26 GENERAL CHEMISTRY Structures of Organic Compounds

26 GENERAL CHEMISTRY Structures of Organic Compounds
Chemistry 140 Fall 2002 TENTH EDITION GENERAL CHEMISTRY Principles and Modern Applications PETRUCCI HERRING MADURA BISSONNETTE 26 Structures of Organic Compounds In this chapter, we will focus on the structures and properties of organic compounds, and we will consider the preparation and some uses of these compounds. In the next chapter, we will turn our attention to reactions that interconvert these compounds. PHILIP DUTTON UNIVERSITY OF WINDSOR DEPARTMENT OF CHEMISTRY AND BIOCHEMISTRY General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Structures of Organic Compounds
Chemistry 140 Fall 2002 Contents 26-1 Organic Compounds and Structures: An Overview 26-2 Alkanes 26-3 Cycloalkanes 26-4 Stereoisomerism in Organic Compounds 26-5 Alkenes and Alkynes 26-6 Aromatic Hydrocarbons 26-7 Organic Compounds Containing Functional Groups 26-8 From Molecular Formula to Molecular Structure Structures of Organic Compounds Coffee beans contain an alkaloid compound commonly known as caffeine. Caffeine is a stimulant of the central nervous system producing alertness and heightened concentration. (Greg Vaughn/PacificStock.com) General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

26-1 Organic Compounds and Structures: An Overview
Chemistry 140 Fall 2002 26-1 Organic Compounds and Structures: An Overview The simplest organic compounds are those of carbon and hydrogen— hydrocarbons—and the simplest hydrocarbon is methane, the chief constituent of natural gas. Tetrahedral structure showing bond angle. Dashed-wedged line structure convention used to suggest a three-dimensional structure through a structural formula. The solid lines represent bonds in the plane of the page. The dashed wedge projects away from the viewer (behind the plane of the page), and the heavy wedge projects toward the viewer (out of the page). Ball-and-stick model. Space-filling model. FIGURE 26-1 Representations of the methane molecule General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

The propane molecule, C3H8
Chemistry 140 Fall 2002 (a) Structural formula. (b) Condensed structural formula. (c) Ball-and-stick model. (d) Space-filling model. (e) Dashed-wedged line notation. FIGURE 26-3 The propane molecule, C3H8 General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Constitutional isomers
Chemistry 140 Fall 2002 Constitutional isomers pentane and 2-methylbutane General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Line-angle formulas or line structures
General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Select the longest continuous carbon chain.
Nomenclature Select the longest continuous carbon chain. This determines the base name. Consider each branch and name similarly except change the name from –ane to –yl. Number the C atoms so that substituents have the lowest possible number.

General Chemistry: Chapter 26
Nomenclature Name each substituent according to its identity and the number of the C atom to which it is attached. Use di, tri, tetra as appropriate. Separate numbers from one another by commas. List substituents alphabetically by name. General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

A carbon atom attached to only one other carbon is called a primary (1º) carbon.
A carbon attached to two other carbons is known as a secondary (2º) carbon. If it attaches to three other carbons it is said to be tertiary (3º) and if attached to four carbons it is quaternary (4º).

(CH3)3CH CH3(CH2)2CH3 CH3CH(CH3)2 8/3/2018

Pentanes Draw Lewis structures and provide IUPAC names for all isomers of pentane, C5H12. Find and circle the isomer which does not contain any 2º or 3º carbons. 8/3/2018

• the shapes of molecules also influences the magnitudes of dispersion forces. The increased surface area in n-pentane leads to higher polarizability and hence stronger dispersion forces.

Functional groups

Functional Groups Common connections of atoms in a molecule lead to similar geometry, polarity and other similar properties. This similarity in chemical nature makes it convenient to group “functional” groups together as they have similar chemistries and related properties. For example;

Alcohols are alkane like compound where one or more hydrogen atoms have been replaced by an OH group. Alcohols can be classified as primary, secondary or tertiary in the same manner as alkane hydrogens. The potential hydrogen bonding due to the OH group results in large intermolecular forces. To name, add an “ol” suffix in place of the “e” in the alkane name.

Ethers are compounds that have two alkyl groups bound together by an oxygen atom. Common names for ethers are the two alkyl group names, in alphabetical order followed by the word ether.

IUPAC nomenclature: treats them as alkanes with an alkoxy substituent.
The smaller substituent is considered part of the alkoxy group; the larger substituent defines the stem.

Determine the number of possible constitutional isomers of C3H8 including propane to the right.
1. 1 2. 2 3. 3 4. 4 5. 5 Slide 23 of 43

Determine the number of possible constitutional isomers of C3H8 including propane to the right.
1. 1 2. 2 3. 3 4. 4 5. 5 Slide 24 of 43

The IUPAC name for the hydrocarbon depicted to the right is:
1. 4,5-dimethylheptane 2. 2-ethyl-3-methylhexane 3. 4-methyl-3-methylheptane 4. 3-methyl-4-methylheptane 5. 3,4-dimethylheptane Slide 25 of 43

The IUPAC name for the hydrocarbon depicted to the right is:
1. 4,5-dimethylheptane 2. 2-ethyl-3-methylhexane 3. 4-methyl-3-methylheptane 4. 3-methyl-4-methylheptane 5. 3,4-dimethylheptane Slide 26 of 43

According to the IUPAC conventions, the name of the alkene depicted to the right is:
1. 2,3-dimethyl-2-pentene 2. cis-2,3-dimethyl-2-pentene 3. trans-3,4-dimethyl-2-pentene 4. 2-ethyl-3-methyl-2-butene 5. 1-ethyl-1,2,2-trimethyl-1-ethene Slide 27 of 43

2. cis-2,3-dimethyl-2-pentene
According to the IUPAC conventions, the name of the alkene depicted to the right is: 1. 2,3-dimethyl-2-pentene 2. cis-2,3-dimethyl-2-pentene 3. trans-3,4-dimethyl-2-pentene 4. 2-ethyl-3-methyl-2-butene 5. 1-ethyl-1,2,2-trimethyl-1-ethene Slide 28 of 43

Which of the structures below is not a constitutional isomer of 3-ethyl-2-pentanol?
Slide 29 of 43

Which of the structures below is not a constitutional isomer of 3-ethyl-2-pentanol?
Slide 30 of 43

The IUPAC-accepted name of the alcohol depicted to the right is:
1. 2,3-dimethyl-4-pentanol 2. 3,4-dimethyl-2-pentanol 3. 3-isopropyl-2-butanol 4. 1,2,3-dimethyl-1-butanol 5. 1-methyl-2-isopropyl-1-propanol Slide 31 of 43

The IUPAC-accepted name of the alcohol depicted to the right is:
1. 2,3-dimethyl-4-pentanol 2. 3,4-dimethyl-2-pentanol 3. 3-isopropyl-2-butanol 4. 1,2,3-dimethyl-1-butanol 5. 1-methyl-2-isopropyl-1-propanol Slide 32 of 43

The correct order of increasing boiling points for the compounds to the right is:
MM g mol-1 a D 1. A < B < C 46 1.7 2. B < C < A 46 1.3 3. C < B < A 4. B < A < C 44 0.08 5. A < C < B a: dipole moment Slide 33 of 43

The correct order of increasing boiling points for the compounds to the right is:
MM g mol-1 a D bpt oC 1. A < B < C 46 1.7 78 2. B < C < A 46 1.3 -24 3. C < B < A 4. B < A < C 44 0.08 -42 5. A < C < B a: dipole moment * is the trend in boiling points due simply to dipole moments? Slide 34 of 43

Sildenafil, or Viagra, was originally developed to treat a form of cardiovascular disease. All of the atoms depicted numerically below are sp2 hybridized except for the atom designated with which number? 1 2 3 4 5 Slide 35 of 43

Sildenafil, or Viagra, was originally developed to treat a form of cardiovascular disease. All of the atoms depicted numerically below are sp2 hybridized except for the atom designated with which number? 1 2 3 4 5 Slide 36 of 43

In each of the following groups of substances, circle the one that has the given property. For each species you must describe the appropriate attractive force and estimate its strength. (4 marks) Largest vapor pressure at 100 K: SiH4, CH4, GeH4, SnH4 Lowest boiling point: CBr4, CCl4, CI4, CF4 General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Nitrogen (N2) and carbon monoxide (CO) are isoelectronic. What substance should have the higher melting point? The phosphine (PH3) and hydrogen sulfide (H2S) molecules each contain 18 electrons. Their dipole moments are 0.58 and 0.97 D, respectively. Which substance should have the higher boiling point ? General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Consider the compound ("Peteane") below and answer the following questions. Note; all of the hydrogen atoms are omitted for clarity. How many quaternary carbons does Peteane have (1 mark)? How many tertiary hydrogens does Peteane have (1 mark)? How many primary carbons does Peteane have (1 mark)? How many secondary carbons does Peteane have (1 mark)? General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Functional Group Isomerism:
1-propanol and ethyl methyl ether are both isomers of C3H8O but they possess different functional groups. How many more isomers are possible for C3H8O? Provide Lewis structure(s) and name(s) for any additional isomers.

Conformations In alkane, groups can rotate freely with respect to one another about the sigma bond connecting them. Conformations are different spatial arrangements that are possible in a molecule. One conformation can be converted into another by rotation about sigma bonds.

Conformations we use Newman projection to represent different spatial arrangement of atoms that result from rotations about a sigma bond. General Chemistry: Chapter 26

Staggered and eclipsed conformations of ethane
12 kJ/mol FIGURE 26-6 Staggered and eclipsed conformations of ethane General Chemistry: Chapter 26

Chemistry 140 Fall 2002 The barrier to rotation is small enough a room FIGURE 26-7 Potential energy diagram for the internal rotation of the methyl groups in ethane

Alkenes can be classified by the geometry of the items bound to the double bond. Since double bonds do not permit free rotation, geometric structural isomers exist. For Example:

The cis and trans naming of alkenes is valid when at least two substituents are the same. But what about when all are different?

The E (entgegen), Z (zusammen) System sets priority using the following rules: 1. Based on atom to sp2 Carbon. Greater atomic number, higher the priority. 2. For ties, move out to next bound atoms 3. Consider double bonds to be connected to 2 of the atoms bonded and triple to three. 4. Higher mass isotopes have a higher priority than lower isotopes.

The E,Z system.

Preparation and Uses of Alkenes and Alkynes
Elimination reaction General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Addition of bromine to an alkene

Preparation of alkanes

Cycloalkanes are similar to alkanes in physical properties, but they have higher boiling points, melting points, and densities. This is due to stronger London forces. The ring shape allows for a larger area of contact. Cycloalkanes exhibit almost the same degree of unreactivity as alkanes, due to their containing only unreactive C-C and C-H bonds; however, ring strain can cause increased reactivity.

The reason for the ring strain is the fact that the models show carbon as being sp3 hybridized and consequently preferring a tetrahedral geometry of atoms attached to the carbons. With tetrahedral geometry the ideal angle between C-C-C bonds is 109.5o. In cyclopropane the angles between C-C-C bonds are 60o and therefore these bonds are forced to be much closer together than in an open chain alkane. In cyclobutane the angle is 90o, so there is less ring strain than in cyclopropane, but it is still significant.

There is also another factor that makes small rings unstable, which is known as eclipsing (torsional) strain.

In the absence of UV light, cyclopropane can undergo addition reactions in which the ring is broken. For example, with bromine, cyclopropane gives 1,3-dibromopropane.

If a cycloalkane ring is flat, then all the hydrogens on the ring will have an eclipsing relationship to each other and the resulting repulsion of all the eclipsing interactions forces the ring to want to twist into a conformation where some of this interaction is reduced or even eliminated.

In the case of cyclobutane, it can twist two opposite carbons in the ring “up” or “down”.
The result is a more stable conformation that relieves some of the eclipsing hydrogen atoms interaction. Consequently, the most stable conformation of cyclobutane is not a flat ring, but a conformation known as “puckered” or sometimes it is called the “butterfly” conformation. The cyclobutane ring rapidly flips from one puckered conformation to the other.

Cis-Trans Isomerism in Disubstituted Cycloalkanes
Chemistry 140 Fall 2002 Cyclohexane rings can adopt a conformation in which any C-C-C angle is almost 109.5o and consequently they have no ring strain. The C-C-C bond angles in a cyclopentane can also be made to be close to 109.5o and therefore there is only a small amount of ring strain. Boat form. (b) Chair form. The H atoms extending laterally from the ring— equatorial H atoms—are shown in blue. The H atoms projecting above and below the ring—axial H atoms—are in red. Cis-Trans Isomerism in Disubstituted Cycloalkanes

If an Alkene has two double bonds use the suffix “diene”. For Example;

β-Carotene Being highly conjugated, it is deeply colored, and as a hydrocarbon lacking functional groups, it is very lipophilic.

We will now introduce a third type of isomer - STEREOISOMERS
If any two compounds are stereoisomers, then they have the same bond connectivity (the same atoms are attached to the same atoms). BUT, the fixed relative arrangement of the atoms in 3D space is not the same. “Fixed” means that, unless the stereoisomers undergo some type of chemical reaction, they cannot interconvert.

STEREOCHEMISTRY

Mirror-image related stereoisomers (ENANTIOMERS) are said to be left-handed and right-handed and occur when a molecule and its mirror image are non-superimposable.

Solutions containing Chiral Molecules distort or “twist” the plane of polarization of the light either to the left or right. This effect is called optical rotation and the molecule is said to be optically active. Optical activity is measured using a polarimeter.

Polarimeter Light is electromagnetic radiation that oscillates perpendicular to its direction of motion. A polarizer can “filter” all but one “plane” of light. This light is referred to as plane-polarized light. When plane-polarized light interacts with an “optically active” solution, the “plane” gets twisted or altered.

When plane-polarized light is passed through a sample of one of enantiomers, the plane of polarization is rotated either clockwise or counterclockwise. The other enantiomer would rotate the light is an equal amount, but in the opposite direction. If facing the light source: • Clockwise rotation: enantiomer is dextrorotary (+) • Counterclockwise rotation: enantiomer is levorotary (-) This interaction with light is called optical activity and enantiomers are often called optical isomers.

Enantiomers cannot be distinguished on the basis of their physical properties, such as boiling points, melting points, and densities. Enantiomers interact differently with plane-polarized light and can be distinguished in this manner.

Compounds whose mirror images are superimposable are called ACHIRAL
Compounds whose mirror images are superimposable are called ACHIRAL. Examples of chiral and achiral molecules: Above all, the chiral examples contain an atom that is connected to 4 different substituent groups. This atom is called an asymmetric atom or a stereocenter (also called a stereogenic carbon atom) stereocenter are always chiral.

(CH3)2CHCHClCH3 CH3CH2CH(CH3)CH2Cl CH3CHClCH2CH2CH3 (CH3)3CCH2Cl CH3CH2CHClCH2CH3 CH3CH2CCl(CH3)2

The absolute configuration of an enantiomer is the actual spatial arrangement of the substituent groups around the chiral centers.

R-S SEQUENCE RULES FOR STEREOCENTERS A stereogenic carbon has one of two possible arrangements (configurations) of its four different groups and they are mirror images of each other. One form of the stereogenic carbon will be labelled as having “R” configuration and its mirror image will then be called “S” configuration.

There is a set of rules, called the Cahn-Ingold-Prelog sequence rules, described on the following slides, which we use in order to find out which of the two possible absolute configurations of a stereocenter is R (for Rectus, Latin for right) and which is S (for Sinister, Latin for left).

We look first at the atoms attached to the stereocenter
We look first at the atoms attached to the stereocenter. An atom of higher atomic number than another has a higher precedence. If you are comparing isotopes, the one with the higher atomic mass has the higher priority. We would assign the bromine as priority 1, the chlorine as priority 2, the CH3 group as priority 3 and the hydrogen as priority 4 (your text uses the letters a,b,c,d)

We then look at the molecule placing the lowest (4th) priority group away from us.

We then consider the sense of direction from highest to lowest priority of the three groups facing us. In this case it is anticlockwise, which we Label as “S” stereochemistry. A clockwise arrangement would be labelled as being of “R” configuration. For naming the molecule, the “S” goes in brackets at the beginning and is italicized: (S)-1-bromo-1-chloroethane. The mirror image will then be (R)-1-bromo-1-chloroethane.

Rules for Assigning Priorities to Substituents
Rule 1: Substituent of higher atomic number takes precedence over one of lower atomic number: General Chemistry: Chapter 26

Nonsuperimposable mirror images of 3-methylhexane
Chemistry 140 Fall 2002 Notice in the diagram that we have adopted the convention used by organic chemists in drawing chiral centers: The groups attached to the central carbon in the plane of the paper are connected with solid lines, the group in front of the plane of the paper is attached by a solid wedge, and the group behind the plane of the paper is indicated by a dashed wedge. Nonsuperimposable mirror images of 3-methylhexane General Chemistry: Chapter 26

Naming Enantiomers: The R, S System of Nomenclature
Chemistry 140 Fall 2002 Naming Enantiomers: The R, S System of Nomenclature The group of lowest priority is placed as far away from the viewer as possible. Assignemnt of R and S configuration at a tetrahedral stereocentre General Chemistry: Chapter 26

Rule 2: If two substituents attached to the stereocenter have the same priority, proceed along the chains to the first point of difference: General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Ring Strain in Cycloalkanes
Chemistry 140 Fall 2002 Ring Strain in Cycloalkanes (a) Ball-and-stick model. (b) Valence-bond picture of cyclopropane using sp3 hybrid orbitals, which leads to poor overlap of the orbitals and hence weak bonds. FIGURE 26-10 Ring strain in cyclopropane General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Chemistry 140 Fall 2002 Boat form. (b) Chair form. The H atoms extending laterally from the ring— equatorial H atoms—are shown in blue. The H atoms projecting above and below the ring—axial H atoms—are in red. FIGURE 26-12 Cis-Trans Isomerism in Disubstituted Cycloalkanes General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

The interconversion of two chair conformations in cyclohexane
Chemistry 140 Fall 2002 In the figure, two of the H atoms are shown in red to emphasize that when the cyclohexane ring converts from one chair conformation to another, the equatorial hydrogen atoms are converted into axial hydrogen atoms and vice versa. The interconversion of the two chair forms proceeds through other conformations, including the boat form. FIGURE 26-13 The interconversion of two chair conformations in cyclohexane General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Nonsuperimposable mirror images of 3-methylhexane
Chemistry 140 Fall 2002 Notice in the diagram that we have adopted the convention used by organic chemists in drawing chiral centers: The groups attached to the central carbon in the plane of the paper are connected with solid lines, the group in front of the plane of the paper is attached by a solid wedge, and the group behind the plane of the paper is indicated by a dashed wedge. FIGURE 26-15 Nonsuperimposable mirror images of 3-methylhexane General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Naming Enantiomers: The R, S System of Nomenclature
Chemistry 140 Fall 2002 Naming Enantiomers: The R, S System of Nomenclature The group of lowest priority is placed as far away from the viewer as possible. FIGURE 26-16 Assignemnt of R and S configuration at a tetrahedral stereocentre General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Rules for Assigning Priorities to Substituents
Rule 1: Substituent of higher atomic number takes precedence over one of lower atomic number: General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Rule 2: If two substituents attached to the stereocenter have the same priority, proceed along the chains to the first point of difference: General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Rule 3: Double and triple bonds count as if they were single and the atoms are duplicated or triplicated at the other end of the double or triple bond: General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Chemistry 140 Fall 2002 Naming the Stereoisomers of Highly Substituted Alkenes: The E,Z System of Nomenclature IUPAC: the E, Z system. The Cahn–Ingold–Prelog rules, discussed previously, are used systematically to assign priorities to the substituents on the carbon atoms of the double bond. The stereochemistry about the double bond is assigned Z (from the German word zusammen, meaning “together”) if the two groups of higher priority at each end of the double bond are on the same side of the molecule. If the two groups of higher priority are on opposite sides of the double bond, the configuration is denoted by an E (from the German word entgegen, meaning “opposite”). These ideas are summarized in the diagram zed iz on ze zame zide General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

26-6 Aromatic Hydrocarbons

Characteristics of Aromatic Hydrocarbons
Planar (flat) cyclic molecules. Conjugated п systems (4n + 2) August Kekulé ( ) who proposed the hexagonal ring structure for benzene in His representation of this molecule is still widely used. General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Naming Aromatic Hydrocarbons

Preparation and Use of Alcohols

Chemistry 140 Fall 2002 Ethers Preparation and Uses of Ethers Diethyl ether has been used extensively as a general anesthetic. It is easy to administer and produces excellent relaxation of the muscles. Also, it affects the pulse rate, rate of respiration, and blood pressure only slightly. However, it is somewhat irritating to the respiratory passages and produces nausea. Methyl propyl ether (neothyl) is also used as an anesthetic and is less irritating to the respiratory passages. Dimethyl ether, a gas at room temperatures, is used as a propellant for aerosol sprays. Higher molecular mass ethers are used as solvents for varnishes and lacquers. Methyl tert-butyl ether, an unsymmetrical ether, marketed under the name MTBE, has been used as an octane enhancer in gasoline. However, because of its relatively high solubility in water in some localities it has caused rather extensive groundwater pollution through leakage from underground storage tanks; it is currently being phased out of use. conc H2SO4 CH3CH2OH + HOCH2CH3  CH3CH2OCH2CH3 + H2O 140 C General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Extract from natural sources

Carboxylic Acids Oh My Such Good Apple Pie, Sweet As Sugar oxalic, malonic, succinic, glutaric, adipic, pimelic, suberic, azelaic, sebacic General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

Preparation and Uses of Carboxylic Acids

Preparation and Uses of Amides

Quaternary ammonium ions

26-7 Heterocyclic Compounds
Chemistry 140 Fall 2002 26-7 Heterocyclic Compounds In the pyridine molecule, a nitrogen atom replaces onecof the CH units of benzene. Its formula isC5H5 N (a) Structural formula (b) Space-filling model FIGURE 26-17 Pyridine General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

26-8 From Molecular Formula to Molecular Structure
C5H10 If an organic molecule with n carbon atoms has fewer than 2n+2 hydrogen atoms, then it contains elements of unsaturation, such as π bonds or ring structures General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

End of Chapter Questions
Take your knowledge and skills with you. Any area of expertise requires a base on which to build. Your first year courses are that base of knowledge. Your developing problem solving skills are a base on which to build more advanced skills. Chemistry is a field that uses advanced problem solving. General Chemistry: Chapter 26 Copyright © 2011 Pearson Canada Inc.

26 GENERAL CHEMISTRY Structures of Organic Compounds

Similar presentations

Presentation on theme: "26 GENERAL CHEMISTRY Structures of Organic Compounds"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

26 GENERAL CHEMISTRY Structures of Organic Compounds

Similar presentations

Presentation on theme: "26 GENERAL CHEMISTRY Structures of Organic Compounds"— Presentation transcript:

Similar presentations

About project

Feedback