PAPER-1 (B.E./B.TECH.) JEE MAIN 2019...5. Aniline react with Sodium nitrate and dil. HCl and then the solution is added to equimolar mixture of phenol and aniline drop by drop. What

PAPER-1 (B.E./B.TECH.)

JEE MAIN 2019 Computer Based Test

Solutions of Memory Based Questions

Date: 9th April 2019 (Shift-1)

Time: 09:30 A.M. to 12:30 P.M.

Durations: 3 Hours | Max. Marks: 360

Subject: Chemistry

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JEE MAIN 9 APRIL 2019 SHIFT-1 CHEMISTRY

1. Which compound gives the following qualitative test:

(i) Decolourisation of Bromine water.

(ii) Insoluble in HCl

(iii) Soluble in NaOH

(A) Phenol

(B) Aniline

(C) Cyclohexanol

(D) All of these

Solution: (A)

(i) Unsaturated compounds dicolourise bromine water.

(ii) Since the compound is insoluble in HCl, it indicates that the compound is acidic.

(iii) As the compound is soluble in NaOH, this again indicates that the compound is acidic.

The above three conclusion indicate that the compound is unsaturated and acidic.

Hence, the answer is phenol as it is aromatic and slightly acidic in nature.

2. What happened when CO2 increase in Atmosphere?

(A) Global Warming

(B) Decrease in Temperature

(C) Smoke

(D) None of these

Solution: (A)

The increased amount of CO2 in air is mainly responsible for global warming.

3. Choose correct option corresponding the order of following graph. Where [R] is concentration of Reactant

(A) 1, 1

ln [R] [R]

time time


(B) 0, 2

(C) 1, 0

(D) 0, 1

Solution: (C)

First order R ⟶ P differential rate law

d[R]

dt= −K[R]

(zero order time)

R ⟶ P Differential rate low d[R]/dt = −K

4.

(A)

(B)

(C)

(D)

ln [R]

time

[R]

time

H C–C—C–H3

D Cl

D I

H C–C—C–H3

D Cl

DI

H C–C—C–H3

Cl D

DI

H C–C—C–H3

D D

Cl I


Solution: (C)

Addition according to Markovnikov’s Rule

5. Aniline react with Sodium nitrate and dil. HCl and then the solution is added to equimolar mixture of phenol

and aniline drop by drop. What is the major product formed.

(A) P- HydropyAzobenzene and P- Amino Azo benzene.

(B) O- HydroxyAzobenzene and P-Amino Azo benzene.

(C) P- HydroxyAzobenzene and O-Amino Azo benzene.

(D) O- HydroxyAzobenzene and O-Amino Azo benzene.

Solution:

NH2 N Cl2

+ NaNO2 + 2HCl + NaCl + 2H O2

Benzene dizonium salt


6. Arrange in increasing order of oxidation state of following

N2O, NO, NO2, N2O3 , N2O5

(A) N2O > NO > N2O3 > NO2 > N2O5

(B) N2O < NO < N2O3 < NO2 < N2O5

(C) NO > N2O > N2O3 > NO2 > N2O5

(D) NO > N2O > NO2 > N2O3 > N2O5

Solution: (B)

Oxidation States of following

N2O ⇒ 2x + 1(−2) = 0 ⇒ x = 1

NO ⇒ x + 1(−2) = 0 ⇒ x = 2

N2O3 ⇒ 2x + 3(−2) = 0 ⇒ x = 3

NO2 ⇒ x + 2(−2) = 0 ⇒ x = 4

N2O5 ⇒ 2x + 5(−2) = 0 ⇒ x = 5

7.

(A)


(B)

(C)

(D)

Solution: (B)

8. Aerosol is a colloidal solution of

(A) Gas in Liquid

(B) Gas in Solid

(C) Solid in Gas

(D) Liquid in water

Solution: (C)

Aerosol is a colloidal solution of solid in air; for example smoke, dust etc. are consider as Aerosol.

9. In fullerene the structure consist?

(A) 5 Pentagonal & 12 Hexagonal Ring

(B) 20 Pentagonal & 12 Hexagonal Ring

(C) 20 Hexagonal & 12Pentagonal Rings

O

O

OH

Br

O

OH

Br

O O O

OH BrPBr3 alc. KOH

Substitution reactionS 2N

Elimination

acidic Hydrogen


(D) 12 Pentagonal & 5 Hexagonal Rings

Solution: (C)

Fullerene is an Isotope of Carbon which consist 20 Hexagonal and 12 Pentagonal Rings, it has𝑠𝑝2hybridisation

and form cage like structure.

10. Find ∆Go for the following reaction:

Cu2+ + Zn ⟶ Zn2+ + Cu

(Eo = 2 volt, F = 96500 c)

(A) −193 kJ

(B) 193 kJ

(C) 386 kJ

(D) −386 kJ

Solution: (D)

For the reaction

Cu2+ + Zn ⟶ Zn2+ + Cu

n = 2 (two electrons are exchanged)

∆Go = −nFEo

= −2 × 96500 × 2

= −386 kJ

11. Which of the following metal has large difference in its first and second ionization enthalpy.

(A) Ba

(B) K

(C) Sr

(D) Ca

Solution: (B)

Among the given option, except for potassium rest others are alkaline earth metals with outer configuration ns2

In case of Potassium, the outer configuration is 1s2 2s2 2p6 3s2 3p6 4s1


First ionization enthalpy is low, but once an electron is removed, it attain noble gas status and become Stable.

Hence, removing second electron is difficult and requires large amount of energy.

12. The linkage present in the units of a sucrose molecules is:

(A) (1, 2)

(B) (1, 3)

(C) (1, 4)

(D) (1, 6)

Solution: ()

Sucrose is a disaccharide which is formed due to glycosidic linkage between C1 of α − glucose and C2 of β −

fructose.

13. Reactivity towards eletrophilic substitution reaction of following is

(A) a > b > c > d

(B) d > c > b > a

(C) c > d > a > b

(D) b > c > a > d

Solution: (C)

Reactivity of E+ substitution ∝ e− rich πe− cloud present on aromatic component & their πe− exposed.

electrophilic substitution ∝ +m ∝1

−m

Cl CN OMe Me(a) (b) (c) (d)


∝ +HC ∝1

−HC

∝ +I ∝1

−I

due to – I effect of halogen it is ring deactivating but o/p directory.

14. Number of water molecules not directly coordinated with Cu in CuSO4. 5H2O

(A) 1

(B) 2

(C) 4

(D) 5

Solution: (A)

In CuSO4 . 5H2O , 4 H2O molecules are directly attached to the Cu atom while one is present as water of

crystallisation. According to structure it is considered to be distorted octahedral. It form an orthorhombic Crystal

lattice.

15. Which orbital involved in [Cr(H2O)6]+3

(A) dx2 − y2 , dz2

(B) dxy, dyz

(C) dxz, dxz

(D) dxy, dx2 − y2

Solution: (A)

[Cr(H2O)6]+3it is

[Cr(H2O)6]+3 ⇒ Crin+3 oxidation state Cr+3 = ⋯ 3d3 4S0

H2Ois weak field legand its hybridisation d2Sp3 square biprymidal So d- orbital involved is dz2 , dx2 − y2

16. If vapour pressure of M is 450 mm of Hg& of N is 700 mm of Hg , then which is true?

Cl CNOMe Me

+me effect of –OCH

+HyperConjugation

– > + mI –m

> > >


(A) xm

xn>

ym

yn

(B) xm

xn<

ym

yn

(C) xm

xn=

ym

yn

(D) (xm − xn)(ym − yn) > 0

Solution: (A)

Pmo is given as = 450 mm of Hg.

Pnois given as = 700 mm of Hg.

We know that Pm = xm × P mO → Raoult’s law.

Pn = xn × P nO → Raoult’s law.

From dalton’s law Pm = ym × PT

Pn = ym × PT

So,

xm × P mo = ym × PT … (i)

xm × P no = yn × PT … (ii)

(i) / (ii) xm

xn×

P Mo

P No =

ym

yn

𝑥𝑚

𝑥𝑛=

𝑦𝑚

𝑦𝑛×

750

450

𝑥𝑚

𝑥𝑛>

𝑦𝑚

𝑦𝑛

17. Which of the following show optical isomerism

(A)

(B)

A

A

B

B

B

B

B

B

enA

A


(C)

(D)

Solution: (D)

(A)

Symmetry not optically active.

(B)

Symmetry exist, Hence no optical isomerism.

(C)

Symmetry exist, Hence no optical isomerism.

(D)

This has no plane of symmetry.

Hence can rotate plane polarized light and will show optical isomerism.

18. AB ionic having osmotic pressure 4 times of BaCl2 having conc. 0.01m. Find the concentration of AB.

(A) 0.12

A

A

en

en

A

A

en

en

B B

B B

A

A

A

A

B

B

en

A

A

en en

A

A

en

en


(B) 0.04

(C) 0.06

(D) 0.03

Solution: (C)

Osmotic pressure(A)π = i × CRT

𝑖 = Vant's half factor

According to question, for AB i = 2, BaCl2

i = 3

πAB = 4 × πBaCl2

2 × C × RT = 4 × 3 × 0.01 × RT

C = 6 × 0.01 = 0.06 m.

19. Write the correct name of the following compound

(A) 3-Chloro-4-methyl nitrobenzene

(B) 1-Chloro-3-nitro-6 methylbenzene

(C) 1-Chloro-6-methyl-3-nitrobenzene

(D) 1-Chloro-2-methyl-5-nitrobenzene

Solution: (D)

Priority −Cl > −𝑁O2 > −𝐶H3


1-Chloro-2-methyl-5 nitro-benzene

20. Match the following

Column-I Column-II

(a)V2O5 (i) NH3

(b) zeigler natta (ii) H2SO4

(c)PdCl2 (iii)Polyethene

(d) Irom oxide (iv) ethanol

(A) (a) ii, (b) iv, (c) i, (d) iii

(B) (a) ii, (b) i, (c) iii, (d) iv

(C) (a) ii, (b) iii, (c) iv, (d) i

(D) (a) iii, (b) ii, (c) iv, (d) i

Solution: (C)

(i) Contact process : Industrial method to prepare H2SO4 in presence of V2O5

(ii) Wacker oxidation : Conversion of ethylene to acetaldehyde by an acidic, aqueous solution of PdCl2.

(iii) Ziegler-Natta catalyst used to polymerize terminal alkenes (ethylene and alkenes with the vinyl double bond)

n(CH2 = CHR) → [CH2 − CH2R]n

(iii) Haber process : Haber process is used to form NH3in presence of Iron oxide.

21.

(A) H3C − CH = CH − CH2 − OH

(B) H3C − CH2 − CH2 − CH2 − OH


(C) O ||

CH3 − CH = CH − C − O − CH3

(D) H3C − CH = CH − CH2 − O − CH3

Solution (A)

22.

(A)

(B)

(C)

(D)

Solution: (B)

Alkaline KMnO4 oxidised Benzylic carbon to carboxylic Acid.


23.

(A)

(B)

(C)

(D)

Solution: (C)


24. Find the ratio of difference of maximum and minimum Frequency of lyman to Balmer.

(A) 9

4

(B) 3

4

(C) 9

2

(D) 3

2

Solution: (A)

=(vmax−vmin)lyman

(vmax−vmin)Balmar ∵ 𝜈 =

𝑐

𝜆=

𝐸

ℎ

=(Emax−Emin)lyman

(Emax−Emin)Balmar 𝐸 = ℎ𝜈, 𝜈 =

𝐸

ℎ

=𝐸1→∞− 𝐸1→2

𝐸2→∞− 𝐸2→3

=13.6. (

1

1−

1

∞) − 13.6 (

1

1−

1

4)

13.6. (1

4−

1

∞) − 13.6 (

1

4−

1

9)

=1−0−1+

1

41

4−0−

1

4+

1

9

=1/4

1/9=

9

4

25. For the given reaction : N2 + 2H2 → 2NH3

In which of the following option. H2 will be limiting Reagent. (L.R.)?

(A) 14g N2 , 6g H2


(B) 35g N2, 8g H2

(C) 28g N2, 6g H2

(D) 56g N2, 4g H2

Solution: (D)

N2 + 3H2 → 2NH3

=mole(

wt

mol.wt)

statio.coeff which is less that is L.R.

N2 + 3H2 → 2NH3

(A) 14

28

1=

1

2

6

2

3= 1 → In this N2 is L.R.

(B) 35

28=

5

4

8

2

3=

4

3 → In this N2 is L. R.

(C) 28

28

1= 1

6

2

3= 1 → Both are L.R.

(D) 56

58

1= 2

4

2

3= 0.667 → In this H2 is L.R.

26. Vander wall constants a and b are given for following noble gas.

Noble gas a b

Ne 2.8 0.7

Ar 4.2 0.9

Kr 5.1 1.0

Xe 5 1.2

which noble gas will have highest critical temp.

(A) Kr

(B) Ne

(C) Ar

(D) Xe

Solution: (A)

𝑇𝐶 =8𝑎

27𝑅𝑏

for Kr the value of (𝑎

𝑏) is highest so 𝑇𝑐 . for Kr will be highest

27. Which of the given or has fluoride ion?


(A) Cryolite

(B) Sphalerite

(C) Manetite

(D) Malachite

Solution: (A)

Cryolite ⇒ Na3[AlF6]

Sphalerite ⇒ Zns

Manetite ⇒ Fe3O4

Malachite ⇒ CuCO3. Cu(OH)2

28. Which will stabilize by forming an anion.

(A) NO

(B) C2

(C) O2

(D) F2

Solution: (B)

𝐶2 ⇒ 𝜎1𝑠2 𝜎∗1𝑠 𝜎2𝑠 𝜎∗2𝑠 𝜋2𝑃𝑥 = 𝜋2𝑃𝑦 𝜎2𝑃2 (BMO)

In, F2, O2 and NO

𝜎1𝑠 𝜎∗1𝑠 𝜎2𝑠 𝜎∗2𝑠 𝜎2𝑃2 𝜋2𝑃𝑥 = 𝜋2𝑃𝑦 𝜋∗2𝑃𝑥 = 𝜋∗2𝑃𝑦 𝜋2𝑃2

For these 𝑒− are came in ABMO

SO In C2 𝑒− added in BMO is stable.

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PAPER-1 (B.E./B.TECH.) JEE MAIN 2019...5. Aniline react with Sodium nitrate and dil. HCl and then the solution is added to equimolar mixture of phenol and aniline drop by drop. What