10 Introduction to organic chemistry

Transcription

1 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

2 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. 7 The electrophile is the I δ+ 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.

3 [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 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 ( ).

4 ii) B ( ) iii) A ( ) 2 a) i) Free radical substitution ( ) ii) Electrophilic addition ( ) b) i) 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 ( ).

5 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 H = = 100 kj mol 1 ( ) b) i) Homolytic fission is when the bonding pair of electrons ( ) gives one electron to each ii) atom in the bond, forming radicals ( ). iii)

6 c) i) The question is about the reaction between propene and hydrogen bromide. ii) 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 ( ).