Solution

CHEMISTRY SP 6

Class 12 - Chemistry
Section A
1.

(d)

Explanation:

y
The products obtained can be explained by Markonikov’s rule. According to the rule, the hydrogen from HCl will get added to

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the carbon directly bonded to the most hydrogen atoms(addition reaction), and the –Cl will get bonded to the carbon directly
bonded to the least hydrogens.

2.
(b) Amino acids
d
ca
Explanation:
Proteins are made up of amino acids containing mostly hydrogen, carbon, nitrogen and oxygen, as linked together in chains.
dA

3.
(c) C6H5CHO
Explanation:
C6H5CHO
vin

4.
(c) Williamsons reaction
Ar

Explanation:
The reaction of an alkyl halide with sodium alkoxide to give ether (alkoxy alkane), is known as Williamson's synthesis. In this
reaction, an ether (anisole) is prepared by the action of alkyl halide (methyl iodide) on sodium alkoxide (sodium phenate), so it
is an example of Williamson's synthesis.

5.
(c) increases with increase in temperature.
Explanation:
Value of henry constant increases with increase in temperature.

6.
(d) (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)
Explanation:
(a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)

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 7.
(d) b, c
Explanation:
Chiral molecules(the object which is non-superimposable on their mirror image) contains one carbon atom surrounded by four
non-identical species. All straight-chain molecules cannot be chiral because of the presence of two or more identical groups
like hydrogens. Even the carbons with double or triple bonds to a group cannot be considered as a chiral carbon. An
asymmetric carbon needs to be surrounded by four species different from each other through covalent bonds. Hence, the atoms
b and c are asymmetric.

8.
(d) Tyrosine
Explanation:
Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine

9.
(c) quadruple

y
Explanation:

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The rate of the reaction is quadruple.

10. (a) CH3CH2CHO

Explanation: d
CH3CH2CHO will give aldol reaction because of the presence of alpha hydrogen in it.
ca
11. (a) CH3-CH2-CH2-OH
Explanation:
dA

CH3-CH2-CH2-OH
A compound 'X' with molecular formula C3H8O can be oxidised to a compound 'Y' with the molecular formula C3H6O2 'X' is
most likely to be Primary alcohol.
12.
vin

(c) NO+2
Explanation:
NO+2 Nitronium ion electrophile initiates the process of nitration of benzene. It is obtained as:
Ar

+ −
H2 SO4 (conc.) ⟶ H + HSO
4

+ +
H + HNO 3 ⟶ H2 NO
3
+ +
H2 NO ⟶ NO + H2 O
3 2

13.
(d) If both Assertion and Reason are false statements.
Explanation:
Glycosides are formed by treating glucose with methanol in the presence of dry HCl gas. They cannot be hydrolyzed in acidic
medium. Also, they are hemiacetals and not acetals.

14.
(c) A is true but R is false.
Explanation:
A is true but R is false.

15.
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

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 Explanation:
Both (A) and (R) are true, but (R) is not the correct explanation of (A).

16.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
Being a strong nucleophile sodium methoxide can cause an elimination reaction with a tertiary halide.

Section B
nB
17. π = i V
RT

WB
6.5 = i × MB
×
1000
× 0.0821
V

6⋅1 1000
6.5 = i ×
122
×
100L
× 0.0821 × 300 K
6⋅5×122
i =
6⋅1×0⋅0821×300×10
= 0.528
1−i 1−0.528
α =
1
=
1
= percentage association= 0.944 or 94.4%
1− 1−
n 2

18. The acidified solution of (potassium permanganate) KMnO4 acts as an oxidising agent. It oxidises oxalic acid into CO2 and itself

y
changes to Mn2+ ion which is colourless.

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19. Answer the following:
d
(i) As the number increases the chance of simultaneous collision decrease that means the probability of more than three
molecules colliding simultaneously is very small. Therefore, the possibility of molecularity being three is very low.
ca
(ii) Slope = - k
20. PH = -Log[H+]
dA

The cell reaction is-

+ −
H + e → H2 (g)

According to Nearest Equation

0 0.059 1
E = E − log
n +
[H ]
vin

0.03 V = 0 + 0.059

1
(−log
1

+
)
[H ]

= 0 + 0.059PH
PH = 0.03V

0.059
= 5.07 V
Ar

OR
Q = It
= 5 × 193 = 965C

96500 C = 1 mol of electrons = 6.022 × 10 23

mo l
−1

965
965 C = 6.022 × 10 × 23

96500

= 6.022 × 10
21
electrons

21. i.

ii.

Section C
22. The cell reaction is Zn(s) + 2Ag+ (aq) ⇌ Zn2+(aq) + 2Ag(s).
; Eo (Zn2+|Zn) = -0.76V and Eo (Ag+|Ag) = 0.80 V.
+
The cell is represented as Zn(s) ∣∣Zn 2+
(aq)∥Ag (aq)∣
∣ Ag(s)

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 i. Since zinc (Zn) is more reactive than silver (Ag). So, Zn electrode is negatively charged.
ii. At anode : Oxidation will takes place; Zn(s) → Zn2+(aq) + 2e- (oxidation)
At cathode: Reduction will takes place; Ag+ (aq) + e- → Ag(s) (reduction)
iii. Ions are the carriers of current within the cell.
23. i. IUPAC name of the given complex [Cr(NH3)4Cl2]Cl is tetraamminedichlorido chromium(III) chloride.
ii. [Co(en)3]Cl3: It exhibits optical isomerism and exists in two forms: dextro (rotate the plane polarized light towards right) and
laevo (rotate the plane polarized light towards left) that can be represented as:

iii. In the complex [NiCl4]2-, Ni is in +2 oxidation state and the outer electronic configuration of Ni in +2 oxidation state is

[Ar] 3d8 4s0. The Cl- ion being a weak field ligand cannot pair up the two unpaired electrons present in 3d-orbitals. That
means 3d-orbitals are not involved in hybridization. Thus, the complex is sp3-hybridised tetrahedral and is paramagnetic in

y
nature due to the unpaired electrons.

d em
In the complex [Ni(CO)4], the oxidation state of nickel is zero and outer electronic configuration of Ni is [Ar] 3d8 4s2. In the
ca
presence of strong field ligand CO, the 4s-electrons shifts to the two half-filled 3d-orbitals and make all the electrons paired.
The valence 4s and 4p-orbitals are involved in hybridisation i.e sp3 hybridization Thus, the complex is tetrahedral but
dA

diamagnetic in nature due to the all paired electrons.

.
vin

24. a. C H 3 − C H2 − C H2 C H2 − OH
Butan - 1 - ol

b. C H 3 − C H − C H3
|
Ar

OH

Propan - 2 - ol
CH3

c. C H 3 − C − OH
|

CH3

2−Methylpropan−2−ol

OR
i. 2-Chloropropane to 1-propanol
a. alc. KOH heat,
b. HBr peroxide
c. alc. KOH, heat
ii. Isopropyl alcohol to iodoform
I2/NaOH heat
iii. Chlorobenzene to p-nitrophenol
a. Conc. HNO3+ conc. H2SO4
b. aq NaOH (15%), 433 K (c) dil. HCl
25. 1. Rosenmund reduction: Acyl chloride is hydrogenated over a catalyst, palladium on barium sulphate. This reaction is called
Rosenmund Reaction.

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 O O

|| P d−BaSO4 ||

C H3 − C − C l + H2 −−−−−−→ C H3 − C − H + HC l

Ethanoyl chloride quinoline Ethanal

2. Tollen's reagent: On warming an aldehyde with freshly prepared ammonical silver nitrate solution (Tollen's reagent), a bright
silver mirror is produced due to the formation of silver metal. The aldehydes are oxidised to corresponding carboxylate anion.
The reaction occurs in alkaline medium.
− −
+
RC HO + 2[Ag(N H3 )2 ] + 3 OH → RC O O + 2Ag + 2H2 O + 4N H3

26. The standard electrode potential of Mg2+ | Mg can be measured with respect to the standard hydrogen electrode, represented by
H+(aq, 1 M)|Pt(s)|H2(g), 1 bar.

A cell, consisting of Mg2+ |Mg(ag, 1 M) as the anode and the standard hydrogen electrode H+(aq, 1 M)|Pt(s)|H2(g), 1 bar as the
cathode.
The cell represented as
Mg|mg(aq, 1 M)||H+(aq, 1 M)|Pt(s)|H2(g), 1 bar
Thus, emf of cell, E ∘
= E
∘
− E
∘

cell right left

The potential of standard hydrogen electrode is always zero.
Therefore,

y
∘
E = o
right

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Hence,
∘ ∘
E = −E
cell Mg
2+
∣Mg

∘ ∘
E = −E
Mg
2+
∣Mg cell ad
27. Alkyl halides can undergo both beta-elimination as well as nucleophilic substitution reactions, the desired product can be obtained
using an appropriate reagent and environmental conditions of the system. Consider the case of CH3CH2Br, which can undergo β -
elimination as well as nucleophilic substitution reaction. On treatment with alcoholic KOH, at 473-523°K, ethyl bromide
Ac
undergoes elimination to give ethane(C2H4). On the other hand, treatment with aqueous KOH at a lower temperature of 373°K
gives ethyl alcohol(C2H5OH) by a nucleophilic substitution reaction.
∘
473− 523 K

C H3 C H2 Br + alc. K OH −−−−−−→ C H2 = C H2 + HBr

∘
373 K
d

C H3 C H2 Br + aq. K OH −−−→ C H3 C H2 OH + K Br

28. It is known that,

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2.303 [R]0
t = log
k [R]

2.303 1 2.303
= log = log 16
−1 1 −1
60 s 60 s
16

= 4.6 × 10 (approximately)
−2
s
Ar

Hence, the required time is 4.6 × 10 −2

s.

Section D
+

29. 1. H N C HRC O O
3
−

In aqueous solutions, amino acids mostly exist as zwitterion or dipolar ions.

R

+ |
−
N H3 − C H− C O O
Z witter ion

2. Amino acids are least soluble at their isoelectric points. At a specific pH, called isoelectric point, the positive and negative
charges balance each other and the net charge becomes zero. If there is a charge, the amino acid prefers to interact with water,
rather than other amino acid molecules, this charge makes it more soluble.
2.3+9.7
3. Isoelectric point = 2
=6
OR
X>Z>Y
+

Carboxylic acids are stronger acids than - N H , therefore X is the strongest acid. Since -COOH has -I effect which decreases
3

with distance therefore, effect is more pronounced on Z than on Y. As a result Z is more acidic than Y, therefore, the overall
order of increasing acid strength is X > Z > Y.

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30. i. Geometrical Isomerism.

ii. Solvate isomerism

iii. Linkage isomerism is the existence of coordination compounds that have the same composition differing with the connectivity
of the metal to a ligand Typical ligands that give rise to linkage isomers are thiocyanate, SCN, isothiocyanate, NCS
OR
Linkage isomers.
Section E
31. Attempt any five of the following:
(i) a. 2MnO + H2O + I- → 2MnO2 + 2OH- + IO
+
3
−
4

b. 2MnO + 10I- + 16H+ → 2Mn2+ + 8H2O + 5I2

+

(ii) Cr (Z=24) : [Ar]3d54s1

y
(iii)When the particular oxidation state becomes less stable relative to the other oxidation states, one lower and one higher, it

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is said to undergo a disproportionation reaction.
For example, 3M n VI
O
2−

4
+ 4 H+ ⟶ 2 M n V II
O
−

4
+ MnIVO2 + 2H2O
(iv)Cr2+ is a strong reducing agent because after the loss of one electron Cr2+ becomes Cr3+ which has more stable t g
2 3
(half-filled) configuration. d
(v) Group 6 has Cr element in its 3d series. Like Chromium other elements of this group shows +3 and +6 as their common
ca
and stable oxidation state.
Stability of +3 comes from fully filled t2g configuration and stability of +6 comes from nobel gas configuration.
dA

(vi)+2 and +4 oxidation states are due to high stabilities of f0, f7 and f14 configuration.
(vii)Reactivity of an element is dependent on the value of ionization enthalpy. In moving from Sc, the first element to Cu, the
ionization enthalpy increases regularly. Therefore, the reactivity decreases as we move from Sc to Cu.
32. i. mass of solute = 19.5g
vin

molar mass of solute (F- Cl2 - COOH) = 78 g mol-1

mass of solvent i.e water = 500g; kf value for water = 1.86 k kg mol-1
depression in freezing point = 1 C
Ar

degree of dissociation of solute = ?

No. of moles solute = = 0.2519.5

78

molality is the no. of moles of solute in 1 kg solvent

0.25
molality = 500
= 0.50 m
1000

Calculated depression in freezing point;

ΔTf = Kf × m

= 1.86 × 0.50 = 0.93 K

Observed freezing point
i=
Calculated freezing point
i= 1.0

0.93
= 1.0753
Let, C is the initial conc. of bluoroacetic acid and α be its degreee of dissociation.
− +
C H2 F C OOH → C H3 F C O O + H
C(1−α) Cα Cε

Total number of moles = C(1 - α ) + Cα + Cα

= C(1 + 2)
1.0753 = 1 + α
α degree of dissociation of solute= 0.0753

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 ii. a. KCl dissociates in the solution and forms ions K+ and Cl- and glucose does not dissociate. Since the boiling point is a
colligative property and depends on the number of particles. Therefore, 0.1 M KCl has a higher boiling point than the 0.1
M glucose.
b. Meat is preserved for a longer time by salting so that it can be protected against bacterial action for longer.
OR
i. Mole fraction:
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total
number of moles of all the components in the mixture.
Number of moles of the component
i.e., Mole fraction of a component = Total number of moles of all components

Mole fraction is denoted by 'x'.

If in a binary solution, the number of moles of the solute and the solvent are nA and nB respectively, then the mole fraction of
the solute in the solution is given by,
nA
xA =
nA + nB

Similarly, the mole fraction of the solvent in the solution is given as:
nB
xB =
nA + nB

ii. Molality

y
Molality (m) is defined as the number of moles of the solute dissolved per kilogram of the solvent. It is expressed as:
Molality (m) = It is expressed as mol/Kg
Moles of solute

em
Mass of solvent in kg

iii. Molarity
Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of the solution.
It is expressed as: ad
Molarity (M) = It is expressed as mol/L
Moles of solute

Volume of solvent in Litre

iv. Mass percentage:

The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the
Ac
solution. It is expressed as:
Mass of component in solution
Mass % of a component = Total mass of solution
× 100

33. i.
d
vin

ii.
Ar

HN O2 N H3 Br2 /KOH
CH3 NH 2
iii. C 2 H5 C H2 N H2 −−−−→ CH3CH2OH −−−−−−−−−−→ CH3CHOOH −−−→ CH3CH2NOH2 −−−−−−→
Ethyla min e K2 C r2 O7 /H2 S O4 Δ Methylamine
OR
Sn/HCl

i. C6H5-NO2 −−−−→ C6H5.NH2 (Aniline)

ii.

Br2 /KOH

iii. CH3CONH2 −−−−−−−−−−−−−−−−−−−→ CH3NH2+ K2CO3 + 2KBr + 2H2O

Hof mann bromamide degradation

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