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Transcript
10 Introduction to organic chemistry Answers to end-of-chapter questions Page 195 Questions 1 a) CH 3 CH(OH)CH 2 Br is called 1-bromopropan-2-ol. b) CH 2 ClCH 2 COOH is called 3-chloropropanoic acid. [e] There are three carbon atoms in the chain, so the stem name is prop-. c) CH 2 =CHC(CH 3 ) 3 is called 3,3-dimethylbut-1-ene. [e] There are four carbon atoms in the chain, as can be seen from the skeletal formula: 2 a) The formula of 1,2-dichloro-1,2-difluoroethene is F(Cl)C=C(Cl)F. b) The formula of 1-hydroxybutanone is CH 3 CH 2 COCH 2 OH. c) The formula of 2-amino-3-chloropropanoic acid is CH 2 ClCH(NH 2 )COOH. 3 a) The functional groups in CH 2 OHCOCH(NH 2 )COOH are: • –OH (alcohol) • C=O (in a ketone) • –NH 2 (amino) • –COOH (carboxylic acid) b) The functional groups in CH 2 =CHCH(OH)CHO are: • C=C (alkene) • –OH (alcohol) • C=O (in an aldehyde) 4 1,1-dibromopropane 1,2-dibromopropane 1,3-dibromopropane © Hodder & Stoughton Limited 2015 10 Introduction to organic chemistry Answers to end-of-chapter questions 2,2-dibromopropane [e] 2,3-dibromopropane is identical to 1,2-dibromopropane. 5 There are three pentene isomers: • Pent-1-ene CH 2 =CHCH 2 CH 2 CH 3 • (Z)-pent-2-ene cis-pent-2-ene • (E)-pent-2-ene trans-pent-2-ene and there are three butene isomers: • 3-methylbut-1-ene CH 2 =CHCH(CH 3 )CH 3 • 2-methylbut-1-ene CH 2 =C(CH 3 )CH 2 CH 3 • 2-methylbut-2-ene (CH 3 ) 2 C=CHCH 3 [e] A chain of three is impossible because to achieve this the central carbon atom would have to have five bonds (one double and three single). 6 The major organic product is hexachloroethane, CCl 3 CCl 3 . δ+ 7 The electrophile is the I in ICl and the mechanism is: 8 The purple colour of the potassium manganate(VII) solution would be replaced by a brown precipitate. The organic product is butane-2,3-diol, CH 3 CH(OH)CH(OH)CH 3 . © Hodder & Stoughton Limited 2015 10 Introduction to organic chemistry Answers to end-of-chapter questions [e] Do not forget to state the original colour as well as the final appearance. The brown precipitate is manganese(IV) oxide, MnO 2 . 9 In the reaction between ethene and bromine, bromine attacks the electron-rich π-bond. In ethane, all the bonds are σ-bonds, so there is no centre of high-electron density. The propagation step of the photochemical substitution reaction with ethane involves the reaction of a bromine radical with an ethane molecule. This is a slow reaction. It is energetically unfavourable because of the relatively weak C–Br bond that is formed. [e] The mechanism of the reaction between ethene and bromine is electrophilic addition; between ethane and bromine it is free-radical substitution. 10 a) b) The rotation means that the π-bond has to break and this requires energy which is equal to that of a photon of visible light. 11 There are not two identical groups, one on each C atom. Thus the H is cis to the CH 3 group but trans to the C 2 H 5 group, so cis/trans naming will not work. The CH 2 OHCH 2 group has a higher priority than H and C 2 H 5 has a higher priority to CH 3 . If the two higher priority groups are on the same side of the double bond it is the Z- isomer and if on opposite sides it is the E- isomer. Pages 196–197 Exam practice questions 1 a) B () b) But-2-ene exhibits geometric isomerism whereas but-1-ene does not (printed incorrectly in the first printing of this book). The two CH 3 groups in but-2-ene can be either on the same side of the double bond or on opposite sides (), but but-1-ene has two H atoms on one of the double-bonded carbon atoms (). c) D () d) C () e) i) An electrophile is a species that accepts a pair of electrons () from an electron-rich site in another species (). © Hodder & Stoughton Limited 2015 10 Introduction to organic chemistry ii) B () iii) A () 2 a) i) ii) b) i) Answers to end-of-chapter questions Free radical substitution () Electrophilic addition () A homologous series is a series of compounds with the same functional group (), the same general formula () and where one member differs from the next by CH 2 (). ii) CH 3 CH 2 CH=CH 2 + Br 2 → CH 3 CH 2 CHBrCH 2 Br () iii) iv) CH 3 CH 2 CH(OH)CH 2 Br () v) The purple solution () forms a brown precipitate () (of MnO 2 ). The product is CH 2 CH 2 CH(OH)CH 2 (OH) () vi) 3 a) b) The intermediate cation () is stabilised by the electron pushing /inductive effect of the CH 3 group (). + 4 a) The intermediate cation CH 2 BrC H 2 can be attacked by a lone pair of electrons () on either − the O of H 2 O (allow from the O of OH ) forming CH 2 BrCH 2 OH () or from the lone pair on Br − forming CH 2 BrCH 2 Br (). © Hodder & Stoughton Limited 2015 10 Introduction to organic chemistry Answers to end-of-chapter questions + + b) The first step could be the addition of H forming the intermediate CH 3 C H 2 (). This is then + followed by the addition of H 2 O and the subsequent loss of H forming ethanol () or the − addition of Br forming bromoethane (). 5 a) b) Once the HC=CH bond has broken (), the σ-bond can rotate and so cis-/trans- isomerism is no longer possible () c) One possible mechanism is: 6 a) Bonds broken Bonds made Cl–Cl +243 H–Cl −432 C–H +435 () C–Cl −346 () Total +678 – 778 –1 ∆H = +678 – 778 = –100 kJ mol () b) i) Homolytic fission is when the bonding pair of electrons () gives one electron to each atom in the bond, forming radicals (). ii) iii) © Hodder & Stoughton Limited 2015 10 Introduction to organic chemistry c) i) ii) Answers to end-of-chapter questions The question is about the reaction between propene and hydrogen bromide. + The secondary carbocation, CH 3 C HCH 3 (), is stabilised by the electron pushing effect of the two neighbouring CH 3 groups (). The minor product would require the primary + carbocation, CH 3 CH 2 C H 2 , to be formed and this is less stabilised (). d) Iodine is not electronegative enough () to draw the π electrons towards itself (). © Hodder & Stoughton Limited 2015