P 100 % P+1 10 % Compound 1 148/13 = 11.38;.38*13 =5 C 11 H 18.

Slides:

Advertisements

Similar presentations

INTRODUCTION TO SPECTROSCOPY G. RAM KUMAR BY Department of Chemistry

Advertisements

Spectroscopy Structural Analysis

WOODWARD-FEISER RULES FOR CALCULATING λmax

Degree of Unsaturation

IUPAC Nomenclature Organic Compounds (Part 2) of.

Organic Reactions Chemical reactions involving organic compounds.

One Upon a Time Organic compounds – compounds obtained from living organisms Inorganic compounds – compounds obtained from non living things.

Calculate the Empirical Formula for a compound with the following composition: 46.16% carbon; 53.84% nitrogen 1)Change % to grams (if needed) 2)Convert.

Structure Determination: MS, IR, NMR (A review)

Lecture 5 HDI More Sample Problems (NMR & IR) Due: Lecture Problem 3.

Chapter 41 Alkenes. Chapter 4. Chapter 42 Contents of Chapter 3 General Formulae and Nomenclature of Alkenes General Formulae and Nomenclature of Alkenes.

Electronic Excitation by UV/Vis Spectroscopy :

Problem Day 2 CH343 Bruce A. Hathaway. Compound #1: IR sp 2 C-H, sp 3 C-H, C=O, benzene C=C, possible monsubstituted benzene.

Hydrocarbons. Organic Chemistry Carbon compounds (usually C+H+?) If all were removed from earth? –Look like moon with water If all were removed from our.

Check your Unknown #3 mass spectrum

Nuclear Magnetic Resonance (NMR) Spectroscopy

Experiment 14: IR AND NMR IDENTIFICATION OF AN UNKNOWN.

Naming Rings, Alkenes and Alkynes. Alkenes Alkenes: Double bonded carbon Ending: ene Name: Use the same four steps Changes: Carbon 1 is closest to the.

Alkenes Properties Nomenclature Stability Addition Reactions.

1 Chapter 12 Alkenes and Alkynes Geometric Isomers of Alkenes.

1) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound. First,

Empirical Formulas. Empirical formula tells the relative number of atoms of each element in a compound Mole concept provides a way of calculating the.

Alkene: Structure and Reactivity Chapter 6. Alkenes An alkene (also called an olefin) is a hydrocarbon with a carbon-carbon double bond. Alkenes are present.

Alkenes and Alkynes Nanoplasmonic Research Group Organic Chemistry Chapter 3.

Alkenes and alkynes Saturated organic compounds: bonds between carbons are single bonds (share 1 pair of electron; alkanes) Unsaturated organic compounds:

Chapter 2 Ultraviolet and Visible Absorption Spectroscopy (UV-Vis)

Common 1 H NMR Patterns 1. triplet (3H) + quartet (2H) -CH 2 CH 3 2. doublet (1H) + doublet (1H) -CH-CH- 3. large singlet (9H) t-butyl group 4. singlet.

Reactions of Unsaturated Hydrocarbons Combustion Complete combustion C 3 H 6 + O 2 → CO 2 + H 2 O Incomplete combustion C 3 H 6 + O 2 → C + CO + CO 2.

1) Draw a structure consistent with the following data:

Short Answer: 1) What type of electromagnetic radiation is used in nuclear magnetic resonance? radio 2) What is the most abundant peak in a mass spectrum.

1) What is the major product of the following reaction? HBr.

5.1 Alkene Nomenclature. Alkenes Alkenes are hydrocarbons that contain a carbon-carbon double bond also called "olefins" characterized by molecular formula.

Endo-Exo Convention 1) the position of double bonds relative to rings in carbocyclic systems; 2) the relative orientation of substituents of bridged carbocycles;

Empirical Formula Molecular Formula

C 5 H 12 O MW = Solve the structure with no further information.

Background The impact of a stream of high energy electrons causes the molecule to lose an electron forming a radical cation. – A species with a positive.

Naming Hydrocarbons Use your textbook to complete: 1.Naming organic compounds uses prefixes and _________. 2.Alkane names end with the suffix ____. The.

CHAPTER 5 UNSATURATED HYDROCARBONS (ALKENES AND ALKYNES)

ALKYNES  Alkynes are hydrocarbons that contain a carbon – carbon triple bond (with no functional groups, only substituents present).  They are characterized.

1 2.7 Physical Properties of Alkenes. 2 Nonpolar Insoluble in water Soluble in nonpolar organic solvents. Less dense than water: they float on water.

1 Increasing frequency CH 2 =CH-CH=CH 2 Absorption spectrum for 1,3-butadiene.

Section 9.3—Analysis of a Chemical Formula How can we determine a chemical formula?

Alkenes: Structure and Reactivity. Calculating Degree of Unsaturation Relates molecular formula to possible structures Degree of unsaturation: number.

categories of organic reactions There are so many types of organic reactions. We’re going to focus on just a few. There are so many types of organic reactions.

Alkenes : Structure and Reactivity

12. Structure Determination: Mass Spectrometry

Percent Composition, Empirical and Molecular Formulas.

Basic Organic Chemistry

Empirical & Molecular Formulas

Accurate Mass: H ; C ; N ; O How to distinguish CO and C2H4

MKI Instrumentasi Kimia

Chapter 7 Alkenes: Structure and Reactivity

Unsaturated Hydrocarbons

Alkenes, Cycloalkenes and Dienes

The study of carbon and carbon-containing compounds

Ch. 8 – The Mole Empirical formula.

CH 2-2 Hydrogen Deficiency & Constitutional Isomers of Hydrocarbons

Unsaturated Hydrocarbons

UV Spectroscopy problems Y. B. Chavan College of Pharmacy

Electronic Excitation by UV/Vis Spectroscopy :

Organic Compounds (Part 2)

Empirical and Molecular Formulas

Unsaturated Hydrocarbons II: Dienes and Alkynes

Unsaturated Hydrocarbons II: Dienes and Alkynes

Organic Molecules Chapter 20.

10.3 Alcohols These compounds have an -OH attached to the carbon chain. This functional group is called a hydroxyl group. Note: The oxygen is bonded to.

University of California,

Unsaturated Hydrocarbons: Alkenes

Percentage Composition

Presentation On Ultra-violet and Visible Spectroscopy Presentation On Ultra-violet and Visible Spectroscopy 2/22/20191.

Presentation transcript:

P 100 % P+1 10 % Compound 1 148/13 = 11.38;.38*13 =5 C 11 H 18

Compound 1

C 11 H 18 C 10 H 14 O Area : 5:2:2:3

m/e m/e 120 m/e 105 m/e 77

Compound 2 P 100 % P+1: 7.1% P+2 : 33 % = 91 91/13 = 7.0 C 7 H 7 Cl

Compound 2

Area: 5:2

Compound 3 m/e /13 = 8.769; 0.769*13 = 10 C 8 H 18

Compound 3

C 8 H 18 C 7 H 14 O C 6 H 10 O 2

m/e 99P-15 m/e 86P-28 m/e 69P-45 m/e 41 m/e 86 m/e 69 m/e 41

Compound 4 m/e = /13 = 8.077; 0.077*13 = 1 C 8 H 9 Br

Compound 4

Area 5:2:2

Compound 5 C 8 H 18 O 3 mw 162 m/e 75 C 2 H 3 O 3 + C 3 H 7 O 2 + m/e 73 C 2 HO 3 + C 3 H 5 O 2 + C 4 H 9 O + m/e 147 = P - 15 m/e 130 = P - 32 CH 4 O m/e 115 = P – 47CH 3 O 2 ; C 2 H 7 O

Compound 6

Compound 7

A comparison of two EIMS spectra obtained from different instruments and different samples m/e /13 = *13 = 6 C 10 H 16

Compound 7 C 10 H 16

Compound 7 C 10 H 16

Empirical Rules for Dienes (π → π*) Homoannular(cisoid) Heteroannular (transoid) Parent λ = 253 nm λ = 214 nm Increments for: double bond extension 3030 alkyl subst or ring 55 exocyclic double bond55 Alkenes ethylene λ = 175 nm λ = = 273 nm

Compound 8 m/e /13 = *13 = 4 C 8 H 12

Compound 8

d d t

Compound 8 area = 2:2:4:4 left to right

P : m/e 108 m/e 80 = P-28

1 H NMR: no signal 13 C NMR: 4 vinyl carbons

282/13 = ; 0.692*13= 9 C 21 H 30 C 18 H 31 Cl C 15 H 32 Cl 2 C 12 H 33 Cl 3 C 9 H 34 Cl 4 C 6 H 35 Cl 5 C 6 Cl 6 35 Cl = Cl = n = n = n = n = n = n = 6 n = 5 (0.76) 5 = 0.254; 5(0.76) 4 (0.24)= % 157% n = 6 (0.76) 6 = 0.193; 6(0.76) 5 (0.24) = % 189%

146/13 = 11.23; 0.23 *13 =3 C 11 H 14 Compound 9 P-28

C 11 H 14 C 10 H 10 O Compound 9

2 1 C 10 H 10 O C9H6O2C9H6O2 Compound 9

S S D D D D S D D C9H6O2C9H6O2 Compound 9

136/13 = ; 0.462*13 = 6 C 10 H 16 Compound 10

Neat liquid

Compound 10 C 10 H 16 C 9 H 12 O CCl 4 solution

Compound 10 Area: 1:2:2:3 C 9 H 12 O C8H8O2C8H8O2

D S D S D Q C8H8O2C8H8O2

Compound /13 = 8.615; 0.615*13 = 8 C 8 H 16

Compound 11 C 8 H 16 Degrees of unsaturation: 1

Area: 3:2:3 C 8 H 16

S D D D T Q C 8 H 16 C 7 H 12 O C 6 H 8 O 2 Exact mass: Exact mass of C 6 H 8 O 2 :