Chemistry, Canadian Edition
Chapter 10: Student Study Guide
Chapter 10: Organic Chemistry
Learning Objectives
Upon completion of this chapter you should be able to:
• draw and name hydrocarbons using the IUPAC system
• draw and name aromatic compounds using the IUPAC system
• draw and name compounds containing common functional groups
• recognize, name, and draw stereoisomers
• predict the products from SN1and SN2reactions
• write the mechanisms of SN1 and SN2 reactions
• predict the products from E2 and E1 reactions
• write the mechanisms of E2 and E1 reactions
• predict the products of addition reactions
• write the mechanisms of electrophilic addition of hydrogen halides, water, or halogens to alkenes
Practical Aspects
This chapter focuses on substances based on the element carbon. Structures and names of organic
compounds are explored, starting from the simpler hydrocarbons through to molecules that contain atoms
other than only carbon and hydrogen. Stereochemistry is concerned with the spatial arrangement of atoms
in molecules, and these subtleties require special drawing and naming techniques. Three classes of
reactions in organic chemistry are explored: substitution, elimination, and addition.
10.1 HYDROCARBONS
Skills to Master:
 Naming alkanes and alkyl groups.
 Naming branched chain alkanes.
 Naming cylcoalkanes, alkenes, and alkynes.
 Naming other substituents.
Key Concept:
 The IUPAC system of nomenclature gives each compound a unique name.
Key Terms:
 Hydrocarbon – compound that contains only carbon and hydrogen atoms.
 Alkane – hydrocarbon with only single bonds between the carbon atoms.
 Alkene – hydrocarbon with at least one carbon-carbon double bond.
 Alkyne – hydrocarbon with at least one carbon-carbon triple bond.
 Alkyl group – hydrocarbon fragment attached onto a larger hydrocarbon chain.
Helpful Hint
 Reviewing the line-bond drawing technique described in chapter 1 will be helpful.
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Naming Hydrocarbons
Following is a brief summary of rules for naming hydrocarbons. You must also learn the terms in tables
10-1 and 10-2 in the text.
Naming Alkanes
1. Find and name the longest continuous carbon chain.
2. Identify and name groups attached to this chain.
3. Number the chain consecutively, starting at the end nearest a substituent group.
4. Designate the location of each substituent group by an appropriate number and name.
5. Assemble the name, listing groups in alphabetical order. The prefixes di-, tri-, tetra- etc., used to
designate several groups of the same kind, are not considered when alphabetizing.
Naming Cycloalkanes
1. For a monosubstituted cycloalkane the ring supplies the root name (table above) and the substituent
group is named as usual. A location number is unnecessary.
2. If the alkyl substituent has more carbons, the ring is named as a substituent group on an alkane.
3. If two different substituents are present on the ring, they are listed in alphabetical order, and the first
cited substituent is assigned to carbon #1. The numbering of ring carbons then continues in a direction
(clockwise or counter-clockwise) that affords the second substituent the lower possible location number.
4. If several substituents are present on the ring, they are listed in alphabetical order. Location numbers
are assigned to the substituents so that one of them is at carbon #1 and the other locations have the lowest
possible numbers, counting in either a clockwise or counter-clockwise direction.
5. The name is assembled, listing groups in alphabetical order and giving each group (if there are two or
more) a location number. The prefixes di, tri, tetra etc., used to designate several groups of the same kind,
are not considered when alphabetizing.
Naming Alkenes (similar logic applies to Alkynes)
1. The ene suffix (ending) indicates an alkene or cycloalkene.
2. The longest chain chosen for the root name must include both carbon atoms of the double bond.
3. The root chain must be numbered from the end nearest a double bond carbon atom. If the double bond
is in the center of the chain, the nearest substituent rule is used to determine the end where numbering
starts.
4. The smaller of the two numbers designating the carbon atoms of the double bond is used as the double
bond locator. If more than one double bond is present the compound is named as a diene, triene or
equivalent prefix indicating the number of double bonds, and each double bond is assigned a locator
number.
5. In cycloalkenes the double bond carbons are assigned ring locations #1 and #2. Which of the two is #1
may be determined by the nearest substituent rule.
Other Substituents
Atoms other than carbon and hydrogen are commonly part of organic compounds. For example, if
halogens are attached, their substituent names are as follows:
-F
-Cl
-Br
-I
fluoro
chloro
bromo
iodo
These are incorporated into names in the same manner as alkyl groups.
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EXERCISE 1: Draw a structure for each of the following compounds: a) 2,3-dimethylpentane,
b) 1-ethyl-2-methylcylohexane, c) 1-chloro-2-butene.
STRATEGY: Deconstruct the parts of the name and represent each part in the structure. It might be
useful to start looking at the right of the name and work left, adding parts as you go.
For part a):
1.
2.
3.
4.
-ane: all single C-C bonds
–pent-: five carbons
–dimethyl-: there are two (di), one carbon (meth), substituents (yl)
2,3-: the substituents are on carbons 2 and 3
SOLUTION:
For part b):
1.
2.
3.
4.
5.
-ane: all single C-C bonds
–hex-: six carbons
–cyclo-: the six carbons are in a ring
-2-methyl-: a one-carbon substituent on carbon 2
-1-ethyl-: a two-carbon substituent on carbon 1
SOLUTION:
(Yours may be rotated or flipped relative to this one. It doesn’t matter.)
For part c):
1.
2.
3.
4.
-ene: a C-C double bond
–but-: four carbons
2-: the double bond is between carbons 2 and 3
1-chloro-: a chlorine substituent is on carbon 1
SOLUTION:
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EXERCISE 2: Provide a name for each of the following compounds:
a)
b)
STRATEGY: Construct the parts of the name that represent each part in the structure. Again, it might be
useful to start working at the right of the name and continue left, adding parts as you go.
For part a):
1.
2.
3.
4.
5.
there is a double bond so the ending will be “ene”
there are six “continuous” carbons so the “root” will be “hex”
the six carbons are in a ring, so “cyclo” will be added
there is a one-carbon substituent; “methyl”
for cyclic compounds, the double bond must be between carbons 1 and 2. Keeping the
carbon number for the methyl group as small as possible puts it on carbon “3”
SOLUTION: The name of this compound is 3-methylcyclohexene.
For part b):
1.
2.
3.
4.
all C-C bonds are single so the ending will be “ane”
there are four “continuous” carbons so the “root” will be “but”
there are two fluorine atoms; “difluoro”
numbering the chain so as to have the first substituent on the lower number we would
start from the right of the compound as it is drawn above
5. both substituents are on carbon 2 therefore we specify “2,2”
SOLUTION: The name of this compound is 2,2-difluorobutane
Try It #1: Supply the name or structure as required.
a)
b) 2-Chloro-3-methyl-1-pentene
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10.2 AROMATIC COMPOUNDS
Skills to Master:
 Naming benzene compounds using numbers or the o-, m-, p- designations to denote the positions
of substituents.
Key Concepts:
 Parts of molecules with six carbons in a ring with alternating single and double bonds possess special
structural and chemical properties.
 Special naming aspects may be used for benzene compounds
Naming Benzene Compounds
Benzene compounds can be named using the same principles of naming cylcoalkanes, that is, the carbons
of the ring are assigned numbers for the purposes of identifying the locations of substituents.
Alternatively, the prefixes ortho-, meta- and para- (abbreviated o-, m-, p-) can be used to denote the
1,2 disubstituted, the 1,3 disubstituted, and the 1,4 disubstituted cases, respectively.
If the benzene ring is attached to some other group of carbon atoms, the benzene becomes a substituent
called phenyl.
EXERCISE 3: Provide the name (using both naming styles if possible) or structure as required.
a)
b)
c) 1-bromo-3-propylbenzene
STRATEGY:
For part a):
1) The benzene ring contains the only carbon atoms, so it is not a substituent.
2) There are two fluorine atoms; “difluoro”
3) A carbon with one of the fluorine substituents is designated as carbon “1”. Relative to this,
the other is on carbon “4”; “1,4-“
4) The relative positions of the fluorine atoms could also be described as “para”.
SOLUTION: The name of this molecule is 1,4-difluorobenzene, or, p-difluorobenzene.
For part b):
1) The benzene can be considered a phenyl substituent on the four carbon chain.
2) The four carbon chain has all single bonds; “butane”.
3) Numbering the carbons from one end and choosing the end so as to put the substituent on the
lowest number, generates a “2” for the carbon bearing the phenyl substituent.
SOLUTION: The name of this molecule is 2-phenylbutane.
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For part c):
1) The name indicates that there is a benzene ring with a bromine atom on carbon “1”, and a 3carbon substituent (propyl) on carbon “3”.
SOLUTION:
Try It #2: Supply the name or structure as required.
a)
b) 1,2,3-trimethylbenzene
10.3 FUNCTIONAL GROUPS
Skills to Master:
 Naming and drawing the following classes of organic compounds: alcohols, ethers, amines,
aldehydes, ketones, carboxylic acids, esters, and amides
Key Concepts:
 When atoms of oxygen or nitrogen are added to organic molecules, the resulting compounds have
properties very different from the simple hydrocarbons.
 There are common ways in which oxygen and nitrogen atoms occur in organic molecules. These
common atom groupings are called functional groups.
 The naming of these compounds is based on the identity of the functional group.
Helpful Hint
 Table 10-3 in the text summarizes the functional group structures and provides named examples. It is
critical to be able to accurately recognize the functional groups.
Naming Alcohols
Alcohols are named following a procedure similar to that for branched chain alkanes. First, select the
longest continuous carbon chain to which the hydroxyl is directly attached. Change the name of the
alkane corresponding to this chain by dropping the final “e” and adding the suffix -ol. Then number the
longest continuous carbon chain so as to give the carbon atom bearing the hydroxyl group the lower
number. Indicate the positions of other substituents (as prefixes) by using the numbers corresponding
to their positions along the carbon chain. Exercise 4 demonstrates this procedure.
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EXERCISE 4: Provide a name or structure as required.
a)
b) cyclohexanol
STRATEGY:
For part a):
1. The longest continuous carbon chain that contains the –OH contains 4 carbons. The
hydrocarbon would be butane, so the alcohol is butanol.
2. There are two methyl substituents; “dimethyl”.
3. Numbering the carbons from one end and choosing the end so as to put the hydroxyl on the
lowest number, generates a “1” for the –OH and a “2” and “3” for the carbons bearing the
methyl substituents.
SOLUTION: The name of this molecule is 2,3-dimethyl-1-butanol.
For part b):
1. –ol: an alcohol.
2. cyclohexan-: six carbons in a ring with all single bonds
SOLUTION:
Naming Ethers and Amines
One way of naming ethers is to name the two alkyl groups attached to the oxygen in alphabetical order
and attach the word ether. If the two alkyl groups are the same, we use the prefix di-, as in dimethyl ether.
Similarly, amines can be named by naming the groups attached to the nitrogen atom (in alphabetical
order), and adding amine. Exercise 5 provides examples.
EXERCISE 5: Provide a name or structure as required.
a)
b) methylethylamine
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STRATEGY:
For part a):
1. We recognize the characteristic structure of an ether.
2. The benzene rings here are substituents and there are two of them; “diphenyl”.
SOLUTION: The name of this molecule is diphenyl ether.
For part b):
1. –amine: a nitrogen atom with alkyl substituents.
2. methylethyl-: a one-carbon and a two- carbon substituent.
SOLUTION:
Naming Aldehydes and Ketones
Aldehydes are named by naming the corresponding alkane, dropping the final “e” and adding –al. The
carbonyl carbon (the one with the double bonded oxygen attached) is number “1” by convention, so it is
not necessary to specify. With ketones, we take the name of the alkane, drop the “e” and add –one. Here it
may be necessary to specify the position of the carbonyl carbon. Exercise 6 applies these procedures.
EXERCISE 6: Provide a name for each of the following.
a)
b)
STRATEGY:
For part a):
1. The carbon bearing the carbonyl group is not at the beginning of a chain, so this is a ketone.
We must number the chain to have the carbonyl carbon with the lowest number. From either
end, this will put the carbonyl on carbon “3”. The name ending will be “3-one”.
2. There are five carbons in the continuous chain, and all the bonds are single; “pentan-”
3. There is a “chloro” and a “methyl” substituent.
4. Numbering the carbons from one end and choosing the end so as to have the first substituent
on the lowest number yields two possibilities (both put the substituent on carbon “2”). We
choose the option so the alphabetically lower substituent has the lower number in this case.
SOLUTION: The name of this molecule is 2-chloro-4-methylpentan-3-one.
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For part b):
1. The carbon bearing the carbonyl group is at the beginning of a chain, so this is an aldehyde
and the name will end with “-al”. The carbonyl carbon is carbon “1”.
2. There are four carbons and all single bonds; “butan-”
3. There is a “bromo” substituent on carbon “3”.
SOLUTION: The name of this molecule is 3-bromobutanal.
Naming Carboxylic Acids, Esters, and Amides
Naming carboxylic acids consists of naming the corresponding alkane, dropping the –ane and adding –oic
acid. Esters are named by first naming the group that is part of the alkoxyl group (singly bonded to the
oxygen atom), and then the carboxylic acid part of the molecule is named according to the number of
carbons in it, and adding –ate. Amides are named in a fashion similar to esters, but N- and N-,N- are used
to indicate substituents that are attached to the nitrogen atom.
EXERCISE 7: Provide a name or structure as required.
a)
b)
c) N-methyl-N-ethylbutanamide
STRATEGY:
For part a):
1. We recognize the carboxylic acid functional group so the name will end in “–oic acid”.
Carbon “1” is the one bearing the oxygen atoms.
2. There are three carbons all with single bonds; “propan-”
3. There are three fluorine atoms, all on carbon “3”; “3,3,3-trifluoro-”
SOLUTION: The name of this molecule is 3,3,3-trifluoropropanoic acid.
For part b):
1. The alkoxy group has six carbons in a ring; “cyclohexyl”
2. There are two carbons in the carboxylic acid part; “ethanoate”
SOLUTION: The name of this molecule is cyclohexyl ethanoate.
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For part c):
1. N-methyl-N-ethyl-: A one-carbon and a two-carbon substituent attached onto the nitrogen
atom.
2. –butanamide: a four-carbon chain with all single bonds constitutes the “carboxylic acid”
part.
SOLUTION:
Try It #3: Supply a name for each of the following compounds.
a)
b)
10.4 STEREOCHEMISTRY
Skills to Master:
 Naming cis and trans isomers using the E/Z system
 Drawing structures with the wedged and dashed bonds
 Recognizing enantiomers and diastereomers
 Determining from the line structure whether a molecule is chiral or achiral
 Identifying R and S enantiomers
Key Concepts:
 Stereoisomers are molecules that differ only in the spatial orientation of the atoms.
 Enantiomers are stereoisomers that are non-superimposable mirror images of each other. Typically a
molecule with a carbon bonded to four different groups exists as an enantiomeric pair.
 Diastereomers are stereoisomers that are not enantiomers. Typically a molecule with a double bond,
in which there are two different substituents bonded to each of the carbons of the double bond, exists
as a diastereomeric (cis/trans or E/Z) pair.
 Special naming elements are required to describe stereoisomers.
Helpful Hint
 Learn the Cahn, Ingold, and Prelog Priority Rules presented on page 564 of the text.
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The E/Z System for Naming Diastereomeric (cis/trans) Compounds
In certain cases, molecules with a double bond can have two isomeric forms. To differentiate the two
versions, an additional naming component is required: the E/Z system. To determine if a molecule is
labelled E or Z, consider the two carbon atoms participating in the double bond. Use the priority rules and
assign a priority for the two groups bonded to one of the carbon atoms. Do the same for the other. Then
observe if the higher priority group on the first carbon is on the same side (or opposite side) of the double
bond as the higher priority group of the second carbon. If they are on the same side the molecules is Z
(Zee Zame Zide!).If they are on opposite sides the molecule is E (Epposite!).
EXERCISE 8: Assign E or Z configurations to the following molecules.
a)
b)
STRATEGY:
For part a):
1. Considering the carbon of the double bond that is on the left: the two atoms that are bonded
to it are chlorine and carbon. Chlorine has a higher atomic number so it has higher priority.
2. Considering the carbon of the double bond that is on the right: the two atoms that are bonded
to it are both carbons, and we must proceed further along to find differentiation. The lower
chain comes to an oxygen atom which has higher priority than the hydrogens of the upper
chain. The lower chain has priority.
3. The two high priority groups are on opposite side of the double bond.
SOLUTION: This molecule is E.
For part b):
1. Considering the carbon of the double bond that is on the right: The two groups bonded to this
carbon are identical, and no priority assignment can be made. Indeed, this molecule does not
have a stereoisomer!
SOLUTION: This molecule is neither E nor Z as there is no stereoisomer of this compound.
The R/S System for Naming Chiral Compounds
When a molecule has a carbon atom with four different groups attached to it, the molecule typically has
two stereoisomeric forms that are mirror images of each other. These two isomers are called enantiomers;
one will have R configuration, one will have S configuration. The three step process for assigning R or S
is outline in the text on page 569. Consider an alternative method. Assign priorities to the four groups of
the chiral centre. Imagine the thumb of your hand along the bond from the chiral centre pointing towards
the lowest priority group. If the fingers of your right hand curl from priority 1 to priority 2 to priority 3,
the molecule has R configuration. If the fingers of your left hand curl from priority 1 to priority 2 to
priority 3, the molecule has S configuration. Try this with physical models first.
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EXERCISE 9: Assign R or S configurations to the following molecules.
a)
b)
STRATEGY:
For part a):
1. The priorities are, from lowest to highest: H < CH3 < CH2CH3 < OH.
2. With the thumb of your right hand along the bond pointing towards the hydrogen, the fingers
of your hand curl from OH to CH2CH3 and on towards CH3.
SOLUTION: This molecule is R.
For part b):
1. The priorities are, from lowest to highest: H < CH2 < CHCl < Br
2. In this depiction the lowest priority group is pointing away from you. Tracing from Br to
CHCl, to CH2 goes in a counter-clockwise direction.
SOLUTION: This molecule is S.
Try It #4: Supply a name for each of the following compounds, including the stereochemical designation.
a)
b)
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10.5 SUBSTITUTION REACTIONS
Skills to Master:
 Recognizing the nucleophile, substrate, product, and leaving group in a nucleophilic substitution
reaction.
 Stating the factors that make a good nucleophile and a good leaving group.
 Predicting the products and the likely mechanism of nucleophilic substitution reactions.
Key Concepts:
 The nucleophile is the reactant having a negative charge or at least one unshared electron pair. The
substrate contains the positive centre created by the withdrawal of electrons from a carbon atom by
an electronegative atom.
 The SN2 reaction proceeds in a single step and results in inversion of stereochemical configuration.
The SN1 reaction proceeds with two steps and includes a carbocation intermediate.
 A nucleophile is better if it: 1) is negatively charged rather than neutral, 2) has a larger nucleophilic
atom and therefore more polarizable, and 3) is smaller and therefore less “sterically hindered”.
 Tertiary alkyl halides are substituted via the SN1 mechanism, while primary alkyl halides proceed via
the SN2 mechanism.
Predicting the Products and Mechanisms of Substitution Reactions
To predict the outcome given reactants, you must identify the nucleophile and the leaving group. Replace
the leaving group with the nucleophile. Ensure charge balance. To determine the likely mechanism,
consider the carbon to which the leaving group is attached. If it is tertiary (attached to 3 other carbons) it
will likely proceed via SN1. If it is primary or methyl (attached to 1 or no other carbon) it will likely
proceed via SN2. If it is secondary, it is difficult to predict the mechanism without considering several
other factors.
EXERCISE 10: For each of the following, predict the products and the likely mechanism.
a)
b)
STRATEGY:
For part a):
1. Because the sulphur atom is negatively charged, it is likely the nucleophile. As is common,
the halogen (bromine in this case) will act as the leaving group.
2. The bromine is bonded to a tertiary carbon so this will likely proceed via SN1
3. We must ensure the products have a total charge of 1- as that is the sum of the charges of the
reactants.
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SOLUTION: This reaction will likely proceed via an SN1 mechanism and yield the following products:
For part b):
1. Here it is important to consider NaOH as the ions Na1+ and OH1-. Then we can see the oxygen
bearing the negative charge will likely act as the nucleophile. Again, the halogen (chlorine in
this case) will act as the leaving group.
2. The chlorine is bonded to a “methyl” carbon so this will almost certainly proceed via SN2
3. We could restate the reactant as OH-1, and the products would have a total charge of 1- to
reflect charge balance (the “net ionic” reaction). If we leave the reactant as NaOH, then we
should have a net charge of zero in our products, again to reflect charge balance. Let us
proceed with the latter approach.
SOLUTION: This reaction will proceed via an SN2 mechanism and yield the following products:
Try It #5: Given the following reactants, predict the products and the likely mechanism of the reaction.
10.6 ELIMINATION REACTIONS
Skills to Master:
 Predicting the products and likely mechanisms of elimination reactions
Key Concepts:
 In elimination reactions, two atoms on adjacent carbon atoms of the substrate molecule are removed
(eliminated) which results in a multiple bond in the substrate molecule.
 The E2 mechanism proceed in a single concerted step, while the E1 mechanism proceeds with two
steps and involves a carbocation intermediate.
 Primary alkyl halides can only undergo E2 eliminations. Tertiary and secondary ones are more likely
to undergo E1 eliminations.
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Predicting the Products and Mechanisms of Elimination Reactions
A common introductory elimination reaction is dehydrohaolgenation, that is, removal of a halogen and an
adjacent hydrogen atom. This is facilitated by a Lewis base, often an alkoxide ion RO1-. The alkoxide ion
collects the hydrogen atom and the halogen leaves as the halide ion.
EXERCISE 11: For each of the following, predict the elimination products and the likely mechanism.
a)
b)
STRATEGY:
For part a):
1. The alkoxide is shown with the Na1+ counter ion. We must recognize that the CH3O1- is the
operative agent, and the Na1+ is a spectator ion.
2. The bromine is bonded to a primary carbon so this will proceed via E2.
3. While the bromine will leave as bromide ion, we should show it with the Na1+ counter ion for
charge balance.
SOLUTION: This reaction will proceed via an E2 mechanism and yield the following products:
For part b):
1. Here the alkoxide is shown without its counter ion.
2. The chlorine is bonded to a secondary carbon so this will likely proceed via E1.
3. The chlorine will leave as chloride ion.
SOLUTION: This reaction will proceed via an E1 mechanism and yield the following products:
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10.7 ADDITION REACTIONS
Skill to Master:
 Applying Markovnikov’s Rule.
Key Concepts:
 Addition reactions convert a carbon-carbon double bond to a single bond and add an atom or group of
atoms to each of the carbons.
 Markovnikov’s Rule states that in the addition of HX to an alkene, the hydrogen atom adds to the
carbon atom that already has the greater number of hydrogen atoms.
Addition reactions can generically be represented as an alkene to which HX is added. One of the carbons
of the double bond will get the H and one will get the “X”. The identity of “X” varies; for example it
could be a halogen, an -OH group (so HX is water), or another hydrogen atom (so HX is H2).
Because the addition proceeds via a carbocation intermediate, when the alkene is not symmetrical often
one carbocation will be more stable and thus one product will predominate. The outcome is predicted by
Markovnikov’s Rule.
EXERCISE 12: For each of the following, predict the addition products.
a)
b)
STRATEGY:
For part a):
1. We will be adding an H to one of the atoms of the double bond and a bromine atom to the
other.
2. Examining the two carbons of the double bond, we count one hydrogen on the lower one and
zero on the upper one. Applying Markovnikov’s Rule we expect the lower carbon to accept
the hydrogen and the upper to take the bromine atom.
SOLUTION: The following product is expected:
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For part b):
1. We will be adding an H to one of the atoms of the double bond and a –OH group to the other.
2. Examining the two carbons of the double bond, we count one hydrogen on the left one and
two on the right one. Applying Markovnikov’s Rule we expect the right carbon to accept the
hydrogen and the left to take the –OH group.
3. The term over the arrow “cat HCl” refers to the catalyst which is required to make this
reaction proceed. It does not appear in the products.
SOLUTION: The following product is expected:
Try It #6: Given the following reactants, predict the addition product of the reaction.
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Chapter 10 Self-Test
1. Provide a name of the following molecules.
a)
b)
c)
2. For each of the following, provide the stereochemical designation E/Z or R/S as appropriate.
a)
b)
3. Predict the products and the mechanism of the following reaction.
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4. Predict the products for the following elimination reaction.
5.
Predict the product for the following addition reaction.
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Answers to Try Its:
1. a) 4-Ethyl-2-methylheptane
b)
2. a) 1-Bromo-2-methylbenzene, or o-bromomethylbenzene
b)
3. a) 2,3-Dimethylcyclopentanol
b) 2,3-dichloro-3-phenylpropanoic acid
4. a) R-2-Bromobutane
b) E-Pent-2-enoic acid
5.
6.
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Chapter 10: Student Study Guide
Answers to Self-Test
1. a) 1,2,3-Trichloro-2-fluoropropane
b) 3-Methylbutanone
c) 1,2,4-triethylbenzene
2. a) The molecule is R.
b) The molecule is Z.
3. The reaction likely proceeds via a SN1 mechanism. Products:
4.
5.
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