LAB 5: WORKSHOP ON FUNCTIONAL GROUPSAND MOLECULAR MODELSObjectives: By the end of this lab session, you should:1. Be able to recognize the various oxidation states of carbon atoms in organic molecules.2. Be able to recognize various functional groups in organic molecules.3. Understand the tetrahedral structure of carbon atoms and why there is rotation about single bonds.4. Understand why double bonds are rigid, planar bonds, and what cis and trans configurations represent.5. Appreciate the linear nature of C-C triple bonds.6. Know what asymmetric (chiral) carbon atoms are, and understand isomers and enantiomers.7. Be able to understand the three-dimensional structures of organic compounds.Note: Bring your organic chemistry lecture .ppt slide sets along for this exercise. They will be useful.1. OXIDATION STATES OF ORGANIC COMPOUNDS: Note - the R group in the structures shown below represents the remainder of an organic molecule. R can be almost any organic structure (even just H), but the R-C bond shown is almost always a carbon-carbon (C-C) bond.R-CH3 (e.g., propane CH3-CH2-CH3) – saturated hydrocarbon, completely reduced, full of H, energy-rich.CO2 (carbon dioxide) – completely oxidized, no H, energy-poor.*Carboxyl groups are always ionized (proton lost into solution) at physiological pH.Remember that a saturated hydrocarbon R-(CH2)nH is the most reduced form of carbon possible. Notice that a saturated hydrocarbon is full of hydrogen atoms. This property will be important when we encounter carbohydrate (metabolism) oxidation. Animals oxidize reduced organic compounds to obtain the energy required to sustain life (energy is stored as electrons of the hydrogen atoms in reduced organic foodstuffs). The more reduced an organic compound is, the more energy it contains, and we derive some of this energy by oxidation. The further we oxidize organic compounds, the more energy we obtain. In general, the more oxygens you see on an organic molecule, the more oxidized it is; the more hydrogens you see, the more reduced it is. A CC double bond is more oxidized than a C–C single bond because the CC double bond has fewer H.Biochem 107L5-2Drawing organic structures: Remember that each C must have 4 bonds, O must have 2, and H has 1 (and only 1) bond. N always has 3 bonds, and S has 2. Four compounds we oxidize to obtain metabolic energy are glucose, oxaloacetate, fatty acids, and acetate. Condensed structural formulas for these compounds are given below. Write out the completestructural formulas of each (i.e., show ALL C-C, C-O, C-H, and O-H bonds). Then circle the various functional groups they contain and identify these groups. On a per-carbon-atom basis, do you think a fatty acid (e.g., palmitic acid) or a carbohydrate (e.g., glucose) would yield more energy upon complete oxidation? Why? CHO COOH (HCOH)4CO H2COH CH2CH3(CH2)16COOH CH3COOHCOOH palmitic acid glucose oxaloacetic acid (fatty acid) acetic acid2. BIOCHEMICALLY IMPORTANT FUNCTIONAL GROUPS: Biochemistry is about the series of chemical reactions that are required for life, and these reactions are all based on interactions between various functional groups in organic molecules. The most important functional groups are shown below.Biochem 107L5-33. Draw the complete structures of the following compounds, i.e., draw EVERY bond. In each structure, circle any functional groups (i.e., alcohol, aldehyde, ketone, amine, etc.) and identify the groups. Then write the condensed structural (or short-hand) formula below each. a) methane n) diethylketoneb) ethane o) 1,5-dihydroxy diethylketonec) propane p) methanal (formaldehyde)d) butane q) ethanal (ethylaldehyde)e) methanol (methyl alcohol) r) ammoniaf) ethanol (ethyl alcohol) s) methylamineg) 1-butanol (butyl alcohol) t) ethylamineh) 2-propanol (isopropyl alcohol) u) 2-aminobutanei) 2-aminoethanol(ethanolamine) v) benzenej) acetic acid w) toluene(methylbenzene)k) butanoic acid x) 2,4,6-trinitro toluene (TNT)*l) dimethylketone (acetone)y) aminoacetic acid (glycine)m) 1,3-dihydroxyacetone z) 2-amino propionic acid (alanine)*nitro = NO2 – just write it as such4. MOLECULAR MODELS: The three dimensional shapes of molecules determine, in part, their physicaland chemical properties. However, it can be difficult to visualize and work with three dimensional structures on paper, thus kits have been developed for making molecular models. Colored balls, drilled with holes, represent atoms (the number of holes indicates the number of bonds that can form). Black usually represents carbon, yellow hydrogen, and red oxygen. Sticks and springs fit into the holes to represent the bonds. Short sticks are used for bonds from hydrogen, and longer sticks are used for all other single covalent bonds. Springs are used for double or triple bonds. Using these kits, one can prepare models to aid in visualizing the three-dimensional structures of organic molecules.Examine the components of your set of molecular models. The black balls each have 4 tetrahedrally oriented holes. They are used to represent carbon atoms, which can form four bonds in a tetrahedral configuration.Insert a bond (stick) into each hole. Notice how each of the four valence bonds is equidistant from all the other. Your carbon atom will look like the figure below.Biochem 107L5-4The yellow balls each contain one hole. They are used to represent hydrogen atoms, which can form one bond (1 valence electron). If you place a yellow ball at the end of each bond, you will have CH4 (methane), the gas we burn in our kitchen stoves. The red balls contain two holes and can be used to represent oxygen atoms, which can form two bonds (6 valence electrons – needs 2 more). If you insert a stick (bond) in each end of a red ball and place yellow balls on the ends of the sticks, you will have water (H2O). Make a model for n-pentane (CH3-CH2-CH2-CH2-CH3), a linear 5-carbon molecule with 12 hydrogens and no double bonds. Notice how the molecule can rotate around each single bond to form many different conformations, even though the actual structure doesn't change. Now make a molecule of cyclopentane from your n-pentane molecule. What did you have to remove to form this new molecule? Is this molecule an isomer of pentane? Now make a molecule of 2-methylbutane (CH3-CH-CH2-CH3). l CH3This compound is also known as isopentane. Note that