Chemistry 122 Introductory Organic Chemistry Fall Quarter 2015 Dr. Thomas H. Schultz.

Презентация:



Advertisements
Похожие презентации
HPC Pipelining Parallelism is achieved by starting to execute one instruction before the previous one is finished. The simplest kind overlaps the execution.
Advertisements

© The McGraw-Hill Companies, Inc., Chapter 4 Counting Techniques.
Here are multiplication tables written in a code. The tables are not in the correct order. Find the digit, represented by each letter.
Diffraction and Interference. Interference and Diffraction Distinguish Waves from Particles O The key to understanding why light behaves like waves is.
Benford Benford's law, also called the first-digit law, states that in lists of numbers from many (but not all) real-life sources of data, the leading.
Business Statistics 1-1 Chapter Two Describing Data: Frequency Distributions and Graphic Presentation GOALS When you have completed this chapter, you will.
Sequences Sequences are patterns. Each pattern or number in a sequence is called a term. The number at the start is called the first term. The term-to-term.
INVOLUTES An involute is a curve that is traced by a point on a taut cord unwinding from a circle or regular polygon, which is called a base or (plane.
How can we measure distances in open space. Distances in open space.
In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act of permuting (rearranging) objects.
11 BASIC DRESS-UP FEATURES. LESSON II : DRESS UP FEATURES 12.
SPLAY TREE The basic idea of the splay tree is that every time a node is accessed, it is pushed to the root by a series of tree rotations. This series.
Time-Series Analysis and Forecasting – Part IV To read at home.
© 2009 Avaya Inc. All rights reserved.1 Chapter Two, Voic Pro Components Module Two – Actions, Variables & Conditions.
A S ANY LANGUAGE IN THE WORLD A SIGN LANGUAGE HAS MANY ADVANTAGES. F IRST OF ALL, IT IS QUITE RICH TO SHOW THE MOST IMPORTANT MEANINGS THAT EXIST IN ALL.
University of Cambridge University of Oxford Government and Society State schools Private schools Nursery Education Primary Education Secondary Education.
Ideal Family Were prepared by Iryna Molokova and Ilona Synytsia.
Knot theory. In topology, knot theory is the study of mathematical knots. While inspired by knots which appear in daily life in shoelaces and rope, a.
S12-1 NAS122, Section 12, August 2005 Copyright 2005 MSC.Software Corporation SECTION 12 RESIDUAL VECTOR METHOD.
ADVANCED DRESS-UP FEATURES 39. Once OK has been selected, your part will appear with the filleted area highlighted by orange lines at the boundaries.
Транксрипт:

Chemistry 122 Introductory Organic Chemistry Fall Quarter 2015 Dr. Thomas H. Schultz

What is Organic chemistry?

The study of carbon and its compounds.

Что такое органическая химия? Изучение углерода и его соединений. Сначала мы будем говорить о соединений просто содержащие углерода и водорода, эти соединения называются углеводородов.

Что такое органическая химия? Изучение углерода и его соединений. Сначала мы будем концентрироваться на соединений просто содержащие углерода и водорода, эти соединения называются углеводородов. Hydrocarbon Classification Hydrocarbons Alkanes Alkenes CycloalkanesAlkynes Cycloalkenes

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a. CH 4 b. C 2 H 6 c. C 3 H ?

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a. CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H ?

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4 C 2 H 6

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4 C 2 H 6 C 3 H 8

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4 C 2 H 6 C 3 H 8 D. Polarity? Polar or nonpolar?

1.Alkanes (saturated) hydrocarbons, or aliphatic hydrocarbons) A.General formula of C n H 2n+2 B.Examples a.CH 4 b. C 2 H 6 c. C 3 H 8 d. C 4 H 10 C. Draw Lewis Structures CH 4 C 2 H 6 C 3 H 8 D. Polarity? Polar or nonpolar?Nonpolar

E. Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10

E. Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = ? Secondary carbon = Tertiary carbon =

E. Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = 2 Secondary carbon = ? Tertiary carbon =

E. Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = 2 Secondary carbon = 2 Tertiary carbon =

E. Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = 2 Secondary carbon = 2 Tertiary carbon = ?

G.Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = 2 Secondary carbon = 2 Tertiary carbon = 0

E. Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = ? Secondary carbon = Tertiary carbon = Isobutane C 4 H 10

G.Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = 3 Secondary carbon = ? Tertiary carbon = Isobutane C 4 H 10

E. Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = 3 Secondary carbon = 0 Tertiary carbon = ? Isobutane C 4 H 10

G.Types of carbon 1. Primary (1 ) Carbon connected to one carbon atoms. 2. Secondary (2 ) Carbon connected to two carbon atoms. 3. Tertiary (3 ) Carbon connected to three carbon atoms. 4. How many primary, secondary, and tertiary carbons in the two different structures of C 4 H 10 Primary carbon = 3 Secondary carbon = 0 Tertiary carbon = 1 Isobutane C 4 H 10

1.Alkanes (Continued) F. There are two different structures for C 4 H 10 called isomers, because they contain different types of carbon. Structure 1 Structure 2

Constitutional Isomers (Structural Isomers) are different compounds of the same formula. The different structures from the previous slide for the formula C 4 H 10 is an example of Constitutional isomers. Isomerism

Constitutional Isomers (Structural Isomers) are different compounds of the same formula. The different structures from the previous slide for the formula C 4 H 10 is an example of Constitutional isomers. How many isomers are there of an alkane containing five carbons (C 5 H 12 )? Isomerism Isomer Strategy – Draw Lewis possible different length chains of carbons atoms connected with a covalent bond.

Constitutional Isomers (Structural Isomers) are different compounds of the same formula. The different structures from the previous slide for the formula C 4 H 10 is an example of Constitutional isomers. How many isomers are there of an alkane containing five carbons (C 5 H 12 )? Isomerism Isomer Strategy – Draw Lewis possible different length chains of carbons atoms connected with a covalent bond. C C C C C Chains of 5 carbon atoms H H H H H H H H H H H H

Isomerism Chains of 4 carbon atoms C C H H H H C H H H H H H H H

Isomerism Chains of 4 carbon atoms C H H H H C H H H H H H H H Chains of 3 carbon atoms C C C C H H H H H H CH H H H H H C C C

Isomerism Chains of 4 carbon atoms C C H H H H C H H H H H H H H Chains of 3 carbon atoms There are three isomers of C 5 H 12 C C C C H H H H H H CH H H H H H

NOMENCLATURE 1.Common system a.Works best for low molecular weight hydrocarbons b.Steps to give a hydrocarbon a common name: 1.Count the total number of carbon atoms in the molecule. 2.Use the Latin root from the following slide that corresponds to the number of carbon atoms followed by the suffix ane. 3.Unbranced hydrocarbons use the prefix normal, or n-, 4.Branched hydrocarbons use specific prefixes, as shown on a subsequent slide

NOMENCLATURE Common system Examples 1. Give a name for the following compound Step #1, count the number of carbons and write down the memorized Latin name for that number (next slide) Step #2, since this structure fits the alkane general formula, use the ane suffix propane Three carbon Latin root Alkane suffix

Latin Hydrocarbon Roots Number of Carbons Latin Root 1meth 2eth 3prop 4but 5pent 6hex 7hept 8oct 9non 10dec 11undec Latin Hydrocarbon Roots Number of Carbons Latin Root 12dodec 13tridec 14tetradec 15pentadec 16hexadec 17heptadec 18octadec 19nonadec 20eicos 21unicos 22doicos H H n-butane isobutane H C C C H H C H H C H H H H H H H Examples neopentane

2. Systematic System of Nomenclature (IUPAC) Find the longest continuous chain of carbon atoms. Use a Latin root corresponding to the number of carbons in the longest chain of carbons. Follow the root with the suffix of ane for alkanes Carbon atoms not included in the chain are named as substituents preceding the root name with Latin root followed by yl suffix. Number the carbons, starting closest to the first branch. Name the substituent's attached to the chain, using the carbon number as the locator in alphabetical order. Use di-, tri-, etc., for multiples of same substituent. If there are two possible chains with the same number of carbons, use the chain with the most substituent's.

Substituent Names (Alkyl groups)

Which one? Systematic Nomenclature continued.

Which one? Systematic Nomenclature continued. The one with the most number of substituent's

Which one? Systematic Nomenclature continued. The one with the least number of substituent's The top structure has four substituent's and the bottom has three substituent's.

Which one? Systematic Nomenclature continued. The one with the least number of substituent's The top structure has four substituent's and the bottom has three substituent's. Name = ?

Which one? Systematic Nomenclature continued. The one with the most number of substituent's The top structure has four substituent's and the bottom has three substituent's. Name = ? heptane

Which one? Systematic Nomenclature continued. The one with the least number of substituent's The top structure has four substituent's and the bottom has three substituent's. Name = 3,3,5-trimethyl-4-propylheptane

Another Example: Name = 3-ethyl-2,6-dimethylheptane

Another Example: Name = 3-ethyl-2,6-dimethylheptane Notice substituent's are in alphabetical order; di, tri, etc. do not participate in the alphabetical order

Line Structures A quicker way to write structures' (Condensed Structure) (A line structure of the above condensed structure) ethyl methyl

Complex Substituent's If the branch has a branch, number the carbons from the point of attachment. Name the branch off the branch using a locator number. Parentheses are used around the complex branch name. 1-methyl-3-(1,2-dimethylpropyl)cyclohexane

Alkane Physical Properties Solubility: hydrophobic (not water soluble) Density: less than 1 g/mL (floats on water) Boiling points increase with increasing carbons (little less for branched chains) due to dispersion forces being larger. Melting points increase with increasing carbons (less for odd-number of carbons).

Boiling Points of Alkanes Branched alkanes have less surface area contact, so weaker intermolecular forces.

Melting Points of Alkanes Branched alkanes pack more efficiently into a crystalline structure, so have higher m.p.

Reactions of Alkanes I. Combustion reaction II. Cracking reaction III. Halogenation reaction (substitution reaction)

Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 12 Step 2 react each isomer with chlorine Step 3 count the products

Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products

Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products

Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products

Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products

Sample problem: Which isomer of C 5 H 12 has the most monochloro isomers? Problem solving process: Step 1 draw the isomers of C 5 H 10 Step 2 react each isomer with chlorine Step 3 count the products Winner!

Conformers of Alkanes Structures resulting from the free rotation of a C-C single bond May differ in energy. The lowest-energy conformer is most prevalent. Molecules constantly rotate through all the possible conformations.

Ethane Conformers Staggered conformer has lowest energy. Dihedral angle = 60 degrees model H H H H HH Newman projection sawhorse Dihedral angle

Ethane Conformers (2) Eclipsed conformer has highest energy Dihedral angle = 0 degrees =>

Conformational Analysis Torsional strain: resistance to rotation. For ethane, only 12.6 kJ/mol =>

Propane Conformers Note slight increase in torsional strain due to the more bulky methyl group.

Butane Conformers C2-C3 Highest energy has methyl groups eclipsed. Steric hindrance Dihedral angle = 0 degrees => totally eclipsed (methyl groups)

Butane Conformers (2) Lowest energy has methyl groups anti. Dihedral angle = 180 degrees => Staggered-anti

Butane Conformers (3) Methyl groups eclipsed with hydrogens Higher energy than staggered conformer Dihedral angle = 120 degrees => Eclipsed (hydrogen and methyl)

Butane Conformers (4) Gauche, staggered conformer Methyls closer than in anti conformer Dihedral angle = 60 degrees => Staggered-gauche

Conformational Analysis

Cycloalkanes Rings of carbon atoms (-CH 2 - groups) Formula: C n H 2n Nonpolar, insoluble in water Compact shape Melting and boiling points similar to branched alkanes with same number of carbons Slightly unsaturated compared to alkanes

Naming Cycloalkanes Count the number of carbons in the cycle If the bonds are single then use the suffix ane First substituent in alphabet gets lowest number. May be cycloalkyl attachment to chain.

Cis-Trans Isomerism (a type of stereoisomerism) Cis: like groups on same side of ring Trans: like groups on opposite sides of ring

Cycloalkane Stability 6-membered rings most stable Bond angle closest to Angle (Baeyer) strain Measured by heats of combustion per -CH 2 -

Heats of Combustion/CH 2 Alkane + O 2 CO 2 + H 2 O kJ/mol kJ Long-chain

Cyclopropane Large ring strain due to angle compression Very reactive, weak bonds =>

Cyclopropane (2) Torsional strain because of eclipsed hydrogens

Cyclobutane Angle strain due to compression Torsional strain partially relieved by ring puckering =>

Cyclopentane If planar, angles would be 108, but all hydrogens would be eclipsed. Puckered conformer reduces torsional strain.

Cyclohexane Combustion data shows its unstrained. Angles would be 120, if planar. The chair conformer has bond angles and all hydrogen's are staggered. No angle strain and no torsional strain.

Chair Conformer

Boat Conformer

Conformational Energy

Axial and Equatorial Positions

Monosubstituted Cyclohexanes

1,3-Diaxial Interactions

Disubstituted Cyclohexanes

Cis-Trans Isomers Bonds that are cis, alternate axial-equatorial around the ring. => One axial, one equatorial

Bulky Groups Groups like t-butyl cause a large energy difference between the axial and equatoria l conformer. Most stable conformer puts t-butyl equatorial regardless of other substituents. =>

End of Chapter 2