Contents · PDF file · 2015-10-03Topic Page No. Theory 01 - 08 Exercise - 1 09 -...

Topic Page No.

Theory 01 - 08

Exercise - 1 09 - 18




Answer Key 33 - 34

Contents

Syllabus

STRUCTURAL IDENTIFICATION & POC-1

Structural Identification

Degree of unsaturation, Catalytic hydrogenation, Monochlorination, Ozonolysis

Test of hydrocarbons and test of functional groups.

Name : ____________________________ Contact No. __________________

POC-1 & STRUCTURAL IDENTIFICATION # 1A-479 Indra vihar, kotaPh. - 9982433693 (NV Sir) 9462729791(VKP Sir)

Physical & Inorganic By

NV SirB.Tech. IIT Delhi

Organic Chemistry By

VKP SirM.Sc. IT-BHU

KEY CONCEPTSTRUCTURAL IDENTIFICATION & POC-1

STRUCTURAL IDENTIFICATIONCalculation of Degree of Unsaturation (DU) :

It is the hydrogen deficiency index (HDI) or Double Bond Equivalence (DBE)

That means deficiency of 2H is equivalent to 1 DU

(i) 1 DU = Presence of 1 double bond or presence of 1 ring closure(ii) 2DU = Presence of 2 double bond or 1 triple bond or two ring closure or 1 double bond + 1 ring closure

G.F. D.U.

(i) CxHy

y(x 1)2

(ii) CxHyOz

y 0(x 1)2

(iii) CxHyXs

y s(x 1)2

(iv) CxHyNw

y w(x 1)2

(v) CxHyOzXsNw

y s w(x 1)2

MONOCHLORINATION :When an alkane or a cycloalkane is treated with halogen (Cl2, Br2, F2, I2), a photochemical reaction takesplace and a C – H bond cleaves and a C – Cl bond is formed. So, one H-atom is substituted by one halogenatom. This is known as monohalogenation reaction.

Ex 1. Ex 2.

Ex 3. Ex 4.

Ex 5.

Application : If a molecule has more than one type of H-atom, then on monochlorination, it forms amixture of monochloroisomers. All these isomers are position isomers.

Conclusion : Hence, it can be concluded that the total number of position isomers (structural) of monochlorocompounds is equal to the number of different types of H-atoms present in the reactant. The different typeof H-atoms are also known as non-identical Hydrogens or non-equivalent Hydrogens or chemically differentHydrogens.





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Ex 6. (structural)

Ex 7. (structural)

Ex 8. (structural)

Ex 9. (structural)

Ex 10. (structural)

Remark : In aromatic hydrocarbons, the hydrogen atoms of the side-chain are chlorinated but H-atoms ofbenzene ring are stable. In pure benzene, no monochlorination occurs.

Ex 11.

Catalytic Hydrogenation of C = C; C C(i) Alkenes, Alkynes, polyenes or polyynes can be hydrogenated by using catalysts Ni/Pt/Pd at room

temperature.

(ii) All C – C bonds(C = C, C C) are hydrogenated. The reaction can't be stopped at any intermediatestage.

(iii) Aromatic bonds which are stable at room temperature but can be hydrogenated at high temperature.

(iv) It can be concluded that the hydrogenation product of an alkene or alkyne or any unsaturated compoundis always a saturated compound.

(v) The number of moles of H2 consumed by 1 mole of compound is equal to the number of bondspresents.

(vi) All positional isomers of alkenes or alkynes (due to multiple bond) always give sarne product onhydrogenation.

(vii) During catalytic hydroqenation no rearrangement in carbon skeleton takes place.

General reaction :





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[Reaction cannot be stopped at any intermediate stage]

Ex. 12

Ex. 13 5 Monochloro product (structural)

Ex. 14 5 Monochloro product (structural)

Ozonolysis :(i) It tells about position of unsaturation.

(ii) Alkene and polyalkene on ozonolysis undergo oxidative cleavage.

(iii) (a) The reagent of reductive ozonolysis is

(i) O3 (ozone) (ii) Zn and H2 or Zn and CH3COOH or (CH3)2S

(b) The reagent of oxidative ozonolysis is O3 and H2O2.

(iv) The products are carbonyl compounds (aldehydes or ketones). This type of ozonolysis is known asreductive ozonolysis.

(v) Ozonolysis does not interfere with other functional groups.

General Reaction :

Ex. 15

Ex. 16

Ex. 17

Ex. 18





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Applications :The process is used determine the position of C = C in a molecule.

If the products are rejoined, the position of C = C can be determined in the reactant molecule. All C = C(except aromatic ones) undergo oxidative cleavage under normal conditions.

At higher temperature, the aromatic double bonds can also undergo ozonolysis.

Ex. 19

Ex. 20

Ex. 21

Ex. 22

POC-I :IntroductionThe main objective of an organic chemist is the determination of the structure of a new organic compoundwhich has been obtained in pure state either from a natural source or synthesised in the laboratory.

In order to establish the correct structure of an organic compound, it is necessary to detect element andfunctional group present in the organic compound.

Detection of elements (Qualitative Analysis) :Most of the organic compounds contain 2 to 5 different elementsThe principal elements present are carbon, hydrogen and oxygen.Less commonly present elements are nitrogen, sulphur and halogens.

In few organic compounds, phosphorus and metal may also be present.The order of abundance of these elements in organic compounds is given below.

Carbon Always presentHydrogen Nearly always presentOxygen Generally presentNitrogen, halogen, sulphur Less commonly presentPhosphcrus and metal Rarely present

Detection of nitrogen, sulphur and halogens are tested in an organic compounds by lassaigne's test.

The organic compound (N,S or halogen) is fused with sodium metal as to convert these elements inionisable inorganic substance i.e. nitrogen into sodium cyanide, sulphur into sodium sulphide and halogensinto sodium halides.

These cyanide, sulphide or halide ions can be confirmed in the aqueous solution by usual test. The aqueoussolution (fused sodium extract) is called lassaigne's filterate.





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Table : Identification of Elements in Organic Compounds





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Tabl

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PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.

Section (A) : Degree of Unsaturation and hydrogenationA-1. Which of the following pair of hydrocarbon will give same product after hydrogenation.

(A)

(B)

(C) CH3–CH2–CH=C(CH3)2,(CH3)2C–CH=CH–CH3

(D) (B) and (C) both

A-2. Which of the following hydrocarbons give same product on hydrogenation.(A) 2-Methyl hex-1-ene & 3-Methyl hex-3-ene.(B) 3-Ethyl hex-1-en-4-yne & 2-Methylhept-2-en-4-yne.(C) 3-Ethylcycloprop-1-ene & 1,2-Dimethylcycloprop-1-ene.(D) 2-Methylbut-2-ene & 3-Methylbut-1-ene.

A-3. Which of the following is correct DU of the given structures :Column I Column IICompound Degree of unsaturation

(A) 2

(B) 4

(C) 4

(D) 0

A-4. Calculate the DU of following compound C8H12O, C3H5N, C4H8O respectively :(A) 4,3,2 (B) 3,2,1 (C) 2,1,3 (D) 2,2,3





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A-5. Which of the following compound have same degree of unsaturation :

(A) I, II (B) I and III (C) I, II and III (D) I, II and IV

Section (B) : MonochlorinationB-1. Which of the following is not correct option for (P) in the given reaction sequence

(P) Unsaturated Hydrocarbon 2H /Ni

(excess) (Q) Saturated Hydrocarbon 2Cl /h 3-monochloro structural

product :

(A) (B) (C) (D)

B-2. Identify E in the following sequence of reaction.

(A) (B) (C) (D)

B-3.

Number of structural monochloroination product of above compound is :(A) 5 (B) 4 (C) 6 (D) 3

B-4. Which of the following compound will give four monochloro (structural) product on monochlorination.

(A) (B) (C) (D)

B-5. Which of the following alkenes give four monochloro (structural isomer) product after hydrogenation(A) Pent-2-ene (B) 2-Methylbut-2-ene (C) 3-Methylhex-2-ene (D) 2, 3-Dimethylbut-2-ene

B-6. 'M' = C8H16 shows geometrical isomerism and on monochlorination shows 3 monochloro structural products.Which can be the structure of M.

(A) (B) (C) (D)





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Section (C) : OzonolysisC-1.* Which of the following unsaturated hydrocarbon(s) give only ketones upon reductive ozonolysis.

(A) 2, 3-Dimethyl but-2-ene (B) 1, 2, 3, 4-tetramethylcyclopenta-1,3-diene(C) Out-2-yne (D) 2-Methylbut-2-ene

C-2. An alkene (A)

(A) (B) (C) (D)

C-3. A hydrocarbon C6H4 gives C3H2O3 on ozonolysis. The hydrocarbon is :

(A) (B) (C) (D)

C-4. Carbon chain splits into two or more parts by ozonolysis in the following except in :(A) 1, 3-Cyclobutadiene (B) 1, 2-Dimethylcyclohex-1-ene(C) Benzene (D) Isobutene

C-5. A compound P (C7H10) by ozonolysis forms :

The structure and degree of unsaturation of compound P are :

(A) & 3 (B) & 4 (C) & 3 (D) & 3

C-6.* Which of the following compounds form single product in each of the following reactions?(i) Ozonolysis (ii) Catalytic hydrogenation (iii) Hydrogenation followed by monochlorination

(A) (B) (C) (D)

C-7. A conjugated alkadiene having molecular formula C13H22 on ozonolysis yields butanone, ethanedial &cyclohexanecarbaldehyde. Which of the following is the correct structural formula of the alkadiene.

(A) (B)

(C) (D)





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Section (D) : Test for unsaturated compoundsD-1. What is the structure of a optical active compound (C3H2D2) which decolourise bromine water.

(A) (B) (C) CHD = C = CHD (D) CHD2 – C CH

D-2.

Compound X is :

(A) H3C–C=C–CH3 (B) CH2=CH–CH=CH2 (C) CH3–CH2–C=CH (D)

D-3.

Identify X :(A) CH3–CH2–CC–CH2–CH3 (B) CH3–CC–CH2–CH2–CH3

(C) (D)

D-4. Which one of the following will not give white precipitate with ammonical silver nitrate solution

(A) CH3 – C C – CH3 (B)

(C) CH3 – CH2 – CH = CH2 (D) All of these

Section (E) : Test for alcohol and phenol compounds

E-1.

Identify x :

(A) (B)

(C) (D)

E-2. Alcohol X having four carbon atoms. It gives Lucas test in 10 min. Which cannot be the alcohol X.

(A) (B) (C) (D)





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Section (F) : Test for aldehyde and ketoneF-1. Which of the following compound will give black or silver ppt. with Tollen’s reagent.

(A) (B)

(C) CH3 – C C – CH2 – CHO (D)

F-2. The compound 'A' gives following reactions.

Its structure can be :

(A) (B) OHC – (CH2)2 – CH = CH – COOH

(C) (D)

F-3. Which of the following compound will not give positive iodoform test.

(A) (B)

(C) (D) CH3 – CHO





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F-4.

Identify ‘P’

(A) (B) CH3–CH2–CH2–CHO

(C) (D)

F-5. A compound which can give iodoform test, bromine water test, FeCI3 test but not Tollen's test.

(A) (B) (C) (D)

F-6. Tollen's reagent (AgNO3 + NH4OH) can be distinguish between :

(A) (B)

(C) (D)

Section (G) : Test for acids, esters, amides and amines

G-1. Which of the following will not give +ve test with CHCI3/KOH.(A) CH3 – CH2 – NH – CH3 (B) CH3 – CH2 – CH2 – NH2

(C) (D)





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G-2.

Identify 'X'

(A) CH3 – CH2 – COOH (B)

(C) CH3 – CO – CH2 – OH (D) CH2 = CH – CH2 – OH

G-3. An organic compound X (C4H8O2) gives positive test with NaOH and phenopthalein. Structure of X will be:

(A) (B)

(C) (D)

G-4. Which of the following compound will not give positive test with NaHCO3.

(A) (B) (C) (D)

G-5. An organic compound X (C6H10) gives positive test with Na metal and after oxidative ozonolysis gives Ywhich is optically active and gives positive test with NaHCO3. Compound X can be :

(A) CH3 – C C – CH2 – CH2 – CH3 (B)

(C) CH3 – (CH2)2 – CH2 – C CH (D)

G-6. Which of the following compound will give smell of NH3 with conc. NaOH.

(A) (B)

(C) (D)





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PART - II : MISCELLANEOUS QUESTIONS

Comprehensions :

Comprehension # 1Observe the following diagram and answer questions.

1. R and S are :(A) Chain isomers (B) Positional isomers(C) Geometrical isomers (D) Functional isomers

2. R can be :

(A) (B) (C) (D)

3. The correct statement about 'T' is :(A) It has all 2º carbon atoms(B) It has six 1º and four 2º and four 3º hydrogen atoms(C) It has three types of chemically different (nonidentical) hydrogen atoms(D) Its all 2º hydrogen atoms are identical in nature

Comprehension # 2

'X'(C4H6) on catalytic hydrogenation by H2/Ni produces 'Y'(C4H8). 'Y' on dichlorination by CI2/h forms'Z'(C4H6CI2), mixture of total six isomers (all dichloro isomers). Reductive ozonolysis of X produces asingle compound W (C4H6O2)·

4. The compound X can be :

(A) (B) (C) (D)

5. The number of fractions obtained by fractional distillation of mixture Z is :(A) 3 (B) 4 (C) 5 (D) 6

6. The number of dioxime isomers formed by 'W' en reaction with NH2OH is :(A) 1 (B) 2 (C) 3 (D) 4





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Comprehension # 3

The compound 'X' is an optically inactive nonresolvable cyclic compound, with two chiral centres. Onheating it undergoes decarboxylation to give two isomeric products Y and Z. Where X, Y and Z all liberateCO2 with NaHCO3·

7. The compound X can be :

(A) (B) (C) (D)

8. Compound ‘Y’ and ‘Z’ are :(A) Position isomers (B) Geometrical isomers (chiral)(C) Functional isomers (D) Geometrical isomers (Achiral, meso)

9. If X and Y react with excess of NaHCO3 then number of moles of CO2 gas liberated are respectively :(A) 1 , 2 (B) 1 , 1 (C) 2 , 1 (D) 2 , 2

Match the column10. Match the compounds of column-I with the reagent of column-II which can distinguish the compounds of

column-I.Column -I Column -II

(A) (p) Tollen's reagent

(B) (q) I2/NaOH

(C) (r) Lucas reagent

(D) (s) Neutral FeCI3

(t) 2,-DNP





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11. Column-I Column-II

(A) (p) Bromine water solution decolourised

(B) (q) precipitate obtained with AgNO3 + NH4OH

(C) (r) CO2 gas liberated by NaHCO3

(D) (s) Yellow precipitate by 2, 4-DNP

Assertion / ReasoningDIRECTIONS :

Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False.(D) Statement-1 is False, Statement-2 is True.(E) Statement-1 and Statement-2 both are False.

12. Statement-1 :

Statement-2 : Products (a) and (b) can be distinguish by Tollen's reagent.

13. Statement-1 : But-2-yne forms white precipitate with Totten's reagent.Statement-2 : But-2-yne does not react with Tollen's reagent because it has no acidic hydrogen.

14. Statement-1 : 2,4-Dinitrophenol gives NaHCO3 test positive.Statement-2 : Nitrophenols are acidic in water.

TRUE I FALSE15. Br2 water can be used to distinguish ethyne and ethene.

16. Oecolourization of Baeyer's reagent is sure test of alkenes.

17. But-2-yne on oxidative ozonolysis forms formic acid and propanoic acid.

18. Neopentane yields one monochloro product.

FILL IN THE BLANKS

19. Unsaturation of alkenes can be tested with ..............................

20. A white precipitate is obtained by passing prop-1-yne through ammonical ..............................solution.

21. The ozonolysis product of 2,3-Dimethylbut-2-ene is ...............................

22. An olefinic 'X' on ozonolysis gives CH3CHO and formaldehyde. The IUPAC name of the olifin is .......................

23. Chlorination of butane give rise to ..............................monochloro structural products.





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PART - I : MIXED OBJECTIVE

Single Correct Answer Type1. Quinine is the most important alkaloid obtained from cinchona bark. It’s molecular formula is C20H24N2O2.

It may contain(A) 5 double bond & 6 ring (B) 6 double bond & 4 ring(C) 6 double bond & 3 ring (D) 7 double bond & 5 ring

2.

Which of the following is the correct statement about P & Q.(A) Product will be 1-methyl-3-(2-methylpropyl)cyclohexane.(B) Product will be 3-methyl-1-(2-methylpropyl)cyclohexane.(C) DU of reactant P is 3.(D) DU of product Q is zero.

3. An alkane E is produced by hydrogenation of only one alkyne D. Identify D.

(A) HC º C – CH2 – CH2 – CH3 (B)

(C) (D) CH3 – CH2 – C C – CH2 – CH3

4. 3 monochloro structural isomers

for compound C5H10 how many pairs of geometrical isomers are possible :(A) 1 pair (B) 2 pair (C) 3 pair (D) No pair

5. How many alkynes will produce 3-Chloro-3-ethylpentane in the following sequence of reactions :

(A) 1 (B) 2 (C) 3 (D) 4





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6. Three aromatic isomers X, Y, Z have molecular formula C6H4Br2. On mononitration 'X' gives one, 'Y' givestwo and 'Z' gives three isomeric products of molecular formula C6H3Br2NO2. Identify X, Y and Z.

X Y Z

(A)

(B)

(C)

(D)

7. An organic compound X(C5H10) on ozonolysis gives Y & Z. The product mixture Y and Z on reaction withNH2 – OH gives four oximes. The structure of X is :

(A) (B) CH3 – CH = CH – CH2 – CH3

(C) (D)

8. For the following reactions sequence

The structure consistent with X and Y are : X Y X Y

(A) (B)

(C) (D)





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9. The chemical reactions of an unsaturated compound 'M' are given below. Determine the possible structuralformula of 'M'

(A) (B) (C) (D)

10. Compound 'A'(C8H12) does not show stereoisomerism. It adds only one mole of H2. On ozonolysis it givesa symmetrical diketone B(C8H12O2). Identify A.

(A) (B) (C) (D)

11. What are the reductive ozonolysis products of

(A) (B)

(C) (D)

12. Compound 'A' (C3H6O), decolourizes Br2 water. It liberates colourless, odourless gas on addition of sodiummetal. On ozonolysis, it gives B and compound 'C' (C2H4O2). Identify ‘A’ :

(A) (B) CH2 = CH – CH2OH (C) (D) CH3 – CH2 – CHO

13. A compound P(C5H6) gives Baeyers test positive and on hydrogenation from a hydrocarbon B(C6H10) whichgives only one monochloro product. The compound 'P" is :

(A) (B) (C) (D) CH C – CH2 – CH = CH2

14. Compound A (C3H5N) gives precipitate with Tollen's reagent and H2 gas is also evolved on addition of Limetal. Compound A can be :(A) CH3 – CH2 – C N (B) H – C C – NH – CH3

(C) CH3 – CH2 – N (D) CH2 – C = N – CH3





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15. An aromatic compound A (C7H8O) gives following tests with the given reagents.

Identify ‘A’

(A) (B) (C) (D)

16. An organic compound A(C4H8O) gives following test and on monochlorination it gives four structural isomers.

Identify 'A' :

(A) (B) (C) (D)

17.

Compound is :

(A) (B) (C) (D)

18. Compound 'A' (C16H16) on ozonolysis gives only one product 'B', (C8H8O). 'B' gives positive Iodoform testand forms sodium benzoate as one of the product. Identify the structure of ‘A’.

(A) Ph – CH2 – CH = CH – CH2 – Ph (B)

(C) (D)





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19. Compound 'x'(C3H6O) gives negative tests with the following reagents. (a) Br2 (b) 2, 4-Dinitrophenylhydrazine(c) Na metal. It gives two monochloro structural isomers. Identify 'x'.(A) CH2 = CH – O – CH3 (B) CH3 – CH2 – CHO

(C) (D)

20. Observe the following compound and select + ve & – ve test respectively :

(A) + + + – (B) + + + + (C) + – + – (D) + – – +

21. Consider following compounds and decide as to which of the following statement will be true?

CH3O – C C – H

(i) (ii) (iii) (iv)(A) (ii) gives no reaction with Na metal, however, 1 mole of (iv) on reaction with Na metal will liberate

22.4 litres of H2 gas at STP(B) (i) will give brisk effervescence on addition of NaHCO3 but will not bring any change in the colour of Br2

water(C) (iii) liberates H2 gas with Na metal, gives white precipitate with Tollen's reagent but does not respond

towards lucas reagent or 2, 4-DNP test.(D) (iv) gives turbidity with anhydrous ZnCI2

22. The following two compounds I and II can be distinguished by using reagent :

(1) aq. NaHCO3

(2) Neutral FeCI3 (aq.) (FeCI3 + NH4OH + H2O)(3) Blue litmus solution(4) Na metal(5) HCI (ZnCI2 anhydrous)(A) 1 or 3 (B) 2 or 5 (C) 4 or 5 (D) 3 or 4





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23. Identify the structure of X.

(A) (B) (C) (D)

24. Consider the ozonolysis of trans-4,5-Dimethylcyclohexene having the configuration shown below :

(A) (B) (C) (D) None

25. Ozonolysis of Agathene dicarboxylic acid gives

Structure of Agathene dicarboxylic acid should be ?

(A) (B)

(C) (D)





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26. Farnesene is a compound found in the waxy coating of apples. On hydrogenation it gives 2, 6,10-trimethyldodecane. On ozonolysis it gives one mole acetone, one mole of formaldehyde, one mole of2- methylpentanedial and one mole of 4-oxopentanal. The structure proposed for farnesene may be :

(A) (B)

(C) (D)

More than one choice type27. Select the correct statement(s) about the following reaction

(A) The reactant (X) has four positional isomers (including 'X' itself)(B) The hydrogenated product Y has three positional isomers(C) Four monochloro products are formed which are all positional isomers(D) Only one monochloro positional isomer shows geometrical isomerism

28. The correct statement(s) about the products of the following reaction is / are

(A) Three structural isomers are formed(B) All the formed structural isomers are optically active(C) Total number of isomers formed are four(D) Total optically active isomers are two

29. An organic compound with molecular formula C6H8, on reductive ozonolysis gives 2 moles of 2-oxopropanal.The structure of the compound will be :

(A) (B) (C) (D)

30. Which statement(s) is/are correct.(A) DU tells about H-deficiency from any molecule in the form of multiple bond or ring.(B) Hydrogenation tells about carbon skeleton.(C) Monohalogenation tells about type of chemically different H.(D) Ozonolysis tells about position of double or triple bond in molecule.

31. Which will give colour with FeCI3 ?

(A) (B)

(C) (D)





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32. Which will perform iodoform reaction with I2/OH–?(A) CH3COCH2CH3 (B) CH3CONH2 (C) C6H5COCH3 (D) CH3CHO

33. Correct statement about :

(A) It gives coloured solution with neutral FeCI3 solution(B) It liberates one mole H2 gas with Na metal(C) It gives CO2 gas with NaHCO3.(D) It forms sweet smelling compound with alcohols.

34. Correct statement about is

(A) It gives coloured solution with neutral FeCI3 solution(B) It liberates half mole H2 gas with Na metal(C) It gives +ve Iodoform test.(D) It forms sweet smelling compound with alcohols.

35. Correct statement about is :

(A) liberate 32

mole of H2 on treatment with Na.

(B) + test with FeCI3

(C) + test with NaHCO3

(D) + test with tollen's reagent

36. Which of the following will give positive iodoform test.

(A) (B)

(C) (D)

37. Which of the following compounds after complete hydrogenation will form three monochloro structuralisomeric products.

(A) (B)

(C) (D)





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PART - II : SUBJECTIVE QUESTIONS

1. Two optically active straight chain acyclic compounds X and Y (molecular formula C5H9Br) give followingreactions

Identify X, Y, W and S.

2.

Calculate sum of number of products formed in the reaction a, b and c.

3. C5H12O2 gives instant turbidity (white thin precipitate) with anhydrous ZnCI2 / HCI, with sodium metal 1 moleof compound liberates 11.2 litre H2 gas at STP. write the structure of molecule?

4. Compound 'A' (C5H10) on ozonolysis gives two compounds D and E. Both D, E give positive Iodoform test.E responds tc Tollen's test but D does not. Identify the structure of A, D and E.

5. Find the structure of 'X'

6. Compound A (C6H12) exists in two geometrically isomeric form. It decolourises the pink colour of Baeyer'sreagent. On ozonolysis it gives B (C4H6O). B gives a silver mirror with ammonical silver nitrate solution.Write the structural formula of compounds A and B.

7. Calculate the lowest molecular mass of optically active compound (X) which gives the following reactions(a) It Liberates N2 on treating with NaNO2 and HCI(b) It gives positive test with 2,4-DNP(c) It gives positive iodoform test(d) It gives white precipitate with Tollen's reagent.

8. Calculate the DU of following compounds :(i) C6H6 (ii) C5H9N

9. Two compounds X and Y (Molecular formula = C3H6) give following reactions. Identify the structure of Xand Y.

10. Write the products of following reactions :





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11. How many isomeric alkenes gives 2, 3, 4-Trimethylpentane after hydrogenation.

12. P (Total number of monochloro structural products).

13. Q (Total number of monochloro structural products).

14.

Sum of total number of products of P & Q.

15. Write total number of monochloro (Structural) products of following compounds :

(i) (ii) (iii) (iv)

16. Write the product of following reactions :

(i) (ii)

(iii) (iv)

17.

Write the structure of Q.

18.

Write the structure of (P).

19. After ozonolysis of benzene how many products are formed.

20.

Write the structural of (X).

21.

Write the structure of (Y)





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22. An organic compound (X) after hydrogenation gives 2-Methylbutane. After reaction of 'X' with Na metal it

gives 12

mol of H2 gas. 'X' also gives +ve test with Br2 water. Identify the structure of 'X'.

23. Write the lowest molecular weight optically active compound which gives turbidity with lucas reagent after30 min.

24.

Identify the structure of X :

25.

Identify the structure of Y :

26.

Identify the structure of Y :

27.

Identify the structure of P.

28.

Identify the structure of A.

29.

Identify the structure of P.

30.

Identify the structure of Q.





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PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions are having more than one correct option.

1. Five isomeric para-disubstituted aromatic compounds A to E with molecular formula C8H8O2 were given foridentification. Based on the following observations, give structures of the compounds. [JEE 2002, 5/60](i) Both A and B form a silver mirror with Tollen's reagent; also B gives a positive test with FeCI3 solution.(ii) C gives positive iodoform test.(iii) D is readily extracted in aqueous NaHCO3 solution.(iv) E on acid hydrolysis gives 1, 4-dihydroxybenzene.

2. The number of chiral compounds produced upon monochlorination of 2-methylbutane is : [JEE 2004, 3/84](A) 2 (B) 4 (C) 6 (D) 8

3. In conversion of 2-butanone to propanoic acid which reagent is used. [JEE 2005, 3/84](A) NaOH, Nal / H (B) Fehling solution (C) NaOH, I2 / H

(D) Tollen's reagent

4. [JEE 2006, 2/184]

The number of possible isomers of [N] and number of fractions of [P] are(A) (6, 6) (B) (6, 4) (C) (4, 4) (D) (3, 3)

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)

1. On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only onemonochloroalkane this alkane could be : [AIEEE 2003](1) propane (2) pentane (3) isopentane (4) neopentane.

2. Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is:[AIEEE 2005](1) n-Hexane (2) 2,3-Dimethylbutane (3) 2,2-Dimethylbutane (4) 2-Methylpentane

3. Among the following the one that gives positive iodoform test upon reaction with I2 and NaOH is :(1) CH3CH2CH(OH)CH2CH3 (2) C6H5CH2CH2OH

(3) (4) PhCHOHCH3

4. In the following sequence of reactions, the alkene affords the compound 'B' [AIEEE 2008, 3/105]

(1) CH3CH3CHO (2) CH3COCH3 (3) CH3CH2COCH3 (4) CH3CHO





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NCERT QUESTIONS

1. Observe the following flow chart and answer the number of X, Y, Z.

2. An organic compound (A) adds two moles of Br2 . (A) gives positive test with 2,4-DNP, while on treating withNaOI solution gives product which on oxidative ozonolysis gives lowest molecular mass optically activedibasic acid along with a monobasic acid of molecular mass 74. what is the molecular mass of (A).

3. Observe the following reaction

(y) Total number of fraction obtained (x) total number of monochloro productss

If total number of optically active isomers = z in the end product and total number of diastereomeric pairs

= w in the end product. Then write your answer in the form of .





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4. In the following reaction

The number of stereoisomers of B and number of optically active stereoisomers of B formed respectivelyare :If X = number of stereoisomers of B

Y = number of optically active stereoisomers of BZ = total isomers of C produced

Then given your answer in the form of

5. A compound “M” (C10H20) gives a product “P” on hydrogenation which has only 2 monochloro structuralisomeric products when treated with Cl2/h. If “M” is ozonolysed in presence of zinc,. “W” is the onlyproduct obtained.Now if the following statements are true write “1” (one), if false use “0” (zero) to fill code key.I. “M” exists as 2 diastereomers.II. One out of 2 monochloro structural of “P” is chiral.III. “W” when treated with hydroxylamine gives two products.IV. “W” gives Tollen’s test positive.





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Exercise # 1PART - I

A-1. (D) A-2. (D) A-3. (B) A-4. (B) A-5. (D) B-1. (B) B-2. (A)B-3. (A) B-4. (D) B-5. (B) B-6. (A) C-1.* (ABC) C-2. (C) C-3. (C)C-4. (B) C-5. (C) C-6.* (AC) C-7. (B) D-1. (C) D-2. (C) D-3. (A)D-4. (D) E-1. (B) E-2. (A) F-1. (C) F-2. (C) F-3. (C) F-4. (A)F-5. (D) F-6. (C) G-1. (A) G-2. (A) G-3. (C) G-4. (D) G-5. (D)G-6. (A)

PART - II1. (C) 2. (C) 3. (C) 4. (D) 5. (C) 6. (C) 7. (C)8. (D) 9. (C) 10. (A) p, q, t (B) s (C) r, t (D) q, r, s 11. (A) p, s (B) p, q (C) q, r, s (D) r12. (C) 13. (D) 14. (B) 15. F 16. F 17. F 18. T19. Baeyer’s reagents 20. silver nitrate 21. acetone22. propene 23. two


1. (B) 2. (A) 3. (B) 4. (A) 5. (A) 6. (B) 7. (B)8. (D) 9. (C) 10. (C) 11. (C) 12. (B) 13. (C) 14. (B)15. (A) 16. (D) 17. (B) 18. (C) 19. (D) 20. (A) 21. (C)22. (B) 23. (D) 24. (B) 25. (C) 26. (C) 27. (ACD)28. (ACD) 29. (ACD) 30. (ABCD) 31. (ABC) 32. (ACD) 33. (ABCD)34. (ABC) 35. (ABCD) 36. (BCD) 37. (CD)

PART - II

1. 2. 6

3. 4.

5.

6. 7. 97

8. (i) 4 (ii) 2

9. Y = CH3 – CH = CH2





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10. 11. 3 12. 4 13. 2

14. 7 15. (i) 3 (ii) 4 (iii) 5 (iv) 3

16.

17. CH3 – CH = CH – CH3 18. 19. 1 20. CH3–CH=CH2

21. 22. 23.

24. 25. 26.

27. 28. 29.

30. CH3 – CH2 – COOH


1.

2. (B) 3. (C) 4. (B)PART - II

1. (4) 2. (2) 3. (4) 4. (4)

Exercise 41. 2. Molecular mass = 138 3. 4.5. 1 1 1 1

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