Textbook notes and problems

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CHAPTER 14

Steps in Naming Alkanes

BLM 14.2.1

OVERHEAD

• Identify the root.

– Find the longest continuous chain. 6 carbon atoms.

– Find the root name for the number of carbon atoms in the chain.

Root is hex-.

• Identify the suffix.

– For alkanes the suffix is –ane. Compound is an alkane.

hex- plus –ane is hexane

• Identify the prefix.

– Find the number of carbon atoms in each side group.

– Find the root name for each side group according to the number of carbon atoms.

Add –yl to the root to name the side group.

One side group has one carbon so it is a methyl

group. The other side group has two carbon

atoms so it is an ethyl group.

– Determine the alphabetical order of the side

groups (if there is more than one). Alphabetical order is ethyl methyl.

– Find the position of each side group.

– Precede each side group name with the number of the carbon atom to which it is attached on the main chain.

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– There are hyphens between numbers and words but no hyphen or space between the last prefix and the root.

The name of the compound is 3-ethyl-2-methylhexane.

CHAPTER 14

Drawing Alkanes

BLM 14.2.2

OVERHEAD

Step 1

Identify the root and the suffix of the name:

This step gives you the number

of carbons in the main chain and the type of functional group present. For

3-ethyl-3-methylpentane, the root and suffix are −pentane. The root, pent, tells you there are five

carbons in the main chain. The suffix, –ane, tells you that the chain has only single

carbon–carbon bonds.

Step 2

Draw the main chain first:

Draw a straight chain containing five carbon

atoms with single bonds between the atoms. Do not include hydrogen atoms yet.

Step 3

Choose one end of your carbon chain to be carbon number 1 and number

the rest of the chain in sequence. Add the indicated branches to the appropriate

carbon:

In this case, add a methyl group and an ethyl group both at carbon three. You

may place the branches on either side of the main chain.

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CHAPTER 14

Naming Alkenes

BLM 14.2.3

OVERHEAD

The IUPAC name of alkenes has the same three basic parts: a prefix, a root, and a suffix.

Determining the prefix, root, and suffix for alkenes is very similar to the procedure for alkanes, with a few modifications as follows:

• Identify the root:

Find the longest continuous chain that contains the double bond. The root name for a given chain of carbon atoms is the same as the root name for alkanes.

• Identify the suffix:

The suffix for all alkenes is −ene. However, the suffix of an alkene must also indicate the location of the double bond.

Assign position numbers:

The numbering of the main chain must begin at the end of the chain nearest the double bond and continue through the double bond. The position of the double bond is indicated by the number of the carbon atom that precedes the double bond. The suffix consists of a hyphen, a number, a hyphen, and −ene.

• Identify the prefix:

The rules for naming alkyl groups as side groups on alkenes are the same as they are for alkanes.

Sample Problem

Name the following alkene:

Identify the root: The longest chain that contains the double bond has six carbon atoms. The root is -hex-.

Identify the suffix: The molecule has a double bond so the suffix ends with –ene.

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Identify the prefix: There is one methyl group on carbon atom number 4 and another on carbon atom number 5. There is also an ethyl group on carbon atom number 4. The prefix is 4-ethyl-4,5-dimethyl-.

Solution: The full name of the compound is 4-ethyl-4,5- dimethylhex-1-ene.

CHAPTER 14

Drawing Alkenes

BLM 14.2.4

OVERHEAD

When drawing alkenes

The double bond appears after the carbon atom with the number stated in the suffix. The two carbon atoms sharing the double bond each have only two other bonds. Sample Problem

Draw a structural formula for 2-methylbut-2-ene.

Solution

Step 1The root is but- so there are four carbon atoms in the main chain.

Step 2The suffix is -2-ene so it has a double bond after carbon atom number two.

Step 3The prefix is 2-methyl- so there is a −CH3 group of the second carbon atom.

Step 4All carbon atoms must be bonded to enough hydrogen atoms to give them exactly four bonds.

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CHAPTER 14

Naming Cyclic Hydrocarbons

BLM 14.2.5

OVERHEAD

Determine the number of carbon atoms in the ring. The root name is the same as the straight chain alkane, alkene, or alkyne, with the same number of carbons atoms, preceded by cyclo.

Determine whether the molecule has all single bonds, at least one double bond, or at least one triple bond. The suffix is −ane, −ene, or −yne, respectively. There are no numbers to indicate the location of the double or triple bonds because they are always assumed to be between carbon atoms number one and two.

The names for the alkyl prefixes are the same as they are for the straight-chain

hydrocarbons. They are given in alphabetical order. However, there are a special set of rules for numbering the carbon atoms in the ring to which side groups are attached.

– If there are no side groups or only one side group on an alkane, the carbon atoms are not numbered.

– If the molecule is a cycloalkane and there are two or more side groups, the numbering must result in the lowest possible numbers. The carbon atom to which one of the side groups is attached is carbon atom number one. Numbering goes in the direction to make the numbers of any other side groups as small as possible. You can choose any carbon atom with a side group to start numbering.

– If the molecule is an alkene or alkyne, the multiple bond takes highest priority. The carbon atom on one side of the multiple bond is carbon atom number one and the one on the other side is number two. If there is a side group, the numbering starts in the position that will make the number of the carbon atom with the side group as small as possible.

Sample Problem

Identify the root: The ring has six carbon atoms so the root is cyclohex−.

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Assign the position numbers: The carbon atoms beside the double bond must be numbered 1 and 2. The numbering must proceed in the direction of the side group. Therefore, the lower left carbon atom must be number one and the numbering then goes clockwise giving the side group the number 3. Because the

numbering of carbon atoms one and two is mandatory, the numbers are not included in the name.

Identify the prefix: The side group has one carbon atoms so the prefix is methyl.

Solution: The name of the compound is 3-methylcyclohexene.

DATE: NAME: CLASS:

CHAPTER 14

Aliphatic Hydrocarbons Quiz

BLM 14.2.6

ASSESSMENT

Test your understanding of naming and drawing alkanes, alkenes, and alkynes.

1. Name the following alkanes:

(a) CH3CH2CH2CH2CH3

(b)

2. Name the following alkenes:

(a) CH3CH2CH2CH2CH2CH═CHCH2CH3

(b)

3. Name the following alkynes:

(a) CH3CH2CH2CH2CH2CH2CCH

(b)

4. Draw the following hydrocarbons:

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(b) oct-2-ene

CHAPTER 14

Aliphatic Hydrocarbons Quiz

Answer Key

BLM 14.2.6A ANSWER KEY

1. (a)pentane

(b) 3,3-dimethylhexane

2. (a)non-3-ene

(b) 3-methylbut-1-ene

3. (a)oct-1-yne

(b) 5-ethyl-3,4,6-trimethylhept-1-yne

4. (a)CH3—CH2—CH2—CH2—CH2—CH3

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DATE: NAME: CLASS: CHAPTER 14

Naming and Drawing of

Hydrocarbons Quiz

BLM 14.2.7 ASSESSMENT

Practice naming and drawing hydrocarbons.

1. Draw condensed structural diagrams for the following hydrocarbons:

(a) methylpropane

(b)3-ethyl-2-methylpentane

(c) 4-ethylhex-2-yne

(d)Cyclohept-1-ene

(e) 1-butyl-2,4-diethylbenzene

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(a) cyclopentane

(b)2,3,4,5-tetramethylhexane

(c) isopropylcyclohexane

(d)5,6,7-triethyl-4-methylnon-2-ene

(e) 3, 4-dimethyl-4-propyloct-1-yne

CHAPTER 14

Naming and Drawing of

Hydrocarbons Quiz

(continued)

BLM 14.2.7 ASSESSMENT

3. Name the following hydrocarbons:

(a)

(b)CH3CH2CH═CHCH2CH2CH3

(c)

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(e)

(f)

CHAPTER 14

Naming and Drawing of

Hydrocarbons Quiz Answer Key

BLM 14.2.7A ANSWER KEY

1. (a)

(b)

(c)

(d)

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2. (a)

(b)

(c)

(d)

(e)

3. (a) 1,4-diethylbenzene

(b) hept-3-ene

(c) 2-methylhexane

(d) 3-ethyl-2,4-dimethyloctane

(e) 3-ethyl-2-methylhex-2-ene

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CHAPTER 14

Naming and Drawing Alkyl

Halides

BLM 14.3.1 OVERHEAD

Locate the longest chain that includes the halogen atom(s). Name the parent alkane. • Identify the prefix:

– Number the parent carbon chain starting at the end nearest the halogen atom(s). If the compound is a cycloalkane, numbering starts at the carbon atom bonded to the halogen atom.

– Name and number any alkyl side groups on the main chain.

– Insert the number(s) of the carbon atom(s) bonded to the halogen(s).

– Use the prefix(es) that identify the specific halogen(s) (chloro-, fluoro-, bromo-, iodo-) – If there are two or more of the same type of halogen, use a prefix to indicate the number. – If there is more than one type of halogen present, write them alphabetically. Any prefixes

(di-, tri-) are not considered when alphabetizing the terms.

Sample Problem

Name the following alkyl halides.

(a) (b)

Solution

(a) Identify the root: The chain has six carbon atoms, therefore, the parent compound is hexane.

Identify the prefix: The carbon atom on the right end is nearest the first halogen so numbering starts at the right end. There is a bromine atom on carbon number two and carbon number four. The prefix will include 2,4-dibromo-. There is a chlorine atom on carbon atom number three so the prefix will include 3-chloro-. Alphabetically, bromo- is before chloro-. The name of the compound is 2,4-dibromo-3-chlorohexane.

(b) Identify the root: The parent compound is a cyclohexane.

Identify the prefix: Two bromine atoms are present as well as a methyl group.

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CHAPTER 14

Naming and Drawing

Carboxylic Acids

Remember that



The carboxylic acid group must always be on the end of a chain.



The carbon atom of the carboxyl group must be given the number one.

• Identify the root:

Locate the longest chain that includes the carboxyl group. Name the

parent alkane.

• Identify the suffix:

Drop the

−e

at the end of the name of the parent alkane and

replace it with

−oic acid

.

• Identify the prefix:

Name and number any alkyl side groups on the main chain. The

carbon atom of the carboxyl group is always number one.

Sample Problem

Name the following carboxylic acid:

Identify the root:

The longest chain that includes the carboxyl group has five

carbon atoms including the carbon atom in the carboxyl group.

The parent alkane is pentane.

Identify the suffix: Replace the

−e

at the end of pentane with

−oic acid

. The name

includes pentanoic acid.

Identify the prefix: There is a methyl group on carbon atom number three, so the

prefix is 3-methyl.

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CHAPTER 14

Naming and Drawing Esters

BLM 14.3.3

OVERHEAD

Identify the part of the ester that contains the C–O group. This is the part of the ester that came from the acid. The root of the name of the ester is based on the name of the acid. Determine the name of the parent acid.

Remove the −oic acid from the name of the parent acid and replace it with −oate.

To form the prefix, consider the part of the ester that is associated with the alcohol. Ignore the oxygen atom and use only the alkyl group. Identify the name of the alkyl group. The name of the alkyl group is the prefix for the name of the ester. There is always a space between the name of the alkyl group and the root.

Sample Problem

Name the following ester:

Identify the root: The C–O is part of a three carbon group making the parent acid a propanoic acid.

Identify the suffix:Remove the −oic acid from the name of the parent acid and replace it with

−oate. The root plus the suffix is now propanoate.

Identify the prefix:The part of the ester that is associated with an alcohol has one carbon atom therefore the prefix is methyl.

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CHAPTER 14

Common Chemical Families and

Their Functional Groups

Family name Suffix or_prefix Functional_group Example Generic_formula

alkene -ene

hept-3-ene

alkyne -yne

pent-2-yne

alcohol -ol _{(commonly called}

the hydroxyl group)

propan-2-ol

alkyl halide _{with halogen}prefix varies

X = a halogen 2-chlorobutane

carboxylic acid -oic acid _{(commonly called}

the carboxyl

group) propanoic acid

ester -oate

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CHAPTER 14

Fractional Distillation Towers

BLM 14.4.1

OVERHEAD

Fractional distillation involves successive heating, evaporation, cooling, and

condensation. Fractional distillation towers can be as high as 60 m.

Kerosene, Fuel oil 15 – 18 carbon atoms

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CHAPTER 15

Addition, Substitution,

Elimination, and Esterification

Reactions

BLM 15.1.2

OVERHEAD

Sample Problem

Identify each type of reaction and then complete the equation.

1.

2.

3.

4.

Solution

1. The alcohol is heated in the presence of a strong acid. This reaction is an elimination reaction. An

alkene and a small second product are formed in elimination reactions:

2. A small molecule reacts with an alkene in such a way that two parts of the small molecule

become bonded to two adjacent carbons in the organic molecule. This reaction is an addition reaction. The addition reaction gives a single product, as follows:

3. A small molecule reacts with an alkane in the presence of ultraviolet light in such a way that one

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4. A carboxylic acid reacts with an alcohol. This is an esterification reaction. An ester and water will be formed as products. The equation is completed as follows:

DATE: NAME: CLASS:

CHAPTER 15

Identifying Organic Reactions

BLM 15.1.5

ASSESSMENT

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CHAPTER 15

Identifying Organic Reactions

(continued) BLM 15.1.5 ASSESSMENT A dd it io n : D oe s th e or ga ni c pr od uc t ha ve f ew er d ou bl e or tr ip le b on ds th an th e re ac ta nt ? E lim in at io n : D oe s th e or ga ni c pr od uc t ha ve m or e do ub le o r tr ip le b on ds th an th e re ac ta nt ? S u bs ti tu ti on : H as a n at om o r gr ou p of at om s be en s ub st itu te d fo r a di ff er en t at om o r gr ou p of a to m s? C on d en sa ti on : D o tw o or ga ni c m ol ec ul es c om bi ne to f or m a la rg e or ga ni c m ol ec ul e, s uc h as a n es te r or am id e? H yd ro ly si s: D oe s a la rg e or ga ni c m ol ec ul e, s uc h as a n es te r or a m id e, sp lit in h al f to f or m tw o sm al le r or ga ni c m ol ec ul es ?

Type of reaction

Reaction 1 yes no no no no addition

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CHAPTER 15

Identifying Organic Reactions

Answer Key

BLM 15.1.5A ANSWER KEY A dd it io n : D oe s th e or ga ni c pr od uc t ha ve f ew er d ou bl e or tr ip le b on ds th an th e re ac ta nt ? E lim in at io n : D oe s th e or ga ni c pr od uc t ha ve m or e do ub le o r tr ip le b on ds th an th e re ac ta nt ? S u bs ti tu ti on : H as a n at om o r gr ou p of at om s be en s ub st itu te d fo r a di ff er en t at om o r gr ou p of a to m s? C on d en sa ti on : D o tw o or ga ni c m ol ec ul es c om bi ne to f or m a la rg e or ga ni c m ol ec ul e, s uc h as a n es te r or am id e? H yd ro ly si s: D oe s a la rg e or ga ni c m ol ec ul e, s uc h as a n es te r or a m id e, sp lit in h al f to f or m tw o sm al le r or ga ni c m ol ec ul es ?

Type of reaction

Reaction 3 no yes no no no elimination

Reaction 4 no no yes no yes hydrolysis (substitution)

Reaction 5 no no yes yes no condensation (substitution)

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CHAPTER 15

Addition and Condensation Polymers

OVERHEAD

Examples of Addition Polymers

Name

Structure of

_monomer

Structure of polymer

Uses

polystyrene __{styrene and Styrofoam}

insulation packaging

polyvinylchloride

(PVC, vinyl) _construction materials_{sewage pipes}

medical equipment

Examples of Condensation Polymers

Name

Structure

Dacron™ (a polyester)

synthetic fibres used to make fabric for clothing and surgery

Nylon-6 (a polyamide)

tires