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Innisfail High School Chemistry 30 Ms. Littke Spring 2014 Unit A: Organic Chemistry.

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1 Innisfail High School Chemistry 30 Ms. Littke Spring 2014 Unit A: Organic Chemistry

2 Assessment  Formative  Students must participate in lessons - complete daily notes and examples.  Students are expected to self assess and seek help if necessary Complete daily activities and review new concepts regularly  F/S - Quizzes will be given periodically throughout the unit  You should use these to your advantage as they test smaller sections of the curriculum and help prepare you for the Unit Exam  Labs – 2 labs per unit  Summative Unit Exam – test format will be given to you later

3 Unit A: Chapters 9 and 10 O rganic C hemistry Today’s Objectives: 1)Define organic compounds as compounds containing carbon, recognizing inorganic exceptions such as carbonates, cyanides and carbides 2)Identify and describe significant organic compounds in daily life, demonstrating generalized knowledge of their origins and applications 3)STS: Demonstrate an understanding that science and technology are developed to meet societal needs and expand human capability

4 Section 9.1 (pg ) Organic Introduction Today’s Agenda: 1)Introduce organic chemistry and review the origins and applications of some major organic compounds 2)“Carbon – The element of life” video 3)Are You Ready pg. 354 #1-6 – due tomorrow

5 What is Organic Chemistry?  The early definition related to compounds obtained only from living things.  Today, it is a major branch of chemistry that deals with compounds of carbon, called ORGANIC compounds*.  *Carbon compounds that are exceptions and considered INORGANIC are compounds like: Oxides carbon monoxide (CO (g) ) and carbon dioxide (CO 2(g) ), and Ionic compounds of carbon-based ions, such as carbonate CO 3 2-, cyanide CN -, and carbide ions, SiC (silicon carbide)  The major source of carbon compounds is still living or previously living things, such as plants, animals and all types of fossil fuels.

6 Organic or Inorganic?? FormulaOrganic or Inorganic? CaCO 3(s) Inorganic (carbonate ion) C 25 H 52(s) Organic Ca 2 C (s) Inorganic (carbide ion) CCl 4(l) Organic CH 3 COOH (l) Organic CO 2(g) Inorganic (oxide) KCN (s) Inorganic (cyanide) C 12 H 22 O 11(s) Organic

7 Why is carbon special?  There are millions or organic compounds and only a thousand inorganic compounds. WHY?  Carbon has a bonding capacity of 4 Remember Lewis Dot Diagrams from Chem 20??Lewis Dot Diagrams from Chem 20??  This means carbon can bond extensively and can bond together to form chains effectively = called PolymerismPolymerism  Carbon covalently bonds by sharing 4 pairs of electrons. These bonds may be single, double or triple, all producing stable compounds  Compounds can form with same number of each type of atom but different structures = IsomerismIsomerism

8  So why do we care about bonding capacity?  If we know how many bonding e - ’s an atom has, we can predict what structure a molecular compound will have Determining Lewis Formulas AtomNumber of valence electrons Number of bonding electrons Bonding capacity carbon444 nitrogen533 oxygen622 halogens711 hydrogen111 H I.e. Carbon can form 4 single bonds, 2 double bonds, 1 triple and 1 single, or 1 double and 2 singles

9 Polymers  Examples of repeating carbon chains:carbon

10 Isomers  Compounds with the same number of each type of atom but different structures (C 4 H 10 ) Compounds  We will talk about this in more detail later

11 Importance of Organic Chemistry  Building units of all living matter: carbohydrates, proteins, fats  All foods are organic compounds  Photosynthesis is a reaction that makes carbon a part of our food. Carbon is passed along through food chains and sugar from photosynthesis is modified and combined with other materials.  Dead organisms are food for other organisms, or are buried in the earth and converted to fossil fuels like peat, coal and petroleum  Petroleum is the source of fuel and starting material for plastics, fabrics and industrial chemicals

12  The carbon cycle is an illustration of the interrelationship of all living things with the environment and with technologies that refine and use fossil fuels  We will continually outline the importance of organic compounds in our daily lives

13 Carbon: The Element of Life Complete the worksheet provided as you watch the following video (20 min)

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15 Today’s homework  Ensure video worksheet is complete and in your notes to review later  Work on Are you Ready pg # 1- 6 – due tomorrow  What is coming up tomorrow?  Naming alkanes, branched alkanes and cycloalkanes

16 Section 9.2 (pg ) Naming Organic Compounds Today’s Objectives: 1)Name and draw structural, condensed structural, and line diagrams and formulas for saturated and unsaturated aliphatic (including cyclic) Containing up to 10 carbon atoms in the parent chain/cyclic structure Containing only one type of a functional group or multiple bond Using the IUPAC nomenclature guidelines 2)Identify types of compounds from the functional groups, given the structural formula 3)Define structural isomerism and relate to variations in properties

17 Four Types of Formulas 1. Molecular FormulasC 5 H 10(g) Not very useful for organic compounds because so many isomers can exist 2. Structural Formulas 1. Condensed Structural Formulas 2. Line Diagrams – end of line segment represents carbon – it is assumed to satisfy each carbon’s octet

18 Naming Organic Compounds  Aliphatic Hydrocarbons – contains only hydrogen and carbon atoms  Straight line chains of carbon atoms  Alicyclic hydrocarbons have carbon atoms forming a closed ring. Still considered aliphatic AlkanesAlkenesAlkynes Only single C-C bonds Double C-C Bond present Triple C-C bond present General formula C n H 2n+2 General formula: C n H 2n General formula: C n H 2n-2 SaturatedUnsaturated

19  In organic chemistry, names have a root and a suffix.  The root describes the number of carbons present in the chain or ring.  The suffix describes the type of compound it is. Naming Organic Compounds

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21 Naming ALKANES 1. Find the parent chain (the longest continuous chain of carbon atoms). Use the appropriate root and the suffix-ane.

22 Naming ALKANES 1. Find the parent chain. Use the appropriate root and suffix. 2. Number the carbon atoms, starting from the end closest to the branch(es) so that the numbers are the lowest possible 3. Identify any branches and their location number on the parent chain (us the suffix –yl for branches) 4. Write the complete IUPAC name, following the format: (number of location, if necessary) – (branch name) (parent chain) 2-methylheptane

23  If more than one of the same branch exist, use a multiplier to show this (di, tri). Remember to include all numbers  Draw 2,4,6-trimethylheptane Naming ALKANES

24  If different branches exist, name them in alphabetical order ethyl before methyl (e before m in the alphabet) Naming ALKANES

25  If there is more than one branch of the same type, a locating number is given to each branch and a prefix indicating the number of that type of branch is attached to the name.  This numbering prefix does not affect the alphabetical order of the branches  Draw the structural formula for 3,4-dimethylhexane Naming ALKANES

26 Summary of Naming Alkanes 1. Find the parent chain. Use the appropriate root and suffix. 2. Number the parent chain carbon atoms, starting from the end closest to the branch(es) so that the numbers are the lowest possible 3. Identify any branches and their location number on the parent chain (us the suffix –yl for branches) 4. If more than one of the same branch exist, use a multiplier (di, tri) to show this. Remember to include all numbers 5. If different branches exist, name them in alphabetical order 6. Separate numbers from numbers using commas, and numbers from words using dashes (no extra spaces)

27 Don’t forget  Questions will specificallyask about structural, condensed structural or line structural formulas.  You must be comfortabledrawing any of the three

28 Practice  Write the IUPAC name for the following 2,5-dimethyl-4-propyloctane

29 Correct the following names:  4-ethyl-2-methylpentane  ACTUALLY 2,4-dimethylhexane  ALWAYS LOOK FOR LONGEST CHAIN!!

30 Correct the following name:  4,5-dimethylhexane  Actually 2,3-dimethylhexane  **Want branch numbers to be as low as possible

31 CYCLOALKANES  Based on evidence, chemists believe that organic carbon compounds sometimes take the form of cyclic hydrocarbons:  Cycloalkanes: Alkanes that form a closed ring  General Formula C n H 2n Two less hydrogens are present than in straight chain alkanes because the two ends of the molecule are joined Are these considered saturated?? Yes, because they have only single bonds and the max amount of hydrogen's bonded to the carbons Cyclo-compounds will have a higher boiling point than their straight chain partners (because there is an additional bond present)

32  Cycloalkanes are named by placing the prefix cyclo in front of the alkane name, as in cyclopropane and cyclobutane  If branches are present, treat the cycloalkane as the parent chain and identify the branches.  Since there is no end at which to start the numbering, use the lowest numbers possible Naming CYCLOALKANES

33 Name the following: ,2-dimethylcyclopentane ethylcyclohexane **Why don’t we need a number?

34 Today’s homework  Pg. 370 – 71 #7 – 11  Pg. 372 #5-6  What is coming up tomorrow?  Naming Alkenes, Alkynes, Cycloalkenes and Cycloalkynes  Comparing properties of Isomers

35 Section 9.3 (pg ) Naming Organic Compounds: Alkenes and Alkynes Today’s Objectives: 1)Name and draw structural, condensed structural, and line diagrams and formulas for saturated and unsaturated aliphatic (including cyclic Containing up to 10 carbon atoms in the parent chain/cyclic structure Containing only one type of a functional group or multiple bond Using the IUPAC nomenclature guidelines 2)Identify types of compounds from the functional groups, given the structural formula 3)Define structural isomerism and relate to variations in properties

36 Review:  Find and name all of the isomers of pentane (C 5 H 12(l) )

37 Structural Isomerism  Compound with the same molecular formula but different structures  They will have different chemical and physical properties – based on their different structures

38 Alkenes and Alkynes  Alkenes – hydrocarbons containing a double C-C bond  General formula (C n H 2n ) - (like cycloalkanes)  Alkynes – hydrocarbons containing a triple C-C bond  General formula (C n H 2n-2 ) – (like cycloalkenes)  Alkenes and Alkynes are considered unsaturated compounds. They do not have the maximum number of hydrogen atoms surrounding each carbon.  Reactivity: Alkynes (highest), Alkenes, Alkanes (lowest)

39 Naming Alkenes and Alkynes 1. Find the parent chain. It MUST contain the multiple bond.  If the bond is a double, the suffix for the parent chain will be -ene  If the bond is a triple, the suffix for the parent chain will be –yne 2. Count carbon atoms so that the multiple bond will be on the lowest possible number. Indicate the number that the multiple bond falls on directly before the suffix 3. Name branches as before

40 Naming Alkenes and Alkynes  Draw the following as condensed structural formulas: 4-methylpent-2-yne methylpropene (why don’t we need a number?)

41 Naming Alkenes and Alkynes  Name the following: 3-methylbut-1-ene 5-methylhex-2-ene

42 Naming Alkenes and Alkynes 4. It is possible for a molecule to have more than one double bond. These are called alkadienes and have the same general formula as alkynes (C n H 2n-2 )  If this is the case, indicate both numbers where the double bond is formed, and change the suffix to –diene. a) Draw buta-1,3-diene: b) What is the IUPAC name for the following: buta-1,2-diene

43 Cycloalkenes and -ynes  The rules for naming cycloalkenes and cycloalkynes are the same as naming cycloalkanes  The numbering for the carbon atoms begins with the double bond; the carbons of the double bond are carbons 1 and 2; lowest numbers possible  Draw 3-methylcyclohexene as a condensed structural formula

44 Today’s homework  Pg. 377 #1-5  Pg. 380 #6, 7, 11  What is coming up tomorrow?  Hydrocarbon Quiz #1  Naming Aromatics  Boiling point and Chemical Properties Analysis

45 Section 9.4 (pg ) Naming Organic Compounds: Aromatics Today’s Objectives: 1)Name and draw structural, condensed structural, and line diagrams and formulas for aromatic carbon compounds Using the IUPAC nomenclature guidelines 2)Identify types of compounds from the functional groups, given the structural formula 3)Define structural isomerism and relate to variations in properties

46 Hydrocarbon Quiz #1

47 Aromatics  Originally, organic compounds with an aroma or odour were called aromatic compounds  Now, aromatics refer to compounds containing a benzene-ring structure  Benzene’s formula is C 6 H 6, which would suggest a highly unsaturated and reactive compound  Benzene is actually quite unreactive and is considered more stable than alkenes and alkynes  Did You Know?? Benzene is a carcinogen and is found naturally in petroleum – why would this be a problem?

48  What do we know about benzene?  Formula is C 6 H 6 (3D link)3D link  Unreactive – so no true double or triple bonds  Carbon-carbon bonds are the same length and strength  Each carbon is bonded to a hydrogen  So what does benzene look like?? The three double bonds resonate resulting in an overall bond length somewhere in between a single and a double bond, explaining benzene’s stability We will use this line structural formula to represent benzene in compounds

49 Common Aromatic Compounds  Include Aspirin and Vanillin (one of the flavour molecules in vanilla)  You will notice many aromatic molecules are often depicted using a condensed structural formula except for the benzene ring, which is shown as a line structural formula.  This combination is commonly used by chemists, and we will use this method when drawing aromatics.

50 Naming Aromatics 1. If an alkyl branch is attached to a benzene ring, the compound is named as an alkylbenzene.  Alternatively, the benzene ring may be considered as a branch of a large molecule: in this case, the benzene ring is called a phenyl branch. Which has a phenyl branch? An alkylbenzeneContains a phenyl branch

51 Naming Aromatics 2. If more than one alkyl branch is attached to a benzene ring, the branches are numbered using the lowest numbers possible, starting with one of the branches.  Given the choice between two sets of lowest numbers, choose the set that is in both numerical and alphabetical order 1-ethyl-2,4-dimethylbenzene3-phenyl-4-propyloctane

52 Practice Naming Aromatics  Draw 1,2-dimethylbenzene  Are there any isomers of this compound?  There is also classical way of naming these isomers. The arrangements are denoted by the prefixe:s ortho (o), meta (m) and para (p). These names are still used in industry so you may encounter them in other references. 1,2-dimethylbenzene1,3-dimethylbenzene1,4-dimethylbenzene o-dimethylbenzenem-dimethylbenzenep-dimethylbenzene

53 Practice Naming Aromatics  Draw the line structural formula for 1-ethyl-3-methylbenzene  Draw the line structural formula for 2-phenylpentane

54 Practice Naming Aromatics  Draw 3-phenylpent-2-ene  Name the following propylbenzene Why is no number needed? Are the hydrogen’s wrong??

55 Summary  We have now learned about both aliphatic and aromatic hydrocarbons. You will need to be comfortable naming all of the following:

56 Teacher Note:  End if not enough time for two dry labs  Can use during the Station Lab Periods

57 Review of Intermolecular Forces  London Forces – temporary dipoles resulting from an uneven distribution of e - in all molecules  Temporary (-) end will repel e - in neighbouring molecules and so on  Depends on size of molecule (number of e - ’s)  Weakest of the intermolecular forces  Dipole-Dipole – only exists in polar molecules  Attraction between + and – ends of molecule  Hydrogen Bonding – super strong force  Only exists when H-N, H-F, H-O bonds are present

58 Applications  Read pg. 384 Lab Exercise 9.A – complete the parts in red  Purpose: To test the generalization that aromatic hydrocarbons react like saturated rather than unsaturated hydrocarbons  Design: cyclohexane, cyclohexene and benzene are all mixed with potassium permanganate (purple). Evidence for a reaction is a change in the initial purple colour of the solution.  Prediction: Based on your current knowledge, predict the order in which the compounds will react, from least reactive to most reactive. Explain your reasoning.  Analysis: On the basis of the evidence, determine the order of the reaction rate for the three compounds.  Evaluation: Determine if your prediction was verified or falsified. Was the generalization about aromatic hydrocarbons acceptable based on the evidence? Was the purpose of the investigation accomplished?

59 Sample lab report for 9.A

60 Applications  Read pg. 384 Lab Exercise 9.B – complete the parts in red  Purpose: To test the ability of the concept of London forces to predict the relative boiling points of aliphatic and aromatic compounds (Remember: the more electrons in a compound = the stronger the intermolecular forces = the higher the boiling point needed to pull the molecules apart)  Problem: What is the relative order of the boiling points of hexane, hex-1-ene, cyclohexane, cyclohexene, and benzene?  Prediction: Determine the number of electrons in each molecule and use these numbers to determine the order of boiling points  Analysis: On the basis of the evidence given, determine the order of the boiling points. (from lowest to highest)  Evaluation: Determine if your prediction was verified or falsified. Was the predictive power of the concept of London forces judged to be acceptable based on the evidence? Was the purpose of the investigation accomplished?

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62 Extra Practice for Lab Reports  Complete pg. 355 #6  Complete the prediction, analysis and evaluation  Remember lab report guidelines in textbook starting on pg. 790

63 Today’s homework  Pg. 385 #3-8 – due tomorrow  What is coming up tomorrow?  Review Aromatic and Aliphatic Compounds  Refining

64 Section 9.5 and 9.6 (pg ) Properties of Organic Compounds & Crude Oil Refining Today’s Objectives: 1)Compare boiling points and solubility of organic compounds 2)Describe fractional distillation and solvent extraction 3)Describe major reactions for producing energy and economically important compounds from fossil fuels

65 Physical Properties of Simple Hydrocarbons AlkanesNon-polar molecules Only intermolecular forces are London Force Boiling point and melting point increase with number of carbons All insoluble in water (like dissolves like) – nonpolar and polar don’t mix 1-4Cs = gas, 5-16Cs = liquid 17 and up = solid at SATP AlkenesNon-polar molecules, therefore insoluble in water Boiling points slightly lower than alkanes with the same number of carbons due to less electrons (unsaturated), resulting in lower London Forces AlkynesNon-polar molecules, therefore insoluble in water In theory boiling points slightly lower than alkenes with the same number of carbons due to less electrons (unsaturated), resulting in lower London Forces BranchingThe more branching, the less significant the London Force (~lower b.p.) - more surface area in straight chain hydrocarbons allows more separation of charge, resulting in greater London Force - see Table #3 pg. 378 (i.e. pentane (with 5Cs) has a b.p. of 36 o C which is much higher than dimethylpropane (5Cs) -12 o C) = because branching decreased the strength of the London force

66 Sample Question  Predict the relative order of boiling points of the following compounds (lowest to highest). Explain your reasoning. butanol but-1-ene cyclobutane butanoic acid butane Lowest > Highest but-1-ene cyclobutane butane butanol butanoic acid Reasoning: but-1-ene has lower LF’s for than butane because it is unsaturated, cyclobutane has lover LF’s than butane, butanol has H- bonding, butanoic acid has stronger polar forces)  Which would be soluble in water?  Butanol and butanoic acid – because they are the only polar molecules and like dissolves like!

67 Crude Oil Refining  Crude oil is a complex mixture of hundreds of thousands of compounds, all of which have different boiling points  We can take advantage of these different b.p.’s and physically separate the different components using heat  This process is called fractional distillation or fractionation

68  A fractional distillation tower contains trays positioned at various levels.  Heated crude oil enters near the bottom of the tower.  The bottom is kept hot, and the temperature gradually decreases toward the top of the tower.  As compounds cool to their boiling point, they condense in the cooler trays. The streams of liquid (called fractions) are withdrawn from the tower at various heights along the tower. Electronic Visual

69 Fractional Distillation Video

70  A more detailed look…  The vaporized components of the crude oil rise and gradually cool.  To get from one level to the next, the vapours are forced to bubble through the liquid condensed in each tray.  The figure shows the bubble caps used to allow this to happen.  If a gas cools to its boiling point, it will condense and be piped out through the draining tube  Q: How does the number of carbon atoms in a hydrocarbon chain affect its boiling point?  Smaller molecules have fewer electrons, so weaker London forces compared with larger molecules. The fractions with higher boiling points are found to contain much larger molecules (see Table 1 pg. 387) Crude oil is heated in the fractionation tower without air being present to reduce the risk of mixtures starting to burn or explode

71 Cracking  Cracking: large hydrocarbons are broken into smaller fragments  Historically, thermal cracking used extremely high temperatures but created large quantities of solid coke.  Now, catalytic cracking uses a catalyst to speed up the reaction and produce less residual products like tar, asphalt and coke Example: C 17 H 36(l)  C 9 H 20(l) + C 8 H 16(l) + C (s) larger molecules  smaller molecules + carbon  In 1960, hydrocracking was developed, which combines catalytic cracking and hydrogenation and produces no coke. Example: C 17 H 36(l) + H 2(g)  C 9 H 20(l) + C 8 H 16(l) larger molecule + hydrogen  smaller molecules

72 Oil Refining  The refining of crude oil can be divided into two main categories: 1. Physical Processes Fractional Distillation: see previous slides Solvent Extraction: solvent is added to selectively dissolve and remove an impurity or to separate a useful product from a mixture 2. Chemical Processes Cracking – larger molecules are broken down into smaller ones Reforming – large molecules are formed from smaller ones These chemical processes are needed because fractional distillation does not produce enough of the hydrocarbons that are in demand (i.e. gasoline) and produces too much of the heavier fractions

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74 Catalytic Reforming and Alkylation  Catalytic Reforming: improves the quality of the gasoline aliphatic molecule  aromatic molecule + hydrogen  Alkylation : increases the branching; improves the quality of the fuel aliphatic molecule  more branched molecule  (AKA: isomerization because it converts molecules into a branched isomer) FYI pg. 392 on Octane Numbers

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76 Combustion Reactions  Burning of hydrocarbons in the presence of oxygen  Complete Combustion: abundant supply of oxygen; products are carbon dioxide, water vapour and heat Ex. C 3 H 8(l) + 5O 2(g)  3CO 2(g) + 4H 2 O (g)  Incomplete Combustion: limited supply of oxygen; products are carbon monoxide, soot (pure carbon) or any combination of carbon dioxide, carbon monoxide and soot in addition to water vapour and heat Ex. 2C 8 H 18(l) + 17O 2(g)  16CO (g) + 18H 2 O (g) OR 2C 8 H 18(l) + 9O 2(g)  16C (s) + 18H 2 O (g) ** Assume complete combustion unless specified otherwise

77 Balancing FYI Ex. 2C 8 H 18(l) + 17O 2(g)  16CO (g) + 18H 2 O (g) Can also be balanced using a fraction (you need to be comfortable using this method) – divide each number by 2 C 8 H 18(l) + 17/2 O 2(g)  8CO (g) + 9H 2 O (g) Ex. 2C 8 H 18(l) + 9O 2(g)  16C (s) + 18H 2 O (g) can also be balanced as … C 8 H 18(l) + 9/2 O 2(g)  8C (s) + 9H 2 O (g)

78 Simple Distillation

79 Today’s homework  Read pages  Pg. 391 #9,  Pg. 397 #1, 4, 5, 8, 12, 13  Read pages  What is coming up tomorrow?  Chapter 9 Exam Review (Unit A Part 1)  Exam on Friday  Functional Groups and Hydrocarbon Reactions Halides, alcohols, carboxylic acids, esters and polymers Addition, Substitution and Elimination, esterfication and polymerization

80 Aliphatic and Aromatic Compounds Hydrocarbon Quiz #2

81 Section 10.2 and 10.3 (pg ) Organic Chemistry: Organic Halides and Alcohols Today’s Objectives: 1)Name and draw structural, condensed structural and line diagrams and formulas for organic halides and alcohols 2)Identify types of compounds from their functional groups, given the structural formula and name of the functional groups

82 Today’s Agenda: Review  You will have another Hydrocarbon Quiz next class covering all aliphatic and aromatic compounds  Today’s agenda:  Review Quiz – go over common mistakes  Pg. 385 #6-8 **7d) 4-methyl (not 3-methyl) Make the change in your book if not done already

83 Review of Intermolecular Forces  London Forces – temporary dipoles resulting from an uneven distribution of e - in all molecules  Temporary (-) end will repel e - in neighbouring molecules and so on  Depends on size of molecule (number of e - ’s)  Weakest of the intermolecular forces  Dipole-Dipole – only exists in polar molecules  Attraction between + and – ends of molecule  Hydrogen Bonding – super strong force  Only exists when H-N, H-F, H-O bonds are present

84 Organic Halides  Organic compounds where one or more hydrogen has been replaced with halogens (F, Cl, Br, I)  Common example: CFC (chlorofluorocarbons)  Nomenclature is similar to naming branch chains of hydrocarbons, but the branch name used is based on the halogen used  chloro-, fluoro-, bromo-, iodo-

85 Organic Halides  What do you need to know about organic halides?  May by polar or nonpolar molecules or may have a relatively nonpolar (hydrocarbon) end and a polar (halide) end (**Remember Electronegativity differences)  Have higher boiling points than similar hydrocarbons  Have very low solubility in water but higher solubility than similar hydrocarbons  Are typically good solvents for organic materials such as fats, oils, waxes, gums, resins or rubber  Usually toxic or ecologically damaging (DDTs and PCBs)

86 Practice Naming Organic Halides  Draw 1,2-dichloroethane  Draw 2,2,5-tribromo-5-methylhexane

87 Practice Naming Organic Halides  Name the following: CH 2 Cl 2 1,2-dibromoethene Bonus: Try 1,2-dibromo-1,2-dichloroethene chlorobenzene dichloromethane

88 Did You Know?  There is always a new concept to learn that extends what you have already learned.  In addition to the structural isomers that you know about, there are cis and trans isomers

89 Alcohols  An alcohol is an organic compound that contains the –OH functional group (hydroxyl)  General formula is R-OH (R = rest of molecule)  Alcohols are classified as primary, secondary or tertiary depending on the number of carbons bonded to the carbon that contains the hydroxyl group

90 Common Alcohols  Methanol (also called wood alcohol) is extremely toxic, causing death and blindness  Ethanol (also known as grain alcohol) is the alcohol found in alcoholic beverages and is used in the production of vinegar  Gas line antifreeze, windshield de-icer, windshield washer fluid – all contain methanol

91 Naming Alcohols 1. Locate the longest chain that contains an –OH group attached to one of the carbon atoms. Name the parent alkane 2. Replace the –e at the end of the name of the parent alkane with –ol (i.e. butane becomes butanol) 3. Add a position number before the suffix –ol to indicate the location of the –OH group  REMEMBER to number the main chain of the hydrocarbon so that the hydroxyl group has the lowest possible position number propan-1-ol

92 Naming Alcohols 4. If there is more than one –OH group (called polyalcohols), leave the –e in the name of the parent alkane and put the appropriate prefix before the suffix –ol (i.e. diol, triol, tetraol) 4. Name and number any branches on the main chain. Add the names of these branches to the prefix.  Draw 2,3-dimethylbutan-2-ol

93 Practice Naming Alcohols  Draw line structural formulas for: 1. cyclohexanol 2. phenol 1. The three isomers of C 5 H 11 OH that are pentanols These are the only two cyclic or aromatic alcohols you will need to know as they get very complicated

94 Alcohols  What do you need to know about alcohols? Question: Why is the propane used in a barbecue a gas at room temperature, but propan-2-ol (also known as rubbing alcohol) a liquid at room temperature? Answer: Propane is a non-polar hydrocarbon with weak intermolecular forces, thus it has a low boiling point and is a gas at room temperature. Propan-2-ol is an alcohol, with a polar hydroxyl group and strong intermolecular forces, thus it has a higher boiling point than propane and is a liquid at room temperature

95 Alcohols  What do you need to know about alcohols? Question: Glycerol (propane-1,2,3-triol) is more viscous than water, and can lower the freezing point of water. When added to biological samples, it helps to keep the tissues from freezing, thereby reducing damage. From your knowledge of the molecular structure of glycerol, suggest reasons to account for these properties of glycerol. Answer: Each molecule of glycerol contains three hydroxyl groups which can hydrogen-bond with water, interfering with the attractions between water molecules and thus interfering with the freezing of water. When water in tissues does not freeze, there is less damage to the tissues.

96 Application Question  Predict the order of increasing boiling points for the following compounds, and give reasons for your answer. butan-1-ol pentane 1-chlorobutane  Answer: (Lowest b.p.): pentane, 1-chlorobutane, butan-1-ol Why? All molecules have a similar number of electrons. Pentane has the lowest boiling point, because it is non-polar so will only have London forces between the molecules. 1- chlorobutane is polar so will have dipole-dipole forces as well as London forces. Butan-1-ol has the highest boiling point because its molecules will have all three intermolecular forces, most importantly, hydrogen bonding

97 Today’s homework  Pg. 418 #1-5  Pg 424 #2-3  Pg. 430 #5,6,7,8  Due tomorrow  What is coming up tomorrow?  Carboxylic Acids and Esters

98 Section 10.4 (pg ) Organic Chemistry: Carboxylic Acids and Esters Today’s Objectives: 1)Name and draw structural, condensed structural and line diagrams and formulas for carboxylic acids and esters 2)Identify types of compounds from their functional groups, given the structural formula and name of the functional groups

99 Carboxylic Acids  A carboxyl group is composed of a carbon atom double bonded to an oxygen atom and bonded to a hydroxyl group (-COOH)  Note: Because the carboxyl group involves three of the carbon atom’s four bonds, the carboxyl is always at the end of a carbon chain or branch methanoic acid ethanoic acid Examples: Carboxylic acids are weak organic acids

100 Naming Carboxylic Acids 1. Name the parent alkane 2. Replace the –e at the end of the name of than parent alkane with –oic acid 3. The carbon atoms of the carboxyl group is always given position number 1. Name and number the branches that are attached to the compound. Draw 3-methylbutanoic acid HOOC Remember COOH or HOOC can also represent the carboxyl group

101 Naming Carboxylic Acids 1. Draw trichloroethanoic acid (key ingredient in chemical peels) CCl 3 COOH 2. Draw 3-propyloctanoic acid Do you need the CH 3 here? No – but sometimes you will see it written this way. Don’t be confused because it doesn’t change the meaning

102 Esters  The reaction between a carboxylic acid and an alcohol produces an ester molecule and a molecule of water  This reaction is known as a condensation or esterification reaction  The ester functional group –COO– is similar to that of a carboxylic acid, except that the H atom of the carboxyl group has been replaced by a hydrocarbon branch.  Esters are responsible for natural and artificial fragrance and flavourings in plants and fruits.

103  In naming an ester you have to recognize that an ester has 2 distinct parts. The main part contains the C=O group which comes from the parent acid. The second part is the alkyl group. Esters

104 Naming Esters 1. Identify the main part of the ester, which contains the C=O group. This part comes from the parent acid. 2. Begin by naming the parent acid but replace the –oic acid ending of the name with –oate. (propanoic acid becomes propanoate) 3. The second part is the alkyl group that is attached to the single oxygen atom. Name this as you would any other alkyl group (in this case = methyl) 4. Put the names together. Note that esters are named as two words.

105 Naming Esters  Name the following ester and the acid and alcohol from which it can be prepared. ethyl butanoate ethanolbutanoic acid water Tip: The branch attached to the oxygen (of the –COO) comes first in the name, the chain attached to the carbon (of the –COO) comes second A strong acid catalyst, such as H 2 SO 4(aq) is used along with some heating to increase the rate of the organic reaction

106 Naming Esters  Name the following ester and the acid and alcohol from which it can be prepared. ethyl benzoate ethanolbenzoic acidwater

107 Today’s homework  Pg. 438 #1,2  Pg. 441 #3-4  Due tomorrow  What is coming up tomorrow?  Organic Reactions  Summary and Review of all organic molecules  Full Naming Quiz (in two days)

108 Physical Properties of Hydrocarbon Derivatives AlcoholsMuch higher boiling points than hydrocarbons (1-12Cs are liquids at SATP) due to hydrogen bonding between hydroxyl groups of adjacent molecules Small alcohols are totally miscible in water, but the larger the hydrocarbon part of the alcohol (nonpolar part), the more nonpolar the alcohol is Carboxylic Acids Like alcohols they have hydrogen bonding, but is more significant due to the C=O. This means greater bps and solubility than alcohols with same number of Cs. Carboxylic acids with 1-4Cs are completely miscible in water EstersFruity odour in some cases Polar but they lack the –OH bond therefore do not have hydrogen bonding, so lower bps than both alcohols and carboxylic acids Esters with few carbons are polar enough to be soluble in water CompoundBoiling Point ( o C) butane-0.5 butan-1-ol117.2 butanoic acid165.5

109 Summary Table - Organic Complete the summary table of all of the organic compounds we have studied. This will be a good page to reference when studying

110 Section 10.2,10.3 and 10.4 (pg ) Hydrocarbon Reactions: Addition, Substitution and Elimination Today’s Objectives: 1)Define, illustrate and provide examples of simple addition, substituion, elimination and esterification (condensation) 2)Predict products and write and interpret balanced equations for the above reations

111 All Organic Compounds Functional Groups Quiz

112 Hydrocarbon Reactions 1. Addition alkenes and alkynes + H 2(g)  alkanes (hydrogenation) alkenes and alkynes + HX (or X 2 )  organic halides 2. Substitution alkanes and aromatics + X 2  organic halides 3. Elimination alcohols  alkenes + water (dehydration) organic halides + OH -  alkenes + halide ion + water 4. Esterification (already covered, just review) carboxylic acid + alcohol  ester + water 5. Polymerization – monomer + monomer = polymer addition and condensation

113 1. Addition Reactions: reaction of alkenes and alkynes with hydrogen gas, a halogen compound, a hydrogen halide compound or water.  Addition reactions usually occur in the presence of a catalyst a) Addition with H 2(g) (also called hydrogenation)

114 1. Addition Reactions: reaction of alkenes and alkynes with hydrogen gas, a halogen compound, a hydrogen halide compound or water b) Addition of a halogen Since addition reactions involving multiple bonds are very rapid, the alkene product 1,2-dibromoethene can easily undergo a second addition step to produce 1,1,2,2-tetrabromomethane. Excess bromine promotes this second step. 1,2-dichloroethane is a useful solvent and is used to produce chloroethene, the monomer used to make PVC. See Pg. 417 equation

115 1. Addition Reactions: reaction of alkenes and alkynes with hydrogen gas, a halogen compound, a hydrogen halide compound or water c) Addition of an HX (hydrogen halide) molecule Show both possible isomers when predicting the products The addition of hydrogen halides (HF, HCl, HBr or HI) to unsaturated compounds can produce structural isomers, since the hydrogen halide molecules can add in different orientations. However, the type of isomers produced are not always equal. For example only a tiny amount of 1-chloropropane is produced compared to 2-chloropropane in the above reaction.

116 Practice Addition Reactions  For each of the following questions, draw condensed structural diagrams and name all products 1. ethene + bromine  2. hydrogen chloride + ethene  3. 3-methylbut-1-yne + excess hydrogen  1,2-dibromoethane chloroethane __ 2-methylbutane

117 2. Substitution Reactions – breaking of a C-H bond in an alkane or an aromatic ring and replacing it with another atom or group of atoms  Usually occur slowly at room temperature, so light may be necessary as a catalyst  Often substitutes a halogen for a hydrogen  No change in saturation Propane contains hydrogen atoms bonded to end carbons and the middle carbon atom, so two different products (isomers) are formed, in unequal proportions

118 2. Substitution Reactions – breaking of a C-H bond in an alkane or an aromatic ring and replacing it with another atom or group of atoms Benzene rings are stable, like alkanes, so they react slowly with halogens, even in the presence of light. As with alkanes, further substitution can occur in benzene rings, until all hydrogen atoms are replaced by halogen atoms (in the presence of excess halogens)

119 Practice Substitution Reactions  For each of the following questions, draw condensed structural diagrams and name all products 1. propane + fluorine  2. ethane + chlorine  1-fluoropropane + 2-fluoropropane + hydrogen fluoride chloroethane + hydrogen chloride H - F F - F

120 3. Elimination Reactions – involves eliminating atoms or groups of atoms from adjacent carbon atoms; decreases the level of saturation a) Alkane cracked into an alkene (uses high temperatures) b) Alcohol is reacted with a catalyst to produce an alkene and water (dehydration – removes a water molecule from the alcohol) a) Alkyl halide reacts with a hydroxide ion (OH - ) to produce an alkene (dehydrohalogenation – removes a hydrogen and halogen atom)

121 Practice Elimination Reactions  Write a structural formula equation for the preparation of but- 2-ene from chlorobutane, in the presence of a strong base  Write a structural formula equation for the preparation of but- 2-ene from butan-2-ol

122 Reaction type Complete Combustion AdditionEliminationSubstitutionEsteri- fication Reactants Hydrocarbon* + O 2 ene or yne + HOH, X 2, HX, or H 2 alcohol or alkyl halide alkane or aromatic + X 2 alcohol + carboxylic acid ProductsCO 2 (g) + H 2 O(g) or (l) alcohol, alkyl halide*, alkane alkene + HOH or HX alkyl halideester + water Othertriple or double to single single to double slow; needs uv light acid catalyst needed Othermore bonds*fewer bonds*

123 Today’s homework  Pg. 422 #7,8 (Addition and Substitution)  Pg. 433 #18-19 (Elimination)  Pg. 443 #1, 5-8 (Esterfication)  What is coming up tomorrow?  Polymerization  STS Connections

124 Section 10.5 (pg ) Organic Chemistry: Polymerization Reactions Today’s Objectives: 1)Define, illustrate, and give examples of monomers, polymers, and polymerization in living and non-living systems STS: 2) Illustrate how science and technology are developed to meet societal needs and expand human capabilities STS: 3) Illustrate how science and technology have both intended and unintended consequences

125 Polymerization  Polymers are large molecules made of chains of monomers, small molecules that link together.  Polymerization is the formation of polymers from these small units  Polymers can occur naturally (proteins, carbohydrates) and can be synthesized (nylon, Teflon, polyethylene)  They play an integral part in the function of life systems and have revolutionized the way society functions

126 Polymerization Video

127 Addition Polymerization  Many plastics (synthetic polymers) are made by this process  The polymerization process is initiated with a free radical (a species with an unpaired electron). The free radical attacks and breaks the double bond forming a new free radical that attacks another monomer

128 Addition Polymerization always results in one product, the polymer  Requires unsaturated hydrocarbon monomers and bond saturation occurs when the polymer is made  Common polymers produced by addition polymerization:

129 Things to know about addition polymers…  The polymer names end in –ene (i.e. polystyrene, polypropene). Does this mean they have double bonds?  No, the double bonds are saturated by adjacent monomers, as the polymer is made. The name refers to the starting monomer (i.e. polyethene is started by the monomer ethene)  What properties make Teflon a good product for non-stick materials?  Teflon is made up of C-F bonds which are very strong (not C-H bonds). These very strong bonds make the Teflon highly unreactive (non-sticking), it has a high melting point and it has a slippery surface

130 Condensation Polymerization  Monomers combine to form a polymer and a bi-product. Each time a bond forms between monomers, small molecules, such as water, ammonia, or HCl are “condensed” out.  The polymerization of nylon: For condensation polymerization to occur, monomers must be bifunctional, meaning they have at least two functional groups. If they only had one functional group, then only one bond would form.

131 Condensation Polymerization  Condensation polymerization also produces natural polymers, called proteins.  Amino acids (monomers) polymerize to make peptides (short chains of amino acids) or proteins (long chains of amino acids)

132 Comparison of Addition and Condensation Polymerization  Needs a double or triple bond in the monomer  Produces only one product, the polymer  Needs bifunctional monomers (have two functional groups)  Produces two products: the polymer and the biproduct (water, ammonia or HCl) Addition Condensation

133 Polyester  When a carboxylic acid reacts with an alcohol in an esterification reaction, a water molecule is eliminated and a single ester molecule is formed.  This esterification reaction can be repeated so many esters are joined in a long chain… a polyester  This is created using a dicarboxylic acid (an acid with a carboxyl group at each end) and a diol (an alcohol with a hydroxyl group at each end)  The ester linkages are formed end to end between alternating acid and alcohol molecules

134 Polyester: Dacron  Another example of a polyester:  Note the two carboxyl groups in the dicarboxylic acid and the two hydroxyl groups in the polyalcohol that start the chain reaction

135 Natural Polymer Examples  Starch, wood, silk, DNA

136

137 Today’s homework  Pg. 448 #2-4  Pg. 452 #13  Pg. 455 #16, 17  Due tomorrow  What is coming up tomorrow?  Stations Lab – will take two days

138 Chemistry 30 Organic Review Unit Exam Learning Tip Pg. 441


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