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Section 9.5 and 9.6 (pg. 386-400) Properties of Organic Compounds & Crude Oil Refining Today’s Objectives: 1)Compare boiling points and solubility of organic.

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Presentation on theme: "Section 9.5 and 9.6 (pg. 386-400) Properties of Organic Compounds & Crude Oil Refining Today’s Objectives: 1)Compare boiling points and solubility of organic."— Presentation transcript:

1 Section 9.5 and 9.6 (pg. 386-400) 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

2 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 Higher boiling points than alkanes and alkenes with similar C #s Accepted explanation: Linear structure around triple bond allows electrons to come closer together than in alkanes/enes, resulting in greater London Force 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

3 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

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

5 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 butane cyclobutane butanol butanoic acid Reasoning: but-1-ene has lower LF’s than butane because it is unsaturated, cyclobutane has an additional bond because cyclic, butanol has H-bonding, butanoic acid has stronger H-bonding)  Which would be soluble in water?  Butanol and butanoic acid – because they are the only polar molecules and like dissolves like!

6 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

7  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

8  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

9 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

10 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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12 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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14 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

15 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)

16 Today’s homework  Read pages 410-416  Pg. 388 #2  Pg. 391 #11  Pg. 397 #4 b, c, d only  Pg. 430 #11  What is coming up tomorrow?  Hydrocarbon Reactions Addition, Substitution and Elimination


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