NEXT GRAND TEST ON 27.01 - Dr. Sangeeta Khanna · D:\Important Data\2018\+1\Org\Test\Grand Test-6\+1 Grand Test-6 23.12.2018.doc Test Dated: 23.12.2018 Topic: Organic Chemistry READ

$Page 1: NEXT GRAND TEST ON 27.01 - Dr. Sangeeta Khanna · D:\Important Data\2018\+1\Org\Test\Grand Test-6\+1 Grand Test-6 23.12.2018.doc Test Dated: 23.12.2018 Topic: Organic Chemistry READ$
Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2018\+1\Org\Test\Grand Test-6\+1 Grand Test-6 23.12.2018.doc

1. IUPAC Naming 2. Nomenclature

3. General Organic Chemistry 4. Hydrocarbon

(Alkane, Alkene, Alkyne & Benzene) 5. Isomerism

Structural Isomerism

Conformational Isomer of Cyclohexane and

Alkane

Geometrical Isomerism (Alkene)

NEXT GRAND TEST ON 27.01.2019

Dr. Sangeeta Khanna Ph.D


Test Dated: 23.12.2018 Topic: Organic Chemistry

READ THE INSTRUCTIONS CAREFULLY

1. The test is of 2 hour duration. 2. The maximum marks are 265. 3. This test consists of 63 questions. 4. Keep Your mobiles switched off during Test in the Halls.

Level – 1

Section – A (Single Correct Choice Type) Negative Marking

This Section contains 50 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. (Mark only One choice) 50 × 4 = 200 Marks

1. Which of the following compounds will not exhibit enolization?

a. b. c. d.

C

Sol. No - H 2. Which of the following is correct order for bond energy for C – H bonds in these compounds? a. Y > Z > X b. X > Z > Y c. X > Y > Z d. Z > X > Y C Sol. As the stability of radical increases bond energy decreases 3. Which of the following order is correct for hyperconjugation of these radicals?

a. P > Q > R b. R > Q > P c. Q > P > R d. P >R > Q D

Sol. 8 - H in P. 4. The barrier for rotation about indicated bonds will be maximum in which of these compounds?

CH2 – C – H

O

CH3 – C – CH3

O

CH3

O

CH3 H3C

H3C Ph – C – CH3

O

CH3

CH3 CH3

(P) (Q) (R)

CH2 H

X

CH2 H

Y CH2 H

Z

(X) (Y) (Z)



a. X b. Y c. Z d. Same in all

B

Sol.

5. Consider the following reactions:

Productmajor As

SOH/HOH

NaBH )ii( Hg(OAc) )i(

2

|

|3 ]X[

product major as ]Y[

CHCH

CH

CH

CCH 42

4

2

3

3

[X] and [Y] respectively be :

a. 2, 3-Dimethyl-2-butanol and 3,3-Dimethyl-2-butanol

b. 2, 3-Dimethyl-2-butanol and 2,3-Dimethyl-2-butanol

c. 3, 3-Dimethyl-2-butanol and 3,3-Dimethyl-2-butanol

d. 3, 3-Dimethyl-2-butanol and 2,3-Dimethyl-2-butanol

A

Sol.3

|

|

|3

HOH3

||3

shif t2,13

|

|32

|

|3 CHH

CH

C

OH

CH

CCHCHH

CH

C

CH

CCHCHHC

CH

CH

CCHHCHCH

CH

CH

CCH

3333

3

3

3

3

Y is formed by Markovnikoff’s addition without rearrangement

3

||

|3

NaBH )ii(

Hg(OAc) i)(2

|

|3 CHH

OH

C

CH

CH

CCHCHCH

CH

CH

CCH

3

3

4

2

3

3

6. Consider the following reaction sequence

In this B and D respectively are a. b.

Cl2/FeCl3

HNO3/H2SO4

A HNO3/H2SO4

B

C Cl2/FeCl3

D Below 330K

[Both are aromatic]

Θ

Θ

[One ring is antiaromatic]

Θ

[One ring is antiaromatic]

Cl

NO2

and NO2

Cl Cl

NO2

and

Cl

NO2



c. d. A Sol.

Remember : – Cl is o, p-directing while –NO2 is meta directing group. 7. Find out most stable substituted cyclohexane among the following:

a. b. c. d.

C

Sol. Bulky groups should be substituted on equatorial position. 8. Which are isomers?

a. Ethyl alcohol and dimethyl ether b. Acetone and acetaldehyde c. Propanoic acid and propanone d. Methyl alcohol and dimethyl ether A

9. Which of the following is the correct order of stability of different conformations of butane?

a. Staggered > Gauche > Partially eclipsed > Fully eclipsed b. Gauche > Staggered > Partially eclipsed > Fully eclipsed c. Staggered > Fully eclipsed > Partially eclipsed > Gauche d. None A

10. One among the following pairs of compounds is not a pair of isomers.

a. (CH3)2CHOC2H5; CH3(CH2)2OC2H5 b. CH3CH2CO2H; CH3COOCH3 c. (CH3)2CHCH2CH3; CH3CH2CH(CH3)2 d. CH3CH2NO2; CH2(NH2)COOH C

Sol. Both are same 11. Removal of a proton (H+, deprotonation) from cyclopropene is more difficult than from propene

because

a. the cyclopropenyl cation is produced b. the cyclopropenyl anion is anti-aromatic c. The allyl cation is not resonance –stabilized d. the allyl anion is not resonance-stabilized B

Cl

NO2

and

NO2

Cl NO2

Cl2/FeCl3 HNO3 Below 330K

Cl NO2

HNO3/H2SO4 Cl2/FeCl3

Cl NO2

NO2

Cl

o & p-directing (meta-directing)

o –isomer +

Cl

SO3H

and

Cl

NO2



Sol.

12. Identify the least stable carbocation.

a. b. c. d.

D

Sol. Electron withdrawing groups decreases stability of carbocation.

13. Phenol exists in 100% enol form. The reason is

a. Phenol is more stable than its keto form as phenol is aromatic

b. Phenol has high b.p.

c. Phenol is stabilized by H-bonding.

d. Keto form is non planar

A

14. The ortho/para directing group for benzene, among the following is:

a. –COOH b. –CN c. –COCH3 d. –NHCOCH3

D

Sol. NHCOCH3 is o, p – directing while all others are m – directing.

15. The Z-isomer from the following is

(i)

5237

3

HCCCH||

HCCH

(ii)

7325

3

HCCCH||

HCCH

(iii) FCH

||BrCCl

(iv)

HCF||

BrCCl

a. only (i) b. only (iii) c. Both (i) & (iii) d. both (i) & (iv)

C

16. The type of isomerism exhibited by the compounds with formula C4H10O is/are

a. chain and position b. functional and position

c. metamerism d. chain, position, functional and metamerism

D

Sol. Alcohol & ether are functional isomer

17. The functional groups present in Cortisone are:

a. Ether, alkene, alcohol b. Alcohol, ketone, alkene, ether

c. Alcohol, ketone, ester d. Ether, amine, ketone

B

CH2 CH2

OCH3

CH2

NO2

CH3

HOH2C

O

O

OCH3

Cortisone

CH2

CH3

H H

H+ +

Θ

(Antiaromatic)



18. The IUPAC name of the compound is : a. Toluene carbonyl chloride b. 3-Methyl benzene carbonyl chloride c. 5-Methyl benzoyl chloride d. 5-Methyl benzene carbonyl chloride B

19.

a. position isomers b. chain isomers c. functional isomers d. metamers D

20. Which of the following is not the correct order as indicated:

a. )(Stability HCCHCHHCCHHCCHCH 233322

b.

c.

d. (CH3)2CH = CH2 > CH3 – CH = CH2 > CH2 = CH2 (Stability) C

Sol. (a) = allylic free Radical is more stable than alkyl free Radical

(b) =

(c) Benzoic acid is stronger acid than alkyl carboxylic acid 21. What is the end product in the following reaction:

CBACaC3

3

AlCl

ClCH

tube glass

hot dReHOH2

a. Cyclohexane b. Benzene c. Toluene d. 2-Butene C

Sol.

22. In a mixture of Isooctane and n-heptane, the percentage of n-heptane is 60, the octane number of the

fuel is:

a. 60 b. 40 c. 20 d. 4

COCl

CH3

C O

O

and C O H

O

Θ

is aromatic and

Θ

is antiaromatic and least stable

Θ

>

Θ

>

Θ

(Stability)

>

>

(Stability)

CaC2

3

3

AlCl

ClCHHOHBenzeneCHHC

CH3



B

Sol.Octane number is volume of Isoctane in a mixture of Isooctane & heptane, which have same knocking

as that of sample petrol

23. Which of the following acid will form Ethane on decarboxylation and butane on electrolysis of aqueous

solution of its sodium salt

a. CH3COOH b. CH3CH2COOH

c. COOHH

CH

CCH

3|

3 d. COOHH

CHCH

CCHCH

32

|23

B

Sol. CH3CH2 COOH CH3CH3

3223iselectroly s s'Kolbe

23NaOH

CHCHCHCHNaOCOCHCH

24. Which of the following on ozonolysis will form two molecule of ethanal. a. 1-Butene b. 2-Butene c. 2-Butyne d. Butane B

Sol. 33oznolysis

33 CHOCHCHOCHCHCHCHCH

25. Which is the major product (P) in the following reaction:

a. b. c. d.

B

Sol.

26. Hydroxylation of 2-butene with Baeyer’s Reagent will form a. Butan-2-ol b. Butanone c. Butane-2, 3-diol d. Butane-2,3-dione C

Sol. 3||

3agentRe

s'Baey er33 CHH

OH

CH

OH

CCHCHCHCHCH

27. Which of the following is correct order of stability of alkene?

a.

b. c.

P 1. CH3Cl, Anhyd. AlCl3

2. Oxidation 3. Cl2/FeCl3

1. CH3Cl + AlCl3

Friedal craft Reaction

CH3

Oxidation

COOH

(meta directing)

Cl2/FeCl3

COOH

Cl

CH3 – CH = CH2 < < <

CH3 – CH = CH2 < < <

< < < CH3 – CH = CH2

CH3

Cl

COOH

Cl

COOH

Cl

+

COOH

Cl

CH2Cl



d. B

Sol. As the number of - H increases stability of alkene increases; Trans alkene is more stable than cis-alkene

28. :

Correct order of stability

a. P > Q > R > S b. P > S > Q > R c. P > R > Q > S d. S > R > Q > P C Sol. 29. In which of the following benzene ring is most electron rich.

a. b. c. d. A

Sol. 30. Bromobenzene is treated with magnesium in dry ether to form compound ‘X’ which is then treated with

ethanol to form compound ‘Y’. The compound y is

a. Benzene b. Phenol c. Ethylbenzene d. ethyl phenyl ether A

Sol. 31. Which of the following alkene has highest value of heat of hydrogenation?

a. b. c. d. D

< CH3 – CH = CH2 < <

CH2

CH3

>

CH2

CH3

(H.C & +I effect maximum)

(H.C & less +I)

>

CH2

CH3

>

CH2

(Only +I)

CH3

CH2CH3 CH

CH3 H3C

H3C – C – CH3

CH3

H C H

H

3 - H

due to maximum hyperconjugation

Br

Mg

MgBr

C2H5OH

H

+ Mg

Br

OC2H5

CH2 CH2 CH2 CH2

CH3

(Q) (R) (S)

CH3

CH3

(P)



Sol. Compound (d) is least stable as it contain minimum no. of - H.

alkene of stability

1H ionhy drogenat

32. 2-Bromopentane COH)CH/(COK)CH( 3333 P

The major product (P) formed in the above reaction is

a. pent-1-ene b. cis pent-2-ene

c. trans pent-2-ene d. 33|

223 )CH(OCH

CH

CCHCHCH

3

A Sol. With Bulky base major product is

Hoffmann’s product is less stable alkene

3222Bulky Base

322|

3 CHCHCHCHCHCHCHCHH

Br

CCH

33. Which of the following alkene will be most stable?

a. (CH3)2CH = CH2 b. CH3CH = CHCH3 c. (CH3)2C = C(CH3)2 d. CH3CH = CH2 C

Sol. Alkene with maximum hyperconjugation is most stable. 34. Ethylbenzene with bromine in presence of FeBr3, predominantly gives

a. b. c. d. A Sol. Ethyl group is o & p-directing; In presence of FeBr3, reaction is electrophilic substitution. 35. Which alkane (molecular mass 72) would yield three different monochloro derivatives? (excluding

stereoisomer)

a. n-Pentane b. Isopentane c. neopentane d. Isohexane A

Sol. CnH2n + 2 = 72 n = 5; CH3CH2CH2CH2CH3 give three mono chloropentane 1-Chloropentane + 2-chloropentane + 3 –chloropentane & isopentane give four isomer

3|

|

33|

|

332

|

|332

|

2|

CHH

Cl

CH

CH

CCH ;CHH

Cl

CH

CH

CCH;CHCH

CH

Cl

CCH;CHCHH

CH

CH

Cl

C

3333

36. In which of the following reactions Markownikov’s rule is not observed

a. peroxide Organic

23 HClCHCHCH b. peroxide Organic

23 HBrCHCHCH

c. peroxide Organic

223 OHCHCHCH d. CH3CH = CH2 + H2SO4

B Sol. It is anti-markovnikoff’s addition is presence of peroxide. No peroxide effect with HCl, H2SO4 & H2O

CH2CH2Br Br

Br

CH2CH3 +

CH2CH3

Br

CH2CH3

Br

+

CHCH3

Br

Br

+ CH2CH2Br CH2CH3

Br



37. In the following sequence of the reactions:

CH3CH2CH2I BA2

2

NaNH )ii(

Br )i((Alc.) KOH

The end product B is a. Alkene b. Alkanol c. Alkyne d. Alkyl amine C

Sol. )B(

3)ii(

NaNH2

||3

)i(23 CHCCHH

Br

CH

Br

CCHCHCHCHA 2

38. be can X ;XAIHCK 770

)g(Cl(alc.) KOH73

2

a. 1-chloropropene b. 3-chloropropene c. Propyl alcohol d Propene B

Sol. 22

|

onChlorinati K Ally lic 770

Cl 23

KOH .alc73 CHCHH

Cl

CCHCHCHIHC 2

39. Propene BABoil/OH)ii(

Mg/ether )I(HBr

2

In the above sequence of reactions B is

a. Propane b. Butane c. Propene d. Ethane A

Sol. 323HOH

3|

3Mg

3

|

3HBr

23 CHCHCHCHH

MgBr

CCHCHH

Br

CCHCHCHCH

40. Which one of the following will undergo meta-substitution on monochlorination?

a. Ethoxybenzene b. Chlorobenzene c. Methyl benzoate d. Phenol C

Sol. It is meta directing

41. When 3||

3 CHH

Cl

CH

Cl

CCH is treated with NaNH2 the major product formed is:

a. CH3 CH = CH2 b. CH3 – C ≡ C – CH3 c. CH2 = C = CH – CH3 d. CH2 = CH – CH = CH2

D Sol. double dehydro halogenations of dihalide will form alkyne 42. What is X in the following reaction?

a. b. c. d. A Sol.

C – O – CH3

O

CH2CHO CHO COCH3

C ≡ CH Hg

2+/HOH

H2SO4 X

O O

C ≡ CH HOH

C = CH2

OH

C – CH3

O

Hg+2

, H+

Tautomerise to Keto form



43. Which of the following will not produce ethane?

a. Reduction of CH3COOH with HI and Red Phosphorus b. Reduction of CH3COCH3 with HI and Red Phosphorus

c. Soda lime decarboxylation of sodium propionate d. Hydrogenation of ethene in the presence of Ni B

Sol. It will produce propane

22323K 423

P red3

||

3 I2OHCHCHCHHI4CH

O

CCH

44. When CH3[CH2]3 C ≡ CH is oxidized with hot acidic KMnO4 the product is:

a. CH3CH2COOH b. CH3CH2CH2COOH c. CH3CH2CH2COOH & CO2 d. CH3CH2CH2CH2COOH & CO2 D

Sol. Alkyne give lower acid. HCOOH oxidised to give CO2.

OHCOHCOOHCOOHCHCHCHCHCHCCHCHCHCH 222223Oxidation

2223

45. A metallic carbide of gp(II) on treatment with water gives a colourless gas which burns readily in air and gives a precipitate with ammonical silver nitrate. The gas is:

a. Methane b. Ethene c. Ethyne d. Propane C

Sol. Hydrocarbon is terminal alkyne. Only terminal alkyne give white ppt. with AgNO3 & metallic carbide is CaC2.

46. Friedal craft reaction of benzene with isobutyl chloride produces:

a. isobutylbenzene b. tert–Butylbenzene c. n–butylbenzene d. sec–Butylbenzene

B

Sol. 3

|

32

|

3 CH

CH

CCHCHH

CH

CCH

33

47. YXHC3

2

42

3

FeCl

Cl

SOH

HNO66

In the above sequence Y can be a. ortho and p-Chloronitrobenzene

b. 3-Nitrochlorobenzene c. only 4-Nitrochlorobenzene d. equal mixture of all the above products B

Sol.

48. :oductPr)major(

Mole) (1 H2

H3C – C – CH3

CH3

x = NO2

; y =

NO2

Cl

; NO2 group is meta directing due to –M & –I effect



a. b. c. d. None of these B Sol. 49. Which of the most preferred site of electrophilic attack in the following compound.

a. 2 & 3 b. 6 & 4 c. 1 & 3 d. 1, 3 & 5 B

Sol.

HN with lone pair will exert e– donating effect & makes the ring activating at o & p-position 50. An alkyl bromide (A) forms Grignard’s reagent which on treatment with water yields n-hexane . (A) with

sodium in presence of dry ether forms 4,5-diethyloctane. (A) is:

a. CH3[CH2]5Br b. CH3[CH2]3CH(Br)CH3 c. CH3 – [CH2]2 – CH(Br)CH2CH3 d. CH3[CH2]2CH(Br) CH = CH2

C Sol.

H

H

1, 4-addition product is major

NH O

1

3

2

4

5

6

CH3 – CH2 – CH2 – CH – CH2CH3

Br

Mg CH3CH2CH2 – CH – CH2 – CH3

MgBr

HOH Hexane

Na + dry ether

CH3 – CH2 – CH2 – CH – CH – CH2 – CH2 – CH3

CH2

CH3

CH2

CH3

4,5-Diethyloctane



LEVEL – 2

SECTION – D (More than One Answer) No Negative Marking

This Section contains 5 questions. Each question has four choices A), B), C) and D) out of which ONE OR MORE may be correct. (5 × 5 = 25 Marks)

1. Which of the following represent correct matching?

a. OH

O

CCH| |

3 and 3

| |

OCH

O

CH Metamers

b. CH3 – CH2 – C CH and CH3 – C C – CH3 Position isomers

c. CH3CH2CH2NH2 and 3

2

|3 CHH

NH

CCH Tautomers

d. CH3CH2OH and (CH3)2O Functional isomers. B,D

Sol. Metamers have same functional group. 2. Which of the following cycloalkanes will show cis-trans isomerism?

a. b. c. d. C,D 3. Keto-enol tautomerism is observed in:

(a) C6H5 – CHO (b) C6H5COCH3 (c) C6H5COC6H5 (d) C6H5COCH2COCH3 B,D

4. Which of the following alkenes are more stable than ? a. b. c. d. A,B,C,D Sol. A = 10 CH; B = 5 – CH; C = 4CH; D = 4 – CH; given compound have – 3CH 5. Which of the following compound on oxidation with acidic KMnO4 will give Benzoic acid?

a. Ethylbenzene b. Benzaldehyde c. Phenol d. Methylbenzene A,B,D

Sol.

CH3

CH3

CH3

CH3

H3C

CH3 H3C

CH2CH3 COOH CH2OH

CH3

CHO

Toluene



SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. (2 × 8 = 16 Marks)

1. Match the Column – I with Column – II. (More than one Match)

Column- I Column – II a. (p) Structural isomer b. (q) Functional isomers

c. (r) Tautomerism

d. CH3 – CH2COOH & H

O

CCHH

OH

C||

22|

(s) Metamerism

Sol. A p,q,r; B p,q,r; C s,p; D p,q 2. Match Column (I) with Column (II). (More than One Match)

Column – I Column – II

(a) (p) Nucleophilic substitution

(b) (q) Electrophilic substitution

(c) (r) Cation intermediate

(d) (s) Free radical substitution Sol. A – (q), (r); B – (q), (r) ; C – (p); D – (s)

(a) In electrophilic substitution, Intermediate is cation.l

Et

Me

Et

N and

Me

Pr

N Me

CHO and

OH

NH HN

O

N O O

H

and

OH

N

HO OH

N

N

+ Ph – CH2 – Cl AlCl3

CH2 – Ph

H3C – C – CH3

CH3

CH3

Br2, hv CH3 – C – CH2 – Br

CH3

CH3

+ HNO3 H2SO4 NO2

Br

KCN CH3 – CH – CH3

CN

+ KBr



SECTION – F (Integer Type) No Negative Marking This Section contains 6 Questions. The answer to each question is a Single Digit Integer ranging

from 0 to 10. 6 × 4 = 24 Marks

1. The number of geometrical isomers in case of CH3 – CH = CH – CH = CH – C2H5 is Sol. 2n = 4 2. Examine the structural formulas of following compounds and find how many compounds will produce

CO2 on with hot acidic KMnO4

(a) , (b) , (c) , (d)

(e) , (f) H3C – C CH, (g) Ph – CH = CH2 Sol. 5 (b); (c); (e); (f); (g) (terminal alkene & alkyne) 3. From the following compounds/ ions:

(a) 3HC

(b) 4HN

(c) BF3 (d) NH3 (e) AlCl3 (f) F– (g) 2CCl (h) CH2 = CH2

Identify value of X. Where X is the total number of electrophiles. Sol.4 (a, c, g, e) 4. How many trichloroethanes would be produced when 1,1-Dichloroethane reacts with chlorine? Sol. 2;

5. How many of the following are meta directing and deactivating than Benzene Sol. 4; (ii), (iii), (iv), (viii) 6. The total number of structural isomers possible for structure C3H6 are Sol. 2

C – C = C &

CH3 – C – Cl

Cl

Cl

& CH2 – CH Cl

Cl

Cl2 h

CH3 – CH Cl

Cl

Cl

(i)

Cl

; (ii)

NO2

; (iii)

CN

; (iv)

COOH

;

(v)

O – C – CH3

;

O

NH2

; (vi)

C(CH3)3

; (vii)

CF3

(viii)

Documents

NEXT GRAND TEST ON 27.01 - Dr. Sangeeta Khanna · D:\Important Data\2018\+1\Org\Test\Grand Test-6\+1 Grand Test-6 23.12.2018.doc Test Dated: 23.12.2018 Topic: Organic Chemistry READ