Alkanes
ALKANES (a “family” of hydrocarbons) Cn. H 2 n+2 CH 4 C 2 H 6 C 3 H 8 C 4 H 10 etc. C 2 H 6 ethane H H H—C—C—H H H
sp 3, bond angles = 109. 5 o σ-bonds (sigma) rotation about C--C (conformations) representation: “andiron” or “sawhorse”
“staggered” “eclipsed” torsional strain: deviation from staggered. Newman projections:
The barrier to rotation about the carbon-carbon bond in ethane is 3 Kcal/mole. The rotation is ~ “free. ”
Two isomers of butane C 4 H 10: CH 3 CH 2 CH 3 n-butane bp 0 o. C mp – 138 o. C d 0. 622 g/cc CH 3 CHCH 3 bp -12 o. C mp -159 o. C d 0. 604 g/cc isobutane
conformations about C 2 -C 3 in n-butane:
Alkane name isomers CH 4 methane 1 C 2 H 6 ethane 1 C 3 H 8 propane 1 C 4 H 10 butanes 2 C 5 H 12 pentanes 3 C 6 H 14 hexanes 5 C 7 H 16 heptanes 9 C 8 H 18 octanes 18 C 9 H 20 nonanes 35 C 10 H 22 decanes 75 ……. C 20 H 42 eicosanes 366, 319 each new common name requires a new prefix…
hexanes C 6 H 14 common names CH 3 CH 2 CH 2 CH 3 CH 3 CHCH 2 CH 3 n-hexane isohexane CH 3 CH 2 CH 3 CCH 2 CH 3 ? ? ? CH 3 neohexane CH 3 CHCHCH 3 ? ?
IUPAC nomenclature (Geneva, 1920) names of radicals (alkyl groups): CH 3 - “methyl” CH 3 Cl methyl chloride CH 3 OH methyl alcohol, etc. CH 3 CH 2 - “ethyl” CH 3 CH 2 - “n-propyl” CH 3 CHCH 3 “isopropyl” |
CH 3 CH 2 CH 2 - “n-butyl” CH 3 CH 2 CHCH 3 or CH 3 CH 2 CH- “sec-butyl” | CH 3 CHCH 2 CH 3 CCH 3 | “isobutyl” “tert-butyl”
Web problems to help with naming and recognizing alkanes: Sciencegeek. net AP Chemistry Review (scroll to bottom) Naming Alkanes Practice Also try copy/paste this url in the address line in your browser: http: //proton. csudh. edu/structures/butyls/hwbutyls. html
1. Find the longest carbon chain containing the principal functional group and assign it as the parentchain. In case for a presence of two or more functional groups, the highest priority functional group will be chosen for the parent chain and the rest will simply be substituents. 2. Determine the type and number (quantity) of the substituents attached on the parent carbon chain. 3. Use prefixes di, tri, tetra, etc to indicate the number (quantity) of similar substituents. 4. Start assigning numbers for each carbon in the parent chain beginning at the terminal carbon nearest the principal functional group or the first branch point (in alkanes and alkyl halides). Make sure to assign the LOWEST possible numbers for all substituents. In case of ambiguity in numbering (i. e. same number at both terminal ends) assign the lowest number to the substituent which follows the alphabetical order. 5. Determine the position (carbon number) of all the substituents in the main/parent chain. 6. Use comma to separate numbers and use dash/hyphen to separate numbers and words. 7. Arrange the names of the substituents in alphabetical order (excluding the prefixes like di, tri, tetra; sec and tert); only iso and neo are considered in alphabetization and they are not separated by a dash. 8. Write the name of the compound according to the format above.
hexanes C 6 H 14 IUPAC names CH 3 CH 2 CH 2 CH 3 CH 3 CHCH 2 CH 3 (n-hexane) (isohexane) n-hexane 2 -methylpentane CH 3 CH 2 CH 3 CCH 2 CH 3 (no common name) CH 3 3 -methylpentane (neohexane) 2, 2 -dimethylbutane CH 3 CHCHCH 3 (no common name) 2, 3 -dimethylbutane
CH 3 CH 2 CH 2 CHCH 3 CH 3 2, 4 -dimethylheptane CH 3 CH CH 3 CH 2 CHCH 2 CH 3 CH 2 CCH 3 CH 3 6 -isopropyl-2, 2 -dimethylnonane
“classes of carbons” primary carbon (1 o) – a carbon bonded to one carbon secondary carbon (2 o) – a carbon bonded to two carbons tertiary carbon (3 o) – a carbon bonded to three carbons quaternary carbon (4 o) – a carbon bonded to four carbons 1 o 4 o CH 3 CHCH 2 CCH 3 1 o CH 3 3 o 2 o
classification of hydrogens, halides – hydrogens or halides are classified by the carbon to which they are attached. 1 o CH 3 CHCH 2 CH 3 1 o 3 o 2 o 2 o 1 o CH 3 CH 2 CHCH 3 sec-butyl bromide 2 o bromide Br CH 3 CCH 3 Cl tert-butyl chloride 3 o chloride
alkanes, physical properties non-polar or only weakly polar, cannot hydrogen bond relatively weak intermolecular forces lower mp/bp; increase with size; decrease with branching @ room temperature: C 1 – C 4 are gases C 5 – C 17 are liquids > C 17 are solids alkanes are water insoluble
alkane methane propane n-butane n-pentane n-hexane … n-heptadecane n-octadecane branching lowers mp/bp n-pentane isopentane mp o. C -183 -172 -187 -138 -130 -95 bp o. C -162 -89 -42 0 36 69 22 28 292 308 -130 -160 36 28
fossil fuels: natural gas petroleum coal petroleum is a complex mixture of hydrocarbons 1. solvents 2. fuels 3. raw materials for chemical syntheses separated into fractions by fractional distillation in an oil refinery
products from fractional distillation of petroleum: fraction b. range carbons natural gas below 20 o C 1 – C 4 petroleum “ether” 20 – 60 o C 5 – C 6 ligroin 60 – 100 o C 6 – C 7 raw gasoline 40 – 205 o C 5 – C 10 kerosine 175 – 325 o C 12 – C 18 gas oil above 275 o C 12 & up lube oil non-volaltile liquids asphalt non-volatile solids coke solid carbon
syntheses Industrial Laboratory large amounts (tons) small amounts (grams) lowest cost non-profit mixtures often okay pure substances dedicated apparatus flexible apparatus on exams, homework: laboratory syntheses!
Alkanes, syntheses: 1. (to be covered later) 2. Reduction of an alkyl halide a) hydrolysis of a Grignard reagent b) with an active metal and an acid 3. Corey-House synthesis (coupling of an alkyl halide with lithium dialkylcopper)
2. Reduction of an alkyl halide a) hydrolysis of a Grignard reagent (two steps) i) R—X + Mg RMg. X (Grignard reagent) ii) RMg. X + H 2 O RH + Mg(OH)X SB SA WA WB CH 3 CH 2 -Br + Mg CH 3 CH 2 -Mg. Br n-propyl bromide n-propyl magnesium bromide CH 3 CH 2 -Mg. Br + H 2 O CH 3 CH 2 CH 3 + Mg(OH)Br propane
CH 3 CH 3 CH-Br + Mg CH 3 CH-Mg. Br isopropyl bromide isopropyl magnesium bromide CH 3 CH-Mg. Br + H 2 O CH 3 CH 2 CH 3 propane CH 3 CH 2 -Mg. Br + D 2 O CH 3 CH 2 D heavy water CH 3 CH 3 CH-Mg. Br + D 2 O CH 3 CHD
b) with an active metal and an acid R—X + metal/acid RH active metals = Sn, Zn, Fe, etc. acid = HCl, etc. (H+) CH 3 CH 2 CHCH 3 + Sn/HCl CH 3 CH 2 CH 3 + Sn. Cl 2 Cl sec-butyl chloride n-butane CH 3 CH 3 CCH 3 + Zn/H+ CH 3 CHCH 3 + Zn. Br 2 Br tert-butyl bromide isobutane
3. Corey-House synthesis R-X + Li R-Li + Cu. I R 2 Cu. Li + R´-X R—R´ (alkane) (R´-X should be 1 o or methyl) This synthesis is important because it affords a synthesis of a larger alkane from two smaller alkyl halides.
note: the previous equations are not balanced: R-X + 2 Li R-Li + Li. X 2 R-Li + Cu. I R 2 Cu. Li + Li. X R R 2 Cu. Li = R-Cu-, Li+ R 2 Cu. Li + R´X R-R´ + RCu + Li. X
CH 3 CH 3 CH-Br + Li CH 3 CH-Li + Cu. I (CH 3 CH)2 -Cu. Li isopropyl bromide CH 3 (CH 3 CH)2 -Cu. Li + CH 3 CH 2 -Br CH 3 CH-CH 2 CH 3 2 -methylpentane (isohexane) Note: the R´X should be a 1 o or methyl halide for the best yields of the final product.
Alkanes, syntheses: 1. (to be covered later) 2. Reduction of an alkyl halide a) hydrolysis of a Grignard reagent b) with an active metal and an acid 3. Corey-House synthesis (coupling of an alkyl halide with lithium dialkylcopper)
ALKYL HALIDES Mg H 2 O ALKANES Li Sn, HCl Cu. I R’X
Reactions of alkanes: alkane + H 2 SO 4 no reaction (NR) alkane + Na. OH NR alkane + Na NR alkane + KMn. O 4 NR alkane + H 2, Ni NR alkane + Br 2 NR alkane + H 2 O NR (Alkanes are typically non-reactive. They don’t react with acids, bases, active metals, oxidizing agents, reducing agents, halogens, etc. )
Alkane, reactions: 1. Halogenation 2. Combustion (oxidation) 3. Pyrolysis (cracking)
2. Combustion Cn. H 2 n+2 + (xs) O 2, flame n CO 2 + (n+1) H 2 O + heat gasoline, diesel, heating oil… 3. Pyrolyis (cracking) alkane, 400 -600 o. C smaller alkanes + alkenes + H 2 Used to increase the yield of gasoline from petroleum. Higher boiling fractions are “cracked” into lower boiling fractions that are added to the raw gasoline. The alkenes can be separated and used in to make plastics.
1. Halogenation R-H + X 2, heat or hv R-X + HX a) heat or light required for reaction. b) X 2: Cl 2 > Br 2 I 2 c) yields mixtures d) H: 3 o > 2 o > 1 o > CH 4 e) bromine is more selective
CH 3 + Cl 2, hv CH 3 CH 2 -Cl + HCl ethane ethyl chloride CH 3 CH 2 CH 3 + Cl 2, hv CH 3 CH 2 -Cl + CH 3 CHCH 3 propane n-propyl chloride Cl isopropyl chloride 45% 55% gives a mixture of both the possible alkyl halides!
CH 3 CH 2 CH 3 + Cl 2, hv CH 3 CH 2 CH 2 -Cl 28% n-butane n-butyl chloride + CH 3 CH 2 CHCH 3 72% Cl sec-butyl chloride CH 3 CH 3 CHCH 3 + Cl 2, hv CH 3 CHCH 2 -Cl 64% isobutane isobutyl chloride + CH 3 CCH 3 36% Cl tert-butyl chloride
CH 3 + Br 2, hv CH 3 CH 2 -Br + HBr ethane ethyl bromide CH 3 CH 2 CH 3 + Br 2, hv CH 3 CH 2 -Br + CH 3 CHCH 3 propane n-propyl bromide Br isopropyl bromide 3% 97%
CH 3 CH 2 CH 3 + Br 2, hv CH 3 CH 2 CH 2 -Br 2% n-butane n-butyl bromide + CH 3 CH 2 CHCH 3 98% Br sec-butyl bromide CH 3 CH 3 CHCH 3 + Br 2, hv CH 3 CHCH 2 -Br <1% isobutane isobutyl bromide + CH 3 CCH 3 99% Br tert-butyl bromide
In the reaction of alkanes with halogens, bromine is less reactive but more selective. Why? How? mechanism: initiating step: 1) X—X 2 X • propagating steps: 2) X • + R—H H—X + R • 3) R • + X—X R—X + X • 2), 3), 2), 3)… terminating steps: 4) 2 X • X—X 5) R • + X • R—X 6) 2 R • R—R
chlorination of propane, mechanism: 1) Cl—Cl 2 Cl • 2) abstraction of 1 o hydrogen: Cl • + CH 3 CH 2 CH 3 CH 3 CH 2 • + HCl or abstraction of 2 o hydrogen: Cl • + CH 3 CH 2 CH 3 CH 3 CHCH 3 + HCl • 3) CH 3 CH 2 • + Cl 2 CH 3 CH 2 Cl + Cl • or CH 3 CHCH 3 + Cl 2 CH 3 CHCH 3 + Cl • Cl plus terminating steps
2) abstraction of 1 o hydrogen: Cl • + CH 3 CH 2 CH 3 CH 3 CH 2 • + HCl or abstraction of 2 o hydrogen: Cl • + CH 3 CH 2 CH 3 CH 3 CHCH 3 + HCl • The chloride that is produced depends on which hydrogen is abstracted by the chlorine free radical in step 2. The npropyl free radical gives the n-propyl chloride while the isopropyl free radical yields the isopropyl chloride. The relative reactivity in chlorination: H: 3 o : 2 o : 1 o = 5. 0 : 3. 8 : 1. 0
The number of hydrogens (probability factor) may also be important. CH 3 CH 2 CH 3 + Cl 2, hv CH 3 CH 2 CH 2 -Cl n-butane + CH 3 CH 2 CHCH 3 Cl n-butyl chloride = (# of 1 o hydrogens) x (reactivity of 1 o) = 6 x 1. 0 = 6. 0 sec-butyl chloride = (# of 2 o hydrogens) x (reactivity of 2 o) = 4 x 3. 8 = 15. 2 % n-butyl chloride = 6. 0/(6. 0 + 15. 2) x 100% = 28% % sec-butyl chloride = 15. 2/(6. 0 + 15. 2) x 100% = 72%
CH 3 CH 3 CHCH 3 + Cl 2, hv CH 3 CHCH 2 -Cl + CH 3 CCH 3 isobutane Cl isobutyl chloride = (# of 1 o H’s) x (reactivity of 1 o) = 9 x 1. 0 = 9. 0 tert-butyl chloride = (# of 3 o H’s) x (reactivity of 3 o) = 1 x 5. 0 = 5. 0 % isobutyl = (9. 0/(9. 0 + 5. 0)) x 100% = 64% In this case the probability factor outweighs the difference in relative reactivity of 1 o and 3 o hydrogens.
Relative reactivity in bromination: 3 o : 2 o : 1 o = 1600 : 82 : 1 In bromination the relative reactivity differences are much greater than any probability differences. isobutane + Br 2, hv isobutyl bromide + tert-butyl bromide isobutyl bromide = 9 H x 1 = 9 tert-butyl bromide = 1 H x 1600 = 1600 % tert-butyl bromide = (1600/1601) x 100% = >99%
Why is relative reactivity of H: 3 o > 2 o > 1 o ? CH 3—H CH 3 • + H • ΔH = 104 Kcal/mole CH 3 CH 2—H CH 3 CH 2 • + H • ΔH = 98 Kcal/mole 1 o free radical CH 3 CH 2 CH 3 CH 2 • + H • ΔH = 98 Kcal/mole 1 o free radical CH 3 CHCH 3 + H • ΔH = 95 Kcal/mole • 2 o free radical CH 3 CH 3 CHCH 3 CH 3 CCH 3 + H • ΔH = 92 Kcal/mole • 3 o free radical
Relative reactivity in halogenation: Stability of free radicals: Ease of formation of free radicals: Ease of abstraction of H’s: 3 o > 2 o > 1 o > CH 4
1. Halogenation R-H + X 2, heat or hv R-X + HX a) heat or light required for reaction. b) X 2: Cl 2 > Br 2 I 2 c) yields mixtures d) H: 3 o > 2 o > 1 o > CH 4 e) bromine is more selective
Alkane, reactions: 1. Halogenation 2. Combustion (oxidation) 3. Pyrolysis (cracking)
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