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Che 163 Introductory Organic Chemistry ALKANES Summer Quarter 2010
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What is Organic chemistry?
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The study of carbon and its compounds.
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What is Organic chemistry? The study of carbon and its compounds. First we will concentrate on compounds just containing carbon and hydrogen, these compounds are called hydrocarbons.
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What is Organic chemistry? The study of carbon and its compounds. First we will concentrate on compounds just containing carbon and hydrogen, these compounds are called hydrocarbons. Hydrocarbon Classification Hydrocarbons Alkanes Alkenes CycloalkanesAlkynes Cycloalkenes
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1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a. CH 4 b. C 2 H 6 c. C 3 H ?
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1.Alkanes A.General formula of C n H 2n+2 B.Examples a. CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H ?
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1.Alkanes A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4 C 2 H 6 C 3 H 8
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1.Alkanes A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4 C 2 H 6 C 3 H 8 D. Polarity? Polar or nonpolar?
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1.Alkanes A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4 C 2 H 6 C 3 H 8 D. Polarity? Polar or nonpolar?Nonpolar
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1.Alkanes (Continued) E. Draw three dimensional structures, bond angles and hybridization. CH 4 C 2 H 6 C 3 H 8 F. There are two different structures for C 4 H 10 Structure 1 Structure 2
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G.Types of carbon 1. Primary (1 ◦ ) Carbon connected to one carbon atoms. 2. Secondary (2 ◦ ) Carbon connected to two carbon atoms. 3. Tertiary (3 ◦ ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 => Primary = ? Secondary = Tertiary = Primary = Secondary = Tertiary =
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G.Types of carbon 1. Primary (1 ◦ ) Carbon connected to one carbon atoms. 2. Secondary (2 ◦ ) Carbon connected to two carbon atoms. 3. Tertiary (3 ◦ ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 => Primary = 2 Secondary = ? Tertiary = Primary = Secondary = Tertiary =
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G.Types of carbon 1. Primary (1 ◦ ) Carbon connected to one carbon atoms. 2. Secondary (2 ◦ ) Carbon connected to two carbon atoms. 3. Tertirary (3 ◦ ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 => Primary = 2 Secondary = 2 Tertiary = ? Primary = Secondary = Tertiary =
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G.Types of carbon 1. Primary (1 ◦ ) Carbon connected to one carbon atoms. 2. Secondary (2 ◦ ) Carbon connected to two carbon atoms. 3. Tertiary (3 ◦ ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 => Primary = 2 Secondary = 2 Tertiary = 3 Primary = ? Secondary = Tertiary =
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G.Types of carbon 1. Primary (1 ◦ ) Carbon connected to one carbon atoms. 2. Secondary (2 ◦ ) Carbon connected to two carbon atoms. 3. Tertiary (3 ◦ ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary = 2 Secondary = 2 Tertiary = 3 Primary = 3 Secondary = ? Tertiary =
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G.Types of carbon 1. Primary (1 ◦ ) Carbon connected to one carbon atoms. 2. Secondary (2 ◦ ) Carbon connected to two carbon atoms. 3. Tertiary (3 ◦ ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary = 2 Secondary = 2 Tertiary = 3 Primary = 3 Secondary = 0 Tertiary = ?
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G.Types of carbon 1. Primary (1 ◦ ) Carbon connected to one carbon atoms. 2. Secondary (2 ◦ ) Carbon connected to two carbon atoms. 3. Tertiary (3 ◦ ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary = 2 Secondary = 2 Tertiary = 3 Primary = 3 Secondary = 0 Tertiary = 1
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Constitutional Isomers (Structural Isomers) are different compounds of the same formula. The different structures from the previous slide for the formula C 4 H 10 is an example of Constitutional isomers. How many isomers are there of an alkane containing five carbons (C 5 H 10 )?
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NOMENCLATURE 1.Common system a.Works best for low molecular weight hydrocarbons b.Steps to give a hydrocarbon a common name: 1.Count the total number of carbon atoms in the molecule. 2.Use the Latin root from the following slide that corresponds to the number of carbon atoms followed by the suffix “ane”. 3.Unbranced hydrocarbons use the prefix normal, or n-, 4.Branched hydrocarbons use specific prefixes, as shown on a subsequent slide
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Latin Hydrocarbon Roots Number of Carbons Latin Root 1meth 2eth 3prop 4but 5pent 6hex 7hept 8oct 9non 10dec 11undec Latin Hydrocarbon Roots Number of Carbons Latin Root 12dodec 13tridec 14tetradec 15pentadec 16hexadec 17heptadec 18octadec 19nonadec 20eicos 21unicos 22doicos H H n-butane isobutane H C C C H H C H H C H H H H H H H Examples neopentane
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2. Systematic System of Nomenclature (IUPAC) Find the longest continuous chain of carbon atoms. Use a Latin root corresponding to the number of carbons in the longest chain of carbons. Follow the root with the suffix of “ane” for alkanes Carbon atoms not included in the chain are named as substituents preceding the root name with Latin root followed by “yl” suffix. Number the carbons, starting closest to the first branch. Name the substituents attached to the chain, using the carbon number as the locator in alphabetical order. Use di-, tri-, etc., for multiples of same substituent. If there are two possible chains with the same number of carbons, use the chain with the most substituents.
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Substituent Names (Alkyl groups)
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Which one? Systematic Nomenclature continued.
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Which one? Systematic Nomenclature continued. The one with the most number of substituents
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Which one? Systematic Nomenclature continued. The one with the least number of substituents The top structure has four substituents and the bottom has three Substituents.
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Which one? Systematic Nomenclature continued. The one with the least number of substituents The top structure has four substituents and the bottom has three Substituents. Name = ?
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Which one? Systematic Nomenclature continued. The one with the least number of substituents The top structure has four substituents and the bottom has three Substituents. Name = ? heptane
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Which one? Systematic Nomenclature continued. The one with the least number of substituents The top structure has four substituents and the bottom has three Substituents. Name = 3,3,5-trimethyl-4-propylheptane
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Another Example: Name = 3-ethyl-2,6-dimethylheptane
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Another Example: Name = 2,6-dimethyl-3-ethylheptane Notice substituents are in alphabetical order; di, tri, etc. do not participate in the alphabetical order
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Line Structures A quicker way to write sturctures (Condensed Structure) (A line structure of the above condensed structure) ethyl methyl
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Complex Substituents If the branch has a branch, number the carbons from the point of attachment. Name the branch off the branch using a locator number. Parentheses are used around the complex branch name. 1-methyl-3-(1,2-dimethylpropyl)cyclohexane 1 2 1 3
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Alkane Physical Properties Solubility: hydrophobic (not water soluble) Density: less than 1 g/mL (floats on water) Boiling points increase with increasing carbons (little less for branched chains) due to dispersion forces being larger. Melting points increase with increasing carbons (less for odd-number of carbons).
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Boiling Points of Alkanes Branched alkanes have less surface area contact, so weaker intermolecular forces.
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Melting Points of Alkanes Branched alkanes pack more efficiently into a crystalline structure, so have higher m.p.
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Reactions of Alkanes I. Combustion reaction II. Cracking reaction III. Halogenation reaction (substitution reaction)
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Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 12 Step 2 react each isomer with chlorine Step 3 count the products
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Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products
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Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products
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Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products
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Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products
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Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products Winner!
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Conformers of Alkanes Structures resulting from the free rotation of a C-C single bond May differ in energy. The lowest-energy conformer is most prevalent. Molecules constantly rotate through all the possible conformations.
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Ethane Conformers Staggered conformer has lowest energy. Dihedral angle = 60 degrees model H H H H HH Newman projection sawhorse Dihedral angle
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Ethane Conformers (2) Eclipsed conformer has highest energy Dihedral angle = 0 degrees =>
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Conformational Analysis Torsional strain: resistance to rotation. For ethane, only 12.6 kJ/mol =>
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Propane Conformers Note slight increase in torsional strain due to the more bulky methyl group.
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Butane Conformers C2-C3 Highest energy has methyl groups eclipsed. Steric hindrance Dihedral angle = 0 degrees => totally eclipsed (methyl groups)
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Butane Conformers (2) Lowest energy has methyl groups anti. Dihedral angle = 180 degrees => Staggered-anti
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Butane Conformers (3) Methyl groups eclipsed with hydrogens Higher energy than staggered conformer Dihedral angle = 120 degrees => Eclipsed (hydrogen and methyl)
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Butane Conformers (4) Gauche, staggered conformer Methyls closer than in anti conformer Dihedral angle = 60 degrees => Staggered-gauche
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Conformational Analysis
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Cycloalkanes Rings of carbon atoms (-CH 2 - groups) Formula: C n H 2n Nonpolar, insoluble in water Compact shape Melting and boiling points similar to branched alkanes with same number of carbons Slightly unsaturated compared to alkanes
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Naming Cycloalkanes Count the number of carbons in the cycle If the bonds are single then use the suffix “ane” First substituent in alphabet gets lowest number. May be cycloalkyl attachment to chain.
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Cis-Trans Isomerism (a type of stereoisomerism) Cis: like groups on same side of ring Trans: like groups on opposite sides of ring
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Cycloalkane Stability 6-membered rings most stable Bond angle closest to 109.5 Angle (Baeyer) strain Measured by heats of combustion per -CH 2 -
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Heats of Combustion/CH 2 Alkane + O 2 CO 2 + H 2 O 658.6 697.1 686.1 664.0663.6 kJ/mol 662.4658.6 kJ Long-chain
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Cyclopropane Large ring strain due to angle compression Very reactive, weak bonds =>
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Cyclopropane (2) Torsional strain because of eclipsed hydrogens
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Cyclobutane Angle strain due to compression Torsional strain partially relieved by ring puckering =>
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Cyclopentane If planar, angles would be 108 , but all hydrogens would be eclipsed. Puckered conformer reduces torsional strain.
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Cyclohexane Combustion data shows it’s unstrained. Angles would be 120 , if planar. The chair conformer has 109.5 bond angles and all hydrogens are staggered. No angle strain and no torsional strain.
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Chair Conformer
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Boat Conformer
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Conformational Energy
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Axial and Equatorial Positions
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Monosubstituted Cyclohexanes
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1,3-Diaxial Interactions
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Disubstituted Cyclohexanes
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Cis-Trans Isomers Bonds that are cis, alternate axial-equatorial around the ring. => One axial, one equatorial
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Bulky Groups Groups like t-butyl cause a large energy difference between the axial and equatoria l conformer. Most stable conformer puts t-butyl equatorial regardless of other substituents. =>
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End of Chapter 2
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