2.ALKANES (89 - 105)

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ALKANES

ALKANES Alkanes: Physical Properties The alkanes can exist as gases, liquids, or solids at room temperature. The unbranched alkanes methane,ethane, propane, and butane are gases; pentane through hexadecane are liquids; the homologues larger than hexadecane are solids. Branched alkanes normally exhibit lower boiling points than unbranched alkanes of the same carbon content. This occurs because of the greater van der Waals forces that exist between molecules of the unbranched alkanes. These forces can be dipole dipole, dipole induced dipole, or induced dipole induced dipole in nature. The unbranched alkanes have greater van der Waals forces of attraction because of their greater surface areas. Solid alkanes are normally soft, with low melting points. These characteristics are due to strong repulsive forces generated between electrons on neighboring atoms, which are in close proximity in crystalline solids. The strong repulsive forces counterbalance the weak van der Waals forces of attraction. Finally, alkanes are almost completely insoluble in water. For alkanes to dissolve in water, the van der Waals forces of attraction between alkane molecules and water molecules would have to be greater than the dipole dipole forces that exist between water molecules. This is not the case.  Alkanes: Enthalpy of Combustion Alkanes can be oxidized to carbon dioxide and water via a free radical mechanism. The energy released when an alkane is completely oxidized is called the heat of combustion. For example, when propane is oxidized, the heat of combustion is 688 kilocalories per mole. In a homologous series like the straight chain alkanes, the energy liberated during oxidation increases by approximately 157 kilocalories for each additional methylene (CH 2) unit. Heat of combustion data is often used to assess the relative stability of isomeric hydrocarbons.Because the heat of combustion of a compound is the same as the enthalpy of that compound in its standard state, and because potential energy is comparable to enthalpy, the differences in heats of combustion between two alkanes translate directly to differences in their potential energies. The lower the potential energy of a compound, the more stable it is. In the alkanes, the more highly branched isomers are usually more stable than those that are less branched. 



Halogenation Selectivity When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. Four constitutionally isomeric dichlorinated products are possible, and five constitutional isomers exist for the trichlorinated propanes. Can you write structural formulas for the four dichlorinated isomers?

The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.

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







ALKANES

CH3-CH2-CH3 + Cl2 ——>  45% CH3-CH2-CH2Cl + 55% CH3-CHCl-CH3 The results of bromination ( light-induced at 25 ºC ) are even more suprising, with 2-bromopropane accounting for 97% of the mono-bromo product. CH3-CH2-CH3 + Br2  ——>  3% CH3-CH2-CH2Br + 97% CH3-CHBr-CH3 These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3:1. Further experiments showed that 3º-hydrogens are even more reactive toward halogen atoms. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro 2methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule. (CH3)3CH + Cl2 ——>  65% (CH3)3CCl + 35% (CH3)2CHCH2Cl It should be clear from a review of the two steps that make up the free radical chain reaction for halogenation that the first step (hydrogen abstraction) is the product determining step. Once a carbon radical is formed, subsequent bonding to a halogen atom (in the second step) can only occur at the radical site. Consequently, an understanding of the preference for substitution at 2º and 3º-carbon atoms must come from an analysis of this first step. First Step: R3CH + X· ——>  R3C· + H-X Second Step: R3C· + X2 ——>  R3CX + X· R (in R–H) methyl ethyl i-propyl t-butyl B.D.E. 103 98 95 93 phenyl benzyl allyl 110 85 88 B.D.E. = Bond Dissociation Energy(kcal/mole) The difference in C-H bond dissociation energy reported for primary (1º), secondary (2º) and tertiary (3º) sites agrees with the halogenation observations reported above, in that we would expect weaker bonds to be broken more easily than are strong bonds. By this reasoning we would expect benzylic and allylic sites to be exceptionally reactive in free radical halogenation, as experiments have shown. The methyl group of toluene, C6H5CH3, is readily chlorinated or brominated in the presence of free radical initiators (usually peroxides), and ethylbenzene is similarly chlorinated at the benzylic location exclusively. The hydrogens bonded to the aromatic ring (referred to as phenyl hydrogens above) have relatively high bond dissociation energies and are not substituted. h C6 H 5 CH 2 CH 3 + Cl2   C6 H 5 CHClCH 3 + HCl

Since carbon-carbon double bonds add chlorine and bromine rapidly in liquid phase solutions, free radical substitution reactions of alkenes by these halogens must be carried out in the gas phase, or by other halogenating reagents. One such reagent is N-bromosuccinimide (NBS), shown in the second equation below. By using NBS as a brominating agent, allylic brominations are readily achieved in the liquid phase.

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







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The covalent bond homolyses that define the bond dissociation energies listed above may are described by the general equation: R3C-H + energy ——>  R3C· + H· Since the hydrogen atom is common to all the cases cited here, we can attribute the differences in bond dissociation energies to differences in the stability of the alkyl radicals (R3C·) as the carbon substitution changes. This leads us to the conclusion that: alkyl radical stability increases in the order: phenyl < primary (1º) < secondary (2º) < tertiary (3º) < allyl H < benzyl. Because alkyl radicals are important intermediates in many reactions, this stability relationship will prove to be very useful in future discussions. The enhanced stability of allyl and benzyl radicals may be attributed to resonance stabilization. Formulas for the allyl and benzyl radicals are shown below. Draw structural formulas for the chief canonical forms contributing to the resonance hybrid in each case.

The poor stability of phenyl radicals, C6H5·, may in turn be attributed to the different hybridization state of the carbon bearing the unpaired electron (sp2 vs. sp3). Since the H-X product is common to all possible reactions, differences in reactivity can only be attributed to differences in C-H bond dissociation energies. In the case of carbon-hydrogen bonds, there are significant differences, and the specific dissociation energies (energy required to break a bond homolytically) for various kinds of C-H bonds have been measured. These values are given in the following table. alkyl radical stability increases in the order: phenyl < primary (1º) < secondary (2º) < tertiary (3º) < allyl H” benzyl. Because alkyl radicals are important intermediates in many reactions, this stability relationship will prove to be very useful in future discussions. The enhanced stability of allyl and benzyl radicals may be attributed to resonance stabilization.Formulas for the allyl and benzyl radicals are shown below. Draw structural formulas for the chief canonical forms contributing to the resonance hybrid in each case.

The poor stability of phenyl radicals, C6H5·, may in turn be attributed to the different hybridization state of the carbon bearing the unpaired electron (sp2 vs. sp3). Narayana Junior Colleges

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LEVEL-IV STRAIGHT OBJECTIVE QUESTIONS 1.

Observe the following reaction sequence

X Br2 /hv Y Mg/Ether Z

W dil. H2SO4  U O3 / Zn /H2O 

X can be CH 3

a) 2.

b)

c)

d)

Which of the following is correct potential energy diagram for the given chain propagating step. 



CH 3  H  F

 

B . E . 435 kJ / mol

CH 3  H  F

H   32 kcal / mol

B . E . 569 kJ / mol

δ *

δ *

F ........H......C H3 *

CH 4  F

*

C H 3  HF *

C H 3  HF

a)

H  East  32 kcal mol1

δ *

H  East  32 kcal mol 1

b) *

CH 4  F

δ *

δ *

δ *

F .....H..... C H 3

C H 3 ....H..... F *

CH 4  F

Eact  5.0 kJ mol 1

*

C H 3  HF

c)

d) *

CH 4  F

92

Eact  32 kcal mol1

*

C H 3  HF

ΔH  32 kcal mol 1

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3.

a)

b)

c)

d)

The deuterium kinetic isotope effect for chlorination of an alkane is defined as follows : 4.

Which of the following is correct comparison for the above effect ? Deuterium Kineticisotopic effect=

5.

Rate of homolytic clevage of C-H bond by Cl Rate of homolytic clevage of C-D bond by Cl

a) IBr  IC1 b) IBr  IC1 c) IBr  IC1 d) cannot be predicted In the given conformation if C2 is rotated about C2 C3 bond anti clockwise by an angle of 1200 then the conformer obtained is

a) Fully eclipsed

b) Partly elipsed

c) Gauche conformer

d) Anti-confirmer

MULTIPLE CORRECT ANSWER TYPE QUESTIONS

6.

 82 82  Radioactive  Br  Br  adds to 1- bromocyclohexene. The product is   a) is 1,1,2- tribromocyclohexane b) has radioactive bromine at vicinal positions c) has radioactive bromine trans to each other d) has radioactive bromine cis to each other

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7.

ALKANES

CH 4 is not obtained by a) Wurtz reaction c) Reduction of alkylhalide

b) Corey-house synthesis d) kolbe’s electrolysis

LINKED COMPREHENSION TYPE QUESTIONS Passage-1: In the study of chlorination of propane, four products (A,B,C&D) of the formula C3 H 6Cl2 were isolated. Each was further chlorinated to provide trichloro products  C3 H 5Cl3  .

8.

It was found that A provide one trichloro product, B gave two and C&D each gave three. It is found that D is optically active. Formula of the compound A is

a) 9.

c)

d)

Correct formula of the product of chlorination of B is —-

a) 10.

b)

b) ClCH 2CHClCH 2Cl c) Both A and B

d)

Correct formula of the compound D is a)

b)

c)

d)

Passage-2: Wurtz reaction involves the condensation of two molecules of alkyl halides in the presence of sodium and dry ether R - X + 2Na + R - X dry ether R - R + 2NaX

In this reaction small amount of alkene is also formed as by-product.

11.

12.

Tertiary alkyl halides do not give Wurtz reaction. Frankland reaction is similar but has certain advantages over Wurtz reaction. It is useful in the synthesis of symmetrical alkanes. Frankland reaction is shown by primary, secondary as well as tertiary alkyl halide. Among all isomers of hexane which of the following isomer cannot be obtained in a good yield by Wurtz reaction (from symmetrical alkyl halides)? 1) n-hexane 2) iso hexane 3) 2,3,-dimethyl butane 4) Neo hexane a) 1,4 b) 2,4 c) 1,3 d) 3,4 Which among the following compounds will give Wurtz reaction with good yield? a) CH 2  CH  Br b) ph  Br c) CH 2  CH  CH 2  Br

94

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13.

A mixture of ethyl iodide and methyl iodide is subjected to Wurtz reaction. The product that will not be formed is a) Ethane b) Butane c) Propane d) Pentane MATRIX MATCH TYPE QUESTIONS

14.

Column I

Column II

Li CH 3CH 2Cl  ? A) CH 3CH 2Cl  Cul

p) (CH3)2CHCH2CH(CH3)2

CH 3 2 CHCH 2 Br Li B)  CH 3  2 CHI   ? Cul

q) CH 3CH 2CH 2CH 3

Li CH3CH2Br  ? C) CH3CH2C CH3  Br  Cul

r)  CH 3 3 CCH 2CH 3

Li

D) 15.

(C H 3)3C C l

CH3CH2Br

CuI

s) CH 3CH 2C  CH 3  2 CH 2CH 3

Match the Column that are used to get cyclohexane. Column I

Column II

A)

(p) Electrolysis followed by H 2 / Ni

COONa B)

(q) NaOH  CaO  / 

C) Br2CH 2   CH 2  4  CH 2  Br

(r) H 2 / Ni / 

COONa D)

(s) Zn / 

COONa

ASSERTION-REASONING TYPE QUESTION 16.

Statement - 1: Alkanes are monochlorinated with (CH 3 )3 COCl. Statement - 2: The initiating step is the heterolytic cleavage of hypochlorite 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.

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INTEGER ANSWER TYPE QUESTIONS 17.

The number of types of monochloroalkanes formed by chlorination of isobutane is

18.

The number of moles of CH 4 formed by reacting 96 grams of CH 3OH with excess CH 3 MgI is

19 20.

The number of structural isomers of alkane containing six carbon atom is The number of different types of substitution products possible when bromine and ethane react in presence of light is The maximum number of carbon atoms in the expected products of the following reaction

21.

CH 3CH 2 Br + CH 3 - CH - CH 2 Br CH 3 22

Na / Ether   Product

The number of possible enantiomeric pairs that can be produced during monochlorination of 2-methylbutane is CH 3 Cl2 , hv   isomeric products

23

H 3C CH 3

24.

 C5 H11Cl 

(including sterioisomers)

is mono chlorinated through free radical substitution reaction. What is the total number of possible optical isomers that can be formed. SUBJECTIVE TYPE QUESTIONS

25.

26.

27.

96

Chlorination of optically active 2-chlorobutane yields a mixture of isomers with the formula C4H8Cl2 a) How many different isomers would you expect to be produced? What are their structures? b) Which of these fractions would be optically active? Two isomeric organic compounds (A) and (B) of molecular formula C4H8Br2 on hydrolysis gave two compounds (C) and (D) of formula C4H8O. (C) and (D) gave two isomeric acids (E) and (F) of formula C4H8O2 on oxidation. Both the acids on decarboxylation with soda-lime gave propane. Identify (A) to (F). n-Butane is produced by the monobromination of ethane followed by Wurtz reaction. Calculate the volume of ethane at STP required to produce 55 g n-butane, if the bromination takes place with 90% yield and the Wurtz reaction with 85% yield. Narayana Junior Colleges

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KEY LEVEL -IV STRAIGHT OBJECTIVE QUESTIONS 1) A

2) D

3) C

4) A

5) C

MULTI CORRECT ANSWER 6) A,B,C

7) A,B,D LINKED COMPREHENSION TYPE QUESTIONS

8) B

9)C

10) D

11) B

12) C

13) D

MATRIX MATCHING TYPE QUESTIONS 14) (A – q), (B – p), (C – s), (D – r)

15) (A-r) , (B- q) ,(C- s), (D- q)

ASSERTION-REASONING TYPE QUESTIONS 16) C 22) 2

17) 2 23) 6

18) 3 24) 8

19) 5

20) 9

21) 8

HINTS LEVEL-IV STRAIGHT OBJECTIVE QUESTIONS 1.

3

Aromatisation

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Because a bromine radical is less reactive than a chlorine radical , a bromine radical has a greater preference for the more easily broken C-H bond. Bromination , therefore would have a greater deuterium kinetic isotope effect than chlorination Conceptual

5

MULTIPLE CORRECT ANSWER TYPE QUESTIONS 82

6

82

82

82

Br  Br  Br  Br Br

Br

82

Br 

82

Br

82

Br

Trans addition

Br 82 Br

Br H

Conceptual

7

LINKED COMPREHENSION TYPE QUESTIONS

8

(B) (C) (D)

9.

(B) (C) (D)

10.

(B) (C) 98

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(D) 11. 12. 13.

Conceptual Conceptual Conceptual MATRIX MATCHING TYPE QUESTION

15. Conceptual INTEGER ANSWER TYPE QUESTIONS CH 3

CH 3

CH 3 CH 3

CH

17.

CH

Cl 2   h

CH 3

c

Cl + CH 3

CH 2

Cl

CH 3

CH 3

CH 3OH  CH 3 MgI  CH 4  Mg  OH  I

18.

19. 20.

32 g 96 g

1 mole 3 mole CH3

H3C

H3C

,

,

CH3

CH3

CH3, H C 3

H3C

CH3

,

H 3C

CH3

CH3

H 3C

Monosubstituted =1, disubstituted=2, trisubstituted =2 tetrasubstotuted = 2, pentasubstituted=1, hexasubstited=1 Total =9 CH 3

21.

CH3

H3C

H3C

CH 3

2

CH

Br  Na / Ethers 

CH 2

CH 3

c

22.

CH

CH 2

CH 2

CH CH 3

CH 3

CH 3

CH 3

H

CH 3

CH 2

Cl

Cl

CH 2

c C2 H 5

C2 H 5

H

CH3 CH(CH 3 )CH(Cl)CH 3 and its mirror image

CH 3 23.

CH 3

c

Cl

C2 H 5

CH 3

CH 2

Cl

CH

CH 2

CH 3

(d & l)

CH 3 CH3 CH(CH 3 )CH(Cl)CH 3 and its mirror image CH 2 Cl CH CH 2 Total isomeric products = 6 24). 2 - Methyl and 3 - Methyl chloro-cyclohexanes can have optical isomerism CH 3

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SUBJECTIVE TYPE QUESTIONS CH3

CH3

25.

a)H3C

CCl2

Cl  Cl2  

CH2

CH2CH3



CH2Cl

CH3

CHCl

CHCl

CH2

CH3



CHCl

CH3

CH3 CHCl



CH3

CH2 CH2Cl

4 possible isomers

b)The compounds which are having chiral centres (asymmetric carbon) will be optically active Cl

H

CH3

CH3

CH2 Cl ;

C2 H 5

H

Cl

Cl

H

(I)

H

;

Cl CH2

CH3

CH2Cl

(II)

(III)

Three fractions are optically active H3C

Br

26.

A

H3C

H3C

H3C H3C

Br

C

Br

B

O

O H3C

D

OH

H3C

O

E

H3C F

HO

27.

Br

O

H3C

55.57 Lts

LEVEL-V STRAIGHT OBJECTIVE QUESTIONS 1. CH2 Cl Na (Dry ether) CH3 CH2 Cl Product Obtained in above Wurtz reaction is:

CH3 a)

b) CH3

CH3

c)

d) Both (a) and (b) CH3

100

CH3

CH3 Narayana Junior Colleges

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2.

3.

An alkane C7 H16 is produced by the reaction of lithium di(3-pentyl) cuprate with ethyl bromide. The structural formula of the product is a) 3-ethylpentane b) 2-ethylpentane c) 3-methylhexane d) 2-methylhexane What is the IUPAC name for the following compound? H

CH3 CH3

4. 

5. 

H

a) dimethylcyclohexane b) 1,3-dimethylcyclohexane c) cis-1,3-dimethylcyclohexane d) trans-1,3-dimethylcyclohexane The most stable conformational isomer of cis-1-bromo-2-chlorocyclohexane will have... a) both halide atoms in axial positions. b) both halide atoms in equatorial positions. c)the bromine atom in an axial position and the chlorine atom in an equatorial position. d) the bromine atom in an equatorial position and the chlorine atom in an axial position. The most stable conformational isomer of trans-1-ethyl-2-methylcyclohexane will be... Et Et H CH3

a)

H

b)

H

6. 

7. 

H

H

H3C

Et CH3

H

c) H

d)

H

Et

CH3

Which Newman projection shows the most stable conformation of the following compound? Br H

H

Cl

Br

H

Which of the these compounds represents the major monochlorination isomer formed in the following reaction?

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8.  9. 10.

ALKANES

How many dichlorinated isomers can be formed by the halogenation of CH3CH2CH2CH3 with Cl2 in the presence of light? A) 2 B) 3 C) 5 D) 6 Which of the following cycloalkanes has the smallest heat of combustion per CH2 group? A)cyclopropane B)cyclobutane C)cyclopentane D)cyclohexane Which has the greatest molar heat of combustion? A) trans-1,2-dimethylcyclopentane B) cis-1,2-dimethylcyclopentane C) trans-1,3-dimethylcyclopentane D) methylcyclohexane MORE THAN ONE ANSWER QUESTIONS

11.

Which of the following alkanes cannot be synthesized by the Wurtz reaction in good yield?

a)

b)

c)

d)

LINKED COMPRENHESION TYPE QUESTIONS Passage-1: Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon. h R  H  X 2   R  X  HX

Chlorine free radical make 10 , 20 ,30 radicals with almost equal ease, where as bromine free radicals have a clear preference for the formation of tertiary free radicals. So, Bromine is less reactive, and more selective whereas chlorine is less selective and more reactive.

30  20  10 The relative rate of abstraction of hydrogen by Br  is (1600) (82) (1)

12.

30  20  10 The relative rate of abstraction of hydrogen by Cl is (5) (3.8) (1) Consider the above argument and answer the following: 1-halo-2,3-dimethyl butane will be obtained in better yields, if halogen is: a) Br2 b) Cl2 c) I2 d) Can’t be predicted CH3 X

CH3

13.

X 2 / h

Above product will obtained in better yield if X is a) Cl2 b) I2 c) Br2 102

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14.

CH 3 | Cl 2 / h CH 3  CH  CH 3   Product Major product in the above reaction is CH 3 | a) CH 3  CH  CH 2  Cl

CH 3 | CH 3  CH  CH3 b) | Cl

c) CH 3  CH 2  CH 2  Cl

CH 3  CH  CH 2  CH3 | d) Cl

ASSERTION REASONING TYPE QUESTIONS: Directions: In each of the following questions, a statemenet of assertion (A) is given and a corresponding statement of reason (R) is given just below it. Of the two statments, mark the correct answer as. a) If both A and R are true and R is the correct explanation of A b) If both A and R are true and R is not the correct explanation of A c) If A is true but R is false d) If A is false and R is true H H3C

15.

Assertion (A): When

H

Br D H

is subjected to dehydrohalogenation using NaOEt/EtOH,

the product contains no deuterium. Reason (R): E2 follows anti elimination. INTEGER TYPE QUESTIONS 16.

Identify number of chiral centers present in product obtained by following reaction: Br2 CCl4 CH3

H

ADDITIONAL QUESTIONS 1. 2.

The number of carbons in simplest optically active alkane. The number of possible enantiomers of simplest possible optically active alkane of formula Cx H y .

3.

The number of Primary carbons in the branched isomer of first simplest alkane that react with bayer’s reagent to give alcohol.

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4.

ALKANES

The number of dichloro derivatives of propane is equal to number of mono chloro derivatives of C x H y ( straight chain alkane). The number of possible isomers for this compounds including

5.

6.

stereo isomers. Sequence that represents progressive difference between number of hybrid to pure orbitals involved in homologues series of alkanes. a) 1,3,5,7,.... b) 2,4,6,8,.... c) 1,2,3,4... d) 1,4,9,16,...... A( C x H y , alkane where x = 6) undergoes aromatisation in presence of a catalyst to form compound B which on reduction with H 2 / Ni form compound C. Difference in number of hydrogens of C and A. a) 2 b) 4 c) 6 d) 8

7.

Na

 X(major) t-Butylchloride  DryEther

Number of SP 2 hybridised carbons present in the product molecule. Column - II

Column - I h a) CH  Cl   Products 4 2 excess

8.

p) CH3Cl

(Possible)

b) Will not react with amm. AgNO3

q) CH2Cl2

c) Have dipole moment O Cl2 d) Products NaOH

r) CHCl3 s) CCl4

KEY LEVEL-V 1. D 9. D

2.A 10. B

STRAIGHT OBJECTIVE QUESTIONS 3.D 4. D 5.C 6. A 7. B

8.D

MORE THAN ONE CORRECT ANSWER 11. A,C,D 12. B

LINKED COMPREHENSION QUESTIONS 13. C 14.A ASSERTION AND REASON

15.A INTEGER TYPE 16.3 ADDITIONAL QUESTIONS 1. 7 2. 4 3. 3 4. 11 5. B 6. A 8. (A – p, q, r ,s), (B –r, s), (C – p,q,r), (D –r)

104

7. 2

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HINTS LEVEL -V 1.

Wurtz reaction CH3CH2

2.

CH3CH2

CH

CuLi+BrCH2CH3 2

CH 2 CH 3 | CH 3CH 2  CHCH 2 CH 3  (CH3CH 2 )2 CHCu  LiBr 8.

The products are (1,1), (1,2), (1,3), (1,4), (2,2), (2,3) dichlorobutanes. LINKED COMPREHENSION QUESTIONS C C

12.

, Cl2 is less selective x-C-C-C-C  10 halide will be obtained in better yield

13.

To obtain 30 halide Br2 must be used.

14.

CH 3 | CH 3  CH  CH 2  Cl Relative formation is  9 1 % yield  9 /14 100 ASSERTION AND REASON Br H 3C

15.

D

EtO EtOH

CH3

(Hoffmann product due to anti-coplanarity between H and Br) INTEGER TYPE 16. Br Br2 CCl4 H3C H

Narayana Junior Colleges

Br H3C

H

105