Dr. Sangeeta Khanna Ph.D 1 · A chloro compound (X) on treatment with Zn/HCl gave a hydrocarbon with five carbon atoms in the molecule. When (X) is dissolved in ether and treated

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+1\Org\Test\GT-7\Grand Test -7 (organic Chemistry) Level -2.docx

Dr. Sangeeta Khanna Ph.D


LEVEL – II

TOPIC: HYDROCARBON & GENERAL ORGANIC CHEMISTRY

READ INSTRUCTIONS CAREFULLY

1. The test is of 1 hour duration.

2. The maximum marks are 161.

3. This test consists of 37 questions.

SECTION – A (Single Answer Type) Negative Marking [-1]

This Section contains 15 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. (Mark only One choice) 15 × 4 = 60 Marks

1. Cyclohexene when treacted with Br2 in CCl4 gives:

a. Bromocyclohexane b. trans –1, 2 – Dibromocyclohexane

c. cis – 1, 2 – Dibromocyclohexane d. 1, 1-Dibromocyclohexane

B

Sol. Addition of Br2 is Anti-addition; Br2 will add on opposite side to ring.

2. The treatment of benzene with isobutene in the presence of sulphuric acid gives:

a. Isobutyl benzene b. ter–Butyl benzene

c. n–Butyl benzene d. No reaction

B

Sol.

3. A chloro compound (X) on treatment with Zn/HCl gave a hydrocarbon with five carbon atoms in the

molecule. When (X) is dissolved in ether and treated with sodium 2,2,5,5-tetramethylhexane is

obtained. The compound (X) is

a. 2-Chloro-2-methylpropane b. 1-Chloro-2, 2-dimethylpropane

c. 2-Chloro-2-methylbutane d. Isopropyl chloride

B

Sol. 3

|

|22

|

|3

actionReWurtz2

|

|3 CH

CH

CH

CCHCH

CH

CH

CCHClCH

CH

CH

CCH

3

3

3

3

3

3

4. The reaction of one equivalent of HBr with CH2 = CH – C CH gives

a. CH2 = CH – C CBr b. CH2 = CH – CBr = CH2

c. CH3 – CHBr – C CH d. CH2 = CH – CH = CHBr

B Sol. First addition on triple bond due to formation of stable product a conjugated diene.

+ H3C – C = CH2

CH3

H+ C CH3

CH3

CH3



5. The bond dissociation energy of the C – H bond for the compound

)P(

HCH3

)Q(

HCHCH 23

decreases in the order: a. P > Q > R > S b. S > R > Q > P c. S > P > Q > R d. Q > P > S > R C Sol. Bond dissociation energy can be compared by stability of free radical

CH3

radical benzy lstabilise

Resonance2

gationhy perconjuby Stabilise

23 HCPh HCCH

6. :oductPr)major(

Mole) (1 H2

a. b. c. d. None of these B Sol. 7. Which of the following molecules is expected to have the greatest resonance stabilization?

a. b. c. d. B Sol.

8. Consider the following reaction sequence In this B and D respectively are a. b.

Highly destabilise

4n + 2 = 6 Aromatic

4n + 2 = 6 Aromatic

Cl

NO2

and

Cl

NO2

Cl

NO2

and NO2

Cl

Cl2/FeCl3

HNO3/H2SO4

A HNO3/H2SO4

B

C Cl2/FeCl3

D Below 330K

H

(S)

CH2 – H

(R)

H

H

1, 4-addition product is major



c. d. A Sol. Remember : – Cl is o, p-directing while –NO2 is meta directing group. 9. a. R > S > Q > P b. P > S > R > Q c. Q > P >S > R d. S > R > Q > P D Sol. As - H decrease, stability of anion increases.

Stability of carbanion

)gationHy perconju(

effect donating e

1

10. Which of the following acids has lowest pKa value?

a. OH

O

CH

OH

C||

2|

b. OH

O

CCHH

NO

C| |

22

2

|

c. OH

O

CH

Cl

C| |

2|

d. OH

O

C

F

F

CCH|||

|3

D

Cl

SO3H

and

Cl

NO2

Cl

NO2

and

NO2

Cl NO2

CH2

CH3

(P)

CH2

CH2CH3

CH2

CH CH3

CH3

CH2

CH3

H3C – C – CH3

(Q) (R) (S)

Cl2/FeCl3 HNO3 Below 330K

Cl NO2

HNO3/H2SO4 Cl2/FeCl3

Cl NO2

NO2

Cl

o & p-directing (meta-directing)

o –isomer +



Sol. ,COOH

F

F

CCH|

|3 2 (–I) groups ; is most acidic & have lowest pKa & highest Ka

11. In given structure which lone pair is localised a. 1 & 2 b. 2 & 4 c. 1 & 4 d. 2 & 3 C

Sol. Lone pair on doubly bonded atom will not be delocalized. 12. 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

13. Which Benzene ring is most reactive in electrophilic attack a. (1) b. (2) c. (3) (d) Same Reactivity B

Sol.

(+ve charge is delocalized in both the ring. More structure have aromatic ring in carbocation)

14. 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

NH O

1

3

2

4

5

6

1 2 3

E

N

H NH

N

NH H

1

2

4

3



Sol. 15. Treatment of 2-methyl-2-butene with HBr in the presence of a peroxide yields

a. A primary alkyl bromide b. A secondary alkyl bromide c. A tertiary alkyl bromide d. A vicinal dibromide B

Sol. 3|

|

|3

H

Radical f ree in entrearrangem no

3

||

3Peroxide

HBr3

|

3 CHH

Br

C

CH

H

CCHCHH

Br

C

CH

CCHCHCH

CH

CCH

333

SECTION – C (Comprehension Type) Negative Marking [-1]

This Section contains 2 Comprehension. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 9 × 4 = 36 Marks

Passage 1

When (C – H) sigma electrons are in conjugation with pi bond, this conjugation is knwon as (C – H)

conjugation, excess conjugation or hyperconjugation. (i) Compound should have at least one sp2-hybrid carbon of either alkene, alkyl carbocation or alkyl

free radical.

(ii) -carbon with respect to sp2 hybrid carbon should have at least one hydrogen. (iii) Resonating structures due to hyperconjugation may be written involving "no bond” between the

alpha carbon and hydrogen atoms

2

Θ|

2

Θ|

|2

Θ|

|2

|

|HCCH

H

H

CHHCCH

H

H

C HHCCH

H

H

CHCHCH

H

H

CH

In the above resonating structures there is no covalent bond between carbon and hydrogen, and from this pooint of view, hyperconjugation may be regarded as "no bond resonance”. Actually the hydrogen atom is not free from the carbon. These resonating structures only suggest that: (a) there is some ionic character between C – H bond and (b) carbon-carbon double bond acquires some single bond character. We can explain the stability of alkene, carbocation and carbon free radical on the basis of hyperconjugation.

ionhydrogenat of Heat

1H- of number alkene of Stability

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



1. Which of the following statements are correct for C6H5 – CCl3? (a) CCl3 group is electron withdrawing due to the –I effect and reverse hyerpconjugation. (b) CCl3 group is meta directing due to the – M effect. (c) CCl3 group is o, p-directing because it is +R group. (d) CCl3 group can exert +M effect. A

2. Which of the following has the lowest heat of hydrogenation?

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

D Sol. Maximum Hyperconjugation. Means least stable alkene 3. Carbon-carbon double bond length will be maximum in which of the following compounds?

(a) CH3 – CH = CH2 (b) CH3 – CH = CH – CH3

(c) 3

|

|3 CH

CH

C

CH

CCH

3

3

(d) CH2 = CH2

C

Sol. Maximum hyperconjugation will make – C = C –; a single bond

4. The heats of combustion for the four C6H12 isomers shown are (not necessarily in order): 955.3, 953.6, 950.6, and 949.7 (all in kilocalories per mole). Which of these values is most likely the heat of combustion of isomer 1?

a. 955.3 kcal/mol b. 950.6 kcal/mol c. 953.6 kcal/mole d. 949.7 kcal/mol

B Sol. Stable alkene have lowest heat of Hydrogenation. Stability order is 2 > 1 > 3 > 4; 2nd lowest value

for 1 5. The heats of combustion for the four C6H12 isomers shown are (not necessarily in order): 955.3,

953.6, 950.6, and 949.7 (all in kilocalories per mole). What can be said about isomers 3 and 4?

a. 3 is more stable by 1.7 kcal/mol. b. 3 more stable by 3.0 kcal/mol. c. 4 is more stable by 1.7 kcal/mol. d. 3 is more stable by 0.9 kcal/mol. A

Sol. )3(6.950 ;

)4(7.949 ; difference is 1.7 kcal/mol. Isomer 3 is more stable by 1.7.

1 2 3 4

1 2 3 4



Passage – 2

6. Which of the following is compound P? (a) (b) (c) (d) A

Sol. As this compound on dehydrohalogenation will form non terminal alkyne Which don’t react with Na. 7. R and S are:

(a) position isomers (b) enantiomers (c) geometrical isomers (d) functional group isomers C

Sol. R = cis-pent-2-ene; S = trans pent-2-ene 8. Identify structure of compound T:

(a) OH

O

CCH||

3 (b)

(c) OH

O

CCHCH||

23 (d) H

O

CCH||

3

B Sol. Alkyne on ozonolysis form dione;

2|

||||

3OH/Zn .2

O .1323 H

CH

C

O

C

O

CCHCHCHCCCH

3

2

3

O

O

Br

Br Br

P(C5H9Br)

NaNH2

Q Decolourise Br2 water

and cold KMnO4 Q T 1. O3

2. Zn/H2O (No reaction with Na)

Lindlar catalyst Na/NH3()

R (C5H10)

S (C5H10)

(1) O3

(2) (CH3)3S Z + Y

Br



9. Compound z & y are a. CH3CHO + CH3CH2CHO b. CH3COOH + CH3CH2COOH

c. CHOCHCHOH

CH

CCH 3

|

3

3

d. Pentane-2, 3-dione

A

Sol. OS\/

CH

CHCHOCHCHCHOCHCHCHCHCHCH

3

3

233Ozonoly sis

323

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

This Section contains 5 multiple choice 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 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 2. Which of the following compound on oxidation with acidic KMnO4 will give Benzoic acid?

a. Ethylbenzene b. Benzaldehyde c. Phenylmethanol d. Toluene A,B,C,D

Sol.

3. )X(CH

H

C

H

CCH 2Br

)cis(

3||

3 ;

The product (X) is:

a. formed by anti addition of Br2 b. racemic mixture of ‘d’ and ‘’ form c. CH3CHBrCHBrCH3 d. formed by syn addition of Br2 A,B,C

CH2CH3 COOH CH3OH

CH3

CHO



4. Which of the following compound on Reaction with Red P + HI will give 2-Methyl propane a. 2-Methylpropan-2-ol b. 2-methylbutanoic acid c. Isobutyl iodide d. 2-methyl propanal A,C,D

Sol. (A) 3

|

3H P dRe

3

|

|3 CHH

CH

CCHCH

CH

OH

CCH

33

I

(B) etanMethy lbu2

32

|

3H P dRe

2

|

3 CHCHH

CH

CCHCOOHCHH

CH

CCH

33

I

(C) 3

|

3HIP Red

2

|

3 CHH

CH

CCHCHH

CH

CCH

33

I

(D) 3

|

3HIP dRe

|

3 CHH

CH

CCHCHOH

CH

CCH

33

5. Ni/H2

33CH

DCC

CH

D

/\

\

/

Product of above reaction will be:

a. racemic mixture b. optically inactive c. meso d. constitutional isomers B,C

Sol.

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. 8 × 2 = 16 Marks

1. Column (I) Column (II)

(a) 23 HCCH

(p) Mesomeric effect

(b) H3C - 2HC

(q) Hyperconjugation

(c) Cl

Cl

CClΘ

| (r) + I effect

(d) (s) Reverse Hyperconjugation

Sol. A Q, R; B Q, R; C S; D P, R 2. Match Column (I) with Column (II)

(a) (p) Nucleophilic substitution

C O

O

H Θ

+ Ph – CH2 – Cl AlCl3

CH2 – Ph

H3C

D C = C

D

CH3 H2/Ni H3C

D C C

D

CH3

H H

Meso (Plane of symmetry) Syn addition



(b) (q) Electrophilic substitution (c) (r) Cation intermediate (d) (s) Free radical substitution

Sol. A – (q), (r); B – (q), (r) ; C – (p); D – (s)

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 9. (6 × 4 = 24 Marks)

1. 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) 2. How many carbocations undergo rearrangements?

(i) (ii) (iii) 3

|

23 CH

CH

CCHCH

3

(iv) (v) (vi) (vii) (viii) (ix) (x)

Sol. 6; (i), (ii), (iv), (v), (vi); (x) (i) methylshift (ii) ring contraction, (iv) & (v) ring expansion; (vi) H– shift (x) H– shift

H3C – C – CH3

CH3

CH3

Br2, hv CH3 – C – CH2 – Br

CH3

CH3

CH2

Me3C – CH2 – CH2

CH2

CH2 = CH

CH2 = CH – CH2

+ HNO3 H2SO4 NO2

Br

KCN CH3 – CH – CH3

CN

+ KBr

HC – CMe2 Ph

Ph

C – CH3

OH



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 of the following halide on reaction with Mg, dry ether followed by acidification will form 2-

methyl pentane

(a) 32

||

3 CHCHH

Br

CH

CH

CCH

3

(b) (c)

(d) BrH

CH

CCH)CH(

3|

23 (e) 3|

|

23 CHH

Br

CH

CH

CCHCH

3

(f) 3||

|

3 CHH

CH

CH

Br

CH

CH

CCH

3

3

(g) (CH3)3C – CH2Br Sol. 3 (a), (b), (c)

2|

2

|

3H

32

||

3Mg

32

||

3 H

CH

CCH

CH

CCHCHCHH

MgBr

CH

CH

CCHCHCHH

Br

CH

CH

CCH

3

333

5. How many of following reactions are electrophilic addition

(a) 3|

3H

HOH23 CHH

OH

CCHCHCHCH

(b) OH

CH

CH

CCHH

CH

CH

CCH

3

3

4

3

3

|

|3

H

KMnO|

|3

(c) (d)

(e) 3

|

|3

HCl2

|

3 CH

CH

Cl

CCHCH

CH

CCH

33

(f) 3

||||

3Ozonoly sis

33 CH

O

C

O

CCHCHCCCH

Ans. 2 (a, e) (B = oxidation (free Radical); C = free Radical addition; D = electrophilic subsitution 6. How many trichloroethanes would be produced when 1,1-Dichloroethane reacts with chlorine ? Sol. 2;

H3C – CH – CH2 – CH2

CH2 – Br

CH3

+ CH3Cl Anhyd.

AlCl3 + HCl +

CH3

CH3 – C – CH2

CH3

Br

CH2

CH3

+ Cl2 h

Cl

Cl

Cl

Cl

Cl

Cl

CH3 – C – Cl

Cl

Cl

& CH3 – CH Cl

Cl

Cl2 h

CH3 – CH Cl

Cl

Cl

Documents

Dr. Sangeeta Khanna Ph.D 1 · A chloro compound (X) on treatment with Zn/HCl gave a hydrocarbon with five carbon atoms in the molecule. When (X) is dissolved in ether and treated