Time: 3 hrs
Max. Marks: 70
General Instructions
Read the following instructions carefully.
Section
A
(The following questions are
multiple-choice questions with one correct answer. Each question carries 1 mark.
There is no internal choice in this section.)
Question 1.
The electrolytic purification of copper can be carried out in
an apparatus similar to the one shown below. [1]
The half equation for the reaction at the anode is
(a)
Cu(s) + 2e– → Cu2+
(b) Cu(s) → Cu2+ +
2e–
(c) Cu2+ + 2e– → Cu
(d)
Cu2+ → Cu + 2e–
Answer:
(b) Cu(s) → Cu2+
+ 2e–
Impure copper acts as anode. So, the reaction involved at anode is Cu(s) → Cu2+ + 2e–.
Question 2.
The major product formed when ethanol is heated with HI and
red P is [1]
(a) ethyl iodide
(b) ethane
(c) ethylene
(d) ether
Answer:
(b) ethane
When C2H5OH is heated with HI and red R it forms
ethane.
Question 3.
A magnetic moment of 1.73 BM will be shown by which among the
following, [1]
(a) [CU(NH3)4]2+
(b)
[Ni(CN)4]2-
(c) TiCl4
(d)
[CoC]6]4-
Answer:
(a)
[CU(NH3)4]2+
Magnetic moment, µ is related to number of unpaired electrons as
µ =
\(\sqrt{n(n+2)}\) BM ⇒ (1.73)2 = n(n + 2)
On solving, we get n =
1
Thus, the complex/compound having one unpaired electron exhibit a magnetic
moment of 1.73 BM.
(a) In [CU(NH3)4]2+, Cu2+ = [Ar]
3d9
(Although in the presence of strong field ligand
NH3, the unpaired electrons gets excited to higher energy level but
it still remains unpaired.)
(b) In [Ni(CN)4]2-, Ni2+ = [Ar]
3d8
But CN– being strong field ligand pair up the
unpaired electrons and hence, in this complex, number of unpaired electrons =
0
(c) In [TiCl4], Ti2+ = [Ar] no unpaired electron.
(d) In [CoCl6]4-, Co2+ = [Ar]
3d7
It contains three unpaired electrons.
Thus,
[Cu(NH3)4]2+ is the complex that exhibits a
magnetic moment of 1.73 BM.
Question 4.
Which one of the following alkyl halides will undergo
SN1 reaction most readily? [1]
(a)
(CH3)3C—F
(b) (CH3)3 C—Cl
(c)
(CH3)3C—Br
(d) (CH3)3C—I
Answer:
(d) (CH3)3C—I
All the given compounds are tertiary alkyl halides but the bond formed between carbon and iodine (C—I) bond is the weakest bond due to large difference in the size of carbon and iodine. So, (CH3)3C— I gives SN1 reaction most readily. In other words, iodine is a better leaving group.
Question 5.
Consider the following reaction,
2CHCl3 +
O2 → A + B
The products A and B of above reaction respectively are
[1]
(a) CO2 and HCl
(b) COCl2 and HCl
(c) CO and
HCl
(d) None of the above
Answer:
(b) COCl2 and HCl
The products A and B in the given reaction are COCl2 and HCl.
Chloroform is slowly oxidised by air in the presence of light to an extremely
poisonous gas, carbonyl chloride (phosgene).
Question 6.
The maximum oxidation state shown by Mn in its compound is
[1]
(a) +4
(b) + 5
(c) +6
(d) + 7
Answer:
(d) + 7
In excited state,
Hence, maximum oxidation state exhibited by Mn is+7 which is
in KMnO4.
Question 7.
Which of the following statements is not correct for
aldehydes?
(a) Compounds having —CHO group can reduce Fehling solution.
(b) Formaldehyde is the least polar among ethanal and propanal due to lowest
electron density.
(c) Butanal is more polar than ethoxyethane.
(d)
Aldehydes on oxidation gives 1° alcohols. [1]
Answer:
(b) Formaldehyde is
the least polar among ethanal and propanal due to lowest electron density.
The incorrect statement is option (b). It is because formaldehyde is the most polar among ethanal and propanal due to lowest electron density.
Question 8.
Which has the least freezing point?
(a) 1% Sucrose
(b)
1% NaCl
(c) 1% CaCl2
(d) 1% Glucose [1]
Answer:
(c) 1%
CaCl2
Depression of freezing point is a colligative property (depends only upon the
number of particles of solutes). Thus, the compound which produces maximum ions
has the least freezing point.
∵ Concentration is same.
Question 9.
Match the laws with expressions:
Codes
[1]
Answer:
(a)
The correct match is A – 2, B – 4, C – 3, D – 1.
Question 10.
The reaction by which benzaldehyde is converted in benzyl
alcohol, is [1]
(a) Fittig reaction
(b) Cannizzaro reaction
(c) Wurtz
reaction
(d) Aldol condensation
Answer:
(b) Cannizzaro reaction
The given reaction is called Cannizzaro reaction and involves oxidation and reduction of same compound.
Question 11.
Identify (i), (ii) and (iii) in the following diagram.
[1]
Answer:
(c) (i) Frozen solvent – (ii) Liquid solvent –
(iii)Solution
In the given graph, the curve (ii) shows the variation of vapour pressure of
pure liquid solvent with temperature. The addition of non-volatile , solute
decreases the vapour pressure of solution 1 and therefore curve (iii) shows the
variation of vapour pressure of solution with temperature. Curve (i) corresponds
to the vapour pressure of frozen solvent at different temperature.
Thus, (c)
is the correct option.
Question 12.
What is (are) the number(s) of unpaired electrons in the
square planar [Pt(CN)4]2- ion? [1]
(a) Zero
(b)
1
(c) 4
(d) 6
Answer:
(a) Zero
Outer electronic configuration of Pt is 5d9, 6s1.
Outer electronic configuration of Pt2+ is 5d8.
As
CN– is strong field ligand, so pairing will take place.
So, there is no unpaired electron in the square planar
[Pt(CN)4]2- ion.
Hence, it contains zero unpaired
electrons.
Direction (Q. Nos. 13-16) In the following questions as Assertion (A) is
followed by a corresponding Reason (R). Use the following keys to choice the
appropriate answer.
(a) Both (A) and (R) are true, (R) is the correct
explanation of (A).
(b) Both (A) and (R) are true, (R) is not the correct
explanation of (A).
(c) (A) is correct, (R) is false.
(d) (A) is false,
(R) is true.
Question 13.
Assertion (A) MeNH2 is a weaker base than
MeOH.
Reason (R) N is less electronegative than O, lone pair of electrons on
N is more easily available for donation in MeNH2. [1]
Answer:
(d) (A) is false, (R) is true.
MeNH2 is a stronger base than MeOH
because N is less electronegative than O and lone pair of electrons on N is more
easily available for the donation in MeNH2.
Hence, (A) is false
but (R) is true.
Question 14.
Assertion (A) Bond angle in ether is slightly less than the
tetrahedral angle.
Reason (R) There is a repulsion between the two bulky (—R)
groups. [1]
Answer:
(d) (A) is false, (R) is true.
(A) is false but (R)
is true.
Correct Assertion Bond angle in ether is slightly more than the tetrahedral
angle.
Question 15.
Assertion (A) Glucose does not form the hydrogen bisulphite
addition product.
Reason (R) Glucose is not so reactive to form the product
with NaHSO3. [1]
Answer:
(c) (A) is correct, (R) is false.
Glucose does not form the hydrogen bisulphite addition product because it has
cyclic structure in which —CHO group is not free to react. Therefore,
(A) is
true, but (R) is false.
Question 16.
Assertion (A) CO is stronger complexing reagent than
NH3.
Reason (R) CO is a good σ-donor and a good π-acceptor while
NH3 can form only σ-bonds and no π-bonds. [1]
Answer:
(a) Both
(A) and (R) are true, (R) is the correct explanation of (A).
As CO is a good donor and a good π-acceptor ligand, there exists a back
bonding in CO complexes in which CO accepts an appreciable amount of electron
density from the filled d-orbitals of metal atom into their empty π or π*
orbitals.
These π interactions increase the value of ∆0. Whereas
NH3 can form only σ-bonds and no π-bonds with metals, therefore CO is
a better complexing reagent than NH3.
Section
B
(This section contains 5
questions with internal choice in one question. The following questions are very
short answer type and carry 2 marks each.)
Question 17.
(a) Write the reaction when 2-propanone is reduced with zinc
amalgam. [2]
Answer:
(b) Write the name of this reaction. The IUPAC name of the product formed in
above reaction.
Answer:
The above reaction is called as Clemmensen
reduction reaction. The IUPAC name of the product formed in above is
propane.
Question 18.
(a) Name the suitable haloalkane and reagent from which
methyl isocyanide can be prepared. [2]
Answer:
It can be done by treating
the methyl bromide with alcoholic silver cyanide.
(b) Out of ethyl bromide and ethyl iodide, which one undergoes SN2
reaction faster?
Answer:
As I– ion is a better leaving group
than Br– ion, therefore iodides are more reactive than bromides.
Therefore, CH3—CH2—I is more reactive than
CH3—CH2 — Br towards SN2 reaction and hence,
CH3—CH2—I would undergo SN2 reaction faster
than CH3—CH2—Br.
Question 19.
Calculate ∆rG° and log KC for the
following reaction at 298 K. [2]
2Cr(s) + 3Cd2+(aq) →
2Cr3+(aq) + 3Cd(s)
[Given: E°cell] = + 0.34 V, F =
96500 C mol-1]
Answer:
∆rG °= -nFE°cell,
n = 6
= -6 × 96500 C / mol × 0.34 V
= -196860 J/mol or – 196.860
kJ/mol
∵ E°cell = 0.059 V/n × logKc
logKc
= \(\frac{0.34 \mathrm{~V} \times 6}{0.059 \mathrm{~V}}\) = 34.5762
Question 20.
Explain how and why [2]
(a) elevation of boiling point of
1M KCl solution is nearly double than that of 1 M sugar solution?
Answer:
Elevation in boiling point is directly proportional to ‘i’. ∆Tb ∝ i.
Now as given in the question, elevation of boiling point of 1 M KCl solution is
nearly double than that of 1 M sugar solution. It is because KCI
being ionic,
dissociates into K+ and Cl– and therefore, it’s van’t Hoff
factor is 2 whereas for sugar van’t Hoff factor is 1 as it does not undergo such
a dissociation.
(b) Al2(SO4)3 exhibits largest freezing
point depression than KCl and C6H12O6?
Answer:
Al2(SO4)3(aq) \(\rightleftharpoons\)
2 Al3+ + 3\(\mathrm{SO}_4^{2-}\)
(Total ions = thus, i = 5)
Whereas for KCI the value of i is 2 and glucose do not ionise.
Hence, the
value of i is largest for Al2(SO4)3. So,
Al2(SO4)3 will exhibit largest freezing point
depression.
Question 21.
Account for the following. [2]
(a) A solution containing
non-volatile solute have higher boiling point than the pure solvent.
Answer:
Boiling point of a solution containing a non-volatile solute is
higher than the pure solvent because the solute particles decreases the vapour
pressure of the solution since the solute particles have no vapour pressure of
their own. So, the solution is heated to a higher temperature to become equal to
atmospheric pressure for the boiling process to take place.
(b) Elevation of boiling point is a colligative property.
Answer:
Elevation of boiling point is a colligative property because it depends upon the
number of solute particles present in a solution.
Or
What happens when blood cells are placed in [2]
(a) water ?
(b)
concentrated solution ?
Answer:
(a) When blood cells are placed in water,
then water enters the cell by osmosis. Blood cells will swell.
(b) When blood cells are placed in concentrated solution, then water moves out of the cell faster than it comes in. This results in shrinking of blood cells.
Section
C
(This section contains 7
questions with internal choice in one question. The following questions are
short answer type and carry 3 marks each.)
Question 22.
(a) Identify the major product formed when chlorobenzene
undergoes Friedel-Crafts acylation.
Name the reagent which is used to carry
out the reaction. [3]
Answer:
To carry out Friedel-Crafts acylation of
chlorobenzene, reagent anhyd. AlCl3 is used.
(b) Why does the presence of nitro group (—NO2) at o/p-positions
increase the reactivity of haloarenes towards nucleophilic substitution
reaction?
Answer:
The reactivity of haloarenes towards nucleophilic
substitution reactions can be increased by the presence of an electron
withdrawing group (—NO2)at ortho and pa/a-positions. This is because,
—NO2 group at ortho and para-positions withdraw electron density from
the benzene ring and thus, facilitates the nucleophilic attack on
haloarenes.
The carbanion thus formed is stabilised through resonance. The negative charge appeared at ortho and para position with respect to the halogen substituent is stabilised by —NO2 group.
Or
(a) Name the possible haloalkane which will yield n-butane on their reaction
with sodium.
Write the reactions involved. [3]
Answer:
Chloroethane
will yield n-butane on their reaction with sodium which is as follows
(b) Thionyl chloride method is preferred for preparing alkyl chloride from
alcohols. Why?
Answer:
Thionyl chloride (SOCl2)is preferred for
preparation of alkyl chloride from alcohol, because the by-products we get in
this reaction are escapable gases, thus we get good yield of aikyl-chioride
i.e.
Question 23.
Answer the following questions. [3]
(a) What are pseudo first order reaction? Explain the kinetics of acidic
hydrolysis of ester.
Answer:
Pseudo first order reaction The reaction
which is bimolecular but has order one, is called pseudo first order reaction,
e.g. acidic hydrolysis of ester.
(b) In a first order reaction, the concentration of the reactant is reduced
from 0.6 mol L-1 to 0.2 mol L-1 in 5 min. Calculate the
rate constant of the reaction.
Answer:
Rate constant, k =
\(\frac{2.303}{t}\) log \(\frac{[R]_0}{[R]}\).
(c) For a reaction R → P, half-life (t1/2) is observed to be
independent of the concentration of reactants. What is the order of reaction
?
Answer:
As given reaction is independent of initial concentration of
reactants. Thus, it follows first order reaction.
Question 24.
Write the equations for the following reaction. [3]
(a)
Benzyl chloride is treated with aq. NaOH.
Answer:
(b) Methyl magnesium bromide is reacted with propanone followed by
hydrolysis.
Answer:
(c) Propene undergoes acid hydrolysis.
Answer:
Question 25.
Using valence bond theory, explain the following in relation
to the complex K4[Mn(CN)6]. [3]
(a) IUPAC name
(b)
Type of hybridisation
(c) Magnetic behaviour
Answer:
(a)
K4[Mn(CN)6]
IUPAC name Potassium i exacyanomanganate
(II)
(b) Let the oxidation state of Mn be x.
1 × 4 + x + (-1)6 = 0 ⇒ x = +2
Thus, in this complex Mn is present a Mn2+.
Mn2+ =
[Ar]3d5
[CN– being strong field ligand, causes
pairing]
K4[Mn(CN)6]
⇒ Low spin, inner orbital complex.
Thus, the structure of
this complex is octahedral,
(c) Magnetic behaviour Paramagnetic
(as one unpaired electron is present
in the above orbital diagram).
Question 26.
An organic compound A (C3H80) on
treatment with copper at 573 K gives B.
B does not reduce Fehling solution
but gives a yellow ppt. of compound C with I2 / NaOH. Deduce the
structures of A, B, and C. [3]
Answer:
Compound B gives a yellow ppt. with
I2/NaOH, i.e. positive iodoform test and not reduce Fehling solution,
it means that it contains — COCH3 (methyl ketone) group, and is a
ketone. Moreover, B is obtained by the oxidation of A, thus A must be a 2°
alcohol.
(As only 2° alcohol give ketones on oxidation with Cu at 573 K).
Hence, the structure of compound A is
Comparing with the given molecular formula gives R =
CH3. Thus, compound A is
The reactions are as follows
Question 27.
Compound A having molecular formula
C4H10O, is found to be soluble in concentrated sulphuric
acid. It does not react with sodium metal or potassium permanganate.
On
heating with excess of HI, it gives a single alkyl halide. Deduce the structure
of compound A and explain all the reactions. [3]
Answer:
Question 28.
How can reducing and non-reducing sugars be distinguished?
Mention the structural features characterising reducing sugars. [3]
Answer:
Reducing sugars The sugars which reduce Fehling solution and Tollen’s
reagent are called reducing sugars, e.g. all monosaccharides containing
free
groups are reducing sugars. Thus, the presence of free
aldehydic or ketonic group is main feature of reducing sugars.
Non-reducing sugars The sugars which do not reduce Fehling solution or
Tollen’s reagent are called non-reducing sugars, e.g. sucrose.
In
non-reducing sugars, reducing groups of monosaccharides, i.e. aldehydic or
ketonic groups are bonded.
Section
D
(The following questions are
case-based questions. Each question has an internal choice and carries 4(1+1+2)
marks each. Read the passage carefully and answer the questions that
follow)
Question 29.
Polysaccharides may be very large molecules. Starch,
glycogen, cellulose, and chitin are examples of polysaccharides.
Starch is the stored form of sugars in plants and is made up of amylose and amylopectin (both polymers of glucose). Amylose is soluble in water and can be hydrolysed into glucose units breaking glycocidic bonds, by the enzymes a-amylase and β-amylase. It is straight chain polymer. Amylopectin is a branched chain polymer of several D-glucose molecules. 80% of amylopectin is present in starch.
Plants are able to synthesise glucose and the excess glucose is stored as starch in different plant parts, including roots and seeds. The starch that is consumed by animals is broken down into smaller molecules, such as glucose. The cells can then absorb the glucose.
Glycogen is the storage form of glucose in human and other vertebrates, and is made up of monomers of glucose. It is structural support to the cell. Wood and paper are mostly cellulosic in nature.
Like amylose, cellulose is a linear polymer of glucose. Cellulose is made up of glucose monomers that are linked by bonds between particular carbon atoms in the glucose molecule.
Every other glucose monomer in cellulose is flipped over and packed tightly as extended long chains.
Answers the following questions. [4]
(a) Glycogen is known as animal starch. In animals it is stored in which
part?
Answer:
In animals, glycogen is stored in liver.
(b) What type of linkage is present between monosaccharides to form long
chain polysaccharides?
Answer:
Glycosidic linkage joins monosaccharides
and polysaccharides.
(c) What is glycogen and explain how glycogen is different from starch?
Answer:
Glycogen is a polysaccharide carbohydrate which is stored in animal’s
body. It is also known as animal starch. Starch is a polymer of α-glucose and
consists of two components amylose and amylopectin Amylose is a linear polymer
of α-D-glucose. Both glycogen and amylopectin are branched polymers of
α-D-glucose but glycogen is more highly branched than amylopectin. Amylopectin
consists of 20-25 glucose units while glycogen chain consists of 10-14 glucose
units.
Or
What is essentially the difference between the a -form of glucose and β-form
of glucose? Explain. [4]
Answer:
The α-glucose and β-glucose form differ
from each other in orientation of —OH group at C-1.
Moreover, the α-form is
obtained by cystallisation from concentrated solution of glucose at 303 K while
β-form is obtained by crystallisation from hot and saturated solution at 371
K.
Question 30.
Riya performed an experiment on chemical kinetic of a
reaction. She took the A and B reactants and changed their concentrations. The
initial rates for different initial concentrations of the reactants rate are
given below. [4]
The reaction involved is A + B → C + D
Now, answer the
following questions according to the above given reaction and its experimental
data.
(a) What will be the rate constant?
Answer:
Since, r = k[A]
Then,
rate constant,
k = \(\frac{r}{[A]}\) = \(\frac{2 \times 10^{-3}}{1.0}\) = 2 ×
10-3s-1
Or
What is the predicted t1/2 of the reaction with respect to A?
[4]
(b) Why did Riya collect different sets of readings?
(c) What is the
order of reaction with respect to A and B ?
Answer:
t1/2 for
first order reaction,
t1/2 = \(\frac{0.693}{k}\)
=
\(\frac{0.693}{2 \times 10^{-3}}\) [from above answer]
= 3465 s
(b) To maintain the accuracy and to avoid the error in data collection, different sets of readings were taken.
(c) Rate of reaction α [concentration]”, where, n is the order of
reaction.
Given reaction, A + B → C + D
Suppose the given reaction is of
order p with respect to A and q with respect to B.
The rate of formation of C
and D can also be written as
From Eqs. (i) and (ii), we get
\(\frac{4.0 \times 10^{-3}}{2.0 \times
10^{-3}}\) = \(\frac{k(2)^p(1)^p}{k(1)^p(1)^q}\) ⇒ (2)1 =
2p
Hence, p = 1
From Eqs. (i) and (iii), we get
\(\frac{2.0
\times 10^{-3}}{2.0 \times 10^{-3}}\) = \(\frac{k(1)^p(2)^q}{k(1)^p(1)^q}\) ⇒ 1
= 2q ⇒ 20 = 2q
Hence, q = 0
∴ The order
of reaction with respect to A is first order, i.e. r ∝ [A] and with respect to 6
is zero order, i.e. r ∝ [B]0.
Section
E
(The following questions are
long answer type and carry 5 marks each. All questions have an internal
choice.)
Question 31.
(a) Chromium metal is electroplated using an acidic solution
containing CrO3 according to the following equation.
CrO3 (aq) + 6H+ + 6e– → Cr(s) +
3H2O
Calculate how many grams of chromium will be electroplated by
24000 coulombs. How long will it take to electroplate 1.5 g chromium using 12.5
A current?
[Atomic mass of Cr = 52 g mol-1, 1 F = 96500 C
mol-1] [5]
Answer:
Given : Charge (Q) = 24000 C and
current
(i) = 12.5A.
Mass of chromium (Cr) deposited = 1.5 g
Molar mass of (Cr) =
52 g mol-1
and 1 F = 96500 C mol-1
∵ 6 × 96500 C
charge deposit Cr = 52 g
∴ 24000 C charge deposit Cr = \(\frac{52 \times
24000}{6 \times 96500}\)
= \(\frac{1248000}{6 \times 96500}\)
Mass of
chromium deposited = 2.16 g
Time taken to deposit 1.5 g Cu by 12.5 ampere
current.
⇒ m = Zlt
m = \(\frac{E}{96500}\)l.t
1.5 = \(\frac{52}{6
\times 96500}\) × 12.5 × t(sec)
t = \(\frac{1.5 \times 6 \times 96500}{52
\times 12.5}\)
t = 1336.16 s
t =22.2 minutes
(b) Write applications of Kohlrausch’s law.
Answer:
Kohlrausch’s law of
independent migration of ions states that the limiting molar conductivity of an
electrolyte can be expressed as the sum of the individual contributions of the
anion and the cation of the electrolyte, e.g.
Applications of Kohlrausch’s law: Some important applications
of Kohlrausch’s law are as follows
Or
(a) The molar conductivity of A is 46.1 S cm/mol and that of B is 0.025 S
cm/mol. at 0.02 M. Which of the two is most likely to be methanoic acid? [5]
Answer:
Molar conductivity of weak electrolyte is more than the strong
electrolyte. Therefore, A would be methanoic acid.
(b) How many moles of mercury will be produced by electrolysing 10 M
Hg(NO3)2 solution with a current of 2.00 A for three
hours?
[Hg(NO3)2 = 200.6 g mol-1]
Answer:
Given, current = 2A,
Time = 3h = 3 × 60 × 60s
m = Zlt
=
\(\frac{E}{96500}\)
= \(\frac{200.6 \times 2 \times 3 \times 60 \times 60}{2
\times 96500}\)lt
= 22.45g
Number of moles
= \(\frac{22.45}{200.6}\) =
0.112 mol
(c) Following reactions occur at cathode during the electrolysis of aqueous
silver chloride solution.
Ag+ (aq) + e– → Ag(s); E° =
0.80 V
H+(ag) + e– → \(\frac{1}{2}\)H2(g);
E° = 0.00 V
On the basis of their standard reduction electrode potential (E°)
values, which reaction is feasible at the cathode and why?
Answer:
Electrolysis of an aqueous solution of AgCl.
From ∆G°= -nFE°, more negative AG° will be more will be the
feasibility of that reaction.
Since, E° is higher for Ag + (ag) → Ag,
feasibility of occurrence of this reaction at cathode is higher.
Question 32.
An aromatic compound ‘A’ on heating with Br2 and
KOH forms a compound ‘B ‘ of molecular formula C6H7N,
which on reacting with CHCl3 and alcoholic KOH produces a foul
smelling compound ‘C’. [5]
Write the chemical equation involved in the formation of A, B and C. Explain
the reaction of B with Br2/H2O. Also, write the equation
for reduction of A with LiAlH4/THF.
Answer:
Since, compound ‘A’
on heating with Br2 and KOH forms a compound ’S’ which further reacts
with CHCl3 and ale. KOH producing afoul smelling compound
‘O’.
Thus, compound ‘C ’ is an aromatic isocyanide and compound ‘S ’ is
aniline with formula C6H5NH2 and compound ‘A ’
must be benzamide.
Or
(a) A compound (X) has molecular formula as C3H7NO. It
gives the following reactions. Hydrolysis of (X) gives an amine (Y) and
carboxylic acid (Z). Amine (Y) with Hinsberg reagent forms water insoluble
product. [5]
(i) Identify X, Y and Z.
(ii) Write down the reaction for the
formation Z and Y from X.
(iii) Give the test for compound Y which shows it
is insoluble in water.
Answer:
(i) Compound [X]
C3H7NO must be an amide because it gives carboxylic acid
and an amine on hydrolysis.
Amine [Y] must be a secondary amine as it produces an
insoluble compound on treatment with Hinsberg’s reagent, i.e. benzene sulphonyl
chloride.
Also, as only formic acid gives a positive silver mirror test,
hence acid [Z] must be HCOOH.
(ii)The reaction involved in the formation of Y and Z from X is as
follows
(iii) Reaction of compound Y, i.e.
N, N-dimethylamine reacts with benzene
sulphonyl chloride to yield insoluble product as shown.
(b) Account for the following.
(i) Which compound is soluble in water, N,
N-dimethylamine or formic acid? Why?
(ii) Write down the reaction for formic
acid on treatment with Tollen’s reagent giving a positive silver mirror
test.
Answer:
(i) Both the given compounds are soluble in water because
both can form hydrogen bonds with water.
(ii)
Question 33.
Attempt any five of the following. [5]
(a) Which element
of transition series has highest second ionisation enthalpy?
Answer:
Copper has highest second ionisation enthalpy because second electron is to be
removed from fully-filled d-subshell.
Cu = 3d104s1
(b) Why does copper not replace hydrogen from acids?
Answer:
\(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0\) = + 0.34V
Positive standard
reduction potential shows that copper will not displace hydrogen gas on
treatment with mineral acid as Cu is less reactive than hydrogen.
(c) Draw the structure of dichromate ion.
Answer:
Dichromate ion
(Cr2\(\mathrm{O}_7^{2-}\))
(d) How will you obtain Na2Cr2O7 from
Na2CrO4 ?
Answer:
(e) Why scandium (21) is a transition element but not calcium (20)?
Answer:
Sc(21) with electronic configuration [Ar]
3d14s2 have incompletely filled d-orbitals, whereas Ca(20)
with electronic configuration [Ar] 4s2 does not have electrons in d-orbitals.
Thus, Sc(21) is a transition element.
(f) Explain the observation, the second and third rows of transition elements
resemble each other much more than they resemble the first row.
Answer:
Due to lanthanoid contraction, the atomic radii of the second and third rows of
transition elements is almost same. So, they resemble each other much more as
compared to first row elements and show similar characteristics.
(g) What happens when KMn04 is treated with oxalic acid in acidic medium
?
Answer:
KMnO4 oxidises oxalic acid to CO2 and
itself changes to Mn2+ ions which is colourless because
KMnO4 acts as an oxidising agent.