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Organic Chemistry Chemistry of Carbon Compounds. Bonding Capacity H can form only 1 bond halogens (F, Cl, Br, I) form only 1 bond O and S form 2 bonds.

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Presentation on theme: "Organic Chemistry Chemistry of Carbon Compounds. Bonding Capacity H can form only 1 bond halogens (F, Cl, Br, I) form only 1 bond O and S form 2 bonds."— Presentation transcript:

1 Organic Chemistry Chemistry of Carbon Compounds

2 Bonding Capacity H can form only 1 bond halogens (F, Cl, Br, I) form only 1 bond O and S form 2 bonds N and P form 3 bonds

3 Why so many C compounds? covalentlycarbon atoms have the unique ability to covalently bond with other carbon atoms in chains, rings, and networks

4 Lewis Diagram of C C carbon has 4 unpaired electrons so it can form 4 covalent bonds

5 molecular Properties of Covalent Substances (also called molecular substances) low melting points, low boiling points poor conductors of heat & electricity usually soft or brittle generally non-polar –van der Waals forces (dispersion) tend to dissolve in non-polar solvents react more slowly than ionic compounds

6 Chemical Molecular Formulas show kind & # of atoms CH 2 Cl 2 CH 4 CH 3 OH CH 3 Cl

7 Structural Formulas show kind & # of atoms show bond type (single, double or triple) show approximate shapes of molecules 2-D picture of 3-D object so not totally realistic H H–C–H H

8 Structural Formulas single line (–) represents one pair of shared electrons (single bond) double line (=)represents two pairs of shared electrons (double bond) ● triple line (  )represents three pairs of shared electrons (triple bond)

9 Condensed Structural Formula shows kind & # of atoms shows some structural information structural formula: H H H CCC H–C–C–C–H H H H CCC Condensed structural formula: CH 3 CH 2 CH 3

10 Vocabulary hydrocarbons hydrocarbons: organic compounds containing only C and H

11 Homologous Series Homologous Series: group of compounds with related structures & properties – each member of series differs by addition of repeating unit –molecules have fixed, numerical relationships among # of atoms

12 saturated hydrocarbons s saturated hydrocarbons: organic compounds containing only single bonds

13 ALKANES C n H 2n+2

14 Alkanes homologous series of saturated hydrocarbons CH 4 H H–C–H H C 2 H 6 H H H–C–C–H H H

15 Alkanes (C n H 2n+2 ) C 3 H 8 H H H H–C–C–C–H H H H CH 3 CH 2 CH 3 molecular formula structural formula condensed structural formula

16 Alkanes: base unit CH 4 CH 3 CH 3 or C 2 H 6 CH 3 CH 2 CH 3 or C 3 H 8 CH 3 CH 2 CH 2 CH 3 or C 4 H 10 CH 3 CH 2 CH 2 CH 2 CH 3 or C 5 H 12 CH 2 repeating unit: CH 2

17 straight-chain alkane C 4 H 10 H H H–C–C–C–C–H H H butane

18

19 Naming straight-chain Alkanes name describes molecule so can draw it aneaneall alkanes have the suffix –ane prefix: tells # of C ’ s

20 10dec 9non 8oct 7hept 6hex 5pent 4but 3prop 2eth 1meth # of C atomsPrefix

21 C 5 H 12 C 4 H 10 C3H8C3H8 C2H6C2H6 meth + ane CH 4 NameFormulaethanepropane butane pentane

22 Properties of Alkanes change systematically with # of C ’ s as # of C ’ s increases, boiling point increases –molecules get heavier so more energy required to overcome IMF to change from liquid to gas phase

23 Properties of Alkanes low reactivity –except readily undergo combustion (fuels) non-polar – don ’ t dissolve in water low mp & low bp –both increase with molecular mass high VP

24 A.low melting point B.high melting point C.soluble in polar solvents D.insoluble in nonpolar solvents Which property is generally characteristic of an organic compound? correct response = A

25 A.CH 4 B.C 2 H 6 C.C 3 H 8 D.C 4 H 10 Which of the following compounds has the highest boiling point? Correct answer = D Non-polar coval cmpd: bp depends on strength of van der Waals interactions bp depends on strength of van der Waals interactions dispersion forces ↑ as size of molecule ↑ dispersion forces ↑ as size of molecule ↑

26 Branched-chain alkanes beginning with butane, C 4 H 10, there is more than 1 way to arrange the atoms H H–C–H H H–C–C–C–H H H H

27 Branched alkane: can ’ t link all the C ’ s without lifting pencil off paper Methyl propane

28 Isomers compounds with same molecular formula but different structural arrangements the more C atoms there are the more isomers can have (more possible ways to arrange atoms)

29 Isomers different structuresdifferent structures different propertiesdifferent properties Isomers have different chemical and physical properties

30 A note about isomers --- If comparing 2 structural formulas & can superimpose them, they are not isomers – they are same molecule! If can rotate/flip one of structural formulas & then superimpose one on the other, they are not isomers – they are same molecule!

31 ? same yes they are

32 ? same yes they are

33 ?same yes they are

34 ?same no they are not; these are isomers

35 Naming Branched-Chain Alkanes goal of name is to describe molecule so you can draw it

36 Naming branched-chain alkanes longest continuous chain 1.find longest continuous chain of C atoms - Bends don ’ t matter! 2.base name 2.base name derived from # C ’ s in continuous chain 3.branchesfirst 3.branches named first “ yl ” to prefix name - count # C atoms & add “ yl ” to prefix name assign #’s to C on backbone 4.# location of branch: assign #’s to C on backbone - # C's so get lowest number for branch - every branch gets a # 5.more than one of same type of branch - use di, tri, etc.

37 H H–C–H H H–C–C–C–H H H H methyl branch has 1 carbon – methyl branch has to be at C-2 (only give branch # if necessary) methyl propane C 4 H 10 CH 3 CH(CH 3 )CH 3 propane longest continuous chain has 3 carbon atoms – propane

38 H H–C–H H H H H–C–C–C–C–H H H H H–C–H H longest continuous chain has 6 hexane C atoms: hexane methyl branch: 1 carbon = methyl 2 branch located at C-2 2-methyl hexane C 7 H 16 CH 3 CH(CH 3 )CH 2 CH 2 CH 2 CH 31 43 6 5 2

39 A note about branches If several branches with same # C atoms, can condense name a little 2-methyl 3-methyl pentane becomes: –2,3-dimethyl pentane

40 H H – C – H H H H H – C – C – C – C – H H H H–C–H H–C–H H H longest continuous chain: 6 C’s = hexane branch: 1 C = methyl branch located at C-3 3-methyl hexane C 7 H 16 CH 3 CH 2 CH 2 CH(CH 3 )CH 2 CH 3 CH 3 (CH 2 ) 2 CH(CH 3 )CH 2 CH 3 1 234 5 6

41 Unsaturated hydrocarbons Unsaturated hydrocarbons: organic compounds containing one or more double or triple bonds

42 Alkenes hydrocarbonsAnother homologous series of hydrocarbons one unsaturatedEach member contains at least one double covalent bond between C atoms  So alkenes are unsaturated General formula = C n H 2n

43 Naming Alkenes name: prefix tells # of C ’ s in parent chain suffix: -ene 1 st member is C 2 H 4: ethene H H H H C=C H H Alkenes with 4 or more C ’ s: need # to show which C double bond attached to

44 Naming Alkenes # C atoms in backbone & give double bond the lowest possible number H H H C=C–C–C–H H H C 4 H 8 CH 2 CHCH 2 CH 3 H H H H H–C–C=C–C–H H H C 4 H 8 CH 3 CHCHCH 3 1-butene 2-butene

45 Naming Branched-Chain Alkenes parent chain = longest chain that contains the double bond position of double bond, not branches, determines numbering of backbone give 1 st C in double bond lowest possible #

46 Properties of Alkenes non-polar – low solubility in H 2 O fairly low melting and low boiling points more reactive than alkanes : –double bond is site of reactivity

47 Alkynes Homologous series of unsaturated hydrocarbons that contain one triple bond each member contains one C≡C bond –Alkynes are unsaturated general formula = C n H 2n-2

48 Naming Alkynes name: Prefix tells # of C ’ s –suffix: yne if necessary, number 1 st carbon atom where triple bond is found

49 H–C  C–HC 2 H 2 ethyne CHCH H H–C  C–C–H H H H H–C  C–C–C–H H H H–C–C  C–C–H H H C3H4C3H4 propyne CHCCH 3 C4H6C4H6 1-butyne CHCCH 2 CH 3 C4H6C4H6 2-butyne CH 3 CCCH 3

50

51 3 Homologous Series of HC ’ s -yneC n H 2n-2 Alkynes -eneC n H 2n Alkenes -aneC n H 2n+2 Alkanes EndingGeneral FormulaName of Series

52 A.C 2 H 2 B.C 2 H 4 C.C 6 H 6 D.C 6 H 14 Which compound belongs to the alkene series? Correct answer = B alkenes follow the format C n H 2n A & C: C n H n, D: C n H 2n+2

53 A.C 2 H 2, C 2 H 4, C 2 H 6 B.C 2 H 4, C 3 H 4, C 4 H 8 C.C 2 H 4, C 2 H 6, C 3 H 6 D.C 2 H 4, C 3 H 6, C 4 H 8 In which group could the hydrocarbons all belong to the same homologous series? Correct answer = D members of homologous series all have same relationship between atoms; every compound in set D fits formula C n H 2n

54 A.Ethene B.Ethyne C.Propene D.Propane Which of the following is a saturated hydrocarbon? Correct answer = D all alkanes are saturated


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