23 Alkyl Halides

23 Alkyl Halides 1. Heating of 2-chloro-1-phenyl butane with EtOK/EtOH gives X as the major product. Reaction of X with Hg(OAc)2 / H 2O followed by NaBH 4 gives Y as the major product. Y is represents the SN 1reaction? (a) PE OH (a) Ph 5. Which of the following potential energy (PE) diagrams (b) Ph (b) PE Progress of reaction Progress of reaction OH OH (c) Ph (c) (d) Ph 2. Which one of the following is likely to give a precipitate with (b) CCl 4 (d) (CH3 )3 CCl 3. The major product Y in the following reaction is EtONa Heat X HBr (b) (c) (d) HO NaOEt D (a) CH3 CH2C == CH2 ½ CO2CH2CH3 (b) CO2CH2CH3 ½ CH3 C == CHCH3 CH2CH3 4. The major product of the following reaction is CH3 ½ CH OH CH3 ¾ C ¾ CHCH3 ¾ ¾3 ¾ ¾® ½ ½ H Br CH3 ½ (c) CH3 ¾ C ¾ CH 2 CH3 ½ OCH3 Cl COOCH2CH3 Br CH3 ½ (a) CH3 ¾ C == CHCH3 C Y Br (a) Br Progress of reaction 6. The major product of the following reaction is CH2CH3 H3C Cl PE Progress of reaction AgNO3 solution? (a) CH2 == CH ¾ Cl (c) CHCl 3 (d) PE CH3 ½ (b) CH3 ¾ C ¾ CH == CH 2 ½ H CH3 ½ (d) CH3 ¾ C ¾ CHCH3 ½ ½ H OCH3 (c) H3C OCH2CH3 C COOCH2CH3 OCH2CH3 (d) H3C H2C C CO2CH2CH3 CH3 7. The major product in the following conversion is CH3O CH CH CH3 HBr (excess) Heat ? Alkyl Halides CH (a) CH3O CH2 12. The major product of the following reaction is CH3 CH3O Br (b) HO CH2 CH CH2Cl (i) AlCl3 (anhyd.) (ii) H2O CH3 Br (c) CH3O CH2 CH CH3O CH3 (a) (b) Br (d) HO CH CH2 CH3O CH3 OCH3 CH3 CH3O CH3 Br (c) (d) 8. The major product of the following reaction is 13. The major product of the following reaction is (i) KOH alc. ¾® CH3 CH2 CH ¾ CH2 ¾ ¾ ¾ ¾ ¾ (ii) NaNH2 ½ ½ in liq. NH3 Br Br Br (a) CH3CH2 C H ¾ CH2 ½ ½ NH2 NH2 (b) CH3CH == CHCH2NH2 (c) CH3CH == C == CH2 (d) CH3CH2C ºº CH NaOMe MeOH OMe (a) 9. The major product of the following reaction is (b) CH3O OMe (i) Cl2/CCl4 (c) (ii) AlCl3 (anhyd.) Cl (a) 14. The increasing order of reactivity of the following halides for CH3O (b) the SN 1 reaction is CH3O Cl Cl CH3O (d) CH3O (c) Cl 10. Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light? CH3 ¾ CH 2 ¾ CH ==CH 2 d (d) g b (E) (a) b-hydrogen (c) g-hydrogen a (b) d-hydrogen (d) a-hydrogen I. CH3CH(Cl)CH2CH3 II. CH3CH2CH2Cl III. p-H3CO ¾ C6H4 ¾ CH2Cl (a) (III) < (II) < (I) (b) (II) < (I) < (III) (c) (I) < (III) < (II) (d) (II) < (III) < (I) 15. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourise the colour of bromine ? (b) Br Br 11. The major product of the following reaction is Br C6H5 O (a) O KOH alc. (excess) D (d) (c) Ph Br Br Br 16. 3-methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is (a) Ph (b) Ph (c) Ph (d) Ph (a) six (c) two (b) zero (d) four Alkyl Halides 17. The major product obtained in the following reaction is 23. The major product of the following reaction is H C6H5 C6H5 Br H 3C Br tBuOK D F PhSNa dimethyl formamide (a) (± ) C6 H5CH(O Bu)CH2C6 H5 t (b) C6 H5CH == CHC6 H5 (c) (+ ) C6H5CH(Ot Bu)CH2C6 H5 NO2 (d) (- )C6 H5CH(Ot Bu)CH2C6 H5 SPh H3 C 18. 2-chloro-2-methylpentane on methoxide in methanol yields CH3 ½ I. C2 H5 CH2 C ¾ OCH3 ½ CH3 reaction with sodium F F (a) II. C2 H5 CH2 C == CH2 ½ CH3 (b) NO2 NO2 Br H3 C III. C2 H5 CH == C ¾ CH3 ½ CH3 SPh H3 C SPh H3 C SPh SPh (c) (a) Both I and III (c) Both I and II (b) Only III (d) All of these NO2 NO2 24. The following compound on hydrolysis in aqueous acetone 19. The product of the reaction given below is (i) NBS/hn (ii) H2O/K2CO3 (d) will give CH3 CH3 H X —NO2 MeO— OH (b) (a) O CO2H (c) H Cl CH3 CH3 H CH3 (d) —NO2 (K) MeO— 20. The synthesis of alkyl fluorides is best accomplished by (a) free radical fluorination (c) Finkelstein reaction H OH CH3 CH3 H (b) Sandmeyer’s reaction (d) Swarts reaction CH3 —NO2 (L) MeO— 21. In SN 2 reactions, the correct order of reactivity for the following compounds CH3 Cl,CH3 CH2 Cl, (CH3 )2 CHCl and (CH3 )3 CCl is (a) CH3Cl > (CH3 )2 CHCl > CH3CH2Cl > (CH3 )3 CCl (c) CH3CH2Cl > CH3Cl > (CH3 )2 CHCl > (CH3 )3 CCl 22. KI in acetone, undergoes S N 2 reaction with each P, Q, R and S. The rates of the reaction vary as Cl Q (a) P > Q > R > S (c) P > R > Q > S CH3 —NO2 It mainly gives (a) K and L (c) L and M (b) only K (d) only M OH Cl R + C2H5I S (b) S > P > R > Q (d) R > P > S > Q CH3 OH 25. The product of following reaction is O P CH3 H H (d) (CH3 )2 CHCl > CH3CH2Cl > CH3Cl > (CH3 )3CCl —Cl CH3 (M) MeO— (b) CH3Cl > CH3CH2 Cl > (CH3 )2 CHCl > (CH3 )3 CCl H3C—Cl OH H (a) C6H5OC2H5 (c) C6H5OC6H5 C2H5O– anhy. C2H5OH (b) C2H5OC2H5 (d) C6H5I Alkyl Halides 26. Identify the set of reagents/reaction conditions X and Y in the following set of transformations– X Y CH ¾ CH ¾ CH CH3 ¾ CH2 ¾ CH2 Br ¾® Product ¾® 3 3 ½ Br (a) X = dilute aqueous NaOH, 20°C,Y = HBr/acetic acid, 20°C (b) X = concentrated alcoholic NaOH, 80°C, Objective Questions II (One or more than one correct option) 37. For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is(are) CH3 Y = HBr/acetic acid, 20°C Br (c) X = dilute aqueous NaOH, 20°C, Y = Br2 / CHCl 3 , 0°C (d) X = concentrated aqueous NaOH, 80°C, Y = Br2/CHCl 3 , 0°C 27. The compound that will react most readily with NaOH to form methanol is (a) (CH3 )4 N + I – (b) CH3OCH3 + – (c) (CH3 )3 S I (d) (CH3 )3 Cl I Br C Br CH3 III II IV (a) Compound IV undergoes inversion of configuration (b) The order of reactivity for I, III and IV is : IV > I > III (c) I and III follow S N 1 mechanism (d) I and II follow S N 1 mechanism inactive compound (s) is/are always gives H (a) an enantiomer of the substrate (b) a product with opposite optical rotation (c) a mixture of diastereomers (d) a single stereoisomer H Br H3C CH3 H (c) H (d) H2C CH3 SN 2 reaction is– CH3 Br Br H2C Br (b) H2C (a) 29. The order of reactivities of the following alkyl halides for a CH3 CH3 (a) RF > RCl > RBr > RI (c) RCl > RBr > RF > RI (b) RF > RBr > RCl > RI (d) RI > RBr > RCl > RF 39. In the following reaction, the major product is CH3 30. Which of the following has the highest nucleophilicity? (c) CH3- (b) OH - CH2 (b) carbene (d) carbocation (a) (a) Lithium methoxide (c) Lithium dimethylamide (b) Lithium acetate (d) Methyl lithium 33. (CH3 )3 CMgCl on treatment with D2 O produces (a) (CH3 )3 CD (b) (CH3 )3 COD (c) (CD)3 CD (d) (CD)3 COD 34. 1-chlorobutane on reaction with alcoholic potash gives (a) 1-butene (c) 2-butene (b) 1-butanol (d) 2-butanol hydroxide produces (c) propyne (d) propanol 36. The reaction condition leading to the best yield of C2 H5 Cl are– UV light (a) C2H6 (excess) + Cl 2 ¾¾¾® dark room temp. (b) C2H6 + Cl 2 (excess) ¾¾¾® UV light (c) C2H6 + Cl 2 (excess) ¾¾¾® UV light (d) C2H6 + Cl 2 ¾¾¾® H2C (b) H3C Br CH3 H2C CH3 Br (d) H3C Br 40. The compounds used as refrigerant are (a) NH3 (e) CH2F2 (b) CCl 4 (c) CF4 (d) CF2Cl 2 Assertion and Reason 41. An Assertion and a Reason are given below. Choose the correct answer from the following options. 35. n-propyl bromide on treatment with ethanolic potassium (b) propene CH3 Br (c) 32. Which of the following is an organometallic compound? CH3 CH3 racemises slowly in the presence of small amount of SbCl5, due to the formation of (a) carbanion (c) free-radical 1 equivalent HBr H2C (d) NH2- 31. A solution of (+)-2-chloro-2-phenylethane in toluene (a) propane H 3C 38. Compound(s) that on hydrogenation produce(s) optically 28. An SN 2 reaction at an asymmetric carbon of a compound (a) F - Br CH3 Assertion (A) Vinyl halides do not undergo nucleophilic substitution easily. Reason (R) Even though the intermediate carbocation is stabilised by loosely held p-electrons, the cleavage is difficult because of strong bonding. (a) Both (A) and (R) are wrong statements. (b) Both (A) and (R) are correct statements and (R) is correct explanation of (A). (c) Both (A) and (R) are correct statements but (R) is not the correct explanation of (A). (d) (A) is a correct statement but (R) is a wrong statement. Alkyl Halides 44. Match the following : Match the Columns 42. List-I contains reactions and List-II contains major products. List-I P. List-II 1. + Q. OMe R. OH Br ONa + HBr Br p. E1 reaction B. Ph—CHBr—CH3 reacts faster than Ph—CHBr—CD3 q. E2 reaction C. Ph—CH2 ¾ CH2Br on treatment with C2H5 OD /C2H5 O- gives Ph ¾ CD ==CH2 as the major product. r. E1CB reaction D. PhCH2CH2Br and PhCD2CH2Br react with same rate. s. First order reaction 3. + NaOMe OMe 4. + MeBr ONa 5. O Column II A. CH3 ¾ CHBr ¾ CD3 on treatment with alc. KOH gives CH2 == CH ¾ CD3 as a major product. 2. Br S. Column I Fill in the Blanks 45. Vinyl chloride on reaction with the dimethyl copper Match each reaction in List-I with one or more products in List-II and choose the correct option. (a) P ® 1, 5; Q ® 2; R ® 3; S ® 4 (b) P ® 1, 4; Q ® 2; R ® 4; S ® 3 (c) P ® 1, 4; Q ® 1,2; R ® 3,4; S ® 4 (d) P ® 4, 5; Q ® 4; R ® 4; S ® 3,4 43. Match the chemical conversion in Column I with the appropriate reagents in Column II and select the correct answer using the code given below the lists. Column I P. Column II 1. Cl (i) Hg(OAc)2 ; (ii) NaBH4 gives ………. 46. The starting material for the manufacture of polyvinyl chloride is obtained by reacting HCl with …… 47. The halogen which is most reactive in the halogenation of alkanes under sunlight is … True/False 48. Photobromination of 2-methylpropane gives a mixture of 1-bromo-2-methyl propane and 2-bromo-2-methyl propane in the ratio 9 : 1. 49. During S N 1 reactions, the leaving group leaves the molecule before the incoming group is attached to the molecule. 50. The reaction of vinyl chloride with hydrogen iodide to give Q. ONa OEt 2. NaOEt 3. Et-Br 4. (i) BH3 ; (ii)H2 O2 / NaOH OH R. S. OH Codes P (a) 2 (b) 3 (c) 2 (d) 3 1-chloro-1-iodoethane is an example of anti-Markownikoff’s rule. 51. Iodide is better nucleophile than bromide. 52. Carbon tetrachloride is inflammable. 53. Carbon tetrachloride burns in air when lighted to give phosgene. Integer Answer Type Question 54. In the following monobromination reaction, the number of possible chiral product(s) is (are)... CH 2 CH 2 CH 3 Q 3 2 3 2 R 1 1 4 4 S 4 4 1 1 H Br Br 2 (1.0 mole) 300°C CH3 (1.0 mole) (Enantiomerically pure) 55. The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is Alkyl Halides 62. Draw the stereochemical structure of product in the Subjective Questions 56. Identify X, Y and Z in the following synthetic scheme and following reaction CH3 write their structures NaOH Br ¾¾¾ H ¾¾¾® (i) NaNH 2 X CH3 CH2 C ºº CH ¾¾¾¾¾® SN 2 (ii) CH3 CH2 Br H / Pd - BaSO C2 H5 alkaline 2 4 X ¾¾¾¾¾® Y ¾¾¾¾® Z 63. Aryl halides are less reactive than alkyl halides towards KMnO4 nucleophilic reagents. Give reason. 57. What would be the major product in each of the following 64. Identify the major product in the following reaction. reactions? H Alc. KOH CH3 ½ C2 H5 OH (i) CH3 ¾ C ¾ CH2 Br ¾¾¾® D ½ CH3 D Br 65. Write the structures of the major organic product expected from each of the following reactions : CH3 58. Write the structures of the products : Alc. KOH Alc. KOH C6 H5 CH2 CHClC6 H5 ¾¾¾¾® (i) H3 C ¾¾¾ CH2 CH3 ¾¾¾® Cl 59. Which of the following is the correct method for synthesising aq. alkali (ii) CH3 CH2 CHCl 2 ¾¾¾® methyl-t-butyl ether and why? or boil (CH3 )3 CBr + NaOMe ¾® CH3 Br + NaO -t -Bu ¾® 66. Arrange the following in order of their (i) Increasing basicity H2O, OH- , CH3OH, CH3O– 60. Predict the structure of the product in the following reaction Br H Ph MeO H HBr C6 H5 ¾ CH2 ¾¾¾ CH3 ¾¾¾® ? ¾¾® ? (ii) Increasing reactivity in nucleophilic substitution reactions CH3 F, CH3 I, CH3 Br, CH3 Cl NaI acetone 67. Chloroform is stored in dark coloured bottles. Explain in not Ph 61. Optically active 2-iodobutane on treatment with NaI in acetone gives a product which does not show optical activity. Explain briefly. more than two sentences. 68. Show by chemical equations only, how would you prepare the following from the indicated starting materials? Specify the reagents in each step of the synthesis. (i) Chloroform from carbon disulphide. (ii) Hexachloroethane (C2Cl6) from calcium carbide. Answers 1. 5. 9. 13. 17. 21. 25. 29. (c) (b) (c) (b) (b) (b) (a) (d) 2. 6. 10. 14. 18. 22. 26. 30. (d) (b) (c) (b) (d) (b) (b) (c) 3. 7. 11. 15. 19. 23. 27. 31. (b) (d) (d) (a) (a) (a) (a) (d) 4. 8. 12. 16. 20. 24. 28. 32. (c) (d) (d) (d) (d) (a) (d) (d) 33. 37. 41. 44. 45. 49. 53. (a) (a, d) (c) A ® q, B ® propene True False 34. (a) 35. 38. (b,d) 39. 42. (b) 43. q, C ® r, D ® p, s 46. ethyne 47. 50. True 51. 54. (5) 55. (b) (d) (a) 36. (a) 40. (a,d) chlorine True (5) 48. False 52. False Hints & Solutions 1. Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives 3. The given reaction takes place as follows : 1-phenyl but-1-ene(X). Cl Reaction of X with Hg(OAc)2 / H2O followed by NaBH4 gives 1-phenyl butan-1-ol (Y). Reaction involved is as follows: Cl EtONa/D b-H for b-elimination Saytzeff elimination –HCl H CH2 EtOH Ph b-H for Hoffmann elimination H –+ EtOK Markownikov’s centre for Brs Ph + OH 3-methyl but-2-ene major (X) (Saytzeff product) 1. Hg (OAc)2,H2O 2. NaBH4 Ph (Y ) Electrophilic addition (Markownikov’s rule) HBr Mechanism 3-methyl-but-1-ene minor (Hoffmann product) Br OAc Cl s Ph H OEt dehydroPh halogenation 2-bromo-3-methyl butane (minor) medium, CH3OH (as well as a poor nucleophile) is protic in nature, So, the reaction will follow mainly S N1 pathways via the formation of a carbocation intermediate (I). OAc Hg+ H H C Me Me Me H 2-bromo-2-methyl butane (major) (Y) 4. In the given question, the substrate is a 2º-halide (bromide) and the Oxymercuration Ph Br OAc + Hg + H H H C2H5 –Br r rds Me r C Me (I) Me H 2O Ph CH CH OH C2 H5 CH—CH—Me Hg Me OAc OCH3 (Minor) The intermediate, I can be rearranged into the more stable form I¢ (3°) by a-hydride shift. I ¢ will give the major product. H s Me OAc Ph CH CH OH C2H5 HgOAc s r C CH—Me Me a-H shift Me r Me C—CH2—Me CH3OH –Hr I¢ (3° carbocation) H I (2° carbocation) Me Demercuration NaBH4 Ph CH3OH r –H Fast (2°) H Br Me CH CH s CH2 + AcO OH C2H5 (Y) Me OCH3 (Major) 5. The potential energy (PE) diagram for S N1 reaction is 2. (CH3 )3CCl gives a precipitate with AgNO3 solution because it forms stable carbocation. (i.e. tertiary) that readily combines with AgNO3 to give precipitates of AgCl. CH3 ½ + CH3 ¾ C ¾ Cl + AgNO3 ¾® (CH3 )3 C + AgCl (White ppts.) ½ CH3 CH2 == CH ¾ Cl forms unstable carbocation. Hence, it does not readily react with AgNO3. C—CH2—Me Carbocation intermediate PE R++X– – +Nu R¾ X R ¾ Nu Progress of reaction Alkyl Halides S N1 reaction has two steps. In the first step, the carbon halogen bond breaks heterolytically, with the halogen retaining the previously shared pair of electron. In the second step, the nucleophile reacts rapidly with the carbocation formed in the first step. In the above graph, the alkyl halide is the only species that participates in the rate limiting step. Here, the rate of reaction depends on the concentration of the alkyl halide and does not depend on the concentration of nucleophile. Key Idea Both alc. KOH and NaNH2 in liquid NH3 are dehydrohalogenating reagents. On comparative terms NaNH2/liquid NH3 is stronger in action. 8. The reaction proceeds as : CH3 CH2 (i) CH3CH2 C == CH2 ½ COOCH2CH3 Br Br Alc. KOH CH3 CH3 CH2 C CH dihalide. Chlorine adds up to alkene via electrophilic addition reaction involving cyclic chlorinium ion formation. Cl2/CCl4 will be the major product The vicinal dihalide in presence of anhyd. AlCl3 results in the formation of carbocation that rearranges itself to form a cyclic compound. Cl H3CO + AlCl3 CH CH CH3 O CH CH2 CH3O CH3O b g d CH3 CH2 a Br2 / hn C H3 ¾ C H2 ¾ C H == C H2 ¾ ¾ ¾® –HBr Br | C Br2 / hn C H3 ¾CH ¾ C H ==C H2 ¾ ¾ ¾® CH3 d g b a Br | H+ C H3 ¾ Cg ¾ C H == C H2 d d– O CH BF H H H d+ Transition state CH2 CH3 Br CH Br Product Here, BB = Bond breakage BF = Bonf formation | Br b a 3,3- dibromobut-1-ene 11. The reaction follows a, b-elimination mechanism to give a more substituted stable alkene as a major product. As the substrate is a a, g -dibromo (1, 3-) compound it gives a conjugated diene. SN2 Reaction (Zeisel’s Method) CH3Br + HO –HBr 3-bromo-but-1-ene H + BF d BB Cl H reacts with Br2 / hn, it readily undergoes allylic free radical substitution and forms 3, 3-dibromobut-1-ene Br Br– –H+ But-1-ene CH + – r + AlCl 4 Br– O Cl 10. The compound (E ) has two allyl-hydrogen atoms (g ). When E H+ HBr excess/Heat + (Anhyd.) H3CO Cl O H3CO Cl The road map of complete reaction is as follows: Br – CNa+ Thus, option (d) is the correct answer. reaction serves for dual purpose: (i) Hydrolysis of ether via SN 2 mechanism, i.e. Zeisel’s method. (ii) Markownikoff addition at double bond of the branch. d– CH2 9. The given reactant in presence of Cl2/CCl4, given vicinal 7. Key Idea The excess of HBr and high temperature in given CH3 CH3CH2C H+ Cl CH3 CH Cl In accordance with Saytzeff rule CH3 CH2 Br H3CO (ii) CH3 CH == C ¾ CH3 ½ COOCH2CH3 CH3 CH == C ¾ CH3 ½ COOCH2CH3 CH2 NaNH2/liquid NH3 6. Presence of strong base (EtO- ) and heat indicates elimination. Thus, the compound undergo dehydrohalogenation and alkene is produced. As per the position of Cl in substrate, following 2 alkenes are formed as product: CH CH2 CH3 HOs Br H Alc. KOH D Ph –2HBr H Br sOH (It is very stable because the conjugated diene system present. It is further conjugated with the benzene ring.) Alkyl Halides 12. In the given reaction, AlCl3 act as Lewis acid and helps in 15. To show decolourisation, compound must be unsaturated. generation of carbocation. The resulting carbocation (1°) rearranges itself to stable carbocation (3°). MeO O Br O—tBu (Saturated) (cannot decolourise Br2 water) C6H5 C6H5 – + tert–BuONa Cl AlCl3(anhy.) MeO MeO s H r (1°) a-Hs shift r s Cl AlCl3 –AlCl4 Br (Unsaturated) (decolourise Br2 water) MeO O r Br 3° carbocation, more stable Protic medium, H2O supports the rearrangement of carbcation Me O – + tert–BuONa MeO Br –Hr (Unsaturated) (decolourise Br2 water) H 16. The number of stereoisomers in molecules which are not 13. Complete reaction can be represented as Br – Na+OMe MeOH + NaBr + MeOH H Thus, the given reaction is dehydrohalogenation which is a b-elimination proceeding through E2 mechanism. Mechanism The reaction proceeds through the formation of following transition state with simultaneous removal of Br and H atoms. BB Na+ Na+Br– BF (Product) Br (Transition State) divisible into two equal halves and have n number of asymmetric C-atoms = 2n. 3-methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product i.e. 2-bromo-3-methyl pentane. It has two chiral centres. Therefore, 4 stereoisomers are possible Br 2 1 2 4 3 4 1 HBr Peroxide 5 3-methylpent-2-ene 3 5 Anti-Markownikoff's addition Four stereoisomers are possible (As molecule has two chiral centres and asymmetric). 17. An alkyl halide in presence of a bulkier base removes a proton Here, BB = Bond breakage BF = Bond formation – BF from a carbon adjacent to the carbon bonded to the halogen. This reaction is called E2 ( b-elimination reaction). Br BB H OMe BF MeOH (Product) of the alkyl halide. (ii) S N 1reaction proceeds through the formation of carbocation. The reactivity is decided by ease of dissociation of alkyl halide. R ¾ X q RÅ + X È Higher the stability of R + (carbocation), higher would be the reactivity towards S N1 reaction. p - H3CO ¾ C6H4 ¾ CH2Å is the most stable carbocation due to resonance and then CH3 CHCH2CH3 (2° carbocation) while Å CH3CH2 CH2(1°) is least stable. Thus, the correct increasing order of the reactivity of the given halides towards the S N 1reaction is CH3CH2CH2Cl < CH3 CHCH2CH3 < p-H3COC6H4CH2Cl | Cl (II) (I) (III) H t-BuO–K+ C6H5 14. (i) The rate of S N 1 reaction depends only upon the concentration Å (Unsaturated) (decolourise Br2 water) O Intramolecular ArSE2 pathway O (Product) O – + tert–BuONa (H2O) r O – + tert-BuONa H C6H5 C==C H C6H5 H H + tert-BuOH + Br– C6H5 - 18. Key Idea Strong nucleophile ( OMe) in polar solvent (MeOH) gives elimination products over substitution products but all products are possible in different yields. Cl CH3—C—CH2CH2CH3 CH3 -+ MeONa MeOH OCH3 CH3—C—CH2—CH2—CH3 + CH3 (Less yield) CH3 —C == CH—CH2—CH3+ CH2 ==C—CH2—CH2—CH3 CH3 CH3 (More yield) Alkyl Halides 19. 23. Nucleophile PhS- substitute the Br - through SN 2 mechanism NBS hv wih inversion of configuration at a-C. H3C Br H 3C Br HO Br OH H O/K CO 2 2 3 + F + PhS SPh F – S N2 NO2 20. Alkyl fluorides can be prepared by action of mercurous fluoride or antimony trifluorides (inorganic fluorides) on corresponding alkyl halide. This reaction is known as Swarts reaction. 24. Reaction proceed through carbocation intermediate CH3 H CH3Br + AgF ¾® CH3F + AgBr Methyl fluoride —NO2 H Free redical fluorination is highly explosive reaction, so not preferred for the preparation of fluoride. CH3O— I H 2O II H2O Note The order of reactivity towards SN 2 reaction for alkyl halides is Primary > Secondary > Tertiary halides halides halides (2 ° ) (3 ° ) (1° ) PLAN Acetone is an aprotic solvent and can dissolve both the nucleophile and the substrate and thus SN2 reaction is favoured. Also 1° 2° 3° Alkyl halides S N1 S. O Cl Q. 1° alkyl halide but (C ¾ Cl). B.E. is decreased by electron withdrawing [C 6H5CO ] group, (a case of I-effect). Thus, maximum rate in SN 2 reaction 2° alkyl halide, rate is minimum Cl P. CH3 — Cl R. Cl CH3 H CH3 K L CH3O— —NO2 + H CH3 3°(II) stabilised by + R effect of – OCH3 OH OC2H5 (More crowded) As steric hinderance (crowding) increases, rate of S N 2 reaction decreases. SN2 CH3 2°(I) CH3 | CH3Cl > CH3CH2Cl > CH3 ¾ CH ¾ Cl > CH3 ¾ C ¾ Cl | (Less 1° | crowded) CH3 CH3 22. + —NO2 H shift H using which we can arrange the reactant in correct order of their reactivity towards SN 2 reaction. 2° CH3 + 21. Steric hindrance (crowding) is the basis of SN 2 reaction, by – Cl+ Cl CH3 CH3 H NaI C2H5Cl ¾¾® C2H5I + NaCl acetone 1 Steric crowding of 'C' CH3 CH3O— But, when aciton of NaI/acetone takes place on alkyl chloride of bromide, alkyl iodide forms. This reaction is called ‘Finkelstein reaction’ Rate of S N 2 µ NO2 1° alkyl halide 1° allylic halide but allylic 1° carbocation is resonance stabilised in SN 1reaction Thus, reactivity order is S > P > R > Q + C2H5I 25. C2H5O– anhy. C2H5OH SN2 reaction Alcoholic NaOH 80 °C 26. CH3 ¾ CH2 ¾ CH2Br ¾¾¾¾¾® CH3 ¾ CH == CH2 HBr ¾¾¾® CH3 ¾ CH ¾ CH3 ½ Br CH3 ½ 27. H3C ¾ N + ¾ CH3I- ¾® CH3OH + (CH3 )3 N+ I½ CH3 HOLack of b-H on quaternary ammonium iodide leads to SN 2 reaction otherwise E2 elimination usually takes place. 28. SN 2 reaction at asymmetric carbon occur with inversion of configuration and a single steroisomer is formed because the reactant and product are not enthtiomer. Therefore the sign of optical rotation may or may not change. 29. If alkyl groups are same, the order of leaving ability of halides in SN 2 reaction is RI > RBr > RCl > RF Alkyl Halides 30. CH-3 , being the strongest base, has highest nucleophilicity. Cl ½ 31. C6H5 ¾ C ¾ CH3 + SbCl 5 l ½ H + C6H5 ¾ C ¾ CH3 + SbCl6- T ½ H 40. Both NH3 and CF2Cl 2 are used as refrigerant. 41. Vinyl halide (CH2 ==CH ¾ Cl) do not undergo nucleophilic substitution reactions. This is because it forms highly unstable Å carbocation (CH2 == CH). It cannot delocalise its p-electron. In vinyl halide C ¾ Cl bond possess double bond character also. (I) he planar carbocation (I), when return back, forms racemic mixture of the starting compounds. 32. Compound in which metal is directly bonded to carbon, is known are organometallic compound, e.g. CH3Li. CH3 CH3 ½ ½ D O 2 33. CH3 ¾ C - MgCl + ¾¾¾® CH3 ¾ C ¾ D ½ ½ CH3 CH3 s CH2 42. CH For P, i.e. Cl r CH2 –+ ONa + CH Cl Br For this reaction 1 and 4 are probable products. Product 1 i.e., OH product 4 i.e., C2 H5 OH 34. CH3 ¾ CH2 ¾ CH2 ¾ CH2 ¾ Cl + KOH ¾¾¾® is formed due to substitution while is formed due to elimination. A tertiary carbocation i.e, + formed during the reaction. Remember for 3° carbocation ions elimination product predominates. CH3CH2 ¾ CH == CH2 1- butene For Q, i.e. 35. An alkyl halide containing at least one b-H, on treatment with ethanolic KOH, undergoes dehydrohalogenation, giving alkene. + HBr OMe Correctly matched product for this reaction is 2 i.e., The reaction proceeds as Ethanol CH3 ¾ CH2 ¾ CH2 ¾ Br + KOH ¾¾¾® CH3 ¾ CH == CH2 OMe 36. During chlorination of alkane, if excess of alkane is treated with Cl 2 (g ) in presence of light or heat, chance of mono-chlorination predominate. For R i.e., UV-light + HBr MeOH + . Br Br + NaOMe Br C2H6 (excess) + Cl 2 ¾¾¾® CH3CH2Cl + HCl Correctly matched product is 4 i.e., elimination reaction and proceeds as 37. (a) Both I and II are 1° halide, undergos SN 2 reaction. (d) III is a tertiary halide, undergoes SN 2 reaction. I is benzylic bromide, it is very reactive in SN 1also as if produces stable benzylic carbonation. +– + NaOMe Br 38. In both cases, hydrogenation of olefinic bond will render H Br H2 H 2C (a) H3C H Br (b) H2C H 3C CH3 – + + NaBr OMe 43. 3. will be considered and thermodynamically controlled product would be the major product as: CH3 Cl Column II Q. CH3 CH3 Br ONa OEt Explanation NaOEt(2) O Et (strong nucleophile) causes dehydrohalogenation of 3° alkyl halide EtBr (3) 3° butoxide undergoes SN reaction with 1° alkyl halide CH2 CH3 3º-carbocation CH2 P. Column I CH3 + H Br– CH3 CH3 + MeBr CH3 CH3 Chiral 39. Since, there is no mention of temperature, room temperature CH2 OMe . The reaction proceeds as ONa CH3 CH2 + MeBr ONa The correct match is 3 i.e., Br H3C CH3 For S i.e., Br Chiral H H2 (Alkali in alcohol) 3° alkyl halide preferes elimination. compound achiral as two identical ethyl group will come at the a-carbon which was earlier chiral carbon. However, in (a) and (c), chirality will be retained even after hydrogenation. H . It is a normal Alkyl Halides R. (i) Hg(OAC) 2 OH (ii) NaBH4 (1) Mercurationdemercuration adds H2O by Markownikoff’s rule without rearrangement 54. Given compound undergoes free-radical bromination under given conditions, replacing H by Br. C* is chiral carbon. (i) BH3 Hydroboro-oxidation (ii) H2O 2 /OH- adds H2O by anti-Markownikoff’s OH (4) rule S. Thus, P — (2), Q — (3), R — (1), I. Chiral Br C-D bond is stronger than C—H bond, C—H is preferably broken in elimination. B. Ph ¾ CHBr ¾ CH3 reacts faster than Ph ¾ CHBr ¾ CD3 in E2 reaction because in latter case, stronger C—D bond is to be broken in the rate determining step. C2 H5 OD Ph ¾ CH2 ¾ CH2Br ¾¾¾¾® Ph ¾ CD == CH2 C2 H5 O Deuterium incorporation in the product indicates E1CB mechanism l - Ph ¾ CH ¾ CH2Br carbanion C2 H5 OD l CH2Br S — (4) CH2CH2CH3 CH—CH2CH3 Ph ¾ CHD ¾ CH2Br I D ½ I ¾¾¾® Ph ¾ C ¾ CH2 ¾ Br ¾® Ph ¾ C == CH2 ½ D C2 H5 O- Br Br E2 reaction is a single-step reaction in which both deprotonation from b-C and loss of leaving group from a-C occur simultaneously in the rate-determining step. Ph ¾ CH2 ¾ CH2Br CH3 II. Achiral 46. Vinyl chloride is obtained by the reaction of HCl with ethyne. 47. Chlorine is most reactive. 48. 2-bromo-2-methylpropane is formed as major product. 49. In SN 1 reaction, leaving group is detached in the first step forming carbocation intermediate. CH2—CH2CH2CH3 CH2— CH—CH3 52. False 53. CCl 4 is fire retardent, used as fire-extinguisher. Br H Br H CH3 CH3 IV. Chiral V. Chiral (III) has two chiral centres and can have two structures. CH2CH3 CH2CH3 H Br H Br H Br Br H CH3 CH3 (III) A (III) B (IV) has also two chiral centres and can have two structures. CH3 CH3 H Br H Br H Br Br H CH3 (IV)A CH3 (IV)B It has plane of symmetry thus, achiral. Thus, chiral compounds are five. I, III A, III B, IV B and V. 55. The substrate has three different types of B—H, therefore, first three structural isomers of alkenes are expected as : H H Br H KOH 50. True 51. Larger the size of donor atom, greater is its polarisability, III. Chiral Br 45. Propene is produced (CH3 )2 Cu + CH2 == CHCl ¾® CH3 ¾ CH== CH2 Br H CH3 D. Both PhCH2CH2Br and PhCD2CH2Br will react at same rate in E1 reaction because C—H bond is broken in fast non rate determining step. Also E1 reaction follow first order kinetics. stronger is the nucleophile. Br H CH3 Alc. KOH E2 C2 H5 O- Br2 Br H 44. A. CH3 ¾ CHBr ¾ CD3 ¾¾¾® CH2 == CH ¾ CD3 C. CH2CH2CH3 CH2CH2CH3 3 types of B-H + I + II III The last two alkenes II and III are also capable of showing geometrical isomerism, hence two geometrical isomers for each of them will be counted giving a total of five isomers. Alkyl Halides Above equilibrium is established which has equilibrium constant equal to 1. Therefore, equilibrium mixture will have both the enantiomers in equal amount giving racemic mixture. (i) NaNH2 56. CH3CH2C ºº CH ¾¾¾¾¾® CH3CH2 ¾ C ºº C ¾ CH2CH3 (ii) CH3 CH2 Br X C2H5 X H H2/Pd–BaSO4 C ==C H Alk. KMnO H 4 OH H OH C2H5 H5C2 62. SN 2 reactions leads to inversion of configuration at a-C CH3 Br C2H5 Y CH3 NaOH S N2 H C2H5 C2H5 meso-diol product with inverted configuration 57. Unimolecular reaction occur CH3 D ½ CH3 ¾ C ¾ CH2 ¾ Br l ½ CH3 CH3 ½ + CH3 ¾ C ¾ CH2 ½ CH3 63. + I CH3 – Aryl halide Due to the above resonance phenomena, C—X bond acquire partial double bond character and becomes difficult to break in the rate determining step of SN 2 reaction. CH3 SN 1 CH3—C—CH2CH3 OC2H5 ½ Br Alc. KOH E2 Br H MeO I Ph NaI H acetone S N2 Ph 61. Ph H MeO H H5C2 + I – K=1 -H O 2 ¾¾¾® CH3CH2CHO 66. (i) H2O < CH3OH < HO- < CH3O(ii) CH3F < CH3Cl < CH3Br < CH3I 67. Chloroform in presence of air and sunlight, oxidises slowly to form a highly poisonous compound called phosgene hn 2CHCl 3 + O2 ¾¾® 2COCl 2 + 2HCl (Phosgene) To prevent the above oxidation reaction, chloroform is kept in dark bottles. 68. (i) CS2 + Cl 2 ¾® CCl 4 + S2Cl 2 Ph CS2 + 2S2Cl 2 ¾® CCl 4 + 6S CCl 4 + 2 [H] ¾¾¾® CHCl 3 + HCl CH3 C—I OH ½ Aq. NaOH (ii) CH3CH2CHCl 2 ¾¾¾® [CH3CH2 ¾ CH ¾ OH] boil unstable inversion of configuration occur only at a-carbon H3C H Cl E2 CH3 CH3 ½ E2 59. CH3 ¾ C ¾ Br + CH3 O- ¾¾® CH2 == C CH3 ½ CH3 CH3 CH3 ½ ½ SN 2 CH3 ¾ C ¾ O- + CH3 Br ¾¾® CH3 ¾ C ¾ OCH3 ½ ½ CH3 CH3 ¯ HBr C6H5 ¾ CH ¾ CH2 ¾ CH3 ½ Br CH3 CH3 ½ Alc. KOH 65. (i) H3C ¾¾¾ CH2CH3 ¾¾¾® CH3 ¾ C == CHCH3 58. C6H5CH2 ¾ CH ¾ C6H5 ¾¾¾® C6H5 ¾ CH == CH ¾ C6H5 ½ Cl Alc. KOH D 64. C6H5CH2 ¾ CH ¾ CH3 ¾¾¾® C6H5 ¾ CH == CH ¾ CH3 CH3—C==CH—CH3 I 60. X+ X primary carbocation CH3 ½ methyl shift ¾¾¾¾® CH3 ¾ C ¾ CH2 CH3 E1 OH H I—C H C2H5 Fe/H2 O (Chloroform) (ii) CaC2 + H2O ¾® C2H2 + Ca(OH)2 H /Ni Cl 2 (excess) 2 C2H2 ¾¾¾® C2H6 ¾¾¾¾® Cl 3C ¾ CCl 3 hn (Hexachloroethane)