Alkanes (Paraffins) - Pharmaceutical Organic Chemistry I PDF

Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I Alkanes (Paraffins) Prof. Dr. Mostafa El-Miligy Professor of Pharmace...

Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I Alkanes (Paraffins) Prof. Dr. Mostafa El-Miligy Professor of Pharmaceutical Chemistry Faculty of Pharmacy Alexandria University 1 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 2 Alkanes 1 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I 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 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. CH3 H2 H3C CH C CH3 CH C CH H2 CH3 CH3 2,5,6-Trimethylheptane X 2,3,6-Trimethylheptane 4 Alkanes 2 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I 5- 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 2 CH 4 CH 6 CH3 CH CH C H2 CH3 CH2CH2CH3 2,3,5-Trimethyl-4-propylheptane (alphabetical) 5 Classification of Alkyl groups They are classified according to the carbon atom to which they are attached into Primary (1°), secondary (2°), tertiary (3°) 6 Alkanes 3 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I 7 H 3C H 3C C X H 3C tert-butyl 8 Alkanes 4 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I 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 9 10 Alkanes 5 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I Zn / HCl / Hg (Clemmensen reduction) H2 C or NH2NH2 / NaOH R R1 (Wolff-Kishner reduction) RX Zn / HCl RH + ZnXCl e.g. H2 Cl C Zn / HCl H3C H3C C CH3 CH3 + ZnCl2 H Its name is….. 11 B) From organomagnesium halides (Grignard reagent): 12 Alkanes 6 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I C) From compounds containing lower number of carbon atoms: Wurtz Reaction: used for the preparation of symmetrical alkanes 13 Substitution reactions: Free radical halogenation: Halogens are CH4 + Cl2 CH3Cl + HCl Arrange CH4, CH3Cl, CH2Cl2, CHCl3 and CCl4 according to polarity. 14 Alkanes 7 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I Cl Cl h H2O2 2 Cl. Free radical a) Cl. H CH3 HCl +.CH3. b) CH3 + Cl Cl CH3Cl + Cl. 15 16 Alkanes 8 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I Factors affecting rate of free radical halogenation 1- Type of the halogen. 2- Ease of abstraction of hydrogen (rate determining step). 17 ❖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. 18 Alkanes 9 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I H3C H2 H2 C Cl2 C CH Cl H3C CH2Cl + H3C CH3 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 19 20 Alkanes 10 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I 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 21 H2 H2 C X2 C H3C H3C CH3 H3C CH2X + h CH X H3C X= Cl 45% 55% X= Br 2% 98% 22 Alkanes 11 Prof. Dr. Mostafa El-Miligy Pharmaceutical Organic Chemistry I 23 Alkanes 12