Topic 10 Organic Chemistry IB Core Objective 10.1.1

Topic 10 Organic Chemistry IB Core Objective  10.1.1

Topic 10 Organic 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. 10.1.1 Describe the features of a 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. 10.1.1 Describe the features of a homologous series. 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 10.1.1 Describe the features of a homologous series.

Molecular models 1. 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. 10.1.2 Predict and explain the trends in boiling points of members of a homologous series. 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. 10.1.2 Predict and explain the trends in boiling points of members of a homologous series.

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 What is the Molecular and empirical formula? What is the Molecular and empirical formula? Pentanoic Acid Pentanoic Acid H

H H H O Condensed Condensed structural structuralformula formula C C C C OH

H H H H C CH (CH ) COOH 3 2 3 CH3(CH2)3COOH Writing structures R is sometimes used to represent the carbon chain.

ROH represents the functional group alcohol attached to some carbon chain. Rings: Benzene BenzeneRing Ring (Alternating double (Alternating double bonds) 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 Structural Isomers OH OH Propanol Propanol Primary Primaryalcohol alcohol o MP -127 MP -127 CoC

Make a molecular model of both. What is the 3 molecular 7 3 7 formula? CCHHOH OH Iso-propanol Iso-propanoloror2-Propanol 2-Propanol Secondary Secondaryalcohol alcohol o MP

-88 C MP -88oC 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 C 6. Mice Eat Peanut Butter! Or. Methane, Ethane, Propane, Butane The 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 Carbon sub chains have the suffix -yl Methy -CH3 = 1 carbon l = meth =Methyl (functional group) 4 2 4 2

1 5 3 5 1 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 didnt do it this way. Name: 2-Methylpentane Multiple Substiuent/Functional Groups Meth yl 4 1 mono

2 di 3 5 2 3 tri 1 4 tetra Meth yl 5 pent 6 hex In this case there are 2 methyl groups 7 hept on the same carbon. 8 oct 9 non Name: 2,2-Dimethylpentane 10 -dec Multiple Functional Groups Methy

l Methy l 5 3 6 di- is used because it indicates 2 of the same functional groups 4 2 Ethyl In this case you list the sub groups in alphabetical order first. Name: 3-Ethyl-3,4-dimethylhexane

1 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 Cl O OCl O O H

-OH (alcohol) R-CHO (aldehyde) C R H R-COR (Ketone) R-COOH (carboxylic acid) -X (halide) R HO O H

Simplified structure H 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 ALCOHOLS RCH2 OH and HCH2OH H H R COH H COH

H General formula H Methanol CH3OH PRIMARY ALCOHOLS (cont.) Structural formula for Ethanol (Ethyl Alcohol) : H H HC C OH H H 1-butanol: H H H H HC C C C OH H H H H 2-methyl-1-butanol: H H

H H 1 2 3 4 C HO C C C H H HCH H H H SECONDARY ALCOHOLS R R2CHOH R COH H

2-propanol 3-methyl-2-butanol H OH H HC C CH H H C3H7OH H H 4 H OH H

2 3 1 HC C C CH H HCH H H H C5H11OH TERTIARY ALCOHOLS R R3COH R C OH R

2-methyl-2-propanol H H HCH H 1 2 HC C 3 CH H OH H C4H9OH 10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes. ISOMERS of C4H10O

H H C H H C C OH H HC H H H H H H H HC C C C OH H H H H PRIMARY ALCOHOLS H TERTIARY ALCOHOL H HCH H

HC C CH H O H H H H H SECONDARY HC C C C H ALCOHOL H H O H H H H H H H HC CO C CH H H H H

ETHER 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 iodoCl Br Name? Example: H3C CH CH CH3

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. ALDEHYDES Aldehydes have the suffix al. ALKANALS are produced by the OXIDATION OF PRIMARY ALCOHOLS (Primary alkanols) RCH2OH RCHO H RC OH H

O RC H KETONES ALKANONES are produced by the OXIDATION OF SECONDARY ALCOHOLS (Secondary alkanols) R2CHOH R2CO H RC OH R O R C

R CARBOXYLIC ACIDS O RC OH ALDEHYDES are OXIDISED to CARBOXYLIC ACIDS O H3C C H Ethanal O MnO4(aq) H+ H3C C OH + 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 C 6. 10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C 6. 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 CH H2C CH2 CH2 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. H ammonia N H R

H N H H Primary Amine R N H R Secondary Amine R N R

R Tertiary Amine Amines methyl amine CH3NH2 dimethyl amine (CH3)2NH CH3 N primary H H CH3 N

CH3 H secondary CH Benzene HC CH HC CH CH Six sided ring with alternating double bonds Compounds with benzene are often called aromatic.

It is highly flammable with a sweet smell. NH2 OH Phenol Phenylamine Review of Functional Groups OO 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 Cl HO H 1Propanoic Butanone Propanal Benzene Propan-1-ol chloropropane acid ESTERS Alcohol

-ol + Carboxylic Acid -oic ESTER -oate + Water R-COO-R O O R OH +

RC RC O H Alcohol + Carboxylic Acid H OH O R ESTER Water O +

CH3 - OH CH3 C OH Methanol (methyl alcohol) Ethanoic Acid Ester functional group O + CH3 C O - CH3 methyl ethanoate H OH

Water 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 Why where chimney sweep so important in the early Why where chimney sweep so important in the early 1900s? 1900s? Evidence of Incomplete Combustion: Black soot or yellow/orange flame. ... Chim chiminey.. Chim chiminey.. Chim Chim cher-oo

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. Un-bonded e move - Un-bonded move directly to H toe form directly H to form a newtobond

a new bond Cl Cl Cl Cl H 1-Chloroethane Cl Cl Cl Cl

Cl H H C C H H Cl H Substitution With more available chlorine, the reaction will

continue to produce a mix of products. Cl Cl Cl Cl H Cl Cl Cl Cl Cl H

C Cl C H H H C C H H Cl

1,1-Dichloroethane 1,2-Dichloroethane With large excess of Chlorine With large excess of Chlorine H Cl Cl Cl C Cl C

Cl Cl Cl Cl Cl hexachloroethane 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. Oh great!!! He`s Fine!! I`m taking mine Fine!!then I`m too! taking mine then too! Cl Cl Oh great!!! He`s so when Ya,unstable well im taking my

so unstable when Ya, well im taking electron andthis!! goingmy he`s like electron andthis!! going he`shome like home Cl

Cl Hey Ma!! I`m a free Hey Ma!!now!!! I`m a free radical radical now!!! Propagation Two possibilities: 1) Cl radical reacts with an alkane 2) Radical alkane reacts with Cl2(g) H Cl Cl C H

H H H H Cl H Cl Cl Cl Termination Three possibilities:

1) Cl radical reacts with another Cl radical 2) Cl radical reacts Sigh...Guess be Sigh...Guesswe`ll we`llwith be an alkane radical 3) Alkane radical react with another alkane radical hanging out together for a hanging out together forJust a about to ask Your electron

or Just about to ask eh? Your while electron or you the same thing!! while your life mate!eh? you the same thing!! your life mate! Let`s Do This!!!! Let`s Do This!!!! Cl Cl Cl Cl Cl

Cl Tired yet? Tired Bringyet? it!! Bring it!! Cl Cl NOPE! NOPE! Ahhhhhhhhh!!! Ahhhhhhhhh!!! 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 H H H C H H H C H

H H H H H C Cl H Cl H

Cl Cl Cl Cl Cl H 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 reaction Reaction 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: H2C CH2 + H2 Ni H H2C H CH2 10.3.1 Describe, using equations, the reactions of alkenes with hydrogen and halogens. 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 reactionsI 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. H H H

H Margarine is made from oils, by saturating some of the double bonds causing the liquid to become solid at room temp. H C H H H C H H 10.3.1 Describe, using equations, the reactions of

alkenes with hydrogen and halogens. Bromination: Alkanes and alkenes can be determined by bromination. The orange bromine will disappear if there are double bonds present. Note: This reaction also Br Br spontaneously takes place Br Br at room temperature with chlorine and iodine H C H H

H C H H 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). 10.3.2 Describe, using equations, the reactions of

symmetrical alkenes with hydrogen halides and water. A spontaneous reaction can occur with a hydrogen halide (HCl) and an alkene: H2C CH2 + H HCl H2C Cl CH2 10.3.2 Describe, using equations, the reactions of symmetrical alkenes with hydrogen halides and water.

Reaction of Alkene with Water H2C CH2 + H2O H H2C OH CH2 What do you notice about the arrows? At a temperature of 300C, 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.

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 H H Will continue adding from both ends until there are no longer any ethene molecules available C CouldCpotentially add to the other end too H H H H H H C

H H C H H C C C C C H

H C H Radicals formed andhave are formed and highly reactiveare highly reactive H H C

H H H C C H H H C H Radicals H haveH H H C

H H 10.3.4 Outline the polymerization of alkenes. Poly(chloroethene) Better known as PVC (PolyVinyl Chloride) Used in pipes, siding, clothing, upholstery, and inflatable toys/products. 10.3.4 Outline the polymerization of alkenes. Poly(chloroethene) Formed by polymerisation of chloroethene. Repeating unit is [-CH2-CHCl-]n Cl Chloroethene monomer H C C

H H Polymerisation: H C H Cl C H H C Cl C

H H H C H 10.3.4 Outline the polymerization of alkenes. Polypropene Also known as polypropylene. Used in manufacture of packaging, ropes, plastic parts, and laboratory equipment. CH3 Formed by propene monomers: H C Polypropene polymer: H

H C H CH3 H C C H H C H CH3 H C C 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 H H C unreactive H

Fluorocarbons are extremely because the carbon-fluorine bond is so strong. Bromochlorodifluoromethane (commonly called Halon 1211) is used for extinguishing electrical fires and aircraft fires. H C F H Objectives 10.5.1 and 10.5.2 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 Objectives 10.5.1 and 10.5.2 An example of a nucleophilic substitution is bromoethane with hydroxide ion. CH3CH2Br + OH-(aq) CH3CH2OH + Br- ethanol The nucleophile attacks and docks, while Princess Halogen escapes Objectives 10.5.1 and 10.5.2 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 Princess Halogen escapes before the nucleophile attacks! Attack of the nucleophile! O O

H + C HR Cl Cl R H H O

R O RH Objectives 10.5.1 and 10.5.2 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 Princess Halogen

realizes after the nucleophile docks there is no hope! So she leaves.. The nucleophile attacks from the back!! H H O O H + C H

Cl Cl Objectives 10.5.1 and 10.5.2 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. Objectives 10.5.1 and 10.5.2 Primary halogenoalkanes proceed by SN2 mechanism. This is because SN1 is unfavorable because the carbon doesnt like to form carbocationstoo much positive charge. In tertiary halogenoalkanes the positive charge is spread over more atoms, called the

positive inductive effect. Objectives 10.5.1 and 10.5.2 Secondary halogenoalkanes can proceed by a mixture of SN1 and SN2. Review What is the product for the below reaction, and by which mechanism does it form? CH3 H3C C Br + OH- CH3 Now, show the intermediate step for this SN1 mechanism Using Curly Arrows to show movement of electrons OH- ion with 2-bromo,2-methylpropane

(SN1) BrCH3 + CH3 C BrCH3 CH3 Br CH3 C+ CH3 CH3

C OH CH3 OH - SN1 S (substitution) CH3 2-methylpropan-2-ol reaction equation N(nucleophilic) 1(species reacting in the slowest step)

Using Curly Arrows to show movement of electrons hydroxide ion with bromoethane H + C CH3 H Br CH3 H OH C OH Br-

H - ethanol SN2 S (substitution) (SN2) N(nucleophilic) reaction equation 2(species reacting in the slowest step)

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