topic 10 organic chemistry. ib core objective 10.1.1 describe the features of a homologous series

Topic 10Organic Chemistry

IB Core Objective

• 10.1.1 Describe the features of a homologous series.

10.1.1 Describe the features of a homologous series.

• Carbon atoms can form long chains.• These chains can also contain functional groups:

which are other atoms such as oxygen, nitrogen or halogens.

• Some chains may have the same functional group, but only differ by the presence of additional carbon atoms and associated hydrogen atoms.

• These series of compounds that are related are called homologous series.


Homologous Series• Differ from each other by a CH2 unit.• Can be represented by a general formula

(example: alkanes is CnH2n+2)• Compounds have similar chemical properties.• As the number of carbon atoms present

increases, the physical properties will vary in a regular manner.


Alkanes Alkanes are an example of a homologous

series. Alkanes have the general formula CnH2n+2

So when using this formula, what kind of bond (single, double, triple) would form between two or more carbons?

A: Single


Molecular models1. Build a methane (CH4) model.2. In homologous series, successive compounds

differ by a CH2 unit. Build a CH2 unit, and join it with your methane.

3. What kind of C-C bond does your new molecule have? What would the molecular formula be? Does it follow the CnH2n+2 formula?

IB Core Objective

• 10.1.2 Predict and explain the trends in boiling points of members of a homologous series.

10.1.2 Predict and explain the trends in boiling points of members of a homologous series.

• First, lets review van der Waals’ forces. • They are weak forces that attract molecules to

each other.• This is caused by random fluctuations in the

electron clouds, where they can temporarily produce dipoles.

When the overall mass increases, so does the strength of the van der Waals’ forces.


What would happen to the forces of attraction between molecules when the number of carbons in the chain (CH2) are increased?

A: The forces of attraction would increase.What happens to the boiling point when forces

of attraction increase?A: Boiling point increases as well.

IB Core Objective

• 10.1.3 Distinguish between empirical, molecular and structural formulas.

10.1.3 Distinguish between empirical, molecular and structural formulas.

• Structural • Molecular • Empirical

H

C

H

H

H

H

HH

C CC

O

C

HH

O

H

Pentanoic Acid

CH3(CH2)3COOH

Condensed structural formula

What is the Molecular and empirical formula?

Writing structures

• ‘R’ is sometimes used to represent the carbon chain. – ROH represents the functional group alcohol

attached to some carbon chain.

• Rings: Benzene Ring(Alternating double

bonds)

IB Core Objective

• 10.1.4 Describe structural isomers as compounds with the same molecular formula but with different arrangements of atoms.

10.1.4 Describe structural isomers as compounds with the same molecular formula but with different arrangements of atoms.

• Totally different compounds with the same molecular formula– This includes a unique MP and BP– May have totally different chemical properties

Isomers

Structural(Different Bonding)

Stereo(Different shape)

Structural Isomers

OH

OH

Propanol Primary alcohol

MP -127 oC

Iso-propanol or 2-PropanolSecondary alcohol

MP -88oC

Make a molecular model of both. What

is the molecular formula?

C3H7OH

IB Core Objectives

• 10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C6.

• 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6.

10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C6.

10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6.

Mice Eat Peanut Butter!Or….

Methane, Ethane, Propane, ButaneThe first four alkanes!

Alkanes• Prefix: Indicates the

longest carbon chain. • Ex. Meth - 1 carbon• Eth - 2 • Prop - 3• But - 4• Pent - 5 • Hex - 6• Hept- 7• Oct- 8• Non - 9• Dec- 10

Hydrogen atoms are

removed for clarity

Alkanes• Suffix: The ending portion of the name.

• All carbon based organic compounds that contain only single bonds have the ending –ane

Methane CH4

Ethane CH3CH3

Propane CH3CH2CH3

Butane CH3CH2CH2CH3

Pentane CH3CH2CH2CH2CH3

Hexane CH3CH2CH2CH2CH2CH3

Substituents/ Functional groups

• 1st number the carbon chain. In this case no matter which way you go it is a 5 carbon chain, hence pentane.

• Next the subgroup must be named, in this case it is a methyl group off of the 4th carbon. BUT the main chain must be numbered closest to the sub group...so you must re-number if you didn’t do it this way.

• Name: 2-Methylpentane

2

31 4

5

4

3

Methyl

5 2

1

-CH3 = 1 carbon= meth=Methyl

(functional group)

Carbon sub chains have the

suffix -yl

Multiple Substiuent/Functional Groups

• In this case there are 2 methyl groups on the same carbon.

• Name: 2,2-Dimethylpentane

4

3

Methyl

5 21

Methyl

1 –mono2 –di3 –tri4 –tetra5 –pent6 –hex7 –hept8 –oct9 –non10 -dec

di- is used because it indicates 2 of the same

functional groups

Multiple Functional Groups

• In this case you list the sub groups in alphabetical order first.

• Name: 3-Ethyl-3,4-dimethylhexane

Ethyl4

2

Methyl

53 1

Methyl

6

IB Core Objectives

• 10.1.9 Deduce structural formulas for compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.

• 10.1.10 Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.

Functional Groups to Know

• -OH (alcohol)

• R-CHO (aldehyde)

• R-COR (Ketone)

• R-COOH (carboxylic acid)

• -X (halide)

H

H

C

H

HO

Simplified structure

H

C

O

RC

O

R RC

O

R O HH

H

C

H

Cl

ALCOHOLS

If a hydrogen atom of an alkane is replaced by an —OH group, the compound is called an Alcohol.

Alcohols derived from ALKANES are called ALKANOLS.

A functional group is a small set of atoms, held together by covalent bonds in a specific, characteristic arrangement that is responsible for the principal physical and chemical properties of an organic compound.

We interrupt this objective to bring you an all new objective!!

IB Core Objective

• 10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes.

Three CLASSES of alcohols occur according to the position of the HYDROXYL, OH, group. The classes of alkanols are:• PRIMARY • SECONDARY• TERTIARY

PRIMARY ALCOHOLSR–CH2– OH and HCH2–OH

R— C—OH

H

H

—C—OH

H

H

H

MethanolCH3OH

General formula

PRIMARY ALCOHOLS (cont.)Structural formula for Ethanol (Ethyl Alcohol) :

OHH–C–H

H

C–H

H

H–C–H

H

C–H

H

C–H

H

C–H

HOH

1-butanol: 2-methyl-1-butanol:

–C —H

H C —H

C—H

H

C–H

H

1 2 3 4H

H–C–H

H

HO

SECONDARY ALCOHOLS

R2CH—OH —C—OH

H

R´

R

2-propanol

H—C—

H

H

C—

H

OH

C—H

H

H

3-methyl-2-butanol

H—C —

H

H

C —

H

C—

H

C—H

H

HOH

H–C–H

H

1234

C3H7OH C5H11OH

TERTIARY ALCOHOLS

R3COH

2-methyl-2-propanol

H—C —

H

H

C —

H–C–H

H

OH

C—H

H

H

C4H9OH

21 3

R´´

R´

R – C – OH

ISOMERS of C4H10O

H–C–H

H

C–H

H

C–H

H

C–H

HOH

H–C–H

H

C–H

H

C–H

O

C–H

HH

H

H

C–H–C–

H

H

–CH

CH H

HOH

HPRIMARY

ALCOHOLS

SECONDARY

ALCOHOLTERTIARY

ALCOHOL

–C–H–C–

H

HC–O

H

H

H

HC–H

H

H

ETHER

H–C–H

H

C–H

H

C–H

O

C–H

HH

H

H—C —

H

H

C —

H–C–H

H

O

C—H

H

H

H

10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes.

Objectives 10.1.9, 10.1.10

Halogenoalkanes

An alkane with a halogen (fluorine, chlorine, bromine, or iodine).Often represented by an X, or can include the actual halogen (F, Cl, Br, I)

Prefix begins with fluoro-, chloro-, bromo-, or iodo-

Example: Name?CH3 CH CH CH3

Br Cl

A: 2-bromo-3-chlorobutane

10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes.

• Primary, secondary and tertiary structures are the same for halgeonoalkanes as alcohols.

Describe the difference between primary, secondary and tertiary structures of halgonoalkanes:

• Primary: Carbon the halogen is bonded to is bonded to only one other carbon.

• Secondary: Carbon the halogen is bonded to is bonded to two other carbons.

• Tertiary: Carbon the halogen is bonded to is bonded to three other carbons.

Back to our original objectives

• 10.1.9 Deduce structural formulas for compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.

• 10.1.10 Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.

ALDEHYDESAldehydes have the suffix –al.

ALKANALSare produced by the

OXIDATION

OF PRIMARY ALCOHOLS (Primary alkanols)

RCH2OH RCHO

R—C—

H

H

OH R—C

H

O

KETONESALKANONES

are produced by the

OXIDATION

OF SECONDARY ALCOHOLS (Secondary alkanols)

R2CHOH R2CO

R—C—

H

R´

OHR— C

O

R´

CARBOXYLIC ACIDS

R — C

O

O–H

ALDEHYDES are OXIDISED to CARBOXYLIC ACIDS

MnO4¯(aq)

H+H3C — C

O

H

H3C— C

O

O H

Ethanal Ethanoic acid(Acetic acid)

– +

Aldehyde functional group

Carboxyl functional group

IB Core Objectives

• 10.1.7 Deduce structural formulas for the isomers of the straight-chain alkenes up to C6.

• 10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C6.

10.1.7 Deduce structural formulas for the isomers of the straight-chain alkenes up to C6. 10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C6.

Alkenes• General formula: CnH2n

• If there are double bonds involved the suffix changes from –ane to –ene.– Molecules are considered un-saturated if

there are double bonds present.

• After naming the longest chain, number the carbons closest to the double bond

Alkenes

• Location of the double bond is important, anything greater than 3 carbons must include a location–Prefix – location – suffix–Pent-1-ene

CH2CH

CH2CH2

CH3

IB Core Objective

• 10.1.11 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring ( ) and esters (RCOOR).

AMINES

Amines are organic compounds derived from ammonia, NH3.

One or more of the hydrogen atoms are replaced by alkyl groups.

Like alcohols, there can be primary, secondary and tertiary amines.

ammonia

Primary Amine

Secondary AmineTertiary Amine

N

H

H

H

N

R

H

H

N

R

R´

H

N

R

R´

R´´

Amines

methyl amineCH3

N

H

H

dimethyl amine CH3

CH3

N H

primary

secondary

CH3NH2

(CH3)2NH

BenzeneCH

CH

CH

CH

CH

CH

Six sided ring with alternating double bondsCompounds with benzene are often called aromatic.It is highly flammable with a sweet smell.

OH

Phenol

NH2

Phenylamine

Review of Functional Groups

• Alcohol: R-OH (-ol)• Halide: R-X (F, Cl, Br, I)• Ketone: R-CO-R or R2CO• Aldehydes: R-CHO• Benzene: C6H6

• Carboxyl: R-COOH

OH

Propan-1-ol1-

chloropropane

ClO

ButanonePropanal

O

H

Benzene

O

HO

Propanoic acid

ESTERS

Carboxylic Acid ESTER + Water

R — C

O

O

+OHR´–

Alcohol +

–H – R´R — C

O

O

+OHH–

Alcohol Carboxylic Acid ESTER Water

R-COO-R´

-oate-ol -oic

OHCH3 - +

CH3 –

MethanolEthanoic Acid

O

OC

H

O

CH3 –

O

OC

- CH3

methyl ethanoate

+OHH–

Water

Ester functional group

(methyl alcohol)

Esters

Esters have strong sweet smells which are often floral or fruity.

Ester Fragrance ethyl methanoate 3-methylbutyl ethanoate ethyl 2-methylbutanoate phenylmethyl ethanoate

raspberries pears apples jasmine

IB Core Objective

• 10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9.

• Those functional groups are alcohol, aldehyde, ketone, carboxylic acid, and halide.

10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9.

• Aldehydes, ketones, and halogens will give the molecule more polarity, which results in dipole-dipole forces. What would happen to the melting and boiling point when these functional groups are added?

• A: They would be higher.• Alcohol and carboxylic acid give rise to hydrogen bonding.

Would the melting points and boiling points be higher or lower than the aldehydes, ketones and halogens?

• A: They would be higher.• What about solubility?• Like dissolves like. More polar, more likely they will dissolve

in water.

IB Core Objective

• 10.2.1 Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarities.

10.2.1 Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarities.

• Chemically, alkanes are very unreactive.• Weird, when you think that most alkanes are used

for quick burning (propane, butane, etc.)• We use alkanes to store reactive metals such as

sodium.• Because alkanes have strong carbon to carbon and

carbon to hydrogen bonds, they require a high activation energy.

• When this activation energy is provided, the reaction is highly exothermic.

IB Core Objective

• 10.2.2 Describe, using equations, the complete and incomplete combustion of alkanes

10.2.2 Describe, using equations, the complete and incomplete combustion of alkanes

• Complete Combustion (Excess Oxygen)____C8H18 + ____O2 ____CO2 + ____H2O

• Incomplete Combustion (Limited Oxygen)C8H18 + O2 CO2 + CO + C + H2O

Evidence of Incomplete Combustion: Black soot or yellow/orange flame. ... Chim

chiminey.. Chim chiminey.. Chim Chim cher-oo

Why where chimney sweep so important in the early 1900’s?

10.2.3 Describe, using equations, the reactions of methane and ethane with chlorine and bromine.

10.2.4 Explain the reactions of methane and ethane with chlorine and bromine in terms of free-radical mechanism.

Substitution Reactions (10.2.3)

• One atom is kicked off by another more reactive group.

H

CH

H

C

H

H

H

Cl Cl

Cl

Cl

Un-bonded e- move directly to H to form

a new bond

H

Cl

1-Chloroethane

Substitution • With more available chlorine, the reaction will

continue to produce a mix of products.

CH

H

C H

Cl H

Cl

1,1-Dichloroethane

H C

H

C H

H

Cl Cl

1,2-Dichloroethane

C

C

Cl

Cl

Cl

Cl

Cl

Cl

hexachloroethaneWith large excess of Chlorine

Free Radical Substitution

• Radicals: Highly reactive molecules possessing a singly unpaired electron.

• There are three major steps:– Initiation– Propagation– Termination

Initiation

• Light is used to temporarily break the bond between either Bromine or Chlorine to form the free radical.

Cl ClCl Cl

Ya, well im taking my electron and going

home

Fine!! I`m taking mine then too!

Hey Ma!! I`m a free radical now!!!

Oh great!!! He`s so unstable when

he`s like this!!

Propagation

H

H

C

H

H

• Two possibilities:– 1) Cl radical reacts with an alkane– 2) Radical alkane reacts with Cl2(g)

Cl

C

H

H

H

H

H

C

H

Cl Cl

Termination• Three possibilities:– 1) Cl radical reacts with another Cl radical – 2) Cl radical reacts with an alkane radical– 3) Alkane radical react with another alkane radical

Cl ClCl Cl

Your electron or your life mate!

Just about to ask you the same thing!!

Let`s Do This!!!!

Bring it!!Ahhhhhhhhh!!!

Tired yet?NOPE!

Sigh...Guess we`ll be hanging out together for a

while eh?

Termination• Three possibilities:– 1) Cl radical reacts with another Cl radical – 2) Cl radical reacts with an alkane radical– 3) Alkane radical react with another alkane radical

C

H

H

H

C

H

H

H

Cl

C

H

H

H

C

H

H

H

Cl

C

H

H

H

C

H

H

H

Cl Cl

IB Core Objective

• 10.3.1 Describe, using equations, the reactions of alkenes with hydrogen and halogens.

10.3.1 Describe, using equations, the reactions of alkenes with hydrogen and halogens.

Addition reactionReaction in which double bond of alkene is

converted to a single bond. Two new bonds are formed with the molecule it reacted with.

When hydrogen reacts with an alkene and a nickel catalyst:

CH2 CH2 + H2 CH2 CH2

H HNi


• Alkenes are unsaturated, because they can undergo addition reactions across the double bond.

• Alkanes are saturated, because no more addition reactions can occur!

Mmmm, no more addition

reactions…I like the sound of that

Hydrogenation• This reaction requires more energy and a heated

platinum, nickel or palladium catalyst is needed for hydrogenation of the unsaturated bond.– Decreasing the number of unsaturated bonds will cause an

increase in the MP.

– Margarine is made from oils, by saturating some of the double bonds causing the liquid to become solid at room temp.

C

H

C

H

H

H

H

C

H H

H


• Bromination: • Alkanes and alkenes can be determined by

bromination. The orange bromine will disappear if there are double bonds present.

C

H

C

H

H

H

H

C

Br Br

H

Note: This reaction also takes place spontaneously at room temperature with

chlorine and iodine

IB Core Objective

• 10.3.2 Describe, using equations, the reactions of symmetrical alkenes with hydrogen halides and water.

10.3.2 Describe, using equations, the reactions of symmetrical alkenes with hydrogen halides and water.

• Symmetrical: If you cut the molecule in half right across the double bond, both sides would look the same (mirror images).


• A spontaneous reaction can occur with a hydrogen halide (HCl) and an alkene:

CH2 CH2 + ClH CH2CH2

H Cl


Reaction of Alkene with Water

What do you notice about the arrows?At a temperature of 300°C, and high pressure

(7atm), the equilibrium is driven to the right.Or H2SO4 could be used as a catalyst.This process is used to create ethanol.

CH2 CH2

OHHCH2 CH2 + OH2

IB Core Objective

• 10.3.3 Distinguish between alkanes and alkenes using bromine water.

• We learned about bromination previously.• We will be using bromine water to test for

alkenes in an experiment on Sunday, January 31st.

IB Core Objective

• 10.3.4 Outline the polymerization of alkenes.

• Polymers: long chain molecules formed by joining together of monomers.

• Need to know formation of polyethene, polychloroethene, and polypropene.

• Need to identify the repeating unit. • Example –(CH2-CH2-)n- for polyethene.

10.3.4 Outline the polymerization of alkenes.

• Polyethene, also known as polyethylene, is used for the manufacturing of plastics.

• Use ethene monomers to form polyethene polymers.

Addition Polymerization for Polyethene• Poly = many and Mer = Unit• Will continue adding from both ends until there are no

longer any ethene molecules available – Could potentially add to the other end too

H

H

C

H

H

C

H

H H

H

CC Radicals have formed and are highly reactive

H

H

C

H

H

C

H

C

H

H

HH

C CC C

HH H

H

H

H H

H

CC

H

H H

HCC

H


Poly(chloroethene)• Better known as PVC (PolyVinyl Chloride)• Used in pipes, siding, clothing, upholstery, and

inflatable toys/products.


Poly(chloroethene)• Formed by polymerisation of chloroethene.• Repeating unit is [-CH2-CHCl-]n

• Chloroethene monomer

• Polymerisation:

C CCl

H

H

H

C C C C CH H H

H H H H H

Cl Cl


Polypropene• Also known as polypropylene.• Used in manufacture of packaging, ropes,

plastic parts, and laboratory equipment.• Formed by propene monomers:

• Polypropene polymer:C C C C CH H H

H H H H H

CH3 CH3

C CCH3

H

H

H

IB Core Objective

• 10.3.5 Outline the economic importance of the reactions of alkenes.

We have already reviewed:• The hydrogenation of vegetable oils to make

margarine.• The hydration of ethene to manufacture

ethanol.• Polymerization to manufacture plastics.

IB Core Objective

• 10.5.1 Describe, using equations, the substitution reactions of halogenoalkanes with sodium hydroxide.

• 10.5.2 Explain the substitution reactions of halogenoalkanes with sodium hydroxide in terms of SN1 and SN2 mechanisms.

Objectives 10.5.1 and 10.5.2

• Fluorocarbons are extremely unreactive because the carbon-fluorine bond is so strong.

• Bromochlorodifluoromethane (commonly called Halon 1211) is used for extinguishing electrical fires and aircraft fires.

C C FH

HH

H

H

ATTACK OF THE NUCLEOPHILES

A long time ago---in a lab far, farAway, molecules began mixing.Little did some of the molecules knowThat there was a positive pole beingExposed to the dark side.These molecules knew they were Vulnerable to a great evil lead byDarth Hydroxide. The future of the Halogenoalkane forces lay in the handsOf Princess Halogen. However, the Attack of the nucleophiles could beToo great……


• Because halogens are more electronegative than carbon, the carbon has a slight positive charge.

• This positive charge makes it susceptible to attack by the nucleophiles.

+ Nucleophile

-


• An example of a nucleophilic substitution is bromoethane with hydroxide ion.

ethanolCH3CH2Br + OH-(aq) CH3CH2OH + Br-

The nucleophile attacks and docks, while Princess

Halogen escapes


• There are two mechanisms of substitution, SN1, or SN2.

• SN1 : One stands for unimolecular.– 1) Leaving group leaves taking an electron (slow, rate determining

step)

– 2) Carbocation is formed (Positively charged) • Intermediate step

– 3) Nucleophile donates electrons and forms new bond

• HL Understanding: Enantiomers– Products are a racemic mixture. A 50:50 product of right hand and

left hand rotation.

SN1 Substitution

ClC

H

H

H

δ+ δ––

R

–R

OH H C

–

OH

C

R

H

R

Attack of the nucleophile!

Princess Halogen escapes before the nucleophile

attacks!


• SN2 : An ‘attack from the rear’ substitution reaction – 1) Nucleophile attaches to the opposite side of the leaving

group.

– 2) Both are attached at the same time, hence the 2 because it is bimolecular.

– 3) Leaving group leaves and substitution is complete

• HL Objective: Enantiomers and Polarimetry– The product has the opposite rotation of the starting

material.

SN2 Substitution

Clδ+ δ––

H

– –

H

H COH

The nucleophile attacks from the

back!!

Princess Halogen realizes after the

nucleophile docks there is no hope!So she leaves…..


Primary, Secondary, and Tertiary Halogenoalkanes• Tertiary halogenoalkanes proceed via an SN1

mechanism, because it is too bulky to proceed via SN2.

• This “lack of room” for five groups is known as steric hindrance.

http://images.google.ae/imgres?imgurl=http://www.fugly.com/media/IMAGES/Random/crowded_phone_booth.jpg&imgrefurl=http://www.fugly.com/pictures/16577/crowded_phone_booth.html&usg=__MitIbXaW2ue4E-C6stSwNtZCaQc=&h=554&w=450&sz=48&hl=en&start=65&um=1&tbnid=zn_iNVv68VaH5M:&tbnh=133&tbnw=108&prev=/images?q=crowded&ndsp=21&hl=en&sa=N&start=63&um=1


• Primary halogenoalkanes proceed by SN2 mechanism.

• This is because SN1 is unfavorable because the carbon doesn’t like to form carbocations—too much positive charge.

• In tertiary halogenoalkanes the positive charge is spread over more atoms, called the positive inductive effect.


• Secondary halogenoalkanes can proceed by a mixture of SN1 and SN2.

ReviewWhat is the product for the below reaction, and

by which mechanism does it form?

CCH3

CH3CH3

Br + OH- →

Now, show the intermediate step for this SN1 mechanism

+ -CH3

CH3

BrC

CH3

-OH

Br-

OH- ion with 2-bromo,2-methylpropane (SN1)

2-methylpropan-2-ol

reaction equation

1(species reactingin the slowest step)

SN1

S (substitution)N(nucleophilic)

Br-

CH3

CH3

C

CH3

+ CH3

CH3

OHC

CH3

Using Curly Arrows to show movement of electrons

+ -CH3

H

BrC

H

-OH

CH3

H

OHC

H Br-

hydroxide ion with bromoethane (SN2)

ethanol

reaction equation

2(species reacting in the slowest step)

SN2

S (substitution)N(nucleophilic)

Using Curly Arrows to show movement of electrons

topic 10 organic chemistry. ib core objective 10.1.1 describe the features of a homologous series

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