use of incorrect units for pressure
and volume.
Q3
CORRECT ANSWER: C
This question tests knowledge of reaction stochiometry and manipulation of chemical
formulae.
First calculate the number of mols (n) of Ag+.
n(Ag+) = 0.05 mol dm–3 × 0.033 dm–3 = 1.65 x 10–3 mols.
There are an equal number of mols of Ag+ as there are of Cl–.
There are one third the number of mols of MCl3 compared to number of mols of Cl–.
n(MCl3) = 1.65 x 10–3 × 1/3 = 5.5 x 10–4 mols.
The molar mass (g mol–1) of MCl3 = mass (g)/ n (mols) = 0.100 g / 5.5 x 10–4 mols =
181.8 g mol–1.
The molar mass of Cl3 = 3 x 35.45 g mol–1 = 106.35 g mol–1.
Therefore, the molar mass of M = (181.8 – 106.35) g mol–1 = 75.5 g mol–1.
The group 5A element with a molar mass closest to 75.5 g mol–1 is arsenic (74.9 g mol–1).
(A), (B), (D) and (E) are incorrect because they are the other group 5A elements.
Q4
CORRECT ANSWER: E
From the stoichiometry of the equations, one mole of Na2SO4 should ideally yield two
moles of NaHCO3. However, since there is only a 90% yield for each step, for every
mole of Na2SO4, there will only be 0.93×2.0 = 1.458 moles of Na2SO4 yielded.
The molar mass of Na2SO4 is 142g/mol, hence 100kg is equivalent to 704.23 moles of
Na2SO4. This will yield 1.458×704.23 = 1026.76 moles of NaHCO3. Since the molar
mass of NaHCO3 is 84g/mol, the final amount NaHCO3 yielded in mass is
0.084×1026.76 = 86.2 kg.
(A) and (B)
(C)
(D)
Student is making random guesses
Student may have thought that the three steps in total give 90% yield
Student may have though that there is only two steps in the reaction
Q5
CORRECT ANSWER: D
Most of the volume of an atom IS empty space. Alpha particles are only deflected back
if they hit the nucleus.
Q6
CORRECT ANSWER: E
None of the statements are true. The mass of a proton or a neutron is not exactly 1
a.m.u. Hence the mass number would not be equal to the mass of the nucleus, also an
atom of a stable isotope actually weighs less than the sum of the masses of
electrons, protons and neutrons from which it is built. Note also, that there are actually 3
isotopes of the hydrogen atom ( 11 H, 21 H, 31 H ).
Q7
CORRECT ANSWER: A
According to the Aufbau principle, in the process of building up an atom, electrons are
filled in the atomic orbitals according to this order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d,
5p, 6s, 4f, 5d, 6p, 7s…
Q8
CORRECT ANSWER: C
Ionization energies generally increase as we move up and to the right on the periodic
table. Be is above Mg and to the right of Li. Second ionization energies are greater than
the first ionization energies as a rule, since we are removing an electron from an atom
which already has a charge of +1.
(A), (B), (D) and (E) are wrong. The student does not have a clear idea of first and/or
second ionization energies.
Q9
CORRECT ANSWER: A
In an ionic bond, one atom has donated electron(s) to the other. The two resulting ions
have opposite charges and form an ionic bond based on their electrical attraction for each
other.
(B)
In a polar covalent bond, two atoms share electrons, although they share
unequally.
(C)
(D)
(E)
Q10
Electrons are not donated in hydrogen bonds.
In a nonpolar covalent bond, two atoms share electrons equally.
Obviously incorrect.
CORRECT ANSWER: C
3 lone pairs and 2 bond pairs will give rise to a trigonal bipyramidal shape for the
electron geometry.
(A)
Most students will choose this answer. We are talking about the electron
geometry in this question, so the 3 lone pairs of electrons have to be taken into
account. If the question asked for molecular geometry, then it would be linear
shape.
(B)
The student did not take into account 2 bond pairs.
(D)
A total of 5 pairs of electrons for triiodide. Trigonal pyramidal is a shape for
molecules
with a total of 4 pairs of electrons.
(E)
A total of 5 pairs of electrons for triiodidie. Octahedral is a shape for molecules
with a total of 6 pairs of electrons.
Q11
CORRECT ANSWER: D
For H-O-O-H, each O atom has two lone pairs.
Q12
CORRECT ANSWER: A
Using VSEPR, SO3 structure can be drawn as:
Because no lone pair exists, choices (B) and (C) should be ruled out.
The molecule consists of three oxygen and a sulfur atoms, therefore, (D) and (E) cannot
be the answer.
Q13
CORRECT ANSWER: B
CH4 does not have hydrogen bond. All the other molecules have hydrogen bonds.
Q14
CORRECT ANSWER: A
In this compound, the central atom Br has one lone pair and five covalent bonds with the
F atoms. The VSEPR electron pair geometry is therefore octahedral. With one lone
pair, the molecular geometry should be a square pyramid.
(B)
(C)
(D) and (E)
Q15
This will be correct if Br has no lone pairs, and have five covalent bonds
with the F atoms.
Student is not familiar with VSEPR theory. This structure does not exist in
VSEPR theory.
Student is probably making a random choice.
CORRECT ANSWER: B
In this compound, the central atom O has two lone pair and two covalent bonds with the
Cl atoms. The VSEPR electron pair geometry is therefore tetrahedral. Therefore the ClO-Cl bond angle should be near 109.5 degrees.
Q16
CORRECT ANSWER: B
In this compound, the central atom S has one lone pair and four covalent bonds with the
F atoms. The VSEPR electron pair geometry is therefore trigonal bipyramidal. With one
lone pair, the molecular geometry should be a ‘see saw’.
(A)
(C)
(D)
(E)
Student realizes correctly that VSEPR electron pair geometry is trigonal
bipyramidal but does not realize where to place the lone pair. (B) is more
favourable as lone pair takes up more space and is more favourable in the
equatorial position.
Student got the VSEPR electron pair geometry wrong.
Student mistakenly thinks that there is no lone pair at the central atom.
Student is probably making a random guess.
Q17
CORRECT ANSWER: C
Under constant temperature and pressure, the volume of an ideal gas increases linearly
with the number of moles of the gas. This is as given by the ideal gas equation of state,
i.e., PV = nRT. The equation shows also that V is zero for n = 0, so the correct answer
must be line c, not line b. Lines a, d, and e are all answers from pure random guess.
Q18
CORRECT ANSWER: B
This question tests understanding of the gas laws, reaction stochiometry and the correct
use of SI units.
1 mol of NaN3 produces 3/2 mols of N2
From the ideal gas equation:
pV = nRT
nRT
V=
p
3 mol × 0.0820562 dm3 atm K -1 mol-1 × 298.15 K
V= 2
1.00 atm
= 36.7 dm3
(A)
(C)
(D)
(E)
Incorrect because 8.3143 J K−1 mol−1 was used for the gas constant, R.
Incorrect because 1 mol of N2 was used in the calculation.
Incorrect because the temperature was not converted to kelvin, K.
Incorrect because 1.00 × 105 Pa was used for the units of pressure, p, in the
calculation.
Q19
CORRECT ANSWER: A
The volume of an ideal gas is given by V0 = nRT/P. The new volume at 0.5T and 2P is V1
= nR(0.5T)/(2P). The ratio between V0 and V1 gives V1 = 4V0.
Q20
CORRECT ANSWER: B
For a non-ideal gas that we are concerned only with the non-zero molecular volume, one
can represent the gas law with
nRT
P=
V − vm
where vm represent the molecular volume of the gas molecules. Solving the equation with
the given parameters give vm = 8.19 × 10-6 m3. Hence the ratio of the total molecular
volume of gas molecules X to the volume of the container is (8.19 × 10-6 m3)/ (8.0× 10-4
m3) ≈ 0.01.
Q21
CORRECT ANSWER: B
Number the three equations with ΔH’s given as (1), (2), and (3). Equation (1) minus
equation (2) and minus half of equation (3) gives the reaction equation NO(g) + O(g) →
NO2(g). The enthalpy change for NO(g) + O(g) → NO2(g) is, therefore: (-198.9) - (142.3) - ½(495.0) = -304.1 kJ. The method of enthalpy change calculation is based on the
Hess’s Law of Heat of Summation. Answers (A), (C), (D), and (E) are all pure random
guesses with no understanding of the principle.
Q22
CORRECT ANSWER: A
The amount of heat given from the reaction between the nitrogen and hydrogen gases that
produces 5.00 g of ammonia gas is: 0.42 x 32.16 = 13.51 kJ. The molar mass of ammonia
is 17, and hence the number of moles of ammonia gas produced is 5/17 = 0.294. So the
heat of formation of ammonia is: -13.51/0.294 = -46 kJ/mol. The “-“ sign indicates that
heat is given off when ammonia is produced from reaction between nitrogen and
hydrogen gases. Answer (B) is the heat given for the formation of 5.00 g of ammonia, so
is incorrect. Answer (E) has the same incorrect numerical value as (B) and without the
“-“ sign.
Q23
CORRECT ANSWER: C
The lattice energy of CsCl is described in the following equation.
Cs+(g) + Cl–(g) → CsCl(s) (UL: Lattice enthalpy)
By Hess's law:
ΔHf = V + ½B + I + E + UL
UL = ΔHf – V – ½B – I – E = -443 – 78 – 242/2 – 375 –(-349) = -668 (kJ/mol)
(A)
Students may have used a wrong formula, UL = - ΔHf -V - ½B - I – E = 443 –
78 – 121 – 375 – (-349) = 218 (kJ/mol).
(B)
Students may have used a wrong formula, UL = ΔHf +V + ½B + I + E = -443 + 78
+ 121 + 375 + (-349) = 218 (kJ/mol).
(D)
(E)
Students may have forgotten to divide bond dissociation energy by 2 with
formula, UL = ΔHf – V – B – I – E = -443 – 78 – 242 – 375 –(-349) = -789
(kJ/mol).
Students are making a random guess.
Q24
CORRECT ANSWER: B
Cathode is the electrode where reduction occurs. Also remember that the electrolyte is
molten lithium iodide, so there are only Li+1 and I-1 in the system. The only species that
can be reduced is Li+1, and the only product from the electrolysis at cathode is, therefore,
Li(l) from the reduction of Li+1. Other answers are all pure random guesses with no
understanding of what occurs at cathode in the electrolysis process.
Q25
CORRECT ANSWER: A
The -2 change in the oxidation state at a rate of 4.24 × 10-3 mol/h translates to an electron
flow of [(2 × 4.24 × 10-3)/3600] mol/s = 2.356 × 10-6 mol/s. This multiply by the Faraday
constant (constant for magnitude of electric charge per mole of electrons) gives the
current. Current = (2.356 × 10-6 × 96485) A = 0.227 A.
(B) and (C)
(D) and (E)
Q26
Student may have misplaced or omitted in the equation, the -2 oxidation
state change
Student did not understand the molar concept as applied to electronic
charges.
CORRECT ANSWER: A
This question tests understanding of electrochemical potentials and batteries. The
reactions should both first be written as reductions. In the case of Cu/Cu2+, the direction
of the reaction must be switched so the sign of the potential must also be changed.
Co2+(aq) + 2e− → Co(s)
½ Cu2+(aq) + e− → ½ Cu2+(s)
E0 = −0.28 V
E0 = +0.34 V for reduction
In order to obtain the spontaneous direction, the reaction with the more negative half-cell
potential must be subtracted from the reaction with the more positive half-cell potential.
E0cell = +0.34 – (–0.28) = +0.62 V
The E0cell must be positive in order for the reaction to be spontaneous. This is because the
ΔG will be negative (ΔG0 = –nFE0cell).
(B)
(C)
Incorrect because it is the answer that would be obtained if the half-cell potential
for the Co/Co2+ system was divided by the number of electrons (2). The number
of electrons transferred in the reaction does not affect the potential.
Incorrect because it is simply the difference in potential between the two written
reactions (–0.28 – –0.34).
(D)
(E)
Incorrect because it is simply the difference in potential between the two written
reactions and it is negative.
Incorrect because the sign is incorrect.
Q27
CORRECT ANSWER: C
The information that when B is placed in acid solution, gas bubbles form on its surface,
indicates that reaction (E) occurs. Furthermore, since A can reduce B2+ but not C2+, and if
reaction (E) occurs spontaneously as said, then reaction (A) and reaction (D) would also
occur spontaneously. It also follows that reaction (B) would occur spontaneously.
Reaction (C) cannot occur because B cannot reduce C from the information that A can
reduce B2+ but not C2+. Students need to have clear concepts of what oxidation and
reduction are, and be able to work out from a series of information the relative strength
of oxidants and reductants to answer this question correctly.
Q28
CORRECT ANSWER: C
In a heterogeneous equilibrium such as the one given in this question, the equilibrium
constant is equal to the equilibrium partial pressure of CO2 gas. Secondly, since the
equilibrium constant of a reaction system is dependent only on temperature, so as long as
temperature is maintained at 300oC, the equilibrium constant, and the equilibrium partial
pressure of CO2 will remain unchanged. Answer (C) is therefore the correct answer. The
concept as explained is key to the correct answer, and other answers all arise from
incorrect understanding of the concept of chemical equilibrium, particularly equilibrium
in heterogeneous systems.
Q29
CORRECT ANSWER: C
First, the reaction quotient Q is calculated as follows : (1.055)2/(0.127)(0.134) = 65.4.
Comparing Q and Kp : if Q is greater than Kp, the reverse reaction will take place ; if Q is
less than Kp, the forward reaction will occur ; and if Q = Kp, the system is at equilibrium.
Since Q = 65.4 is greater than Kp = 55.2, the reverse reaction will proceed to establish
equilibrium, and that is answer (C). The concept involves here is clear and straight
forward, so other answers all arise from not understanding the concept correctly.
Q30
CORRECT ANSWER: D
From the 1st step (the association of 2 A atoms to form an A2 molecule), we have the
expression K1[A]2 = K2[A2].
Then [A2] = K1[A]2/K2
Substituting this expression into the rate of equation, Rate = K3[A2][B], we will get the
final rate equation: Rate = K[A]2[B] since K1, K2, and K3 are all constants.
Therefore the overall order of reaction is 3, 2nd order w.r.t A and 1st order w.r.t B.
(A)
(B)
Doubling of A will not double the rate of reaction, it will only push the
equilibrium of the reversible reaction to the right, thus forming more X2.
Doubling [B] will double rate of reaction, but by doubling [A] and [B], the rate of
reaction will not increase by 8 times.
(C)
(E)
Q31
There cannot be intermediates (A2) as part of the rate equation.
The students may be guessing the answer.
CORRECT ANSWER: B
Taking 1 mole of N2O4, the amount (50%) dissociated = 0.5 mole N2O4.
Amount of NO2 formed = 1 mole (From stoichometry ratio)
Partial pressure of N2O4 = 0.5/1.5 * 1 = 1/3
Partial pressure of NO2 = 1/1.5 * 1 = 2/3
Therefore equilibrium constant = (2/3)2/(1/3) = 1.33
(A)
(C)
(D)
(E)
The student made a mistake at the end where he forgets to square the partial
pressure of NO2, therefore (2/3)/(1/3) = 2
The student doesn’t know that when 1 mole of N2O4 is 50% dissociated, he will
get 1 mole of NO2. Using 0.5 mol as the amount of NO2, (0.5)2/(0.5) = 0.5
The student balanced the equation to be 1 N2O4 : 2 NO2.
Upon dissociation, he gets 0.5 mole of N2O4 : 3 NO2.
Using the equation for equilibrium constant, he gets (3)2/(0.5) = 18
The student balanced the equation to be 1 N2O4 : 2 NO2.
Upon dissociation, he gets 0.5 N2O4 : 2.5 NO2.
Q32
CORRECT ANSWER: C
Reaction equation (1) with equilibrium constant K1 plus reaction equation (2) with
equilibrium constant K2 gives reaction equation (3) with equilibrium constant K3. The
three equilibrium constants are thus related in the form: K1 x K2 = K3. Thus, K2 can be
calculated from K1 and K3 as follows : K2 = K3/K1 = (1.3 x 10-20)/(1.0 x 10-7) = 1.3 x 10-13,
and that is answer (C). Answer (A) is the result from K1 x K3 and that is incorrect.
Answer (D) is obtained from K1/K3 and that again is incorrect.
Q33
CORRECT ANSWER: D
For a reaction in which reactant A has a reaction order of one means the the rate of
change of reactant A can be described as
d [A]
d [ A]
= −k [ A] →
= − kdt
[A]
dt
Integrating both sides of the equality sign gives
ln[A] = −kt + C
Solving for the initial condition [A] = [A]0 at t = 0, we get
ln
[A]
[A]0
= − kt .
Q34
CORRECT ANSWER: D
When a reaction is described as being zero order with respect to a particular reactant, it
means the rate of reaction is independent of that reactant concentration. Hence choice (D)
is correct.
Q35
CORRECT ANSWER: E
To lower the activation energy and to increase the frequency of collisions of reactant
molecules are the ways to enhance the reaction rate.
I.
Incorrect: Heat (exothermic) is not related to the activation energy that is lowered
by a catalyst.
II.
Incorrect: The activation energy is not related to temperature.
III.
Correct.
IV.
Incorrect: A catalyst increases the rate of a reaction by lowering the activation
energy.
V.
Correct.
Q36
CORRECT ANSWER: D
Replacing the variables in the rate equation with the values given, we have
0.025 mol dm-3s-1 = k × (0.1 mol dm-3) × (0.5 mol dm-3)2
Solving for k gives the correct answer and units.
Q37
CORRECT ANSWER: A
Acid accumulators use sulphuric acid, and table salt is sodium chloride. Students should
already have learnt from school (group VII elements) that mixing these will produce
hydrogen chloride which is acidic. Dry batteries contain manganese dioxide (they should
have learnt that Chemistry or Physics), so addition of this oxidizing agent will produce
chlorine instead of hydrogen chloride (from group VII chemistry as well). They should
also have been taught that chlorine bleaches litmus paper
It is unlikely anyone should choose option (B). Choosing option (C) is likely made by
those who has a “feeling” that the question is to do with oxidation but not sure where,
assuming that the sulphuric acid somehow decomposes differently under differing
conditions. Options (D) and (E) will be by the really clueless.
Q38
CORRECT ANSWER: D
They should know from the Periodic Table that radium is in group II (alkaline earth
metals), so its properties will closely resemble that of barium.
There is little reason for them to choose any of the other options if they know their group
II chemistry and periodic trends. The likely mistake may be choosing option (C), if they
fail to realize that sulphuric acid is also a source of sulphate ions and not just an acid.
Option (E) is possibly another red herring since they would not have encountered the
colour of Ra2+, but they should have been taught that Ca2+, Sr2+ and Ba2+ do impart
colours to a flame.
Q39
CORRECT ANSWER: D
Since the colour is said to be due to a d-d transition, students should be able to rule out
options (A) to (C); MnO4- is Mn(VII) which has no electrons in a d orbital. For the last
two options, either is possible; Ti3+ is d1 while V3+ is d2. Very few will know that the
latter gives rise to 3 different d-d transitions, although they can always make a guess
(one electron one transition)!
Q40
CORRECT ANSWER: B
[Oxidation State of Mn] + 6*(-1) = -4. [Oxidation State of Mn] = +2
Ground State of Mn electronic configuration: [Ar] 4s23d5
Therefore electronic configuration of Mn2+ = [Ar] 3d5
Other answers cannot be provided if students calculate the oxidation state correctly.
Q41
CORRECT ANSWER: C
SbF3 + F2 → SbF5
Redox reaction: Sb +3 → +5, F (in F2) 0 → -1
(A)
(B)
(D)
(E)
2NO2 → N2O4
No oxidation change: N +4, O -2 on each side
6HCl + As2O3 → 2AsCl3 + 3H2O
No oxidation change: H +1, Cl -1, As +3, O -2 on each side
[Cr2O7]2- + 2[OH]- ' 2[CrO4]2- + H2O
No oxidation change: Cr +6, H +1, O -2 on each side
[H3O]+ + [OH]- ' 2 H2O
No oxidation change: O -2, H +1 on each side
Q42
CORRECT ANSWER: E
The Sn2+ ion has a higher charge density than Pb2+ owing to its smaller ionic radius (Sn2+
93pm, Pb2+ 119 pm), so it will polarize the water molecules more strongly, resulting in
increased loss of H+ and formation of complexes with coordinated hydroxide ions.
Q43
CORRECT ANSWER: A
The lone pairs of electrons on adjacent N atoms repel each other owing to the small
atomic size, weakening the N-N bond.
Q44
CORRECT ANSWER: D
The structure of H4P2O6 is stated below:
HO
OH
HO
OH
P
O
P
O
H4P2O6 has a P-P bond, which does not contribute to the oxidation state. The oxidation
state is therefore one less than in phosphorus acid H3PO4, i.e. +4.
Q45
CORRECT ANSWER: B
Na2O forms NaOH; SO3 forms H2SO4; SO2 forms H2SO3; and P4O10 is a potent
dehydrating agent by forming H3PO4. CuO does not react with water; instead, Cu(OH)2
could lose water upon heating to give CuO. Q46
CORRECT ANSWER: C
The balanced reaction should be
4 H+ + 3 MnO42- Æ 2 MnO4- + MnO2 + 2 H2O,
where 2 equivalents of Mn(VI) are oxidized to Mn(VII) and 1 equivalent of Mn(VI) is
reduced to Mn(IV). Balancing this equation is not required to answer this question and so
It will cost some time for students who are not familiar with redox reactions. A student
would choose (A) if he/she does not check the oxidation states of Mn in the three
compounds. Answers (D) and (E) are designed to confuse students who are not familiar
with redox chemistry.
Q47
CORRECT ANSWER: B
The two compounds are enantiomers – both chiral centres are inverted when going from
one structure to another, although this is not immediately obvious as they are drawn in
different ways. As they are enantiomers, they have identical physical properties and only
differ in their interaction with polarized light.
Q48
CORRECT ANSWER: E
The numbering system is as follows:
CH3
Q49
H3 C 4
2 1
CO2H
3
Br
(A)
(B)
(C)
(D)
Incorrect because the double dong is in the 2-position
Incorrect because the double bond is 4Incorrect because it is a hexenoic acid; also the Br is in the 4- position
Incorrect because it is a hexenoic acid
CORRECT ANSWER: C
An alkyl group is electron donating, increasing the electron density on N.
(A)
(B)
Incorrect because molecular weight does not correlate with basicity.
The hydrogen atoms in the ethyl group cannot hydrogen bond as they are on
carbon which is not electronegative – but even if they did, would that affect the
nitrogen lone pair?
(D)
(E)
Q50
There is no delocalization in either molecule – no pi-bonds
This choice is not relevant
CORRECT ANSWER: B
(B) is a naturally occurring amino acid. It is alanine
(A) is an ester, not an acid
(C) is an amide, not an ester
(D) is a β-amino acid, not α
(E) is an α-amino acid, but not naturally occurring (it has a bromine)
Q51
CORRECT ANSWER: A
Statement I and II are correct
I
Will not because the product will have an extra carbon atom
C N
II
III
IV
Q52
Students may be making a wild guess.
Carboxyl acid is not good nucleophile to react with alkyl chloride
Ester is not good electrophile. Benzene is inert toward ester.
Students may be making a wild guess.
CORRECT ANSWER: C
Methylene protons of 1,3-diketones are most acidic.
(A)
(B)
(D)
(E)
Q54
Will not give any product because the ammonia is a nucleophile and cannot react
with benzene; in addition, in the presence of sulfuric acid, the ammonia will be
converted to ammonium sulfate
Will give aminobenzene. This is a standard methjod to reduce nitro groups to
make amino groups
The compound shown will undergo hydrolysis upon heating with sodium
hydroxide to give aminobenzene and sodium acetate.
CORRECT ANSWER: D
Esters are prepared by reacting acyl chlorides with alcohols.
(A)
(B)
(C)
(E)
Q53
CH2NH2
α-proton of ketone is more acidic compared with normal aliphatic protons, but
less than that of 1,3-diketone.
β-proton of ketone is not acidic.
The same solution with Choice A.
Students may be making a wild guess.
CORRECT ANSWER: A
LiAlH4 reduction of lactones give diols.
(B)
A formyl group should be reduced by LiAlH4.
(C)
(D)
(E)
The C-C bond between the benzene ring and the carbonyl carbon could not be
cleaved
LiAlH4 reduction of ester provides alcohol, not ether.
The C-C bond between the benzene ring and the methylene carbon cannot
cleavage.
Q55
CORRECT ANSWER: D
In the presence of an excess amount of NaBH4, both carbonyl groups would be reduced
to their alcohol functionality. Thus, two chiral centers would be formed in the process
and answer A is incorrect. As each chiral centre can be (R) or (S), the possible
combinations are (R),(R); (R),(S); (S),(R); (S),(S). Thus, the answer is D.
Q56
CORRECT ANSWER: D
Aldehydes are more electrophilic than ketones. Added to this is the fact that the α-proton
of the methyl ketone is acidic and can be easily deprotonated by the NaOH base and form
the nucleophile. Thus, this combination results in the β-hydroxy ketone intermediate.
This further undergoes dehydration to give the conjugated ketone.
Q57
CORRECT ANSWER: B
Among all the oxidants provided, only pyridinium chlorochromate is mild enough to prevent the
oxidation process to give the aldehyde product. All the other oxidants would over oxidize the
alcohol to the carboxylic acid compound.
Q58
CORRECT ANSWER: C
The compound is symmetrical and hence, it only has 3 stereoisomers (meso, cis and trans
isomers).
Q59
CORRECT ANSWER: C
The ring bearing the N is deactivated under acidic conditions and hence, the ring without
the N is more reactive. For the other ring, substitution at C5 will give an intermediate that
is resonance stabilized without disrupting the aromaticity of the electron deficient ring.
Q60
CORRECT ANSWER: A
Compound X does not have any alkene functional group as it does not decolourises
aqueous bromine. A single monochlorination product suggests that Compound X is
highly symmetrical.