Organic Chemistry 第二版

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1、Solutions manual to accompanyOrganic ChemistrySecond EditionJonathan Clayden, Nick Greeves, and Stuart WarrenJonathan ClaydenUniversity of ManchesterStuart WarrenUniversity of Cambridge11Great Clarendon Street, Oxford, OX2 6DP,United KingdomOxford University Press is a department of the University o

2、f Oxford.It furthers the Universitys objective of excellence in research, scholarship,and education by publishing worldwide. Oxford is a registered trade mark ofOxford University Press in the UK and in certain other countries Oxford University Press 2013The moral rights of the authors have been asse

3、rtedFirst edition published 2001Impression: 1All rights reserved. No part of this publication may be reproduced, stored ina retrieval system, or transmitted, in any form or by any means, without theprior permission in writing of Oxford University Press, or as expressly permittedby law, by licence or

4、 under terms agreed with the appropriate reprographicsrights organization. Enquiries concerning reproduction outside the scope of theabove should be sent to the Rights Department, Oxford University Press, at theaddress aboveYou must not circulate this work in any other formand you must impose this s

5、ame condition on any acquirerBritish Library Cataloguing in Publication DataData available9780199663347Printed in Great Britain by Ashford Colour Press Ltd, Gosport, HampshireLinks to third party websites are provided by Oxford in good faith andfor information only. Oxford disclaims any responsibili

6、ty for the materialscontained in any third party website referenced in this work.Suggested solutions for Chapter 2 PROBLEM 1 Draw good diagrams of saturated hydrocarbons with seven carbon atoms having (a) linear, (b) branched, and (c) cyclic structures. Draw molecules based on each framework having

7、both ketone and carboxylic acid functional groups in the same molecule. Purpose of the problem To get you drawing simple structures realistically and to steer you away from rules and names towards more creative and helpful ways of representing molecules. Suggested solution There is only one linear h

8、ydrocarbon but there are many branched and cyclic options. We offer some possibilities, but you may have thought of others. some branched hydrocarbonssome cyclic hydrocarbonslinear saturated hydrocarbon (n-heptane) We give you a few examples of keto-carboxylic acids based on these structures. A keto

9、ne has to have a carbonyl group not at the end of a chain; a carboxylic acid functional group by contrast has to be at the end of a chain. You will notice that no carboxylic acid based on the first three cyclic structures is possible without adding another carbon atom. 2 2 Solutions Manual to accomp

10、any Organic Chemistry linear molecules containingketone and carboxylic acidsome branched keto-acidssome cyclic keto-acidsCO2HOCO2HOCO2HCO2HOOCO2HOHO2COCO2HCO2HHO2CHO2COOO O PROBLEM 2 Draw for yourself the structures of amoxicillin and Tamiflu shown on page 10 of the textbook. Identify on your diagra

11、ms the functional groups present in each molecule and the ring sizes. Study the carbon framework: is there a single carbon chain or more than one? Are they linear, branched, or cyclic? for treatment of bacterial infectionsHOHNNSH HCO2HONH2OSmithKline Beechams amoxycillin-lactam antibioticOOONH2HNH3C

12、H3CH3C OCH3Tamiflu (oseltamivir):invented byGilead Sciences;marketed by RochePurpose of the problem To persuade you that functional groups are easy to identify even in complicated structures: an ester is an ester no matter what company it keeps and it can be helpful to look at the nature of the carb

13、on framework too. Suggested solution The functional groups shouldnt have given you any problem except perhaps for the sulfide (or thioether) and the phenol (or alcohol). You should have seen that both molecules have an amide as well as an amine. Solutions for Chapter 2 Organic structures 3 HOHNNSH H

14、CO2HONH2Ocarboxylic acidsulfideamideaminephenol oralcoholamideOOONH2HNH3CH3CH3C OCH3amideesteramineether The ring sizes are easy and we hope you noticed that one bond between the four- and the five-membered ring in the penicillin is shared by both rings. HOHNNSH HCO2HONH2OOOONH2HNH3CH3CH3C OCH3six-m

15、emberedsix-memberedfive-memberedfour-membered The carbon chains are quite varied in length and style and are broken up by N, O, and S atoms. HOHNNSH HCO2HONH2OOOONH2HNH3CH3CH3C OCH3cyclic C6cyclic C3linear C5linear C2branched C5linear C2cyclic C6linear C2 4 Solutions Manual to accompany Organic Chem

16、istry PROBLEM 3 Identify the functional groups in these two molecules NHO OOOOOHOOOOOHOthe heart drug candoxatril a derivativeof the sugar ribosePhPurpose of the problem Identifying functional groups is not just a sterile exercise in classification: spotting the difference between an ester, an ether

17、, an acetal and a hemiacetal is the first stage in understanding their chemistry. Suggested solution The functional groups are marked on the structures below. Particularly important is to identify an acetal and a hemiacetal, in which both ether-like oxygens are bonded to a single carbon, as a single

18、 functional group. NHO OOOOOHOOOOOHOPhcarboxylic acidamideesteretheretheretherhemiacetalacetal Solutions for Chapter 2 Organic structures 5 PROBLEM 4 What is wrong with these structures? Suggest better ways to represent these molecules H2CNCH2H2C CH2HHNHCOCHHHNH2Me HOHHHPurpose of the problem To sho

19、ck you with two dreadful structures and to try to convince you that well drawn realistic structures are more attractive to the eye as well as easier to understand and quicker to draw. Suggested solution The bond angles are grotesque with square planar saturated carbon atoms, bent alkynes with 120 bo

20、nds, linear alkenes with bonds at 90 or 180, bonds coming off a benzene ring at the wrong angles and so on. If properly drawn, the left hand structure will be clearer without the hydrogen atoms. Here are better structures for each compound but you can think of many other possibilities. N NHOOHNH2 PR

21、OBLEM 5 Draw structures for the compounds named systematically here. In each case suggest alternative names that might convey the structure more clearly if you were speaking to someone rather than writing. (a) 1,4-di-(1,1-dimethylethyl)benzene (b) 1-(prop-2-enyloxy)prop-2-ene (c) cyclohexa-1,3,5-tri

22、ene Purpose of the problem To help you appreciate the limitations of systematic names, the usefulness of part structures and, in the case of (c), to amuse. 6 Solutions Manual to accompany Organic Chemistry Suggested solution (a) A more helpful name would be para-di-t-butyl benzene. It is sold as 1,4

23、-di-tert-butyl benzene, an equally helpful name. There are two separate numerical relationships. 12341,4-relationship betweenthe two substituentson the benzene ringthe 1,1-dimethylethyl group12 (b) This name conveys neither the simple symmetrical structure nor the fact that it contains two allyl gro

24、ups. Most chemists would call it diallyl ether though it is sold as allyl ether. Othe allyl group the allyl group (c) This is of course simply benzene! PROBLEM 6 Translate these very poor structural descriptions into something more realistic. Try to get the angles about right and, whatever you do, d

25、ont include any square planar carbon atoms or any other bond angles of 90. (a) C6H5CH(OH)(CH2)4COC2H5 (b) O(CH2CH2)2O (c) (CH3O)2CH=CHCH(OCH3)2 Purpose of the problem An exercise in interpretation and composition. This sort of structure is sometimes used in printed text. It gives no clue to the shap

26、e of the molecule. Suggested solution You probably need a few trial and error drawings first but simply drawing out the carbon chain gives you a good start. The first is straightforwardthe (OH) group is a substituent joined to the chain and not part of it. The second compound must be cyclicit is the

27、 ether solvent commonly known as dioxane. The third gives no hint as to the shape of the alkene and we have chosen trans. It also has two ways of representing a methyl group. Either is fine, but it is better not to mix the two in one structure. Solutions for Chapter 2 Organic structures 7 OHOO(CH2CH

28、2)2OOOMeOOMeOMeOMe(CH3O)2CH=CHCH(OMe)2C6H5CH(OH)(CH2)4COC2H5 PROBLEM 7 Identify the oxidation level of all the carbon atoms of the compounds in problem 6. Purpose of the problem This important exercise is one you will get used to very quickly and, before long, do without thinking. If you do it will

29、save you from many trivial errors. Remember that the oxidation state of all the carbon atoms is +4 or C(IV). The oxidation level of a carbon atom tells you to which oxygen-based functional group it can be converted without oxidation or reduction. Suggested solution Just count the number of bonds bet

30、ween the carbon atom and heteroatoms (atoms which are not H or C). If none, the atom is at the hydrocarbon level ( ), if one, the alcohol level ( ), if two the aldehyde or ketone level, if three the carboxylic acid level ( ) and, if four, the carbon dioxide level. O NOOOcarboxylicacid levelhydrocarb

31、on levelalcohollevelMeOOOOHNNO Why alkenes have the alcohol oxidation level is explained on page 33 of the textbook. 8 Solutions Manual to accompany Organic Chemistry PROBLEM 8 Draw full structures for these compounds, displaying the hydrocarbon framework clearly and showing all the bonds in the fun

32、ctional groups. Name the functional groups. (a) AcO(CH2)3NO2 (b) MeO2CCH2OCOEt (c) CH2=CHCONH(CH2)2CN Purpose of the problem This problem extends the purpose of problem 6 as more thought is needed and you need to check your knowledge of the organic elements such as Ac. Suggested solution For once th

33、e solution can be simply stated as no variation is possible. In the first structure AcO represents an acetate ester and that the nitro group can have only four bonds (not five) to N. The second has two ester groups on the central carbon, but one is joined to it by a CO and the other by a CC bond. Th

34、e last is straightforward. AcO(CH2)3NO2O NOOOesternitroMeO2CCH2OCOEtMeOOOOesteresterCH2=CHCONH(CH2)2CNHNNOnitrilealkene amide PROBLEM 9 Draw structures for the folllowing molecules, and then show them again using at least one organic element symbol in each. (a) ethyl acetate (b) chloromethyl methyl

35、ether (c) pentanenitrile (d) N-acetyl p-aminophenol (e) 2,4,6,-tri-(1,1-dimethylethyl)phenylamine Purpose of the problem Compound names mean nothing unless you can visualize their structures. More practice using organic elements. Solutions for Chapter 2 Organic structures 9 Suggested solution The st

36、ructures are shown belowthings to look out for are the difference between acetyl and acetate, the fact that the carbon atom of the nitrile group is included in the name, and the way that a tert-butyl group can be named as 1,1-dimethylethyl. OOEtOAcethyl acetateCl Ochloromethyl methyl etherCl OMeCNBu

37、CNpentanenitrileHOHNOHONHAcN-acetyl p-aminophenol2,4,6,-tri-(1,1-dimethylethyl)phenylamineNH2t-Bu t-BuNH2t-Bu PROBLEM 10 Suggest at least six different structures that would fit the formula C4H7NO. Make good realistic diagrams of each one and say which functional groups are present. Purpose of the p

38、roblem The identification and naming of functional groups is more important than the naming of compounds, because the names of functional groups tell you about their chemistry. This was your chance to experiment with different groups and different carbon skeletons and to experience the large number

39、of compounds you could make from a formula with few atoms. Suggested solution We give twelve possible structures there are of course many more. You need not have used the names in brackets as they are ones more experienced chemists might use. There is a list of the abbreviations known as organic ele

40、ments on page 42 of the textbook. 10 Solutions Manual to accompany Organic Chemistry HONH2HNOH2N MeOONHNHON OH O NH2 NHOMeOHOMe NH2NNOONMealkyne, primary amineprimary alcohol(cyclic) amide(lactam)ketone, alkene, primary amine (enamine)ether, alkenesecondary amine(cyclic) tertiary aminealdehydealkene

41、, amine, alcohol(cyclic hydroxylamine)(cyclic) ketoneprimary amineoximeimine and alcoholether, nitrile primary alcohol,nitrileimine, ether(isoxazoline)alkene, primary amide Suggested solutions for Chapter 3 PROBLEM 1 Assuming that the molecular ion is the base peak (100% abundance) what peaks would

42、appear in the mass spectrum of each of these molecules: (a) C2H5BrO (b) C60 (c) C6H4BrCl In cases (a) and (c) suggest a possible structure of the molecule. What is (b)? Purpose of the problem To give you some practice with mass spectra and, in particular, at interpreting isotopic peaks. The molecula

43、r ion is the most important ion in the spectrum and often the only one that interests us. Suggested solution Bromine has two isotopes, 79Br and 81Br in about a 1:1 ratio. Chlorine has two isotopes 35Cl and 37Cl in about a 3:1 ratio. There is about 1.1% 13C in normal compounds. (a) C2H5BrO will have

44、two main molecular ions at 124 and 126. There will be very small (2.2%) peaks at 125 and 126 from the 1.1% of 13C at each carbon atom. (b) C60 has a molecular ion at 720 with a strong peak at 721 of 60 x 1.1 = 66%, more than half as strong as the 12C peak at 720. This compound is buckminsterfulleren

45、e. (c) This compound is more complicated. It will have a 1:1 ratio of 79Br and 81Br and a 3:1 ratio of 35Cl and 37Cl in the molecular ion. There are four peaks from these isotopes (ratios in brackets) C6H479Br35Cl (3), C6H481Br35Cl (3), C6H479Br37Cl (1), and C6H481Br37Cl (1), the masses of these pea

46、ks being 190, 192, 192, and 194. So the complete molecular ion will have three main peaks at 190, 192, and 194 in a ratio of 3:4:1 with peaks at 191, 193, and 194 at 6.6% of the peak before it. Compounds (a) and (c) might be isomers of compounds such as these: BrOHBrClBrClBrCl 3 Buckminsterfullerene

47、 is on page 25 of the textbook. 12 Solutions Manual to accompany Organic Chemistry PROBLEM 2 Ethyl benzoate PhCO2Et has these peaks in its 13C NMR spectrum: 17.3, 61.1, 100150 (four peaks) and 166.8 ppm. Which peak belongs to which carbon atom? You are advised to make a good drawing of the molecule

48、before you answer. Purpose of the problem To familiarize you with the four regions of the spectrum. Suggested solution It isnt possible to say which aromatic carbon is which and it doesnt matter. The rest are straightforward. OO 100150 four types of aromatic carbon: 166.8carbonyl 17.3saturated carbo

49、nnot next to oxygen 61.1saturated carbonnext to oxygenipsoorthometaparaorthometa PROBLEM 3 Methoxatin was mentioned on page 44 of the textbook where we said it proved exceptionally difficult to solve the structure by NMR. Why is it so difficult? Could anything be gained from the 13C or 1H NMR? What

50、information could be gained from the mass spectrum and the infra red? Purpose of the problem To convince you that this structure really needs an X-ray solution but also to get you to think about what information is available by the other methods. Certainly mass spectroscopy, NMR, and IR would have b