KVPY-SX 2018 (CHEMISTRY) · 2020. 11. 12. · KVPY-SX_2018 (CHEMISTRY) KVPY-SX_2018 (CHEMISTRY)...
Transcript of KVPY-SX 2018 (CHEMISTRY) · 2020. 11. 12. · KVPY-SX_2018 (CHEMISTRY) KVPY-SX_2018 (CHEMISTRY)...
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 1
Copyright © Think and Learn Pvt. Ltd.
PART-I
1. The amount (in mol) of bromoform (CHBr3) produced when 1.0 mol of acetone
reacts completely with 1.0 mol of bromine in the presence of aqueous NaOH is
(A) 1
3 (B)
2
3 (C) 1 (D) 2
2. The following compound can readily be prepared by Williamson ether synthesis by
reaction between
(A) and
(B) and
(C) and
(D) and
3. X and Y
are:
(A) enantiomers (B) diastereomers
(C) constitutional isomers (D) conformers
4. The higher stabilities of tert-butyl cation over isopropyl cation, and trans-2-butene
over propene, respectively, are due to orbital interactions involving
(A) → and → * (B) → vacant p and → *
(C) → * and → (D) → vacant p and → *
H Cl
CH3
H
Cl
H3C
Cl
Cl H
H H3C
CH3
X Y
OH
I
OH
Cl
I
OH
Cl
OH
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 2
Copyright © Think and Learn Pvt. Ltd.
5. Benzaldehyde can be converted to benzyl alcohol in concentrated aqueous NaOH
solution using
(A) acetone (B) acetaldehyde (C) formic acid (D) formaldehyde
6. The major product of the following reaction
is:
(A)
(B)
(C)
(D)
7. Among the following species, the H–X–H angle (X = B, N or P) follows the order
(A) PH3< NH3 F–> Na+> N3– (B) N3–> Na+> F–> O2–
(C) N3–> O2–> F–> Na+ (D) Na+> F–> O2–> N3–
9. The oxoacid of phosphorus having the strongest reducing property is:
(A) H3PO3 (B) H3PO2 (C) H3PO4 (D) H4P2O7
10. Among C, S and P, the element(s) that produce(s) SO2 on reaction with hot conc.
H2SO4 is/are:
(A) only S (B) only C and S (C) only S and P (D) C, S and P
CO2H OH
O
OH
OH
OH
CO2H
O
CO2H NaBH4
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 3
Copyright © Think and Learn Pvt. Ltd.
11. The complex that can exhibit linkage isomerism is
(A) [Co(NH3)5(H2O)]Cl3 (B) [Co(NH3)5(NO2)]Cl2
(C) [Co(NH3)5(NO3)](NO3)2 (D) [Co(NH3)5Cl]SO4
12. The tendency of X in BX3 (X = F, Cl, OMe, NMe) to form a bond with boron follows
the order
(A) BCl3< BF3
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 4
Copyright © Think and Learn Pvt. Ltd.
15. The correct representation of wavelength-intensity relationship of an ideal
blackbody radiation at two different temperatures T1 and T2 is:
(A)
(B)
(C)
(D)
16. The pressure (P)-volume (V) isotherm of a van der Waals gas, at the temperature at
which it undergoes gas to liquid transition, is correctly represented by
(A)
(B)
(C)
(D)
17. A buffer solution can be prepared by mixing equal volumes of
(A) 0.2 M NH4OH and 0.1 M HCl (B) 0.2 M NH4OH and 0.2 M HCl
(C) 0.2 M NaOH and 0.1 M CH3COOH (D) 0.1 M NH4OH and 0.2 M HCl
V
P
V
P
V
P
V
P
Wavelength
Inte
nsi
ty T1
T2
T2> T1
Wavelength
Inte
nsi
ty T1
T2
T2> T1
Wavelength
Inte
nsi
ty T2
T1
T2> T1
Wavelength
Inte
nsi
ty T2
T1
T2> T1
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 5
Copyright © Think and Learn Pvt. Ltd.
18. The plot of total vapour pressure as a function of mole fraction of the components of
an ideal solution formed by mixing liquids X and Y is:
(A)
(B)
(C)
(D)
19. On complete hydrogenation, natural rubber produces
(A) polyethylene (B) ethylene-propylene copolymer
(C) polyvinyl chloride (D) polypropylene
20. The average energy of each hydrogen bond in A-T pair is x kcal mol–1 and that in G-C
pair is y kcal mol–1. Assuming that no other interaction exists between the
nucleotides, the approximate energy required in kcal mol–1 to split the following
double stranded DNA into two single strands is
[Each dashed line may represent more than one hydrogen bond between the base
pairs]
(A) 10x + 9y (B) 5x + 3y (C) 15x + 6y (D) 5x + 4.5 y
A–T–A–T–G–C–A–G
T–A–T–A–C–G–T–C
Mole fraction of X 0 1 T
ota
l vap
ou
r p
ress
ure
Mole fraction of X 0 1
To
tal v
apo
ur
pre
ssu
re
Mole fraction of X 0 1
To
tal v
apo
ur
pre
ssu
re
Mole fraction of X 0 1
To
tal v
apo
ur
pre
ssu
re
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 6
Copyright © Think and Learn Pvt. Ltd.
PART-II
21. For the electrochemical cell shown below
Pt|H2 (P = 1 atm)|H+(aq., x M)||Cu2+ (aq., 1.0 M)|Cu(s)
The potential is 0.49 V at 298 K. The pH of the solution is closest to
[Given: Standard reduction potential, E° for Cu2+/Cu is 0.34 V
Gas constant, R is 8.31 J K–1mol–1
Faraday constant, F is 9.65 × 104 J V–1mol–1]
(A) 1.2 (B) 8.3 (C) 2.5 (D) 3.2
22. Consider the following reversible first-order reaction of X at an initial concentration
[X]0. The values of the rate constants are kf = 2s–1 and kb = 1s–1
A plot of concentration of X and Y as function of time is
(A)
(B)
(C)
(D)
t
Co
nce
ntr
atio
n
[X]0
[Y]eq
[X]eq
t
Co
nce
ntr
atio
n
[X]0
[Y]eq
[X]eq
t
Co
nce
ntr
atio
n
[X]0
[Y]eq
[X]eq
t
Co
nce
ntr
atio
n
[X]0
[Y]eq
[X]eq
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 7
Copyright © Think and Learn Pvt. Ltd.
23. Nitroglycerine (MW = 227.1) detonates according to the following equation:
2C3H5(NO3)3() ⎯→ 3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g)
The standard molar enthalpies of formation of ofH for all the compounds are given
below:
ofH [C3H5(NO3)3] = –364 kJ/mol
ofH [CO2(g)] = –393.5 kJ/mol
ofH [H2O(g)] = –241.8 kJ/mol
ofH [N2(g)] = 0 kJ/mol
ofH [O2(g)] = 0 kJ/mol
The enthalpy change when 10 g of nitroglycerine is determined is
(A) –100.5 kJ (B) –62.5 kJ (C) –80.3 kJ (D) –74.9 kJ
24. The heating of (NH4)2Cr2O7 produces another chromium compound along with N2
gas. The change of the oxidation state of Cr in the reaction is
(A) +6 to +2 (B) +7 to +4 (C) +8 to +4 (D) +6 to +3
25. The complex having the highest spin-only magnetic moment is
(A) [Fe(CN)6]3– (B) [Fe(H2O)6]2+ (C) [MnF6]4– (D) [NiCl4]2–
26. Among Ce(4f1 5d1 6s2), Nd(4f4 6s2), Eu(4f7 6s2) and Dy(4f10 6s2), the elements having
highest and lowest 3rd ionization energies, respectively, are
(A) Nd and Ce (B) Eu and Ce (C) Eu and Dy (D) Dy and Nd
27. The major product of the following reaction sequence
is:
(A) (B) (C) (D)
Ph
OH Me
Ph Ph
Me
Ph
Me
(i) B2H6 (ii) H2O2/NaOH
Ph
(iii) conc. H2SO4
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 8
Copyright © Think and Learn Pvt. Ltd.
28. Among the following reactions, a mixture of diastereomers is produced from
(A) HBr⎯⎯→ (B) 2H /Pt⎯⎯⎯→
(C) HBr ROOR , h
⎯⎯⎯→ (D) 2 62 2
B H
H O /NaOH⎯⎯⎯→
29. Reaction of phenol with NaOH followed by heating with CO2 under high pressure,
and subsequent acidification gives compound X as the major product, which can be
purified by steam distillation. When reacted with acetic anhydride in the presence of
a trace amount of conc. H2SO4, compound X produces Y as the major product.
Compound Y is
(A) (B) (C) (D)
30. A tetrapeptide is made of naturally occurring alanine, serine, glycine and valine. If the
C-terminal amino acid is alanine and the N-terminal amino acid is chiral, the number
of possible sequences of the tetrapeptide is
(A) 12 (B) 8 (C) 6 (D) 4
OH
O O
O
O
CO2H
O
OH O
O
O O
CO2H
O
Me H Me H
Me H Me H
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 9
Copyright © Think and Learn Pvt. Ltd.
ANSWER KEY 1. (A) 2. (B) 3. (D) 4. (D) 5. (D)
6. (A) 7. (A) 8. (A) 9. (B) 10. (D)
11. (B) 12. (A) 13. (B) 14. (B) 15. (A)
16. (B) 17. (A) 18. (B) 19. (B) 20. (A)
21. (C) 22. (B) 23. (B) 24. (D) 25. (C)
26. (B) 27. (C) 28. (A) 29. (A) 30. (D)
SOLUTIONS PART-I
1. (A) Reaction of acetone with Br2 in the presence of NaOH follow as
CH3 C
CH3+3Br2
O
NaOH⎯⎯⎯→ CH3 C
O
O
Na + CHBr3+ 3HBr Here 3 moles of Br2react with 1 mole acetone, produce 1 mole of CHBr3
1 moles of 2 31
Br moleof CHBr3
=
Hence, the moles of CHBr3 = 1
moles3
Therefore, the correct option is (A). 2. (B) Williamson ether synthesis→
R2–X
R1–O–H+Na or k → R – O–Na
R1–O–R2 Order of reactivity of halide→ R–I > R–Br > R–Cl > R–F (1° > 2° > 3°) Hence, in the following reaction, we can readily prepare the above ether compound.
O|H
+ I 2SN
Reaction⎯⎯⎯⎯→ O
Because, stability of carbocation (C+) Rate of SN2
Therefore, the correct option is (B).
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 10
Copyright © Think and Learn Pvt. Ltd.
3. (D)
(1)
H C H
CH3
CH3
C
Convert to fischer projection⎯⎯⎯⎯⎯⎯⎯⎯→ C
C
Me
Me
H
H
(2)
View side
CH3
⎯⎯⎯→Rotate180°
C
C
Me
Me
H
H
CH3 C
C
H
H
C C H
H
CH3 CH3
Convert into Fischer projection
(1) & (2) are identical →Hence, they are conformers Therefore, the correct option is (D). 4. (D) Hint:- See the key point and definition of hyperconjugation
(1)
CH3 CH
CH3 (6H)
CH3 C
CH3 (9H)
CH3
Hyperconjugation structure = (Number of H + 1) As the Hydrogen increases number of Hyperconjugation ( bond vacant p orbital
overlapping) structure increases and stability of carbocation also increases. Hence, tert-butyl carbocation is more stable compared to isopropyl carbocation.
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 11
Copyright © Think and Learn Pvt. Ltd.
(2) CH3 CH CH2 (3-H)
Or
CH3 CH3 CH CH
(6H)
Hyperconjugation structure = (Number of H + 1)
As the Hydrogen increase in alkene number of Hyperconjugation ( bond * Interaction) structure increase and stability of alkene also increase.
Hence, 2 butene is more stable compare to propene. Therefore, the correct option is (D).
5. (D)
It is an example of cannizzaro reaction
the reaction mechanism follows as
Reactivity order towards Nucleophilic addition→ HCHO >PhCHO
H
O 2– CH – Ph
O
C
H
O
+
OH H
C
H
O
H
PH C
H
O
+ Ph C
O
O
CH2 OH H
C
O
O
+
H
–
H
Therefore, the correct option is (D).
6. (A)
Solution → NaBH4ideal reducing agent for reduction of carbonyl group to
corresponding alcohol without affecting other groups.
Reaction follow as
4NaBH⎯⎯⎯→
O OH
COOH COOH
Therefore, the correct option is (A).
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 12
Copyright © Think and Learn Pvt. Ltd.
7. (A) Bond angle 1. PH3→ no hybridisation (Drago’s rule) 93° 2. NH3→ sp3hybridisation (with one lone pair) 107° Due to lone pair-bond pair repulsion, bond angle will reduce from 109°28’ to
107°. 3.
4NH → sp3hybridisation 109°28’
4. BF3→ sp2hybridisation 120° 8. (A) Ionic size with addition of electron and size with loss of e . Nitride ion has 10 electrons in contrast with only 7 protons due to the above given
reason. It has highest inter electronic repulsions. Hence its size or radius is maximum.
Note:For isoelectronic species. Ionic size negative charge on anion. Hence, correct order follows as –3 –2N O F Na + Therefore, the correct option is (A). 9. (B)
P
O H3PO3
O
O H
H H
Reducing Hydrogen
1 Hydrogen (P–H bounded) →
(A)
H3PO2 → P
O
O–H (2, Hydrogen) H H
(C)
P
HO (0, Hydrogen) OH
OH
H3PO4 →
O
(D)
P
HO (0, Hydrogen)
OH
H4P2 O7 O
P
OH OH
O O →
Reducing Hydrogen P–H (Bounded Hydrogen) → Hence, in H3PO2 have (2) reducing H due to this it has strongest reducing
properties. Therefore, the correct option is (B).
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 13
Copyright © Think and Learn Pvt. Ltd.
10. (D)
Reaction of C, S & P with H2SO4 follows as
C + H2SO4⎯→ CO2 + H2O + SO2
S + H2SO4⎯→ SO2 + H2O
P + H2SO4⎯→ SO2 + H3PO4 + H2O
Hence, here all produced SO2 gas
So, correct options is (D).
11. (B)
Linkage Isomerism→
Linkage isomerism is the existence of coordination compounds that have the same
composition differing with the connectivity of metal to ligand.
Here, in case (B) NO2 is a ambidentate ligand So it can bind with metal as N(NO2)
& O(ONO).
Hence, compound [Co(NH3)5(NO2)]Cl2 show linkage as follow:
[Co (NH3)5 (NO2)]Cl2 [Co (NH3)5 (ONO)]Cl2
Co–N Linkage Co–O
Linkage Isomers
Linkage
Therefore, the correct option is (B).
12. (A)
In All, highest back bonding in the case of B(NMe2)3 due to less E.N & small size and N
have good e density for back bonding due to +I effect of attached group.
E.N back bonding
e density back bonding
Size back bonding
Hence, the overall order of back bonding follow as
B (NMe2)3> B (OMe)3> BF3> BCl3
Therefore, the correct option is (A).
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 14
Copyright © Think and Learn Pvt. Ltd.
13. (B)
According to Langmuir isotherm→
H of adsorption of each binding site is same because surface ishomogenous.
There is a dynamic equilibrium between free gas and adsorbed gas.
The ability of a molecule to get adsorbed at a given site is independent of the
occupation of neighbouring site.
Therefore, the correct option is (B).
14. (B)
→ Reaction of H2SO4 with NH4OH follow as.
2 4 4 4 2 4 2
Salt
H SO NH OH (NH ) SO H O+ ⎯⎯→ +
Initial → H2SO4 is a strong acid so dissociation 100%
Hence due to small size of (H+) conductivity is maximum.
Upon addition of NH4OH in H2SO4, salt [(NH4)2SO4] will form and small H ion
replace by larger Ion (NH4+) so conductivity .
After salt formation, if we add Excess of NH4OH, there is no change in conductivity
because NH4OH is a week base. Hence dissociation is very less
Therefore, the correct option is (B).
15. (A )
From wein’s displacement law
mT =constant
1
i.e,T
[At higher temperature will be small]
Therefore, the correct option is (A).
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 15
Copyright © Think and Learn Pvt. Ltd.
16. (B)
P
TC→ critical + temp
vap
V →
vapour
Vapour → liquid (volume )
At critical temperature ( vapour , in equilibrium)
Therefore, the correct option is (B).
17. (A)
(A) NH4OH + HCl ⎯→NH4Cl + H2O
Initial 0.2 mmol 0.1 mmol
After reaction 0.1 0 0.1
NH4OH & NH4Cl → from basic buffer solution.
(B) NH4OH + HCl ⎯→ NH4Cl + HCl
Initial 0.2 0.2 0
After reaction 0 0 0.2 → No buffer formed (C) NH4OH + CH3COOH ⎯→ CH3COONa + H2O Initial 0.2 0.1 After reaction 0.1 0 0.10 There is no buffer formed (D) NH4OH + HCl⎯→NH4Cl + H2O
Initial 0.1 0.2 0
After reaction 0 0.1 0.1
→ No buffer form
Therefore, the correct option is (A).
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 16
Copyright © Think and Learn Pvt. Ltd.
18. (B) We know that P x(mole fraction of solute) For ideal solution, according to Raoult’s law
→
→
A
T A B
B
P PartialPof AP = P +P
P PartialPof B
0
0 AA A A A A 0
B
PP X P = P X initialP of solvent
P
Similar 0B B BP = P X
A B
A B0 0T A A B B
B A
X + X = 1
if X = 1 X = 0P =P X + P X
X = X = 0
or P x
Therefore, the correct option is (B).
19. (B)
Natural rubber formed by the polymerization of isoprene and on hydrogenation
provide copolymer of ethylene & propylene.
Isoprene →CH–HC CH2
Natural rubber
Polymerisation
CH3
H2C
H2/Ni
Polymerisation
(ethylene) (Propylene)
CH2 = CH2 + CH2 = CH –CH3
Copolymer → Polymer which is synthesized from more than one species of
monomers.
Ex. [A–B–A–B–A–B–A–B --------]n
Therefore, the correct option is (B).
20. (A)
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 17
Copyright © Think and Learn Pvt. Ltd.
Hint → See the hydrogen bounded structure of A–T (Adenine –Thymine) & – G–C(Guanine– Cytosine)
Solution
→ No O H bond between A–T =2
Average energy of each H bond in A–T pair = X Kcal mol–1
Total no of A–T pair in structure = 5
Hence, total energy = 2x ×5 = 10 x
→ No of H bond between G–C =3
Average energy of each H-bond in G–C pair = Y Kcal mol–1
Total no of G–C pair in structure = 3
Hence total energy = 3y×3 = 9y
→Now, total energy required to split all double stranded DNA in to two single strands.
=10 x + 9y
Therefore, the correct option is (A).
Part-II 21. (C)
Reduction potential of 2
o
H /HE + = 0 V & 2
o
Cu /CuE + = 0.34V
Hence, anode & cathode reaction follow as
ocellE = oCE –
oAE = 0.34 – 0 = 0.34 V
We know that
Ecell = ocellE – 0.0591
nlogQ {Q =
2
2
[H ]
[Cu ]
+
+, [H+] = x, [Cu2+] = 1M}
0.49 = 0.34 – 0.06
2log[H+]2
0.15 = +2 × 0.06
2
[–log[H ]]
pH
+
0.15
pH0.06
= {RT
f = 0.0591, n = 2}
pH = 2.5
Therefore, the correct option is (C).
22. (B)
Anode reaction → H2(g)⎯→ 2H+(aq) + 2e–
Cathode reaction → Cu2+(aq) + 2e–⎯→ Cu(s)
H2(g)+ Cu2+(aq)⎯→ 2H+(aq) + Cu(s)
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 18
Copyright © Think and Learn Pvt. Ltd.
Keq = f
b
K
K =
0x –
= 2 {Keq = f
b
K
K =
2
1 = 2}
= 2x0 – 2 3 = 2x0
= 2
3x0
At equilibrium
[x] = ox
3
[y] = 2
3xo
Hence, here concentration ‘y’ equal to concentration ‘x’ So, graph (B) best represents this. Therefore, the correct option is (B). 23. (B) Reaction 2C3H5(NO3)3() ⎯→ 3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g) We know that of( H ) =
of P( H ) – (Hf)Reactant
= 3 × 2
of N( H ) + 2
of O
1( H )
2 +
2
of CO6( H ) + 2
of H O5( H ) – 3 5 3 3
of C H (NO )2( H )
= 0 + 0 + 6 × (–393.5) + 5 × (–241.8) – 2 × (–364) of( H ) = –2842 kJ
For one mole nitroglycerine of( H ) =2812
2
−= –1421 kJ mole–1
Wt of nitroglycerine = 10 g Mw of nitroglycerine = 227
Moles of nitroglycerine = 10
227moles
Hence, of( H ) for 10
227moles = 1421 ×
10
227.1 = 62.5 kJ
Therefore, the correct option is (B).
24. (D)
Kf→ Rate constant for forward reaction Kb→ Rate constant for backward reaction
x y Kf
Kb
x0 0 At t = 0
At eq. x0–
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 19
Copyright © Think and Learn Pvt. Ltd.
Heating reaction of (NH4)2Cr2O7 follow as
Hence, change in oxidation state of Cr → +6 to +3 Therefore, the correct option is (D). 25. (C)
(1)[Fe(CN)6]3– Fe3+→ d5
(d2sp3→Hybridisation) CN→ Strong field ligand, hence, pairing occur. Unpaired electron = 1
MM = n(n 2)+ = 1.732 BM
(2) [Fe(H2O)6]2+ Fe2+→ d6
(sp3d2→ Hybridisation) H2O→Weak field ligand, hence, pairing does not occur. Unpaired electron = 4 MM = 4.90 BM (3) [MnF6]4– Mn2+→ d5
F–→ Weak field ligand Hence, pairing not occur Unpaired electron = 5 MM = 5.92 BM (4) [NiCl4]2– Cl–→ Weak field ligand Hence, paring not occar Ni2+→ d8
MM = 2.83 BM Therefore, the correct option is (C).
3d 4s 4p 4d
3d5 4s 4p
(NH4)2Cr2O7⎯→ N2 + Cr2O3 + 4H2O
+6 +3
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 20
Copyright © Think and Learn Pvt. Ltd.
26. (B)
Ionisation energy→Minimum amount of energy required to remove the most loosely
bonded electron of an isolated neutral gaseous atoms or molecules.
1. Ce
4f15d16s2
Ce2+
4f2
• Least stable
• Hence, need least amount of energy to remove electron
2. Nd
4f46s2
Nd2+
4f4
3. Eu
4f76s2
Eu2+
4f7
• Half field →most stable, hence need more energy to
remove electron
4. Dy
4f106s2
Dy2+
4f10
Therefore, the correct option is (B).
27. (C)
Therefore, the correct option is (C).
Me
(B2H6) Ph
H
B H H CH3
Ph
BH2
H2O2/NaOH
CH3
Ph
OH
Ph 1/2H
conc. H2SO4 [H
+]
H
+ H2O
Shift
Ph
H
Most stable
–H Ph
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 21
Copyright © Think and Learn Pvt. Ltd.
28. (A)
(A)
Diastereomers→ Stereoisomers that are not mirror images, different compounds
with different physical properties.
(B) 2H /Pt⎯⎯⎯→
(C) HBr ROOR , h
⎯⎯⎯→
(D) 2 62 2
B H
H O /NaOH⎯⎯⎯→
Therefore, the correct option is (A).
29. (A)
Therefore, the correct option is (A).
O–H
NaOH acid-base reaction
ONa
(i) CO2 (ii) H+
OH
COOH
(X)
or (CH3CO)2O CH3–C–O–C–CH3
O O O–C–CH3
COOH
(Y)
O
Aspirin
+ CH3COOH
Me H
OH
Me H
Me H
Br
H
H
Me H
Me H Me H
Me H HBr
Me H
Br
+ Me H
Br
Diastereomers
-
KVPY-SX_2018 (CHEMISTRY)
KVPY-SX_2018 (CHEMISTRY) Page | 22
Copyright © Think and Learn Pvt. Ltd.
30. (D)
Serine (S) →
Valine (V) →
Alanine (A) →
Chiral Compounds
H
NH2–C–COOH
CH3
*
H
NH2–C–COOH
CH
*
Me Me
H
NH2–C–COOH
CH2–OH
*
Glycine →
Possible sequence:
Ser – Val – Gly – Ala
Ser – Gly – Val – Ala
Val – Ser – Gly – Ala
Val – Gly – Ser – Ala
Total 4.
Therefore, the correct option is (D).
H
NH2–C–COOH → Achiral compound
H