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10 Introduction to
organic chemistry
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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
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10 Introduction to
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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 .
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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 ().
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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 ().
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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)
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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 ().
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