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Preview Organic Chemistry for the JEE Mains and Advanced Vol II by I. S. S. Raju Preview Organic Chemistry for the JEE Mains and Advanced Vol II by I. S. S. Raju Preview Organic Chemistry for the JEE Mains and Advanced Vol II by I. S. S. Raju Preview Organic Chemistry for the JEE Mains and Advanced Vol II by I. S. S. Raju Preview Organic Chemistry for the JEE Mains and Advanced Vol II by I. S. S. Raju Preview Organic Chemistry for the JEE Mains and Advanced Vol II by I. S. S. Raju Preview Organic Chemistry for the JEE Mains and Advanced Vol II by I. S. S. Raju
Organic Chemistry for the JEE Main and Advanced Volume II I.S.S Raju No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent Copyright © 2017 Trishna Knowledge Systems This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN 978-93-325-7562-2 eISBN 978-93-325-8663-5 First Impression Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128, formerly known as TutorVista Global Pvt Ltd, licensee of Pearson Education in South Asia Head Office: A-8(A), 7th Floor, Knowledge Boulevard, Sector 62, Noida 201 309, Uttar Pradesh, India Registered Office: 4th Floor, Software Block, Elnet Software City, TS-140, Block & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India Fax: 080-30461003, Phone: 080-30461060 www.pearson.co.in, Email: companysecretary.india@pearson.com Contents Preface v Acknowledgment About the Author vii ix Chapter Halogen Derivatives of Hydrocarbons or Halohydrocarbons 1.1–1.132 Chapter Hydroxy Compounds, Ethers and Phenols 2.1–2.230 Chapter Carbonyl Compounds 3.1–3.160 Chapter Carboxylic Acids 4.1–4.88 Chapter Acid Derivatives 5.1–5.80 Chapter Compounds Containing Nitrogen Chapter Carbohydrates 7.1–7.72 Chapter Amino Acids and Proteins 8.1–8.32 Chapter Polymers 9.1–9.40 Chapter 10 Practical Organic Chemistry 6.1–6.134 10.1–10.28 This page is intentionally left blank Preface Organic Chemistry is one of the most fascinating subjects as it deals with the chemicals related to living organisms The subject has grown to such a level that it is not easy for any person to go through it completely The number of organic compounds known has gone to almost 15 million in the past few years It is further increasing every year Understanding organic chemistry becomes difficult due to this reason In the past few years, its presentation also changed in some cases due to this recent finding, but still there are some ambiguities Writing a book of Organic Chemistry for students at the 10+2 level is a challenge as there are no boundaries for it At this level, students have no practical experience Their requirements should be addressed by good books, effectively guided by an able teacher Organic Chemistry has to be presented to such students with crystal clear explanations covering all concepts, while ensuring that the discussion does not digress to topics beyond the scope of 10+2 level The first volume of Organic Chemistry was published conforming to this requirement This second volume of Organic Chemistry adheres to the style followed in the first volume in its delineation of reaction mechanisms and named reactions An effort has been made to keep the explanations simple to enable students understand and apply the concepts effectively to solve questions in IIT entrance examination Questions framed in the book have been modeled on the syllabus of CBSE and NCERT for the same purpose Any error noticed in the book may please be brought to my notice so that it can be rectified in the subsequent editions Suggestions for the book’s improvement are welcome I.S.S Raju This page is intentionally left blank Acknowledgment I thank my family members and friends who encouraged me to write this book at the age of 73 My family members were unfailingly supportive and encouraging during the long months I spent glued into many relevant resources while writing this book I am especially thankful to my great granddaughter Chy Akshara who plays with me all the time, for not disturbing me at the time of writing the manuscript I am obliged to Pearson India Education Services Pvt Ltd for publishing this book In particular, I am grateful to their editors Sri Abhilash Ayyappan and Mrs G Sharmilee whose sincere efforts helped in bringing out this book in record time I.S.S Raju This page is intentionally left blank About the Author I.S.S Raju retired as Principal, with 35 years of experience in teaching graduate and post graduate students the subject of organic chemistry He currently coaches JEE Main and Advanced aspirants, a vocation that he has been actively pursuing since his retirement 15 years ago His rich experience in this subject has assisted many students in gaining better command over this subject as well as helping them well academically His approach in assisting students is methodical and lucid This helps students immensely, especially when it is about understanding the various complex theories and concepts of organic chemistry The author’s tenure as a principal and his post-retirement contributions, in tirelessly coaching IIT aspirants, has brought him into the crux of celebrating the golden jubilee of his academic career 1.118 ■ Chapter (a); Passage (c); The base is a strong base Leaving group is a poor leaving group There is a strong withdrawing group by the side of β-carbon So, it is E1cB (b); It is an example of E2 as base is strong E1 takes place with moderate base It is accompanied by isotopic effect Option (b) is times more easily formed (b); Substrate is tertiary So, E1 elimination takes place CN is not a very strong base Passage (c); Option (a) and (b) are SN2 as in the first reaction there is inversion Second is primary alkyl halide Concentration of reagent does not influence the rate in SN1 only So, it is tertiary butyl chloride (strong base affects elimination, being weak base acetate ion affects substitution) (b); A poor nucleophile CH3OH affects SN1 In other reactions the reagent is a strong base, and high concentration affects SN2 Passage (d); All statements are correct Passage (b); (d); bond energy is less and so it can form Grignard reagent easily (d); Tertiary butoxide being a bulky base affects Hofmann elimination, rather than Zaitsev elimination Passage (c); A SN2 reaction requires high concentration of nucleophile and a polar aprotic solvent (d); Both SN2 and E2 take place in polar aprotic solvents But E2 requires a strong base, whereas SN2 requires a better nucleophile E2 takes place at high temperature (a), (b); Both low concentration of reagent and polar protic solvent encourage SN1 and E1 Additionally high temperature favours E1 Generally both secondary and tertiary alkyl halides prefer to undergo elimination to avoid steric repulsion (e); (a) After the formation of Grignard reagent it interacts with –OH group which is acidic (b) CH2 = CH2 is formed with Mg (c) is formed with Mg (d) Grignard reagent after formation opens the oxirane ring (a); Halogen Derivatives of Hydrocarbons ■ 1.119 Passage (d); (b); Match the Columns Type Questions Column-I Match the following column: Column-I Column-II C R SN2 D S SNi Column-II A C6H5CH2CD2Br on reaction with P C2H5O– gives C6H5CH = CD2 E1 B C6H5CHBrCH3 and C6 H5 − CH − CD3 both react | Br with the same rate Q E2 C C6H5CH2 CH2Br on treatment with C2H5O–/C2H5OD gives C6H5CD = CH2 R E1cB D C6H5CH2CH2Br reacts faster than C6H5CD2CH2Br on reaction on with C2H5O–/ethanol S First order reaction T No bond around asymmetric arbon breaks U Priorities chagne in the product Match the following column: Column-I A CH3 − CH − C − O − H | || Cl O Column-II P No change in configuration of the product Ag O Column-I A CH3CH CH CH Br C H O → OH B Match the following column: − Column-II P E1 → Q Anchimeric effect B CH3CH CH − CH | Br ( CH ) CO − 3 → Q SN1 1.120 ■ Chapter Column-I Column-II R E2 C CH3 − CH − CH CH3 | Cl SH → Match the following column: Column-I A Column-II P cis alkene C2 H5OH S SN2 D CH3 − CH − CH − CH3 | Cl C H 5O − → B Q trans alkene CH3 | CH3O− E CH3 − C − CH3 → CH3OH | Br C R Transition state D S E2 C2 H5OH F CH3 | CH3OH → CH3 − C − CH3 | Br Match the following column: A B C D Column-I E2 E1 E1cB SN2 P Q R S Column-II Carbanion Transition state Carbocation Bimolecular Match the following column: Column-I Answer Keys A–Q; B–P, S; C–R; D–Q Column-II OH A CH3 − CH − CH − CH3 → P | Br Hofmann B CH3CH − CH − CH3 | Br tert butoxide → Q Zaitsev product C CH3 − CH − CH − CH3 | F R E2 A–PQ; B–P, T; C–P, S; D–R A–S; B–R; C–S; D–R; E–P; F–Q A–Q, S; B–R; C–P; D–Q, S A–QR; B–PR; C–PR; D–S A–Q, R, S; B–P, R, S; C–Q, R, S; D–P, R, S Explanations CH3 | C H 5O − → S E1 D CH3 − CH − C − CH3 C H 5OH | Br A–Q; Primary alkyl halide undergoes E2 eliminate B–P, S; E1 and unimolecular C–R; E1cB due to deuterium exchange D–Q; Isotopic effect E2 A–P, Q; –COOH displaces Cl (neighbouring group participation Then OH once again attacks the intermediate Two SN2 reaction No change in configuration Halogen Derivatives of Hydrocarbons ■ 1.121 B–P, T; No bond around a symmetric carbon break C–P, S; SNi D–R; Inversion A–S; Primary alkyl halide without branching gives SN2 product if base is not large sized B–R; Primary alkyl halide large base gives E2 product C–S; Secondary alkyl halide with a weak base gives SN2 product D–R; Secondary alkyl halide gives elimination product with strong base E–P; Base is strong E1 F–Q; It is solvolysis Substrate is 3°—SN1 A–E2–Q1S, B–E1–R, C–E1cB–P, D–SN2–Q, S A–Q, R B–P, R C–P, R D–S A–Q, R, S B–P, R, S C–Q, R, S D–P, R, S Assertion-Reason Type Questions Direction: If assertion is correct and reason is also correct explanation for assertion—answer is (a) If assertion and reason both are correct and reason is not correct explanation for assertion—answer is (b) If assertion is correct, reason is not correct—answer is (c) If assertion is not correct, reason is correct—answer is (d) Assertion: Halobenzene undergoes nucleophilic substitution at room temperature Reason: It is stabilized by resonance Assertion: Optically active 2-iodobutane when treated with NaI/acetone becomes optically inactive Reason: It is converted to meso form Assertion: Benzonitrile is prepared by the action of KCN on bromobenzene Reason: Cyanide ion is soft nucleophile Assertion: p-nitrofluorobenzene is more reactive than p-bromonitrobenzene towards nucleophilic substitution Reason: It is an addition-elimination type of substitution Assertion: In this reac- tion, the configuration of substrate and product is same Reason: No inversion takes place in this reaction Assertion: is a better leaving group as well as better nucleophile Reason: is a strong base Assertion: Polar protic solvents are better for SN1 reaction Reason: They solvate the carbocation formed and stabilize it Assertion: 2-bromobutane gives but-2-ene when treated with alcoholic KOH Reason: But-2-ene is more stable than but-1-ene Assertion: 3-bromo propanol cannot form Grignard reagent Reason: Due to the presence of acidic hydrogen Grignard reagent is converted to alkane 10 Assertion: Rate of SN2 reaction decreases in polar protic solvents Reason: Polar protic solvents solvate nucleophile So, rate of attack of nucleophile decreases 11 Assertion: Addition of AgNO3 increases rate of SN1 reaction Reason: It accelerates formation of carbocation 12 Assertion: Dipole moment of CH3F is greater than that of CH3Cl Reason: F is more electronegative than Cl 13 Assertion: CH3 − CH − C − O − the 1, Ag 2O | || → 2, KOH (aq) Br O reaction is an example of neighbouring group participation Reason: The product has opposite configuration due to inversion 14 Assertion: CH3–CH2–O–CH2Br is more reactive than CH3–O–CH2–CH2Br towards SN2 reaction Reason: Transition state is stabilized by delocalization of electron cloud CH3 | 15 Assertion: CH3 − C C − CH3 is more reactive | || Cl O towards SN1 reaction Reason: Carbocation intermediate is destabilized 1.122 ■ Chapter Answer Keys (d) (c) (d) (b) (c) (c) (a) (a) (a) 10 (a) 11 (a) 12 (d) 13 (c) 14 (a) 15 (d) (c); Assertion is correct as I– being large, can donate the lone pair easily, as it is less tightly held by the nucleus Being a weak base it becomes a better leaving group also Reason is not correct (a); Due to solvation, the cation is stabilized and it cannot recombine with the leaving group Nucleophile can attack this planar carbocation Explanations (d); Halobenzene does not undergo nucleophilic substitution at room temperature It requires high temperature and pressure Assertion is wrong Reason is correct C–X bond acquires partial double bond character due to resonance (a); Both assertion and reason are correct and reason is correct explanation But-2-ene is Zaitsev product (a); Both are correct Reason is correct explanation 10 (a); Both are correct Reason is correct explanation 11 (a); Halogen in alkyl halide is precipitated as AgX which is insoluble So, carbocation is easily formed Nucleophilic attack becomes fast 12 (d); Dipole moment of CH3F is less than that of CH3Cl Reason is correct Assertion is not correct (c); I– from NaI displaces I present in substrate by SN2 reaction followed by inversion The product with opposite configuration may be reconverted to parent The process continues and a racemic mixture is formed after some time and it becomes inactive With a single asymmetric carbon—it does not have meso form Reason is wrong 13 (c); It is an example of neighbouring group participation − C − O − displaces, Br forming a three cen|| O tered cyclic intermediate It is attacks by –OH Another SN2 reaction takes place So, there is no change in configuration Reason is not correct 14 (a); Both assertion and reason are correct Reason is correct explanation 15 (d); Reason is correct Rate of SN1 decreases due to the presence of withdrawing group (d); Assertion is not correct as X in halobenzene cannot be displaced under normal conditions (b); F—increases the positive charge formed on the carbon holding it by exerting –I effect It is small and does not offer any steric repulsion, when nucleophile attacks the carbon holding it Reason is correct but does not explain how assertion is correct (c); Assertion is correct, the product is Both substrate and product have the same configuration ‘S’, even though there is inversion Cl the second priority group in the substrate, becomes first priority group in the product Reason is not correct Inversion took place, but no change in configuration Integer Type Questions The number of isomers with the formula C7H7Cl The number of isomers with the formula C4H9X including stereo isomers is The number of isomers dichlorocyclohexane can have including stereo isomers is 1-methyl cyclohexane is treated with NBS/ peroxide—the number of bromocyclohexanes which may be formed is Total number of isomers including stereo isomers existing with the formula C5H11Br is is treated with NBS The products formed are Halogen Derivatives of Hydrocarbons ■ 1.123 The number of monochloro derivatives formed including stereo isomers from 2-methyl pentane is The number of monochloro derivatives formed from 3-methyl pentane including stereo isomers is How many of the following produce iodoform? 10 In how many of the following reactions configuration does not change? 12 Number of products that can give with CH3O/CH3OH (a) (b) (c) (d) (e) 11 How many of the following give tertiary alcohol when treated with excess Grignard reagent? (a) Ketones (b) Esters of formic acid (c) Esters of other acids (d) Acid chlorides (e) Acid amides (f) Nitriles Answer Keys 8 8 10 11 11 12 1.124 ■ Chapter Explanations 4; 5; 8; (a) cis-1, 2-dichloro cyclohexane—optically inactive as it has plane of symmetry trans 1, 2-dichloro cyclohexane is active (b) cis 1, 3-dichlo cyclohexane is inactive (plane of symmetry) trans-1, 3-dichloro cyclohexane is active (c) cis and trans-1, 4-dichlorocyclohexanes are inactive cis has plane of symmetry both have no trans has centre of symmetry asymmetric carbon also } 6; 4; Halogen Derivatives of Hydrocarbons ■ 1.125 8; (c) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl (d) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl KI in acetone, undergoes SN2 reaction with each of P, Q, R and S The rates of reaction vary as 8; (a) P > Q > R > S (c) P > R > Q > S (b) S > P > R > Q (d) R > P > S > Q In the reaction The compounds A and B, respectively are: (a) benzenediazonium chloride and fluorobenzene (b) nitrobenzene and fluorobenzene (c) nitrobenzene and fluorobenzene (d) phenol and benzene 6; Consider the following bromides: The correct order of SN1, reactivity is: (a) C > B > A (b) A > B > C (c) B > C > A (d) B > A > C 10 4; (a), (c), (d), (e) 11 4; Ketones, esters of acids other than formic acid, acid chlorides, nitriles 12 Previous IIT Questions Alkyl and Aryl Halides In SN2 reactions, the correct order of reactivity for the following compounds: CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3) CCl is (a) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl (b) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl C2H5I and C6H5CH2I both are separately treated with dil HNO3, and AgNO3 Which gives a yellow precipitate? (a) C6H5I (b) C6H5CH2I (c) Both (a) and (b) (d) None of these Compound formed on heating chlorobenzene with chloral in presence of concentrated H2SO4 is (a) hexachloroethane (b) DDT (c) freon (d) gammexane Which of the following is not chiral? (a) 3-chloro-2-methylpentane (b) 2-chloropentane (c) 1-chloro-2-methylpentane (d) 1-chloropentane 1.126 ■ Chapter Alkyl halides react with lithium dialkyl copper to give: (a) alkenes (b) alkyl copper halides (c) alkanes (d) alkyl halides Elimination of HBr from 2-bromobutane gives: (a) equimolar mixture of 1- and 2-butene (b) predominantly 2-butene (c) predominantly 1-butene (d) predominantly 2-butyne 10 The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly (a) benzoyl chloride (b) m-chlorotoluene (c) benzyl chloride (d) o, p-chlorotoluenes 11 The reaction conditions leading to the best yield of C2H5Cl are: U.V light (a) C2 H6 (excess) + Cl → Dark → (b) C2 H6 + Cl Room temperature UV light (c) C2 H6 + Cl (excess) → UV light (d) C2 H6 + Cl → 12 n-propyl bromide on treatment with ethanolic KOH produces: (a) propane (b) propene (c) propyne (d) propanol 13 Chlorination of toluene in presence of light and heat followed by treatment with aqueous NaOH and acidification gives: (a) o-Cresol (b) p-Cresol (c) 2, 4-dihydroxytoluene (d) benzoic acid 14 1-chlorobutane on reaction with alcoholic KOH gives: (a) but-1-ene (b) butan-1-ol (c) but-2-ene (d) butan-2-ol 15 Butane nitrile may be prepared by heating: (a) propyl alcohol with KCN (b) butyl alcohol with KCN (c) butyl chloride with KCN (d) propyl chloride with KCN 16 The number of possible enantiomeric pairs that can be produced during monochlorination of 2-methyl butane is: (a) 2 (b) 3 (c) 4 (d) 1 17 In the reaction of p-chlorotoluene with KNH2 in NH3(l), the major product is (a) o-toluidine (b) m-toluidine (c) p-toluidine (d) p-chloroaniline Halogen Derivatives 18 Which of the following has highest nucleophilicity? (a) F– (b) OH (c) CH3 (d) NH 19 The number of isomers for the compound with the formula C2BrClIF is (a) 3 (b) 4 (c) 5 (d) 6 20 Identify the set of reagents/reaction conditions X and Y in the following set of transformations: X Y CH3 − CH − CH → Product →CH3 − CH − CH3 | | Br Br (a) X = dil aq NaOH, 20°C Y = HBr/acetic acid, 20°C (b) X = conc alcoholic NaOH 80°C Y = HBr/acetic acid, 20°C (c) X = dil aq NaOH, 20°C Y = Br2/CHCl3, 0°C (d) X = conc alcoholic NaOH, 80°C Y = Br2/ChCl2 20°C 21 A compound that gives a positive iodoform test is: (a) pentan-1-ol (b) pentan-2-one (c) pentan-3-one (d) pentane 22 The compound that will not give iodoform on treatment with I2/NaOH is: (a) acetone (b) ethanal (c) diethyl ketone (d) isopropyl alcohol 23 Benzyl chloride C6H5CH2Cl can be prepared from toluene by chlorination with: (a) SO2Cl2 (b) SOCl2 (c) Cl2 (d) NaOCl 24 A solution of 1-chloro-2-phenyl ethane in toluene racemises slowly in the presence of a small amount of SbCl5 due to the formation of: (a) carbanion (b) carbene (c) free radical (d) carbocation 25 Chlorobenzene is prepared by reacting aniline with: (a) hydrochloric acid (b) cyprous chloride (c) Cl2/AlCl3 (d) nitrous acid followed by heating with cuprous chloride Halogen Derivatives of Hydrocarbons ■ 1.127 26 How many structures of F are possible? + Br H CH3CH CH − CH3 → F → C4 H8 Br2 − H 2O CCl products | OH (a) 2 (c) 6 (b) 5 (d) 3 27 The major product of the following reaction is: 31 The products of reaction of alcoholic silver nitrite with ethyl bromide are (a) ethane (b) ethane (c) nitroethane (d) ethyl alcohol (e) ethyl nitrite 32 The order of reactivity of the following alkyl halides (for a given alkyl group) for SN2 reaction is (a) CF > RCl > RBr > RI (b) RF > RBr > RCl > RI (c) RCl > RBr > RF > RI (d) RI > RBr > RCl > RF CH ONa (a) (b) → produces 33 The reaction (CH )3 CBr mainly: (a) (CH3 )3 CONa + CH3 Br (b) (CH3)3COCH3 + NaBr (c) (CH3)3C = CH2 + CH3OH + NaBr (d) (CH3)3CCH2OH + NaBr 34 (CH3)3C MgCl on reaction with D2O gives: (a) (CH3)3CD (b) (CH3)3COD (c) (CD3)3CD (d) (CD3)3COD 35 Isobutyl magnesium bromide on reaction with absolute alcohol gives (a) (CH3)2CHCH2OH, CH3CH2MgBr (b) (CH3)2CHCH2CH2CH3+ (c) (CH3)3CH and CH3CH2OMgBr (c) (d) 28 Toluene when treated with Br2/Fe gives para bromotoluene as the major product because –CH3 group (multiple answer): (a) is para directions (b) is meta directing (c) activates the ring by hyper conjugation (d) deactivates the ring 29 Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to: (a) the formation of less stable carbocation (b) resonance stabilization (c) longer carbon-hydrogen bond (d) sp2 hybridized carbon attached to the halogen 30 The compounds used as refrigerant are (a) NH3 (b) CCl4 (c) CF4 (d) CF2Cl2 (d) (CH3)3CH, CH2 = CH2 36 The product formed in the reaction is: (b) (a) (c) (d) 1.128 ■ Chapter 37 Among the following, the order of leaving group ability is: I –OAC II –OMe III –OSO2Me IV OSO2CF3 (a) I > II > III > IV (b) IV > III > I > II (c) III > II > I > IV (d) II > III > IV > I 38 The molecule with highest dipole moment is (a) CH3Cl (b) CH2Cl2 (c) CHCl3 (d) CCl4 Answer Keys (c) (b) (a) (c) (b) (b) (d) (c) (b) 10 (d) 11 (a) 12 (b) 13 (d) 14 (a) 15 (d) 16 (a) 17 (b) 18 (c) 19 (d) 20 (b) 21 (b) 22 (c) 23 (c) 24 (d) 25 (d) 26 (d) 27 (b) 28 (a), (c) 29 (b), (d) 30 (a), (c) 31 (c), (e) 32 (d) 34 (a) 36 (d) 33 (c) 35 (c) 37 (b) 38 (a) Explanation (c); As the size or number of alkyl groups on the carbon holding halogen (leaving group) increases crowding increases Rear side attack of nucleophile on that carbon becomes slow Rate of SN2 reaction decreases (b); Presence of − C − group in β-position stabilizes || O transition state due to delocalization of electron cloud So, (S) is most reactive Q is a secondary alkyl halide and least reactive (a); (c); The reactivity of SN1 reaction depends on the stability of intermediate carbocation formed Greater the stability, more reactive, the parent compound is (c); Lithium dialkyl cuprate gives alkanes with alkyl halides (R)2 CuLi + R ′X → R − R ′ Alkane alc KOH → (b); CH3 − CH − CH CH3 | Br CH3 − CH = CH − CH3 but-2-ene The carbocation formed from option (b) is stabilised by both resonance and hyperconjugation A is primary alkyl halide–least reactive (b); Only benzyl iodide gives a yellow precipitate of AgI Halogen connected to benzene ring cannot be displaced (b); 10 (d); –CH3 in toluene is o, p-orienting So, o-, p-chlorotoluenes are formed 11 (a); When excess ethane is treated with Cl2 in presence of UV light—formation of polychloroethanes is avoided Only C2H5Cl is formed (d); 1-chloropentane does not have asymmetric carbon atom C2 H6 + Cl → CH3 CH UV light | Cl Halogen Derivatives of Hydrocarbons ■ 1.129 12 (b); 14 (a); alc KOH CH3 − CH − CH → CH3 − CH = CH ∆ | Br 13 (d); 15 (d); CH3CH CH + KCN → CH3CH CH CN | Cl 16 (a); Cl CH3 − CH− CH − CH3 → ± CH − CH− CH CH3 | | | CH3 Cl CH3 ± CH3 − CH− CH− CH3 | | CH3 Cl pairs of enantiomers are formed 17 (b); The reaction takes place via benzyne mechanism is more stable than which may be formed alternatively The –CH3 destabilises the second one due to +I effect exerted by it – 18 (c); Carbon in C H3 is larger than other atoms and less electronegative The lone pair of it is farther away from its nucleus and less tightly held by it and more easily released for donation 19 (d); 1.130 ■ Chapter 26 (d); 20 (b); 21 (b); I NaOH CH3COCH CH CH3 → CHI3 + CH3CH CH COOH Br2 → CH3CH CH− CH 2 isomers X (±) | | Br Br 1, 2-dibromobutane Br 22 (c); C2 H5 − CO − C2 H5 cannot give iodoform as it is not a methyl ketone Y → 23 (c); anti addition to cis isomer gives racemic mixture Br In presence of peroxide SO2Cl2 also gives this product 24 (d); Z → anti addition to trans isomer gives meso isomer Total isomers of C4H8Br2 is 25 (d); It is Sandmeyer reaction 27 (b); NO2 and F are in metaposition So, it cannot be addition—type of substitution and F cannot be displaced Br in side chain is displaced via SN2 reaction as solvent is non-polar (secondary alkyl halides undergo SN2 in non-polar solvent So, answer is Halogen Derivatives of Hydrocarbons ■ 1.131 28 (a), (c); –CH3 group activates the ring by hyperconjugation and it is o, p-orienting 32 (d); Reactivity of these compounds depends on the leaving group Iodides are more reactive as I– is better leaving group (weak base) Fluorides are less reactive as F– is not a good leaving group It is the conjugate base of weak acid HF So, it is a strong base 33 (c); Tertiary alkyl halides undergo elimination with strong base 34 (a); 35 (c); 29 (b), (d); Haloarenes are stabilized by resonance as they are connected to sp2 carbon bond acquires partial double bond character More energy is required to break it and displace it 30 (a), (c); Both ammonia and dichloro difluoro methane are used as refrigerants 31 (c), (e); AgNO CH3 − CH Br → CH3 − CH + CH3 − CH | | O− N = O NO Ethyl nitrile Nitro ethane 36 (d); Configuration at C2 does not change So, it cannot be (b) or (a) At C1 in the substrate where substitution takes place configuration is R So, it should be S So, it is option (d) (in C it is R) 37 (b); Conjugate acid of IV is CF3SO3H strongest acid Conjugate acid of III is CH3SO3H Conjugate acid of I is CH3COOH Conjugate acid of II is CH3OH So, basic strength of conjugate bases is II > I > III > IV A weak base is a better leaving group So, reactivity is IV > III > I > II 38 (a); As the number of Cl atoms is increasing—dipole moment decreases This page is intentionally left blank ... determining stage is proposed That is SN1 It is a two stage mechanism First the substrate ionises and forms an intermediate carbocation It is the rate determining stage As only substrate is involved, it... exits in four stereo isomeric forms cis and its mirror image, trans and its mirror image Halogen derivatives of cycloalkanes also exhibit optical isomerism Both cis and trans isomers of 1, 4-dichlorocyclohexane... is no H atom in antiperiplanar position to Cl On right hand side β-carbon—there is a H atom in antiposition to Cl So, elimination takes place on this side In its cis isomer there is a H-atom in