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Introduction to Organic Chemistry

Introduction to Organic Chemistry. 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. Organic Chemistry.

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Introduction to Organic Chemistry

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  1. Introduction to Organic Chemistry 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.

  2. Organic Chemistry • Organic Chemistry involves the study of Carbon based compounds • Almost all compounds utilized by living organisms involve carbon hence the name “Organic” chemistry • What are some examples of Organic Based compounds? • Petrochemicals - Biochemicals • Dyes - Polymers • Pharmaceuticals - Life

  3. Why so many C compounds? • Carbon atoms can bond with other carbon atoms in chains, rings, and networks. • These Bonds are covalent. • Carbon has 4 unpaired electrons & can form 4 covalent bonds. H can only form 1 bond Halogens only form 1 bond O, S can only form 2 N, P can only form 3 • • C • •

  4. Carbons ability to form 4 covalent bonds, one or more Of which can even be to another carbon atom leads to carbons ability to form millions of different compounds.

  5. Structural Formulas • Show kind & number of atoms. • Also show bonding patterns and approximate shapes of molecules. • They are 2-D drawings of 3-D objects so structural formulas aren’t totally realistic.

  6. Condensed Structural Formulas • These formulas show kind & number of atoms as well as some information regarding what atoms are bonded to what atoms. • They do not depict bonds Examples: CH3CH2CH2CH3

  7. Molecular Formulas • These formulas show kind & number of atoms but not necessarily any other information Examples: CH3CH2CH2CH3 becomes C4H10

  8. Empirical Formula • of a chemical compound is the simplest whole number ratio of atoms of each element present ina compound. • it makes no reference to isomerism, structure, or absolute number of atoms. • The empirical formula is used as standard for most ionic compounds and for macromolecules • Example: C6H14, would be C3H7

  9. Line structural diagram: • the end of the line and points where they meet represent carbon. • Below shows 6 carbons present, and each is filled with its maximum amount of hydrogen

  10. Vocabulary • Hydrocarbons: organic compounds containing only Carbon and Hydrogen. • Straight chain: All C bonded to only 2 other C’s • Ex: c-c-c-c-c-c-c • Branched: One or more C bonded to 3 or 4 other C’s • Ex: c-c-c-c-c-c-c-c c | | | c c c-c-c | c

  11. Homologous Series • a group of compounds with related structures and properties. Each member of the series differs from the one before it by the same additional unit. (Ex. The Alkanes) • For alkanes, each addition of CH2 lengthens the chain • The compounds have similar chemical properties • Successive compounds have physical properties that vary in a regular manner as the number of carbon atoms present increase.

  12. Physical properties Alkane boiling points • The changes are a result in the changes that occur in the strength of van der Waals’ forces with increasing molar mass (increase in number of electrons, increases the size of the temporary dipole)

  13. Explanation • The curve is initally steep because adding the 2nd C, changes the molar mass by 97%, each additional C doesn’t affect such drastic change. • Similar trends for density and viscosity

  14. Naming straight-chain Alkanes • All alkane names have the suffix –ane. • The prefix depends on the number of C’s. Ex: Methane 1 Carbon Ethane 2 Carbon What’s the chemical formula of each?

  15. Alkanes • A Homologous series of saturated hydrocarbons. • Compounds with a related structure differing by the same repeating unit (-CH2-)

  16. Alkanes: base unit CH4 CH3CH3 or C2H6 CH3CH2CH3 or C3H8 CH3CH2CH2CH3 or C4H10 CH3CH2CH2CH2CH3 or C5H12 • Difference between each is one-CH2- • General formula is CnH2n+2 (n= no. of carbon atoms)

  17. Branched alkanes • There is another unit(s) off the longest continuous chain. • You must name what and where it is found. RULES: • Number the longest chain so the lowest possible number combination is in your name. • Branches are added as a prefix and end in –yl • Above example is called 2-methyl pentane, NOT 4-methyl pentane! Both drawings represent the same molecule.

  18. If there is more than one unit you need to place them alphabetically • If there are two methyl groups, its dimethyl, three methyl groups is trimethyl… • Below example: 2,2-dimethyl butane • Butane is the longest chain (4 Cs), and the two methyl groups come off the same carbon (#2).

  19. Isomers 10.1.4 Describe structural isomers as compounds with the same molecular formula but with different arrangements of atoms. 10.1.5 Deduce structural formulas for the isomers of the non-cyclic alkanes up to C6. 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to C6.

  20. Structural isomer • They have the same molecular formula, but different structural formulas. • You know if you’ve drawn an isomer if they all have different names. • Ex: C5H12 can be called pentane, 2-methylbutane, dimethyl propane • Try to draw all isomers of C6H14

  21. Isomers • Isomers have different structures and different chemical and physical properties. Butane2-methylpropane Formula C4H10 C4H10 B.P. 0 C -12 C M.P. -138 C -159 C Density 0.622 0.604 Sol. In 100ml Alcohol 1813 ml 1320 ml

  22. Isomers (continued) 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.

  23. Alkenes • The Alkenes form another homologous hydrocarbon series • Each member contains one double covalent bond between two C atoms. • General formula = CnH2n

  24. Saturation vs. Unsaturation • Simply put, a saturated hydrocarbon has no double bonds between the Carbon atoms (ex. alkanes) • Unsaturated has one or more double or triple bonds between carbons (alkenes, alkynes)

  25. Naming Alkenes • Names are derived from the name of the alkane chain with the same number of C atoms. • Replace the –ane ending of the alkane name with –ene. 1st member is C2H4, ethene. H H C=C H H

  26. Ethene Propene 1-Butene 1-Pentene 1-Hexene 1-Heptene 1-Octene 1-Nonene 1-Decene C=C C-C=C C-C-C=C C-C-C-C=C C-C-C-C-C=C C-C-C-C-C-C=C C-C-C-C-C-C-C=C C-C-C-C-C-C-C-C=C C-C-C-C-C-C-C-C-C=C Alkene Homologous Series(C1-C2 double bond)

  27. Naming Alkenes • Location of double bond is specified by numbering C atoms in backbone. Give bond the lowest possible number. 1-butene (not 4) 2-butene H H H C=C–C–C–H H H H H H H H H H–C–C=C–C–H H H Structural Formula Chemical formula Condensed formula C4H8 CH2CHCH2CH3 C4H8 CH3CHCHCH3

  28. Naming Alkenes • Once double bond is numbered specify substituents alphabetically by number • Use di, tri and tetra for multiple substituents of the same group 2,3-dimethyl-1-butene C=C–C–C C C C6H12 CH2C(CH3)CH(CH3)CH3

  29. Draw the following alkenes: 2-butene 2–methylpropene 4–methyl–2-pentene 3,3-dimethyl-1-butene Problems:

  30. Alkenes = 1 double bond Dienes have 2 double bonds End in -diene Location still denoted by numbers Determine which way to number the C backbone by assigning the lowest possible number to one of the double bonds Example: 1,3 – pentadiene c=c-c=c-c Not 2,4-pentadiene Substituents are named first Dienes Double bonds must all be in the backbone.

  31. Naming organic compounds 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.

  32. Functional Groups Additional atoms (groups of atoms) bonded to the hydrocarbon Will change the chemical and physical properties of the hydrocarbon SL is responsible for … Halide, alcohol, aldehyde, ketone, carboxylic acid, amines, amides and phenols

  33. Organic Halides • One or more of the hydrogen atoms in an alkane is replaced with a halogen • F, Cl, Br, or I • Nothydrocarbons, often called halocarbons.

  34. Properties of Halides Alkyl halides are extremely unreactive • Often used when chemical inertness • is important • Examples: CFC’s (refrigerants, Aerosol • propellants, Teflon (polymer), Brominated • compounds (Fire retardant clothing) • -many of these compounds are now banned • from use for health and environmental • reasons.

  35. c-c-c-c-c Br Br Naming Organic Halides Use prefixes to specify substituent: • fluoro, chloro, bromo, iodo • If more than one, use di, tri, etc. to specify # of substituents. • Give locations by numbering C-atoms in backbone so that the halide has the lowest number. 2,3-dibromopentane

  36. Alcohols • Alcohols contain the polar -OH functional group • OH groups are capable of Hydrogen bonding • This polar functional group affects physical properties • B.P is much higher than the corresponding Alkane • Ethanol (C2H5OH) +78oC Ethane (C2H6) -89oC • Low Mol. Wt. Alcohols are very soluble in water

  37. Hydroxyl groups (alcohol) Alcohols can hydrogen bond because they have a Hydrogen atom that is bonded to oxygen. This results in Much higher B.P.’s And higher water solubulity CH3–CH2–O-H CH3–CH2–O-H H-O-H

  38. Physical properties of Alcohols

  39. Naming Alcohols • Select as the parent structure the longest continuous carbon chain that contains the –OH • Drop the –e from the alkane name of the carbon chain and add –ol • Indicate by a number (if necessary) the position of the –OH group ethanol 2-propanol

  40. Types of Alcohols • Alcohols are also considered to be either - primary, secondary, or tertiary alcohols • Designations correspond to which carbon atom the –OH group is bonded to - Affects Chemical Reactivity/Properties - Tertiary more reactive than Primary Primary alcohol: OH bonded to –CH2OH Secondary alcohol: OH bonded to –CHOH Tertiary alcohol: OH bonded to -COH

  41. Types of Alcohols Primary alcohol Secondary Alcohol Tertiary Alcohol

  42. Even though ethers have an oxygen atom in their structure they are unable to form hydrogen bonds with themselves - alkyl groups are bonded directly to oxygen Ethers are not linear; They have a bent shape similar to water Ethers

  43. Ether’s Properties • BP’s tend to be low • Comparable to similar weighted hydrocarbons • No internal H-bonding • Solubility in water is reasonably high for lower M.W. ethers • H-Bonding to Water is possible

  44. Name the groups (alkyl) on either side and add ether to the end. List attached alkyl groups in order of increasing size If they are the same the side chain is labeled “Di” NO NUMBERS NEEDED to designate location of Oxygen! Naming Ethers

  45. Naming Ethers Name this compound: Dimethyl ether Name this compound: Ethylpropyl ether

  46. Aldehydes • Group is always on a terminal C – no need to specify location by number • Condensed it is symbolized by a -CHO group at the end of the formula • C = O is known as a carbonyl group • Drives the physical and chemical properties of these compounds • Aldehydes have characteristic scents and tastes • Cinnamon is an example

  47. Aldehyde Drop the –e from the end of the alkane and add –al; C=O is C #1 Name these two compounds: Methanal (formaldehyde) Butanal

  48. Ketone • Like the aldehydes the Functional group is the carbonyl group C=O • Contains an alkyl group on either side of the C=O • In aldehydes one side is an alkyl group the other is H • Often used as a solvent • Acetone is one example

  49. Naming Ketones • Number the C chain so C=O has the lowest # • Drop the final e from the alkane name and add -one Name this compound: 3-hexanone

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