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Drugs are mainly chemicals. So there are a lot of topics from Chemistry. So here you can download handwritten notes for this. An unshared electron pair counts as a group. Electronic configuration of carbon: [He] 2s2 2p2 H H H Here the 2s orbital hybridizes with all three 2p orbitals, to give four new sp3 hybrid orbitals in a tetrahedral geometry bond angle about Carbon has sp3 orbitals and tetrahedral geometry in any molecule in which it is bonded to four atoms or groups.
Similarly for nitrogen, oxygen, and other elements. Multiple bonds. Sigma bonds are generally stronger than pi bonds. Electronic Structures [M8 1. A few of these are shown below, using ethanol ethyl alcohol as an example. As complete structural formulas are very time consuming to draw, condensed structural formulas are usually preferable. For complex molecules, line formulas are often used see page Lewis structures often show all outer electrons as dots.
For large molecules this is time consuming and potentially confusing , and it is usually preferable to show bonds as lines, and used dots only for unshared electron pairs. Electronic structures always show unshared electron pairs and formal charges. Formal Charges. When drawing electronic structures it is important to calculate formal charges on all atoms.
The formal charge of an atom is the difference between the number of valence electrons in the neutral atom, and the number of valence electrons "belonging to" the atom in a particular structure. Exercise 1. Draw the detailed structural formula of each of the following compounds.
Show all unshared electron pairs, and indicate hybridization and geometry at each C or N atom. What is the charge on each nitrogen atom in the following structures? In order to understand organic reaction mechanisms, it is essential to know the formal location of charges and electrons.
There are several ways to generate electron dot structures of organic and inorganic compounds. The following is one way that works well for a variety of molecules. Write the atomic skeleton, using lines to indicate covalent single bonds. If you know that a double or triple bond is present, you may introduce it at this point if you wish. If in doubt, leave it as a single bond. Introduce enough unshared electron pairs so that every atom C, N, O, halogen has a complete octet.
In this step, each single bond is counted as two electrons. Calculate formal charges. To avoid confusing negative charges with single bonds, especially in hand- drawn structures, it is helpful to circle all charges. In this step, each of the atoms joined by a single bond is assigned one of the bond electrons. A carbon assigned 4 electrons would have no formal charge; similarly for nitrogen 5, oxygen 6, and halogen 7. Look at the total charge. If the target is a neutral molecule, remove electrons and introduce unsaturation.
This is done by removing two unshared electron pairs from adjacent atoms and inserting another bond a pi bond between the same two atoms. The overall process removes one pair of electrons two negative charges.
Recalculate the formal charge on each atom. Repeat as necessary. If you have a choice, it is better to remove an electron pair from the site with the greater negative charge, or from the less electronegative atom i.
These are examples of resonance structures -- a topic to be covered later in more detail. Note the double-headed arrow which is used only for this purpose. For nitromethane, the two resonance structures are completely equivalent. For acetamide, the left structure is preferred, since there is no charge separation. Neutral carbon atoms generally have four bonds, nitrogen three, and oxygen two.
A nitrogen atom with four bonds and an oxygen atom with three bonds will invariably have a positive charge. An oxygen atom with one bond will almost always have a negative charge.
Bear in mind that multiple bonds are possible. Compounds with the same electronic structure which differ only in the nucleus are said to be isoelectronic.
The bonding, hybridization sp3 , and electron distribution are equivalent. The only difference is within the nucleus: The nucleus of the oxygen contains one more proton. They are isoelectronic. The above chart is arranged so that each vertical column is isoelectronic. Although isoelectronic molecules have the same electronic structure and hybridization, they have different formal charges, and usually very different chemical reactivity. Only the skeleton is supplied; you need to fill in any multiple bonds, unshared electron pairs, or formal charges.
Consider the following compounds, taken from the above exercise. Multiple bonds are shown, unshared electron pairs and formal charges are not. Examples On this and the following two pages are examples of a number of important organic compounds.
You may be familiar with some of them. Isomers [M8 3. By carefully weighing the CO2 and H2O produced, the percentage of carbon and hydrogen can be calculated. Methods for analysis of nitrogen, halogen, and other elements are also available. The balance is generally assumed to be oxygen. Suppose we have an unknown compound with the following composition: C From this we can calculate an empirical formula of C3H8O.
If we can also determine the molecular weight, we can then determine the molecular formula. For example, a molecular weight of 60 would indicate a molecular formula of C3H8O.
But this doesn't tell us the structure, because there are three isomeric compounds with this formula! Compounds having the same molecular formula but different structural formulas are called isomers. For example, there are three different compounds isomers having the molecular formula C3H8O. They have different physical properties and different chemical behavior.
For this reason clear and unambiguous structural formulas are essential. For each of the following molecular formulas several isomers are possible. Draw all possible isomers and identify the functional groups in each one. When we remove two hydrogens from C3H8O forming C3H6O , we need to introduce a double bond or a ring, and there are a variety of ways to do this.
See if you can draw all 10 isomers of C 3H6O. Drawing Structural Formulas [M8 1. Here are four ways to draw 2-propanol isopropyl alcohol.
Additional representations are possible by bending or rotating the structures. All three give the impression that the central carbon is trivalent.
The first one also gives the impression that the oxygen is trivalent, and the right structure also gives the impression that the third carbon is pentavalent.
Line formulas do not show most carbon and hydrogen atoms, and help focus attention on the functional groups the reactive sites. Drawing Line Formulas 1. Write the structural formula. Erase all hydrogen atoms bonded to carbon, and all C-H bonds. Remove all carbon atoms and connect the C-C bonds. Deciphering Line Formulas 1. Insert a C at each intersection or kink , and at the end of every line which is not occupied by an atom. Add enough hydrogen atoms to make each carbon tetravalent. When you are using line formulas, remember that multiple bonds, formal charges, heteroatoms atoms other than carbon or hydrogen , any hydrogens bonded to heteroatoms, and unpaired electrons in free radicals must always be shown.
Carbon atoms, hydrogen atoms bonded to carbon, and unshared electron pairs are usually not shown but may be for clarity or for special emphasis. Reminder: When drawing structural formulas of any kind, remember that for stable organic molecules, where no formal charge is present carbon always has four bonds, nitrogen always three, oxygen always two, and hydrogen and halogens always one.
The following structural formulas are all acceptable representations of 1-nitropropane. What is wrong with each one? Functional Groups [M8 3. The most important classes of organic compounds are listed below. In this list "R" signifies any alkyl or aryl group, unless otherwise indicated. In some cases, R may also be H. In a few cases R must be alkyl sp3 , but not alkenyl or aryl sp2 or alkynyl sp.
Compounds which are not aromatic are sometimes called aliphatic. Hudrlik Organic Chem I: Structure and Bonding Alkanes and cycloalkanes as well as other compounds which lack multiple bonds are sometimes called saturated compounds. Compounds with multiple bonds especially alkenes and alkynes are sometimes called unsaturated.
Cyclic ethers with a three-membered ring are called epoxides. Cyclic esters and amides are called lactones and lactams, respectively. Alcohols, alkyl halides, amines, and amides are sometimes designated as primary, secondary, or tertiary. For alcohols and alkyl halides, the designation is based on the number of organic groups attached to the carbon. For amines and amides, the focus is on the nitrogen. For each of the following, draw as many isomers as you can.
The total number of possible isomers is given in parentheses. Draw structural formulas for each of the following: a an unsaturated alcohol with formula C4H8O b a nitrile with formula C5H9N c an acyclic amide with 3 carbon atoms d a monocyclic ester C5H8O2 1. Nomenclature of Alkanes [M8 3. They are chemically rather inert and uninteresting — their only reactions are halogenation and combustion. They are sometimes considered to be compounds which lack functional groups. Cycloalkanes general formula CnH2n are cyclic saturated hydrocarbons.
Apart from a few cases where a very strained ring is present e. Alkanes and cycloalkanes are sometimes indicated as R-H. R means any alkyl or cycloalkyl group. Systematic names of many aliphatic compounds are based on the names of the alkanes and cycloalkanes. The names of the first 16 alkanes are listed below. You should memorize the names of the first ten.
Alkanes 1. The first four are commonly abbreviated Me, Et, Pr, Bu. There are two isomeric propyl groups, and four isomeric butyl groups, as shown in the following table. Follow the steps outlined below.
Find the longest carbon chain. If possible pick the chain with the most important functional groups or the most substituents. The corresponding alkane is used as the parent name.
For cycloalkanes, one normally selects the ring as the parent name. Number the atoms of the chain, so that the functional groups or substituents have the lowest possible numbers.
Identify and number the substituents. Complex substituents are named in a similar way, except that each substituent is numbered beginning at the point of attachment to the main chain, and each substituent name is generally enclosed in parentheses. Write out the name. How to Interpret Names.
Here are the steps for decoding a name and drawing the structural formula: 1. Find the parent name, usually an alkane or cycloalkane. Draw its structure. Omit the hydrogens temporarily if you wish. Number it if necessary. Attach the substituents indicated by the prefixes. Decipher the names of complex substituents usually in parentheses. Check the structure carefully. Add any necessary hydrogens, erase any numbers inserted in step 1, and check for pentavalent carbons and other mistakes.
Which of the following names are correct? What is wrong with the other names? The following three compounds have the same carbon skeleton. Notice how the presence of the functional group changes the preferred name. OH Cl 2-propylisopropylpentane 1-hydroxypropylisopropylpentane 3-isopropylmethylheptane 1-hydroxypropylethylmethylpentane 2,4-dimethylethylheptane 1-hydroxy 2-ethylmethylpropyl pentane 4,6-dimethylethylheptane 2-methylethyl hydroxymethyl heptane 2-chloroisopropylpropylpentane 2-chloroisopropylmethylheptane 2-methyl 1-chloroethyl methylheptane C.
The following compounds have the same carbon skeleton as the previous examples, except they are cyclic they contain a ring ; consequently we need to name each one as a derivative of a cycloalkane in this case cyclohexane. Write out structures for the following names: 1-bromopropane 2-hydroxymethyloctane methoxycyclopentane 1-phenylisopropylcyclohexane 2-hydroxyt-butyldecane 1-chloronitrophenylhexane 1-hydroxycyano-3,4-difluoro-2,4-dimethylcyclopentane 2,4,4-trimethyl 2-methylpropyl dodecane Exercise 1.
Write structures for each of the following two compounds.
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