ALL CHAPTERS COVERED
SOLUTIONS MANUAL
,Suggested solutions for Chapter 2 2
PROB LE M 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 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 hydrocarbon but there are many branched and
cyclic options. We offer some possibilities, but you may have thought of
others.
linear saturated hydrocarbon (n-heptane)
some branched hydrocarbons
some cyclic hydrocarbons
We give you a few examples of keto-carboxylic acids based on these
structures. A ketone 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 Solutions Manual to accompany Organic Chemistry
linear molecules containing O
ketone and carboxylic acid CO2H
O CO2H
some branched keto-acids
CO2H O CO2 H HO 2C
HO 2C
O
O O
some cyclic keto-acids
O O
CO2H
CO2H
CO2H O
HO 2C O
PROB LE M 2
Draw for yourself the structures of amoxicillin and Tamiflu shown on page 10
of the textbook. Identify on your diagrams 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?
NH2 O
H H H
N S O
H3 C O CH3
O N
HO O H3C HN
CO2H Tamiflu (oseltamivir):
SmithKline Beechamʼs amoxycillin NH2
O invented by
-lactam antibiotic H3 C
Gilead Sciences;
for treatment of bacterial infections marketed by Roche
Purpose 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 carbon framework too.
Suggested solution
The functional groups shouldn’t 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
amine NH2 ether O ester
H H sulfide
O O
HN S
H3C CH3
O N H3C HN
HO O
amide
phenol or amide CO2H H3C O NH amine
alcohol carboxylic acid 2
amide
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.
O
NH2 five-
six-membered
H H H membered O O CH3
S H3C
N
N H3C HN
O
HO O NH six-membered
2
four-membered CO2H H3C O
The carbon chains are quite varied in length and style and are broken up by
N, O, and S atoms.
cyclic C3 O
cyclic C6 NH H H H linear C5 linear C2
2 O O CH3
N S
H3C
cyclic C6
O H3 C
H
N N
HO O linear C
linear C2 CO2H 2 NH2
branched C5 H3 C O
,4 Solutions Manual to accompany Organic Chemistry
PROB LE M 3
Identify the functional groups in these two molecules
O
O
O O Ph O O
OH
NH O
O O
O
OH the heart drug candoxatril a derivative
of the sugar ribose
Purpose of the problem
Identifying functional groups is not just a sterile exercise in classification:
spotting the difference between an ester, an ether, 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 functional group.
ether
ether
O ether
O hemiacetal
Ph O
O O O
amide OH
NH O
ester
O O O acetal
OH
carboxylic acid
, Solutions for Chapter 2 – Organic structures 5
PROB LE M 4
What is wrong with these structures? Suggest better ways to represent these
molecules
H O
H C C NH OH H
H H H Me H
H2C N CH2 H
NH2
H2C CH2
Purpose of the problem
To shock 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° bonds, 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.
O OH
N N
H
NH2
PROB LE M 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-triene
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-
di-tert-butyl benzene, an equally helpful name. There are two separate
numerical relationships.
4
3 1,4-relationship between
the 1,1-dimethyl 1 1 the two substituents
ethyl group 2 2 on the benzene ring
(b) This name conveys neither the simple symmetrical structure nor the fact
that it contains two allyl groups. Most chemists would call it ‘diallyl ether’
though it is sold as ‘allyl ether’.
the allyl group O the allyl group
(c) This is of course simply benzene!
PROB LE M 6
Translate these very poor structural descriptions into something more realistic.
Try to get the angles about right and, whatever you do, don’t 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 shape 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 straightforward—the
(OH) group is a substituent joined to the chain and not part of it. The
second compound must be cyclic—it is the 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
C6H5CH(OH)(CH2)4COC2H5 O(CH2CH2)2O (CH3O)2CH=CHCH(OMe)2
OH O OMe
OMe
MeO OMe
O
O
PROB LE M 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 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 between 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. ■ Why alkenes have the alcohol
oxidation level is explained on page
33 of the textbook.
hydrocarbon level
O O
H
O N Me O N
O O N
O O O
carboxylic alcohol
acid level level
, 8 Solutions Manual to accompany Organic Chemistry
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SOLUTIONS MANUAL
,Suggested solutions for Chapter 2 2
PROB LE M 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 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 hydrocarbon but there are many branched and
cyclic options. We offer some possibilities, but you may have thought of
others.
linear saturated hydrocarbon (n-heptane)
some branched hydrocarbons
some cyclic hydrocarbons
We give you a few examples of keto-carboxylic acids based on these
structures. A ketone 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 Solutions Manual to accompany Organic Chemistry
linear molecules containing O
ketone and carboxylic acid CO2H
O CO2H
some branched keto-acids
CO2H O CO2 H HO 2C
HO 2C
O
O O
some cyclic keto-acids
O O
CO2H
CO2H
CO2H O
HO 2C O
PROB LE M 2
Draw for yourself the structures of amoxicillin and Tamiflu shown on page 10
of the textbook. Identify on your diagrams 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?
NH2 O
H H H
N S O
H3 C O CH3
O N
HO O H3C HN
CO2H Tamiflu (oseltamivir):
SmithKline Beechamʼs amoxycillin NH2
O invented by
-lactam antibiotic H3 C
Gilead Sciences;
for treatment of bacterial infections marketed by Roche
Purpose 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 carbon framework too.
Suggested solution
The functional groups shouldn’t 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
amine NH2 ether O ester
H H sulfide
O O
HN S
H3C CH3
O N H3C HN
HO O
amide
phenol or amide CO2H H3C O NH amine
alcohol carboxylic acid 2
amide
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.
O
NH2 five-
six-membered
H H H membered O O CH3
S H3C
N
N H3C HN
O
HO O NH six-membered
2
four-membered CO2H H3C O
The carbon chains are quite varied in length and style and are broken up by
N, O, and S atoms.
cyclic C3 O
cyclic C6 NH H H H linear C5 linear C2
2 O O CH3
N S
H3C
cyclic C6
O H3 C
H
N N
HO O linear C
linear C2 CO2H 2 NH2
branched C5 H3 C O
,4 Solutions Manual to accompany Organic Chemistry
PROB LE M 3
Identify the functional groups in these two molecules
O
O
O O Ph O O
OH
NH O
O O
O
OH the heart drug candoxatril a derivative
of the sugar ribose
Purpose of the problem
Identifying functional groups is not just a sterile exercise in classification:
spotting the difference between an ester, an ether, 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 functional group.
ether
ether
O ether
O hemiacetal
Ph O
O O O
amide OH
NH O
ester
O O O acetal
OH
carboxylic acid
, Solutions for Chapter 2 – Organic structures 5
PROB LE M 4
What is wrong with these structures? Suggest better ways to represent these
molecules
H O
H C C NH OH H
H H H Me H
H2C N CH2 H
NH2
H2C CH2
Purpose of the problem
To shock 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° bonds, 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.
O OH
N N
H
NH2
PROB LE M 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-triene
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-
di-tert-butyl benzene, an equally helpful name. There are two separate
numerical relationships.
4
3 1,4-relationship between
the 1,1-dimethyl 1 1 the two substituents
ethyl group 2 2 on the benzene ring
(b) This name conveys neither the simple symmetrical structure nor the fact
that it contains two allyl groups. Most chemists would call it ‘diallyl ether’
though it is sold as ‘allyl ether’.
the allyl group O the allyl group
(c) This is of course simply benzene!
PROB LE M 6
Translate these very poor structural descriptions into something more realistic.
Try to get the angles about right and, whatever you do, don’t 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 shape 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 straightforward—the
(OH) group is a substituent joined to the chain and not part of it. The
second compound must be cyclic—it is the 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
C6H5CH(OH)(CH2)4COC2H5 O(CH2CH2)2O (CH3O)2CH=CHCH(OMe)2
OH O OMe
OMe
MeO OMe
O
O
PROB LE M 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 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 between 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. ■ Why alkenes have the alcohol
oxidation level is explained on page
33 of the textbook.
hydrocarbon level
O O
H
O N Me O N
O O N
O O O
carboxylic alcohol
acid level level
, 8 Solutions Manual to accompany Organic Chemistry
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
