Organic Chemistry: Introduction IB Topic 10

10.1 Introduction 10.1.1 Describe the features of a homologous series. 10.1.2 Predict and explain the trends in boiling points of members of a homologous series. 10.1.3 Distinguish between empirical, molecular and structural formulas. 10.1.4 Describe structural isomers as compounds with the same molecular formula but with different arrangement of atoms. 10.1.5 Deduce structural formulas for the isomers of non-cyclic alkanes up to C6. 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6.

What is organic chemistry? The study of carbon, the compounds it makes and the reactions it undergoes. Over 16 million carbon-containing compounds are known.

What is organic chemistry? Carbon Carbon can form multiple bonds to itself and with atoms of other elements. Carbon can only make four bonds since it has 4 valence electrons and most often bonds to H, O, N and S. Because the C-C single bond (348 kJ mol-1) and the C-H bond (412 kJ mol-1) are strong, carbon compounds are stable. Carbon can form chains and rings.

What is organic chemistry? Hydrocarbons Hydrocarbons are organic compounds that only contain carbon and hydrogen Types of hydrocarbons include Alkanes Alkenes Alkynes Aromatic

10.1.1 Describe the features of a homologous series. A homologous series is a series of related compounds that have the same functional group.

Differ from each other by a CH2 unit Can be represented by a general formula examples: CnH2n+2 (alkanes) or CnH2n (alkenes) or…

10.1.1 Describe the features of a homologous series. Watch out!!! Homologous compounds DO NOT have the same empirical formula!

Have similar chemical properties Have physical properties that vary in a regular manner as the number of carbon atoms increases Example: the alkanes

CnH2n+2 CnH2n # C Prefix Alkane (ane) Alkene (ene) 1 meth CH4 methane ethene 3 prop 4 but 5 pent 6 hex

10.1.2 Predict and explain the trends in boiling points of members of a homologous series. What is the trend? Why? Alkane Formula Boiling Pt./oC methane CH4 -162.0 ethane C2H6 -88.6 propane C3H8 -42.2 butane C4H10 -0.5 Intermolecular forces present & molar mass

Trends in boiling points of members of a homologous series (10.1.2) Melting point and boiling point increase with more carbon atoms Why? intermolecular forces increase adding a CH2 adds more electrons this increases the Van der Waal’s forces Alkane Formula Boiling Pt./oC methane CH4 -162.0 ethane C2H6 -88.6 propane C3H8 -42.2 butane C4H10 -0.5 Intermolecular forces present & molar mass

Intermolecular forces present 10.1.2 Predict and explain the trends in boiling points of members of a homologous series. Intermolecular forces present Simple alkanes, alkenes, alkynes → van der Waals’ forces (nonpolar) → lower b.p. Aldehydes, ketones, esters & presence of halogens (polar) → dipole: dipole forces → slightly higher b.p. Alcohol, carboxylic acid & amine → hydrogen bonding (w/ O, N, F) → even higher b.p. Naming song: https://www.youtube.com/watch?v=mAjrnZ-znkY

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

Empirical Formula: Molecular Formula: 10.1.3 Distinguish between empirical, molecular and structural formulas. Empirical Formula: Smallest whole number ratio of atoms in a molecule Molecular Formula: Formula showing the actual numbers of atoms Molecular Formula Empirical Formula CH4 C2H6 CH3 C6H12O6 C4H8 C8H16

10.1.3 Distinguish between empirical, molecular and structural formulas. Bond angles are drawn as though 90o. The true shape around C with 4 single bonds is tetrahedral and the angle is 109.5o. Show every atom and every bond. Can use condensed structural formulas. Hexane: CH3CH2CH2CH2CH2CH3 (condensed s.f.) M.F. = C6H14 E.F. = C3H7

Structural formula Can use condensed structural formulas bonds are omitted, repeated groups put together, side chains put in brackets CH3CH2CH2CH2CH2CH3 or even CH3(CH2)4CH3 CH3CH(CH3)CH3

10.1.3 Distinguish between empirical, molecular and structural formulas.

Skeletal formula Not accepted in the IB for answers but often used in questions… cuz that’s how they do it Every “corner” represents a carbon Hydrogens are implied

10.1.4 Describe structural isomers as compounds with the same molecular formula but with different arrangement of atoms. Structural isomers: compounds with the same molecular formula, but different arrangement of atoms The Fuse School: https://www.youtube.com/watch?v=9SX-iWpi98g

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

Different isomers are completely different compounds Have different physical properties such as melting point and boiling point

Structural Formulas for C4H10O Isomers

10.1.5 Deduce structural formulas for the isomers of non-cyclic alkanes up to C6. You should be able to draw out and write the structural formulas for all isomers that can be formed by: CH4 C2H6 C3H8 C4H10 C5H12 C6H14 Eventually you should be able to name all isomers, as well

10.1.5 Deduce structural formulas for the isomers of non-cyclic alkanes up to C6. If there is a branch off of the main chain, put that formula in parentheses CH3CH(CH3)CH3 CH3CH2CH2CH3

Monkeys Eat Peanut Butter 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6. Determine the longest carbon chain Use the prefix (next slide) to denote the number carbons in the chain Monkeys Eat Peanut Butter 1 Meth- 2 Eth- 3 Prop- 4 But- 5 Pent- 6 Hex-

Use the suffix “-ane” to indicate that the substance is an alkane Number the carbons in the chain consecutively, starting at the end closest to a substituent (groups attached to the main chain/most busy end)

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

1 Meth- 6 Hex- 2 Eth- 7 Hept- 3 Prop- 8 Oct- 4 But- 9 Non- 5 Pent- 10 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6. 1 Meth- 6 Hex- 2 Eth- 7 Hept- 3 Prop- 8 Oct- 4 But- 9 Non- 5 Pent- 10 Dec-

name and number the location of each substituent 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6. For chains longer than 4 carbons with side chains: name and number the location of each substituent the name of the substituent will be written before the main chain and will end with “–yl” (or just memorize the below) CH3 is methyl C2H5 is ethyl C3H7 is propyl

10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6. And with 2 or more side chains: Use prefixes di-, tri-, tetra-, to indicate when there are multiple side chains of the same type. Use commas to separate numbers and hyphens to separate numbers or letters. Name the side chains in alphabetical order. WHEW!!!!!!!!

How about C5H12? The isomers are: Pentane 2-methyl-butane 2,2-dimethyl propane

Nomenclature Practice Name this compound 1 9 carbons = nonane 2 4 3 5 6 7 8 9 Step #1: For a branched hydrocarbon, the longest continuous chain of carbon atoms gives the root name for the hydrocarbon

Nomenclature Practice Name this compound 9 carbons = nonane 1 2 4 3 5 6 CH3 = methyl 7 chlorine = chloro 8 9 Step #2: When alkane groups appear as substituents, they are named by dropping the -ane and adding -yl.

Nomenclature Practice Name this compound 9 carbons = nonane 1 2 4 3 5 6 CH3 = methyl 7 chlorine = chloro 8 9 1 9 NOT 9 1 Step #3: The positions of substituent groups are specified by numbering the longest chain of carbon atoms sequentially, starting at the end closest to the branching.

Nomenclature Practice Name this compound 9 carbons = nonane 1 2 4 3 5 6 CH3 = methyl 7 chlorine = chloro 8 9 2-chloro-3,6-dimethylnonane Step #4: The location and name of each substituent are followed by the root alkane name. The substituents are listed in alphabetical order (irrespective of any prefix), and the prefixes di-, tri-, etc. are used to indicate multiple identical substituents.

What about boiling points of isomers??? Pentane 2-methyl-butane 2,2-dimethyl propane

Magnitude of the force depends on… Number of electrons and size of the electron cloud with more electrons, valence electrons are farther away from the nucleus and can be polarized more easily Shape of molecules molecules with shapes that have more contact area have greater forces between them than those don’t

boiling point increases these round shapes do NOT allow them to stick to one another this flat shape allows it to stick to one another better boiling point increases

10.1 Introduction, cont. 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. 10.1.9 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. 10.1.11 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring ( ) and esters (RCOOR). 10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes. 10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9.

10.1.7 Deduce structural formulas for the isomers of the straight-chain alkenes up to C6. Remember that structural formulas show the relative location of atoms around each carbon Hexane: CH3CH2CH2CH2CH2CH3 (condensed s.f.) M.F. = C6H14 Determine the molecular formulas for the alkenes below. Draw out and write the structural formulas for all isomers that can be formed by each. C2H4 C3H? C4H? C5H? C6H?

10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C6. Alkenes have a double bond between two or more of the carbons CnH2n Draw out and write the structural formulas for all isomers that can be formed by each C2H4 C3H6 C4H8 C5H10 C6H12 45

10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C6. Suffix changes to “-ene” When there are 4 or more carbon atoms in a chain, the location of the double bond is indicated by a number. Begin counting the carbons closest to the end with the C=C bond Numbering the location of the double bond(s) takes precedence over the location of side chains 1-butene 2-butene methylpropene 46

Naming the isomers (IUPAC) of straight chain alkenes up to C6 (10.1.8) suffix changes to “-ene” when there are 4 or more carbon atoms in a chain, the location of the double bond is indicated by a number begin counting the carbons closest to the end with the C=C bond numbering the location of the double bond(s) takes precedence over the location of any substituents 1-butene 2-butene but-1-ene but-2-ene

Count the number of carbons in a chain 10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes up to C6. Breakin’ it down… Count the number of carbons in a chain Determine the ending of the name, based on # of bonds or functional group Determine any side chains, which will be placed at the front of the name 48

choose the correct ending Naming Practice!!! choose the correct ending ene

determine the longest carbon chain with the double bond ene

assign numbers to each carbon ene

assign numbers to each carbon ene

attach prefix (according to # of carbons) ene 1-hexene

determine name for side chains ethyl methyl methyl determine name for side chains 1-hexene 1-hexene

attach name of branches alphabetically ethyl methyl methyl attach name of branches alphabetically 2-ethyl-4-methyl-4-methyl-1-hexene

group similar branches ethyl methyl methyl group similar branches 2-ethyl-4-methyl-4-methyl-1-hexene

group similar branches ethyl methyl methyl group similar branches 2-ethyl-4,4-dimethyl-1-hexene or 2-ethyl-4,4-dimethyl hex-1-ene

propene 2,4-dimethyl-2-pentene 2,4-dimethyl pent-2-tene 2-butene

a) 3,3-dimethyl-1-pentene b) same C H 3 C H C C C H C H C H 3 3 C H 3 c) 4,5 dimethyl-2-hexene

Organic Chemistry Introduction: Functional Groups Topic 10.1.9 – 10.1.13

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. Functional group: a group of atoms that defines the structure of a family and determines its properties The functional group concept explained: The Chemistry Journey: The Fuse School: https://www.youtube.com/watch?v=nMTQKBn2Iss

http://textbookofbacteriology.net/themicrobialworld/chemoc.html

Functional Group Formula Structural Formula 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 Group Formula Structural Formula Alcohol -OH - O – H Aldehyde -COH (on the end of a chain) O - C – H Ketone -CO- (can’t be on end of chain) - C – Carboxylic Acid -COOH (on the end of a chain) - C – O – H Halide -Br, -Cl, -F, -I - X

Functional Group Formula Suffix (or Prefix) 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 Group Formula Suffix (or Prefix) Alcohol -OH -ol Aldehyde -COH (on the end of a chain) -al Ketone -CO- (can’t be on end of chain) -one Carboxylic Acid -COOH (on the end of a chain) -oic acid Halide -Br, -Cl, -F, -I bromo-,chloro-, fluoro-,iodo-

Know these 7, only have to recognize the 3 in the

Alcohols: suffix = “ol” propan-1-ol propan-2-ol 2-methyl propan-2-ol

Aldehydes: suffix = “al” propanal Note: an aldeyhde group is always on an end carbon so don’t need a number butandianal

Ketones: suffix = “one” propanone (don’t need C#, must be in between two carbons) butanone 2-pentanone or pentan-2-one

butandione pentan-3-one

Carboxylic Acids: suffix = “oic acid” butanoic acid Note: a carboxyl is always on an end carbon propandioic acid

Halides: prefixes = “fluoro, chloro, bromo, iodo” 1-bromopropane 2-chlorobutane 1,2-diiodoethane 1,2-difluoroethene 1,1,2-trifluorothene

Only identify the following functional groups in structures: (10.1.11) Formula Amine - NH2 Ester O R – C – O – R Benzene

amino, benzene ring, ester

A few more groups: Functional Group Formula Amine - NH2 Ester O 10.1.11 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring ( ) and esters (RCOOR). A few more groups: Functional Group Formula Amine - NH2 Ester O R – C – O – R Benzene (phenyl-) Ethyl ethanoate

Functional Groups

Identify each functional group by name… This is a possible idea for making flash cards. http://textbookofbacteriology.net/themicrobialworld/chemoc.html

10.1.11 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring ( ) and esters (RCOOR). Esters are used for fragrances and flavoring agents since one of their major properties is smell Benzene is in a family known as the aromatic hydrocarbons… because they smell 

Primary = carbon atom is only bonded to one other carbon 10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes. With reference to the carbon that is directly bonded to an alcohol group or a halogen: Primary = carbon atom is only bonded to one other carbon Secondary = carbon atom is bonded to two other carbons Tertiary = carbon atom is bonded to three other carbons

Types of Carbon Atoms Primary carbon (1o) a carbon bonded to one other carbon Secondary carbon (2o) two other carbons Tertiary carbon (3o) three other carbons

Secondary halogenoalkane Tertiary alcohol 10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes. Draw a… Primary alcohol Secondary halogenoalkane Tertiary alcohol What type are all aldehydes / carboxylic acids? Why? What type are all ketones? Why?

Volatility: how easily a substance turns into a gas 10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9. Volatility: how easily a substance turns into a gas The weaker the intermolecular force, the more volatile it is So, is a nonpolar or polar substance more volatile? ionic › hydrogen bonding › dipole-dipole › van der Wall’s (Fig. 10.35 for reference) Therefore, volatility: vdW › d-d › H alkane › halogenoalkane › aldehyde › ketone › amine › alcohol › carboxylic acid

Intermolecular forces present 10.1.2 Predict and explain the trends in boiling points of members of a homologous series. Intermolecular forces present Simple alkanes, alkenes, alkynes → van der Waals’ forces (nonpolar) → lower b.p. Aldehydes, ketones, esters & presence of halogens (polar) → dipole: dipole forces → slightly higher b.p. Alcohol, carboxylic acid & amine → hydrogen bonding (w/ O, N, F) → even higher b.p. Naming song: https://www.youtube.com/watch?v=mAjrnZ-znkY

Solubility: a solute’s ability to dissolve in a polar solvent (water) 10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9. Solubility: a solute’s ability to dissolve in a polar solvent (water) The more polar a substance is, the more soluble it is Solubility: If the functional group is soluble (hydrogen bonded), it will be more soluble Solubility decreases as chain length increases Smaller alcohols, aldehydes, ketones & carboxylic acids are typically soluble Halogenoalkanes are NOT soluble since they don’t form hydrogen bonds

Compare the boiling points of C2H6, CH3OH and CH3F 10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9. Which substance is most soluble: ethene, propene, prop-1-ene or hex-1-ene? Rank the following substances in order of increasing boiling point: C5H12, CH3CH2CH2CH2OH, CH3OCH2CH2CH3 Compare the boiling points of C2H6, CH3OH and CH3F Explain, at the molecular level, why ethanol is soluble in water, but cholesterol (C27H45OH) and ethane are not.