Organic Chemistry Chapter 2 PDF

Chapter 2 Alkanes and Cycloalkanes Modified by Dr. Mousa AlTarabeen  Chapter 2: Alkanes and Cycloalkanes: Conformational and Geometric Isomers  Types of Hydrocarbons ❖ Hydrocarbons are compounds that only contain C and H atoms. Hydroca...

Chapter 2 Alkanes and Cycloalkanes Modified by Dr. Mousa AlTarabeen  Chapter 2: Alkanes and Cycloalkanes: Conformational and Geometric Isomers  Types of Hydrocarbons ❖ Hydrocarbons are compounds that only contain C and H atoms. Hydrocarbons CnHm alkanes alkenes alkynes arenes H3C CH3 H2C CH2 HC CH Only CC  CC  and  CC  and 2  6 C ring with bonds all C bonds C=C sp2 bonds CC sp alternating  sp3 hybridized hybridization hybridization and  bonds Aliphatic Aromatic  Structure of Alkanes Alkanes are saturated hydrocarbons, that is they contain the maximum number of H atoms possible for the number of C atoms present. The generic formula for an alkane is: CnH2n+2 Ex: CH4; C2H6; and C5H12 H This means every C atom is sp3 hybridized 109.5° C H H H  Structure of Alkanes (cont’d) ❖ Alkanes can come in two forms: I. Normal (or linear)-----> (A) II. Branched: branched alkanes can have branched branches. Therefore the number of isomers possible growths quickly)-----> (B) A B  Structure of Alkanes (cont’d) Molecular Number of Formula Isomers ❖The large number of isomers CH4 1 possible is due to: C2H6 1 ▪ Number of carbon C3H8 1 ▪ carbon’s ability to form strong C- C4H10 2 C5H12 3 C bonds. C6H14 5 ▪ Example of Structural isomer: C7H16 9 Butane (C4H10) C8H18 18 Same chemial formula but different C9H20 35 chemical stucture CH3 C10H22 75 H H H H H H C15H32 4,347 H C C C C H H C C C H C20H42 336,319 H H H H H H H C30H62 4,111,846,763 butane 2-methylpropane  Nomenclature ❖ IUPAC Rules for Alkane Nomenclature Find and name the longest continuous carbon chain. Identify and name groups attached to this chain. Number the chain consecutively, starting at the end nearest a substituent group. Designate the location of each substituent group by an appropriate number and name. Assemble the name, listing groups in alphabetical order. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing. Halogen substituents are easily accommodated, using the names: fluoro (F-), chloro (Cl-), bromo (Br-) and iodo (I-).  Nomenclature Notes: 1. A –CH3 is a methyl group. 2. A –CH2 is a methylene group. 3. A –CH is a methine group. 4. A-CH2CH3 is a ethyl group. 5. A-CH2CH2CH3 is a propyl group. A: The long chain found (CnHn) Examples: 2,3-dimethylpentane 5-ethyl-2-methylheptane 3-methylhexane Cl 2-chloro-4,7-dimethyloctane  Sources of Hydrocarbons Hydrocarbons are found in: ▪ crude oil ▪ natural gas. Both are mixtures of different hydrocarbons They are separated by “fractional distillation” : a process where the different compounds are separated in a long distilling column based on their boiling points (physical property). Heavy fractions can be “cracked” into small lighter, hydrocarbons using heat and/or catalysts. Fractional Distillation  Intermolecular Forces The physical properties of molecules are in part dependent on the type's of intermolecular forces (IMF) present. The strength of the interaction between molecules is also dependent on the mass of the molecule and the overall shape of the molecule. Boiling points (BP) are also dependent on the mass of the molecule and the overall shape of the molecule. Solubility, the ability to dissolve into a solvent is dependent on IMFs.  Intermolecular Forces ❖ There are 3 types of IMFs, by decreasing strength they are: 1) Hydrogen bonding 2) Dipole-dipole 3) Van der Waals or London Dispersion ❖ Hydrogen bonding: is a complex interaction that includes dipole-dipole, as well as orbital interactions and the transfer of electron density between molecules. ❖ These are the strongest of the IMFs and range from 5 – 25 kJ/mol  Hydrogen Bonding Occur primarily between OH, NH and FH. The more EN the atom the stronger the interaction. The atom H is attached to usually has a lone pair of e-) Geometry: -.... O H O: H O: H H + + H H - H H N H + H N: H N: H + H H H +  Dipole-dipole ❖Dipole-dipole forces arise from the attraction of oppositely charged atoms (other than H) in molecules. These molecules may have a permanent dipole moment. Generally in organic molecules they results from the presence of C-X bonds where X is more electronegative that C. These are generally weaker than H-bonding, ranging from about 5-10 kJ/mol.  Dipole-dipole + H3C Cl - H3C Cl H3C Cl + H3C C N - H3C C N H3C C N  Van der Waals (dispersion) ❖ Van der Waals or (London) dispersion forces arise from the movement of electrons within a molecule. This natural motion can produce an uneven distribution of the electrons (polarization of the distribution) resulting in a temporary dipole moment in the molecule. This will induce the movement of electrons in adjacent molecules producing a dipole moment in them. These “induced” dipole moments are very brief as they disappear when the electrons move to new locations within the molecule, so they forces are very brief and weak, only 2- 5 kJ/mol.  Structural Effects on IMFs The strength of the IMFs depend on the amount of contact between the molecules, especially for dispersion forces. Hence the shape of the molecule can affect the surface area of contact, long thin molecules have more surface in contact than spherical molecules.  Effects on Physical Properties (cont’d) Dipole forces: much strong IMFs so the BP is higher when dipole-dipole interactions are present, i.e. O mass = 86 amu mass = 86 amu BP = 101.7 °C BP = 63.3 °C  Effects on Physical Properties (cont’d) O OH Hydrogen mass = 72 amu mass = 74 amu mass = 74 amu bonding: BP = 36.1 35 °C BP =36.3 35 °CoC BP = 117 °C strongest intermolecular CH2CH2CH2CH3 forces so BP are CH3CH2CH2CH2 O H O very high for H equivalent MW compounds, i.e. H O CH2CH2CH2CH3 N NH2 mass = 73 amu mass = 73 amu BP = 36 °C BP = 78 °C  Effects on Physical Properties (cont’d) Solubility: “like dissolves like” solute must have the same types of IMFs as the solvent, i.e. H OH R O H O O O OH vitamin C H hydrogen bonds HO OH H O H OH vitamin A vitamin E HO O  Conformational Isomers Conformational isomers (rotamers or conformers) are compounds with the same constitution (atoms are bonded in the same order) but the atoms are located in different places in space. This is achieved by rotating about C-C single (s) bonds or the dihedral (or torsion) angle (q). In conformational isomers the atoms are located in different location in space due to rotation about C-C single () bonds  A B  Conformational Isomers (cont’d) Three extremes exist for ethane: staggered, and eclipsed, Less energy Higher energy  Conformational Isomers (cont’d) These two extremes represent high and low energy “conformations” of ethane. The “high” E is the eclipsed and low E the staggered.  Conformational Isomers (cont’d) ❖ Example of conformational isomers:  Geometric Isomers Geometric isomers (or configurational isomers, a subset of stereoisomers) are molecules which have the same chemical formula, the atoms are bonded in the same order, but located in different positions in space. Unlike conformational isomers, where the atoms are located in different location in space due to rotation about C-C single () bonds. Geometric isomers are not related by rotation about  bonds. This situation arises with: 1-cyclic structures. 2-compounds with double bond or triple bond.  Geometric Isomers ❖ An example of this are: A. 1-1,2-dibromoethene B. 1-1,2-dibromoethene ❖ This structures can not interconvert without breaking a C-C bond.  Summary of Isomers (to date) different structural (or constitutional) isomers bonding *See slides 27-29 in Chapt.1 pattern interconvertable Isomers conformational (rotamers) isomers by single bond rotation *See slides 24 in Chapt. 2 same stereoisomers bonding pattern NOT interconvertable configurational isomers by single bond rotation *See slide 26 in Chapt. 2  Chemical Reactions of Alkanes In general, because of their strong non-polar covalent bonds alkanes are fairly inert. They do not react with most common acids, bases, oxidizing or reducing reagents. They means they do make good solvents for extraction, recrystallization or as a reaction solvent. They do however have two substances they react with; molecular oxygen and halogens. All chemical processes are redox (oxidation / reduction) reactions. Formally oxidation is the loss of electrons (increase in oxidation sate) and reduction the gain of electrons (decrease in oxidation sate). The two processes are coupled since whatever gains the electrons must gain them from somewhere and whatever loses the electrons must lose them to somewhere.  Oxidation Reactions For the C atom, oxidation involves increasing the number of C-O (or other atoms more EN than C; like C-Cl) bonds and/or decreasing the number of C-H bonds. Reduction will be the opposite, a decrease in the number of C-O bonds or increase in the number of C-H bonds. To determine the oxidation state: – each bond to a C atom counts: 0 – each bond to a H atom counts: +1 – Each (single) bond to a more EN atom count -1, i.e. – Examples: X= -4 X=-2 X= -2 X=-2 0 3+0 + X = 0  Oxidation Reactions (cont’d) More examples: X=-1 X= -2 X= -3 X= +2 Questions  Oxidation Reactions (cont’d) The most important use of alkanes is as a fuel. The light weight ones are gases and the intermediate weight are liquids which makes them handy for storage and transportation. Natural gas is composed primarily of methane with varying amounts of ethane, propane, and butanes. It is commonly used to heat homes. The liquid hydrocarbons are used as gasoline, kerosene and jet fuels. The energy of the hydrocarbon is released when combusted or burned. This is an oxidation process and requires atmospheric oxygen. The final oxidation product (assuming sufficient oxygen) is carbon dioxide. If insufficient oxygen is present then partial oxidation products such as carbon monoxide, formaldehyde or formic acid may be formed.  Halogenation of Alkanes Besides combustion, the only other useful chemical reaction that alkanes undergo is halogenation. This is a free radical process in which heat or light is used to break a halide- halide bond forming two halide free radicals. This type of reaction is a substitution reaction where a halide atom is substituted for a H atom in the alkane. The two common halides used are chlorine and bromine.  Halogenation of Alkanes The process involves three steps: 1. initiation: this is where the free radical is formed: note we generate 2 free radical atoms in this process. 2. Propagation: this is where the halide free radical reacts with the hydrocarbon, extracting a H atom, generating an intermediate C radical, and then adding a halide atom. This process can repeat many times depending on the amount of halide radical available.  Halogenation of Alkanes 3. Termination: this is the final step where the remaining free radicals are consumed. It can involve the halide radical reacting with another halide radical, a carbon radical or two carbon radicals combining. A single alkane molecule can undergo several substitution steps depending on the concentration of the halide radical, i.e.