Pharmaceutical Organic Chemistry 1 (PMC 102) Lecture 2 PDF

Pharmaceutical Organic chemistry 1 (PMC 102) Lecture 2  Alkanes (Paraffins) General formula: CnH2n+2 number of C atoms name Structure 1 methane CH4 2 ethane CH3CH3 3...

Pharmaceutical Organic chemistry 1 (PMC 102) Lecture 2  Alkanes (Paraffins) General formula: CnH2n+2 number of C atoms name Structure 1 methane CH4 2 ethane CH3CH3 3 n- propane CH3CH2CH3 4 n- butane CH3(CH2)2CH3 5 n-pentane CH3(CH2)3CH3 6 hexane CH3(CH2)4CH3 7 heptane CH3(CH2)5CH3 8 octane CH3(CH2)6CH3 9 nonane CH3(CH2)7CH3 10 decane CH3(CH2)8CH3 -H e.g. methane -H methyl Alkane Alkyl  IUPAC nomenclature 1- Select the longest chain of hydrocarbons. 2- Number the longest chain from the end nearest to the branch so that side chain would take lowest possible no. 6 5 4 3 2 1 CH3CH2CH2CH2CHCH3 1 2 3 4 5 CH2CH3 7 6 5 4 3 6 7 2-Ethylhexane X 2 1 5-Methylheptane X 3-Methylheptane 3- When two or more substituents are identical, use the prefixes di-, tri-, tetra- etc. 4- When branching first occurs at an equal distance from either end of the parent chain, choose the name that gives the lower number at the first point of difference 5 CH3 3 7 H2 1 H3C 6 CH 4 C 2 CH 3 CH 3 C 5 CH 7 2 H2 6 4 1 CH3 CH3 2,5,6-Trimethylheptane X 2,3,6-Trimethylheptane When two chains of equal length compete to be parent, choose the chain with the greatest number of substituents CH3 CH3 1 3 7 5 H3C CH 4 CH 6 CH3 2 CH CH C H2 CH3 CH2CH2CH3 2,3,5-Trimethyl-4-propylheptane (alphabetical)  X H3C H3C H2 H3C CH CH C X CH X CH3 H3C H3C sec-butyl isopropyl isobutyl H3C CH3 H3C C X H3C C CH2 H3C CH3 X tert-butyl neopentyl  Physical properties 1) C1-C4 gases, C5-C17 liquids, C18- solids 2) Branched isomers have lower boiling points than straight chain isomers H3C CH3 H2 CH CH2 H3C C CH3 H3C C CH3 CH3 C C H3C CH3 H2 H2 n-pentane isopentane neopentane B.p. = 36 B.p. = 28 B.p. = 9.5  General methods for preparation A) From compounds containing the same number of carbon atoms: 1) Catalytic reduction of unsaturated hydrocarbons: R R H2 H2 / Ni H2 / Ni R R C C R C C C C R H H H2 2) From ketones: Zn / HCl / Hg O (Clemmensen reduction) H2 C C R R1 R R1 or NH2NH2 / NaOH (Wolff-Kishner reduction) 3) From alkyl halides (Reduction using metal and acid): RX Zn / HCl RH + ZnXCl e.g. H2 Cl C Zn / HCl H3C H3C C CH3 CH3 + ZnCl2 H  Zn / HCl / Hg (Clemmensen reduction) or NH2NH2 / NaOH (Wolff-Kishner reduction)  B) From compounds containing lower number of carbon atoms: 1) Wurtz Reaction: used for the preparation of symmetrical alkanes 2 RX + 2 Na RR + 2 NaX  Reactions of alkanes Substitution reactions X2 = F2, Cl2, Br2, I2 Conditions: uv light (h) or heat 250-400 ( temperature) mechanism: Free radical Cl2 CH4 + Cl2 CH3Cl + HCl CH2Cl2 + HCl Cl2 Cl2 CCl4 + HCl CHCl3 + HCl mixture of products  Mechanism: Free radical 1) Initiation: h H2O2 2 Cl. Cl Cl 2) Propagation: a) Cl. H CH3 HCl +.CH3. b) CH3 + Cl Cl CH3Cl + Cl. 3) Termination: Cl. + Cl. Cl2. CH3 +.CH3 CH3CH3. CH3 + Cl. CH3Cl Factors affecting rate of halogenation: 1- Type of the halogen 2- Ease of abstraction of hydrogen (rate determining step)  1. Type of the halogen rate of reactivity: F2 > Cl2 > Br2 > I2 ❖Flourination and iodination are difficult to perfome in the lab (Flourine reacts with alkanes with an explosion while Iodine doesn’t react with alkanes). ❖Chlorination of alkanes is easier than bromination. 2. Ease of abstraction of hydrogen (rate determining step) H3C H2 H2 C Cl2 C + CH Cl H3C CH3 H3C CH2Cl h H3C n-propane n-propyl chloride isopropyl chloride 45% 55% Sec. alkyl halide is major Due to ease of abstraction 3ry H > 2ry H > 1ry H Stability of the free radical formed H3C H3C H3C H3C H < < CH H < H3C C H H2C H H3C H3C Increasing rate of reaction towards X2 radicals formed are more stable H3C H3C H3C.. CH3 <. < CH < H3C C. CH2 H3C H3C Due to hyperconjugation Selective free radical halogenation H2 H2 C X2 C H3C H3C CH3 H3C CH2X + h CH X H3C X= Cl 45% 55% X= Br 2% 98 Bromine is more selective than chlorine because it is less reactive