Organic Chemistry 3 Lecture 1 - Fall 2024
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Ahram Canadian University
2024
Dr. Dalia Sherif El Gamil
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This document is a lecture on UV-Vis spectrophotometry, part of a course called Organic Chemistry 3, taught at Ahram Canadian University in Fall 2024. The lecture covers definitions, principles, applications, and examples. It also includes the course outline, assessment details, and an OSPE exam description. It is a detailed chemistry lecture, not a past exam paper.
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Organic Chemistry 3 Lecture 1 Ultraviolet-Visible Spectrophotometry By: Dr. Dalia Sherif El Gamil Lecturer of Pharmaceutical Chemistry First floor - Room 2110 Fall 2024 Vision and Mission...
Organic Chemistry 3 Lecture 1 Ultraviolet-Visible Spectrophotometry By: Dr. Dalia Sherif El Gamil Lecturer of Pharmaceutical Chemistry First floor - Room 2110 Fall 2024 Vision and Mission Faculty of Pharmacy – Ahram Canadian University Vision: Leadership in the pharmaceutical field on the national level with international recognition of excellence. Mission: The Faculty of Pharmacy - Ahram Canadian University, is committed to provide an up-to-date learning environment within a framework of advanced programs that enable the graduate to acquire professional ethics and have the competencies required for the job market. The Faculty, through distinguished calibers, supports effective collaboration with the community and encourages scientific research. Course Outline Week Date Lecture Lab 1 29/9 – 3/10 UV-Vis spectroscopy Green Chemistry 2 6/10 – 10/10 Mass spectrometry IHD + Azodye Naphthalene / Anthracene 3 13/10 – 17/10 Infrared spectroscopy picrate + Glucosazone 4 20/10 – 24/10 Proton NMR Mass + IR applications Carbon 13 NMR + Structural NMR and structural 5 27/10 – 31/10 elucidation elucidation applications Heterocyclic chemistry (1)* 6 3/11 -7/11 -- Midterm Exams Heterocyclic nomenclature Exercises* 7 10/11 – 14/11 -- Midterm Exams 8 17/11 – 21/11 Heterocyclic chemistry (2) Mono/Diphenylurea Aldol condensation + 9 24/11 – 28/11 Heterocyclic chemistry (3) Aromatic Esters 10 1/12 – 5/12 Heterocyclic chemistry (4) Revision 11 8/12 – 12/12 Heterocyclic chemistry (5) Final Practical Exam 12 15/12 – 19/12 Heterocyclic chemistry (6) -- * Online lectures Course Grades Distribution Assessment Type Grade Midterm Exam 15% Total Practical 25% Lab Evaluation (Participation) 5% Practical Sheet 5% Practical Final Exam 15% Final Exam 50% Oral Exam 10% OSPE - You are going to be evaluated according to OSPE examination system. - OSPE (Objective Structured Practical Examination) is designed to measure your requisite skills during the semester. The exam provides an assessment of your understanding and competency of the practical skills taught throughout the semester. CHARACTERISTICS Objective: Because examiners use a checklist for evaluating the students. Structured: Because every student performs the same tasks in the same time frame. Practical: Because the tasks are representative of those found in real practical situations. Examination: An examination. Purpose: Provide feedback on performance. Evaluate on the basis of Practical skills. Measures minimal competencies. ADVANTAGES 1. Helps the examiner to observe and assess students for different professional and practical skills. 2. Enables the examiner to have an overall view of the student’s performance. 3. Simulations of real-life situations. OSPE You will be provided by a detailed protocol for synthesis of a certain organic compound. You are required to follow the synthetic steps accurately, applying all laboratory standard and safety measures to obtain the required product. Assessment Checklist Item Score Remark(s) Product name /2 Product physical character /2 Quality of product (correct state, color, texture, odor, ….etc.) /6 Lab coat and commitment /1 Tools maintenance /1 Proper chemicals handling and waste disposal /2 Discipline (behavior / not talking or looking around) /1 Total /15 Lecture (1) Intended Learning Outcomes 1. Define spectroscopy 2. Identify the characteristics of electromagnetic radiation (EMR) 3. Differentiate the regions of EMR and their associated molecular processes 4. Comprehend the main principle of UV-Vis spectrophotometry 5. Understand the role of various parts of a UV-Vis spectrophotometer 6. Apply Beer’s Lambert law for qualitative or quantitative assessment of organic samples Fundamentals of Spectroscopy Introduction Spectroscopy is any procedure that uses the interaction of Electromagnetic Radiation (EMR) with matter to identify and/or to estimate an analyte. Electromagnetic radiation (light) As Waves As Particles Electromagnetic radiation consists of discrete Wavelength (): Distance from packets of energy, called photons. one wave peak to the next. Units: m, cm, m, nm or A0 Photons are the particles of light or the quanta of light. Frequency (): Number of peaks that pass a given point per second. Units: Cycles/second or s-1 or All characteristics of light can be related as follows: Hertz (Hz) c E = h υ = h = hc υ Wavenumber Number of waves per cm. λ 1 where h is the Planck’s constant (=6.626x10-34 J.s) υ= cm-1 and c is the speed of light in vacuum λ Regions of Electromagnetic Radiation Introduction high E low change in the spin of Molecular processes Change in nuclear configuration that occur when protons light is absorbed in each region high low low high Ultraviolet 200 nm 800 nm Ultraviolet-Visible Spectrophotometry (Principle) UV-Vis I0 I 500 nm What types of analytes that absorb UV-Vis light? Analytes that have chromophores. A Chromophore is the part of molecule or functional group(s) that absorb light. It is usually a delocalized pi electron (conjugated) system. The higher the conjugation, the longer the wavelength of light absorbed. What happens when a molecule absorbs UV-Vis radiations? The molecule will be promoted to an excited state (of higher energy) where electrons are promoted into higher orbitals. UV-Vis Spectrophotometry (Instrumentation) UV-Vis Components of a Single Beam Spectrophotometer Deuterium lamp (200-400 nm) Light source slits Separates white light Tungsten lamp into various colors (320-2500 nm) Rotating the grating changes the wavelength going through the sample Detector Monochromator (Grating or Prism) detects light & slits measures intensity by converting Sample transmitted Blank is inside cuvette Sample is inside cuvette Sample compartments (cells) photons energy Ablank Asample are made of glass (Vis only) into electrical or quartz (UV-Vis) signal Corrected Absorbance = Asample - Ablank UV-Vis Spectrophotometry (Applications) UV-Vis Beer-Lambert ’s law Acorrected = bc c , is the concentration of the absorbing species in sample. Unit: mol/L (M) b , is the path length of light through sample. Unit: cm , is called the molar absorptivity or molar absorption coefficient. Unit: M-1 cm-1 , is characteristic for each substance at a particular wavelength, . Qualitative applications Quantitative applications Identification of inorganic, organic and biomedical species Estimation of unknown concentration of an analyte Plot a graph of corrected absorbance readings vs. Plot a graph of corrected absorbance readings vs. wavelengths → UV-Vis Spectrum standards concentrations → Calibration Curve max is wavelength where absorbance maximum absorbance occurs Get unknown conc. via direct extrapolation Shape of spectrum and max Remember, here the are characteristic for a slope is b certain analyte UV-Vis Spectrophotometry UV-Vis Train Yourself Example 1 A 3.96 x 10-4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1 cm cuvette; a blank solution containing only solvent had an absorbance of 0.029 at the same wavelength. Find the molar absorptivity of compound A. Solution: ε = Acorrected/c b = (0.624 – 0.029) / (3.96 x 10-4 M)(1 cm) = 1.5 x 103 M-1cm-1. Value of absorbance corrected for blank UV-Vis Spectrophotometry UV-Vis Example 2 A 0.14 mol L-1 solution of chemical (A) in propanol had an absorbance of 0.42. Another solution of (A) in propanol measured under the same conditions had an absorbance of 0.36. Absorbance reading of propanol solvent only was 0.04. What is the concentration of the second solution of (A)? Solution: A1 = 0.42 , A2 = 0.36 , Ablank = 0.04 c1 = 0.14 M 0.42- 0.04 / 0.36 - 0.04 = 0.14 / c2 Value of absorbance corrected for blank c2 = 0.118 M UV-Vis Spectrophotometry UV-Vis Example 3 At 480 nm, a 4.0x10-5 M solution of [FeSCN]2+ has absorbance of 0.3 when measured in 1.0 cm cell. What is the absorbance when measured in 4.0 cm cell? Solution: A1 b 1 c1 b = = 1 A 2 b2 c2 b2 0.30 1.0 = A2 4.0 A 2 = 1.2 How to get prepared for the next lectures? Aliphatic Aromatic Open Chain or Cyclic Cyclic Planar Fully conjugated Alkanes Obeys Huckel’s rule All sp3 Alkenes Olefinic sp2 or Vinyl group Alkynes Acetylinic group sp