Chapter 3

Hydrocarbons: Nomenclature and Reactions

Chapter 3 Problems• Review Section 3.9

• Read Essays on petroleum (p 204-207) and gasoline (p 221-223) for interest.

• I recommend that you do all problems except Prob. 8, 14, 33, 34 and 41

• Skip ethenyl, 2-propenyl etc on p. 200

• Read Section 3.15 and 3-16, but don’t take them too seriously.

I nternational

U nion of

P ure and

A pplied

C hemistry

Sect. 3.1: IUPAC nomenclaturesystematic nomenclature

“eye-you-pac”

colloquially:

Sect. 3.2: the alkanes

• Hydrocarbons

• “Paraffins”

• Alkanes: formula CnH2n+2

The alkanes: table 3-2

Methane (CH4)

Ethane (C2H6)

Propane (C3H8)

Butane (C4H10)

Pentane (C5H12)

Hexane (C6H14)

Heptane (C7H16)

Octane (C8H18)

Nonane (C9H20)

Decane (C10H22)

CH4

CH3 CH3

CH3 CH2 CH3

CH3 CH2 CH2 CH3

CH3 CH2 CH2 CH2 CH3

CH3 CH2 CH2 CH2 CH2 CH3

CH3 CH2 CH2 CH2 CH2 CH2 CH3

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH3

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3

CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3

Sect. 3.4: IUPAC nomenclature of alkanes

1. Find the longest continuous chain of carbon atoms and name it (use linear names).

3. Give the substituent a name based on the number of carbon atoms it has. replace the -ane ending with -yl

2. Number the chain starting from the end nearest a branch.

4. Give the substituent a number determined on its location on the chain.5. Assemble the name.

Single substituent group

Finding the longest continuous Finding the longest continuous chain chain of carbon atoms is not always of carbon atoms is not always simplesimple

C-C-C-C-C-C-C-C-CC-C-C-C-C-C-C-C-C

C

C-C

C-C-CC-C-C

C-CC-C

CC

-C-C-C-C-CC- C-C-C-C-C-CC-C-C-C-C-C

C-CC-C

C-C-C-

C

all possibilitesmust be examined

it won’t always bethe horizontal one as shown here

try these also ……..

99

66 88

Sect. 3.3: Common alkyl groups (C1 through C4): table 3-3

Methyl

Ethyl

Propyl

Butyl

Isopropyl

Isobutyl

(1-Methylethyl)

(2-Methylpropyl)

sec-Butyl

(1-Methylpropyl)

tert-Butyl

(1,1-Dimethylethyl)

CH3

CH3 CH2

CH3 CH2 CH2

CH3 CH2 CH2 CH2

CH3 CH

CH3

CH3 CH CH2

CH3

CH3 CH2 CH

CH3

CH3 C

CH3

CH3

CH3 CH2 CH CH3

CH3

Name this alkane

4 3 2 1

2-methylbutane

CH3 CH2 CH CH3

CH2 CH3

Find the longest continuouscarbon chain

1 2 3

4 5

3-methylpentane

CH3 CH2 CH CH2 CH2 CH3

CH2 CH2 CH3

You must choose the longestcontinuous carbon chain

4 3 2 1

5 6 7

4-ethylheptane

Two different substituents• number chain from end closest to a group,

regardless of alphabetical order• locate where groups are on chain with

numbers• place groups in alphabetical order, with the

appropriate number• assemble the complete name, using

hyphens to separate numbers from “text”

CH3 CH2 CH2 CH CH CH2 CH3

CH3

CH2 CH3

Number from the end nearestthe first substituent

7 6 5 4 3 2 1

4-ethyl-3-methylheptane

CH3 CH2 CH2 CH CH2 CH CH2 CH3

CH2 CH3

CH3

Number from the end nearestthe first substituent

8 7 6 5 4 3 2 1

3-ethyl-5-methyloctane

Two or more identical substituent groups

• for two identical groups, use prefix di with the name of the group: dimethyl, diethyl, etc.

• dimethyl alphabetized as methyl, not dimethyl

• use numbers to locate groups on chain

• use commas to separate numbers

• prefixes: di = 2 tri = 3 tetra = 4 penta = 5

CH3 CH CH CH3

CH3

CH3

Use “di-” with two substituents

1 2 3 4

2,3-dimethylbutane

CH3 CH2 C CH2 CH2 CH3

CH3

CH3

Every substituent must get a number

1 2 3 4 5 6

3,3-dimethylhexane

CH3 CH CH2 C CH3

CH3 CH3

CH3

You need numbers, even though it appears on the same carbon!

5 4 3 2 1

2,2,4-trimethylpentane

CH3 CH2 CH CH CH2 CH2 CH2 CH2 CH CH3

CH3CH3

CH3

Number from the end nearestfirst substituent

10 9 8 7 6 5 4 3 2 1

2,7,8-trimethyldecane

CH3 CH2 CH CH CH2 CH2 CH2 CH CH2 CH3

CH3

CH3

CH3

Number from the end which hasthe “first difference”

1 2 3 4 5 6 7 8 9 10

3,4,8-trimethyldecane

CH3 CH2 CH CH CH2 CH CH2 CH3

CH3

CH3

CH2 CH3

Number from the end nearestthe “first difference”

Dimethyl alphabetized as methyl, not dimethyl

1 2 3 4 5 6 7 8

6-ethyl-3,4-dimethyloctane

If you can name this,you can name almost anything!

1 2 3 4 5 6

7

8 9

4-isopropyl-2,6,6-trimethylnonane

CH3 CH CH2 CH CH2 C CH3

CH3

CH

CH3

CH3

CH3

CH2

CH2 CH3

The isopropyl group can be named as a “complex” substituent

1-methylethyl

CH3 CH

CH3

2 1

Now, rename the isopropyl group. Notice the alphabetical order!

1 2 3 4 5 6

7

8 9

2,6,6-trimethyl-4-(1-methylethyl)nonane

CH3 CH CH2 CH CH2 C CH3

CH3

CH

CH3

CH3

CH3

CH2

CH2 CH3

Deciding on alphabetical order for complex groups

• Complex groups are alphabetized under the first letter of the name

• (1,3-dimethylbutyl) = d

• (1,1,2-trimethylpropyl) = t

• (1-ethyl-1,2-dimethylbutyl) = e

Naming complex substituents -- this one is aphabetized under d

1,3-dimethylbutyl

C H 3 C H C H 2 C H

C H 3C H 3

4 3 2 1

Naming complex substituents

2-ethyl-1,1-dimethylbutyl

CH3 C CH CH2 CH3

CH3 CH2 CH3

1 2 3 4

CH3 CH2 CH2 CH2 C CH2 CH2 CH2 CH3

C

CH2 CH2 CH3

CH3

CH2CH2 CH3CH3

Name this compound!

1 2 3

1 2 3 4 5 6 7 8 9

5-(1-ethyl-1-methylpropyl)-5-propylnonane

CH2CH2

CH2CH

CH2CH3 C

CH2CH2

CH3

CH2CH3

CH3

CCH3 CH3

CH3

Name this two ways -- (the complex group)

7-(1,1-dimethylethyl)-3-ethyl-7-methyldecane

7-tert-butyl-3-ethyl-7-methyldecane

Sect. 3.5: Common names of alkanes

• butane

• isobutane

• pentane

• isopentane

• neopentane

Sect. 3.6: the cycloalkanes

• The names of the cycloalkanes always contain the prefix cyclo

• Cycloalkanes have the general formulaCnH2n

Cyclic molecules

Cyclopropane

Cyclobutane

Cyclopentane

Cyclohexane

CC

C

H H

H

H

H

H

CH2

CH2

CH2

C

C C

C

H

H

H

H

H

H

H

H

CH2

CH2 CH2

CH2

C

C

C

C

C

H HH H

HH

HH

HH

CH2

CH2

CH2

CH2

CH2

C

CC

C

CC

H HH

H

H

H

HH

H

H

H

HCH2

CH2

CH2

CH2

CH2

CH2

Nomenclature of the substituted cycloalkanes

• If there is only one substituent, do not use the “1”. • If there is more than one substituent, you must use all

numbers, including “1”!• Number around the ring in a direction to get from the

first substituent to the second substituent by the shorter path.

• For equivalent degrees of substitution, number in a direction that follows the alphabetical sequence.

• A carbon with greater substitution has precedence in numbering.

CH3

CH3

1,1-dimethylcyclohexane

CH3

CH3

CH2CH3

4-ethyl-1,1-dimethylcyclohexane

CH3

CH3

CH3

CH3CH3

CH2CH3

CH3

CH3

CH3CH2

1,3-dimethylcyclopentane

1-ethyl-4-methylcyclohexane

3-ethyl-1,1-dimethylcyclobutane

Some Some cycloalkanescycloalkanes

The more substituted carbontakes precedence even though E comes before M.

Drawndifferentlybut samename.

=

E before M

1 2

3

1

2

3

1

231

2

3

4

Two ways of naming this

1-isopropyl-2-methylcyclohexane

1-methyl-2-(1-methylethyl)cyclohexane

CH3

CH

CH3

CH3

Numbering starts at the most highly-substituted carbon

Cl CH3

CH3

CH3

2 13 74 6 5

2-chloro-1,1,6-trimethylcycloheptane

Sect. 3.7: cycloalkyl groups

cyclopropyl cyclobutyl

cyclopentyl cyclohexyl

C

CH3

CH2CH2CH3 CH3

3-cyclobutyl-3-methylpentane

CH CH2CH3CH3

(1-methylpropyl)cyclohexane or2-cyclohexylbutane

No locant is needed. With one substituent on a ring , it is automatically on carbon 1.

1-(1-methylpropyl)cyclohexane is overkill, but OK!

Rings with one substitutentRings with one substitutent

Another name of a group

or

Phenyl

or C6H5

CH3 CH2 CH CH CH3

CH3

3-methyl-2-phenylpentane

Sect. 3.8: Degree of Substitution

methyl methylene

methine

CH3 R R CH2 R

R CH R

R

R C R

R

R

primary (1°) secondary (2°)

tertiary (3°) quaternary (4°)

CH3 C

CH3

CH3

CH CH2 CH3

CH3

QUATERNARY

PRIMARY

TERTIARY

SECONDARY

A hydrocarbon containing carbon atoms withdiffering degrees of substitution

All of the methyl groups (CH3) are primary.

ExampleExample

Sect. 3.9 -- review

We already did this in Chapter 1

Sect. 3.10 and 3.11: nomenclature of halides and

nitro compounds

fluoro chloro bromo iodo

nitro

F Cl Br I

NO2

CH 3 CH 2 Br

bromoethane (IUPAC)

ethyl bromide (common)

Br

bromocyclopropane

CH3 C Cl

CH3

CH3

2-chloro-2-methylpropane (IUPAC)

tert-butyl chloride (common)

CH3 CH CH CH2 CH3

Br CH3

2-bromo-3-methylpentane

I

iodocyclohexane (IUPAC)

cyclohexyl iodide (common)

Br

Cl

1-bromo-2-chlorocyclohexane

CH3 CH CH3

NO2

2-nitropropane

LOCANT PREFIX N STEM SUFFIX

numbers substituents number -ALK(AN)- ending

Sect. 3.12: Block diagram for nomenclature

Sect. 3.13: alkene nomenclature

• ending is ene

• identify the longest chain with the C=C

• number from the end closest to the C=C and assign a number - - i.e. 2-pentene

• C=C is more important than groups!

• now number the attached groups and place them in alphabetical order

CH2 CH2 CH3 CH CH2

CH3 CH2 CH CH2 CH3 CH CH CH3

ethene (IUPAC) propene (IUPAC)

1-butene 2-butene

ethylene (common) propylene (common)

C CH CH3

CH3

CH3

2-methyl-2-butene

CH3 CH CH2 CH2 CH CH CH3

CH3

6-methyl-2-heptene

C C

CH3

H

CH2

CH2

CH2CH3

CH2CHCH2

CH3

CH3

trans-6-methyl-3-propyl-2-octene(Don’t worry about “trans” until Chapter 4)

CH3 CH3

4,4-dimethylcyclohexene

CH3

2-methyl-1,3-cyclohexadiene

CH2 CH3

CH2

CH3

2,5-diethyl-1,3-cyclooctadiene

Very important!

benzeneIt is never cyclohexatriene!!!

Sect. 3.14: nomenclature of alkynes

• similar system used as with alkenes

• ending is yne

• identify the longest chain with the triple bond

• everything else is the same as alkenes

H C C H CH3 C C H

CH3 CH2 C CH CH3 C C CH3

ethyne propyne

1-butyne 2-butyne

“acetylene”

ALKYNES ( -YNE )ALKYNES ( -YNE )

C CCH3 CH2CH2CH3

CCH3 C CH

CH3

CH3

2-hexyne

4-methyl-2-pentyne

The suffix hasprecedenceover anysubstituents

The functional group has precedence in numbering.

functionalgroup

CH3 C C

CH3

Cl

C CH3

CH3 CH C

CH3

C CH CH2

Br

CH3

4-chloro-4-methyl-2-pentyne

5-bromo-2-methyl-3-heptyne

Number from the end closest to either the double bondor the triple bond, whichever is closest to the end.

ene vs. yne: which one wins?ene vs. yne: which one wins?

CH3-CH2-C C-CH2-CH=CH-CH3

2-octen-5-yne

8 7 6 5 4 3 2 1

Compounds are named: en-yne.

COMPUTER PROGRAM

““ORGANIC NOMENCLATURE”ORGANIC NOMENCLATURE”

Available in ChemistryComputer Lab - CB280

Go to ChemApps Folder :

optional, but recommended …..

then choose first and then Organic Nomenclature

Chem Apps

Organic

Sect. 3.15: physical properties of hydrocarbons

• the longer the straight chain, the higher the boiling point -- van der Waals forces

• isomers that are branched have lower boiling points

• hydrogen bonding increases boiling points

• Dipole-dipole attractions increase b.p.

Sect. 3.16: Combustion of alkanes

CnH2n+2 + m O2 n CO2 + (2n+2)/2 H2O

where n = number of carbons

+ HEAT!!!

Example:

2 C6H14 + 19 O2 12 CO2 + 14 H2O

light+ +R H Cl Cl R Cl H Cl

chlorination

examples

CH4 + Cl2 CH3Cl + HCl

CH3CH2CH3 + Cl2 CH3CHCH3 + HCl Cl

+ CH3CH2CH2-Cltakes place at a refineryor a chemical plant - noteasy to do in the lab

free-radicalsubstitution reaction

Sect. 3.17: Halogenation of Alkanes

The previous examples given The previous examples given assumed assumed monochlorinationmonochlorination

(one chlorine added)(one chlorine added)

BUT …the reaction can repeat itselfCOMMON NAMES

CH4 + Cl2 CH3Cl + HCl methyl chloride

CH3Cl + Cl2 CH2Cl2 + HCl methylene chloride

CH2Cl2 + Cl2 CHCl3 + HCl chloroform

CHCl3 + Cl2 CCl4 + HCl carbon tetrachloride

fully chlorinated product What are theIUPAC names ?

The reaction must be initiated

It does not occur in the dark.

Exposure to ultraviolet light (sunlight) will startthe reaction.

Heat will also start the reaction.

Once reaction starts, it is exothermic and continuesalmost explosively.

The first step is the dissociation of chlorine :

Cl-Cl 2 Cl....

..

......: : :.h

or chlorineatoms

diatomic molecule(radicals)

Abstraction of hydrogen atomAbstraction of hydrogen atom

Cl.. :...

BY A CHLORINE “FREE RADICAL” (ATOM)

+

C . H-Cl.. :..

+

Chlorine takes thehydrogen and one of its electrons

unpaired electron= “free” radical

HYDROGEN ABSTRACTION..C H

Cl Cllight

2 : Cl..

...

CH3 H + : Cl..

...

..:

..H Cl + . CH3

. CH3 +

..

..

..

..: :Cl Cl

..

..:CH3 Cl + : Cl

...

..

..

..

..

..: :

1. Initiation

2. Chain Propagation (first step)

3. Chain Propagation (second step)

a free radical

methyl radical

feeds back intostep two

REPEAT ING

STEPS

Mechanism of chlorination of methaneMechanism of chlorination of methaneCHAIN REACTION

“hydrogen abstraction”

“dissociation”

4. Termination Steps

Cl Cl2 : Cl..

...

. CH3CH3

..

..

..

..: :

. + CH3CH3

: Cl..

... . CH3+ : Cl

..CH3

..

These steps stopthe chain reaction

“recombinations”

CH3CH2CH3 + Cl2 CH3CHCH3

Cl+ CH3CH2CH2-Cl

limited amount

QUESTION

A B

WHAT ARE THE RELATIVE AMOUNTS OF A AND B ?

IS IT STATISTICAL ( 2 : 6 ) = (1 : 3 ) ?

DOES SOMETHING ELSE CONTROL THE OUTCOME ?

Monochlorination of propane: Monochlorination of propane: Does one isomer predominate?Does one isomer predominate?

CH3CHCH3

Cl CH3CH2CH2-Cl

A B

STATISTICAL PREDICTION 25 % 75 %

CH3-CH2-CH3 = 6 : 2 or 3:1

Monochlorination of propaneMonochlorination of propane STATISTICAL VERSUS EXPERIMENTAL RESULTS

ACTUALLY FOUND 50 % 50 %

Equal amounts

Equalamounts

Experimental results show:

Secondary hydrogens are energenically more easily removed than primary hydrogens

CH3CHCH3

H

CH3CH2CH3

Secondary HPrimary H

more reactive hydrogen

Stability of free radicals explains results!

H3C C CH3

H

secondary radical

More stable and easier to form!

H3C CH2 C

H

H

primary radical

Less stable and harderto form

Stability of radicals: TERTIARY > SECONDARY > PRIMARY

H3C C CH3

H

secondary radical

less stable than tertiary but more stable than primary

H3C CH2 C

H

H

primary radical

Least stable and hardest to form

H3C C CH3

CH3

tertiary radical

Most stable and easiest to form!

H3C C CH2

CH3

H H

H3C C CH2

CH3

H Cl

Cl2

Another example: isobutaneWhich product should form in the largest amount?

There are 9 primary H’sand only 1 tertiary H

Statistically you could predict a 9:1 ratio or a 90% yield of 1-chloro-2-methylpropane!

Wrong!!

H3C C CH3

CH3

tertiary radical - more stable

H3C C CH2

CH3

H H

H3C C C

CH3

H

ClH-Cl

primary radical - less stable

H

H

Cl H-Cl

Isobutane gives only 62% of 2-chloro-2-methylpropane! Why? Look at the stability of the intermediate radical.

The statistical factor predicts a 9:1 ratio (90%)However, the energy factor predicts that the ratiowill be less than 90% and turns out to be 62%.

H3C C CH3

CH3

tertiary radical

H3C C C

CH3

H

primary radical

H

H

Cl2

Cl2

H3C C CH3

CH3

Cl

ClH

H3C C C

CH3

H

H

H

Cl ClH

38%

62%

Cl2

light or heatCl

same as

Cl

1-chloro-2-methylhexane

The two circled methylgroups are equivalent!!!!

Draw the structure of all of the monochlorinated products. There are 6 total products. The next slide shows the remaining 5 products. Only ONE product is shown here!

Cl2

light or heatCl

Cl

Cl

Cl

Cl

1-chloro-5-methylhexane 4-chloro-2-methylhexane

2-chloro-5-methylhexane

2-chloro-2-methylhexane

3-chloro-2-methylhexane

Here are 5 more isomeric products that are formed!

Ethylcyclopentane: monochlorination products

MONOCHLORINATION PRODUCTS

CH2 CH3

CH2 CH2 Cl

CH CH3

Cl

CH2 CH3

Cl

CH2 CH3

Cl

Cl

CH2 CH3

Hydrochlorofluorocarbons(HCFC’s)

H C Cl

F

F

Depletes the ozone in the upper atmosphere

Hydrogenation is covered in more detail in Chap 4 -we’ll cover it there.

Hydrogenation is included in this chapter (briefly) because it is a method of making ALKANES.

Sect. 3.18: hydrogenation of alkenes

+Pd

H2

CH3

HH

CH3

+ H2 CH3CH2CH2CH3

Pt

catalyst = Pt, Pd, Ni

+C C H H C C

H H

catalyst