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Organic Laboratory 1 คม252: ปฏิบัติการเคมีอินทรีย์ 1

สาขาวิชาเคมี คณะวิทยาศาสตร์ มหาวิทยาลัยแม่โจ้ ประจ าปีการศึกษา 1/2562

จัดท าโดย อ.ดร. วชิระ ชุ่มมงคล และ อ. ดร. ปิยธิดา กล าภู่ 2

Contents Laboratory Safety and MSDS Physical Constants and Purification of Organic Substances 1. Determination of Melting Point and Boiling Point 2 2. Acid-Base Extractions, Separations and Drying Agents 12 3. TLC and Column Chromatographic Adsorption 21 4. Reaction of , Alkene and Alkyne 25 5 Stereochemistry: Addition Reaction: Camphor Derivative 6. Solid-Liquid Extraction and Crystallization: Caffeine Part 1 31 7. Sublimation: Caffeine Part 2 and Simple Distillation: Mixed Compound 34 8. Stream Distillation: Eugenol oil / Citronella oil 38 9. Williamson reaction and identification of ether synthesis: Paracetamol derivative 45 10. The Oxidation of 1o-, 2o –Alcohol and Identification of Aldehyde and 41 Ketone with 2,4-DNP

3 หลักสูตรวิทยาศาสตรบัณฑิต สาขาวิชาเคมี คณะวิทยาศาสตร์ มหาวิทยาลัยแม่โจ้ ประมวลการสอนรายวิชา (Course Syllabus) ประจ าปีการศึกษาที่ 2562 ภาคเรียนที่ 1

ชื่อวิชา ปฎิบัติการเคมีอินทรีย์ 1 รหัสวิชา คม 252 จ านวนหน่วยกิตรวม 3 หน่วยกิต [บรรยาย 3 ชั วโมง/สัปดาห์] อาจารย์ผู้สอน อ. ดร.วชิระ ชุ่มมงคล อ.ดร. ปิยธิดา กล าภู่ ผู้ประสานงาน อ. ดร.วชิระ ชุ่มมงคล ค าอธิบายรายวิชา (Course description) การศึกษาสมบัติของการประกอบอินทรีย์ การระบุเอกลักษณ์ของสารประกอบอินทรีย์ การหาจุด เดือด จุดหลอมเหลว การแยกสารให้บริสุทธิ์ด้วยโครมาโทรการฟี การสกัด การตกผลึก การระเหิด การกลั น รวมถึงศึกษาสเตอรีโอเคมี กลไกการการเกิดปฏิกิริยาการแทนที การขจัด การเพิ ม รีดัคชัน ออกซิเดชัน ของ สารอินทรีย์

วัตถุประสงค์รายวิชา 1. นักศึกษาสามารถน าคุณธรรม จริยธรรม จรรยาบรรณในการรวบรวมผล สรุป และอภิปราย ในการ ปฏิบัติการ CLO1 2. นักศึกษามีความรู้ในหลักการ ทฤษฎี และปฏิบัติ รวมถึงการวิเคราะห์และอภิปราย ในสมบัติทางการ ภาพ การแยกสารให้บริสุทธิ์ และปฏิกิริยาเคมีอินทรีย์ CLO2

อาจารย์ผู้สอน วันอังคาร 15.00-18.00 น. section 3 อ.ดร. ปิยธิดา กล าภู่ วันพฤหัสบดี 08.00-11.00 น. section 2 อ.ดร. วชิระ ชุ่มมงคล

สถานที่เรียน วิทย์ 1214

สอนนักศึกษา สาขาวิชาเคมี ชั้นปีที 2 สาขาเคมีอุตสาหกรรมและเทคโนโลยีสิ งทอ ชั้นปีที 2 และสาขาเทคโนโลยียางและพอลิเมอร์ ชั้นปีที 2

4 แผนที แสดงการกระจายความรับผิดชอบต่อของรายวิชา (Curriculum Mapping)

ทักษะระหว่าง ทักษะวิเคราะห์เชิง ความรู้ ทักษะทาง บุคคลและ ตัวเลข การสื อการ คุณธรรมและจริยธรรม รายวิชา ปัญญา ความ และการใช้เทคโนโลยี รับผิดชอบ สารสนเทศ

1 2 3 4 5 1 2 3 4 1 2 3 1 2 3 1 2 3 4 คม252 ปฏิบัติ การเคมีอินทรีย์ 1            

แผนที แสดงการกระจายความรับผิดชอบ PLOs ของหลักสูตรฯ ต่อของรายวิชา Program Learning Outcome (PLOs) รายวิชา PLO1 PLO2 PLO3 PLO4 PLO5 PLO6 PLO7 คม252 ปฏิบัติ U U การเคมีอินทรีย์ 1 Bloom’s Taxonomy: U = Remembering/Understanding, A = Applying/Analyzing

CLO PLO TQF วิธีการสอนที จะใช้ วิธีการประเมินผล พัฒนาการเรียนรู้ CLO1: นักศึกษาสามารถน า PLO6 TQF1 : คุณธรรม จริยธรรมที่ - ปลูกฝังให้นักศึกษามี - ป ร ะ เ มิ น จ า ก คุ ณ ธ ร ร ม จ ริ ย ธ ร ร ม ต้องพัฒนา ระเบียบวินัย โดยเน้น พ ฤ ติ ก ร ร ม ข อ ง จรรยาบรรณในการรวบรวม 1.1 มีความยึดมั นความดีงาม การเข้าชั้นเรียนให้ตรง ผู้เรียนระหว่างร่วม ผล สรุป และอภิปราย ในการ ในทางวิชาการ ซื อสัตย์ เวลาและการส่งงาน กิจกรรมการเรียน ปฏิบัติการ สุจริต เสียสละ และมี ภายในเวลาที ก าหนด การสอน (1) น้ าใจช่วยเหลือผู้อื น -ปลูกในทางวิชาการ -ประเมินจากรายงาน 1.2 มีวินัย ตรงต่อเวลา และ ซื อสัตย์ สุจริตใ นการ ผลการทดลอง การ รับผิดชอบต่อตนเอง เขียนผลการทดลอง การ สรุปและอภิปราย สังคมและสิ งแวดล้อม สรุปและอภิปรายผล ผลการทดลอง (2) การทดลอง -ประเมินจากการให้ ค ะ แ น น ก าร เข้า ห้องเรียนและการ ส่งงานตรงเวลา (3) CLO2: นักศึกษามีความรู้ใน PLO1- TQF2 ความรู้ที่ต้องได้รับ - การท าปฏิบัติการ มีการ - ประเมินจากรายงาน หลักการ ทฤษฎี และปฏิบัติ 5 PLC7 2.1 มีความสามารถอธิบาย เตรียมความพร้อมก่อน มีการเขียนหลักการ รวมถึงการวิเคราะห์และ หลักการและทฤษฎีที ท าปฏิบัติการในการ ทฤษฎี และปฏิบัติ อภิปราย ในสมบัติทางการ เขียนหลัการ ทฤษฎี และการวิเคราะห์ 5 ภาพ การแยกสารให้บริสุทธิ์ ส าคัญในเนื้อหาวิชาที และวิธีการปฏิบัติ ก่อน อภิปราย การ และปฏิกิริยาเคมีอินทรีย์ ศึกษา เข้าชั้นเรียน ค านวณ ที ถูกต้อง 2.2 มีความสามารถในการบูร - ปฏิบัติการทดลองแบบ ตามหลักวิชาการ ณาการเนื้อหาในสาขา กลุ่มฝึกกระบวน การคิด (2) วิชาชีพและสาขาวิชาที วิเคราะห์และวิพากษ์ - ประเมินผลจาก เกี ยวข้อง ทั้งในระดับบุคคลและ รายงาน ในการ กลุ่ม อภิปรายทีมีการ ค้นคว้าเพิ มเติม (3)

6 รายละเอียดเนื้อหาวิชา สัปดาห์ที่ หัวข้อ/รายละเอียด จ านวน ชั่วโมง 1 แนะน าห้องปฏิบัติการ อธิบายลงและข้อควรปฏิบัติใน 3 (1- 5 กรกฎาคม 62) ห้องปฏิบัติการ แนะน าและอธิบายวิธีการเขียนรายงานการ ทดลอง การเก็บข้อมูล การค านวณ การสรุปผล และการ อภิปรายผลการทดลอง 2 เตรียมความพร้อมก่อนเรียนตรวจอุปกรณ์ แบ่งกลุ่ม 3 (8- 12 กรกฎาคม 62) 3 16 วัน อาสาฬหบูชา และ 17 วันเข้าพรรษา 3 (15-19 กรกฎาคม 62) (งดปฏิบัติการ) 4 ปฏิบัติการที 1 3 (22-26 กรกฎาคม 62) Determination of Melting Point and Boiling Point 5 ปฏิบัติการที 2 3 (29 กรกฎาคม -2 สิงหาคม Acid-Base Extractions, Separations and Drying Agents 62) 6 ปฏิบัติการที 3 3 (5-9 สิงหาคม 62) TLC and Column Chromatographic Adsorption 7 ปฏิบัติการที 4 3 (12-16 สิงหาคม 62) Reaction of Alkane, Alkene and Alkyne 8 ปฏิบัติการที 5 3 (19-23 สิงหาคม 62) Stereochemistry: Addition Reaction: Camphor Derivative สอบกลางภาค (งดปฏิบัติการ) วันที่ 26 สิงหาคม - 1 กันยายน 2562 9 ปฏิบัติการที 6 3 (2-6 กันยายน 62) Solid-Liquid Extraction and Crystallization: Caffeine Part 1 10 ปฏิบัติการที 7 3 (9-13 กันยายน 62) Sublimation: Caffeine Part 2 and Simple Distillation: Mixed Compound 7 11 ปฏิบัติการที 8 3 (16-20 กันยายน 62) Stream Distillation: Eugenol oil / Citronella oil 12 ปฏิบัติการที 9 3 (23-27 กันยายน 62) Williamson reaction and identification of ether synthesis: Paracetamol derivative 13 ปฏิบัติการที 10 3 (30-กันยายน -4 ตุลาคม The Oxidation of 1o-, 2o –Alcohol and Identification of 62) Aldehyde and Ketone with 2,4-DNP 14 * เช็คและคืนอุปกรณ์ 3 (7-11 ตุลาคม 62) ** ตรวจเช็คคะแนนรายงาน 15 ทบทวนปฏิบัติการ 3 (14-18 ตุลาคม 62) สอบปลายภาค วันที่ 21 ตุลาคม ถึง 3 พฤศจิกายน 2562 ตามประกาศมหาวิทยาลัย

คะแนนเก็บของอาจารย์ผู้สอนแบ่งเป็น 1. สังเกตพฤติกรรมการเข้าปฏิบัติการ แบบทดสอบก่อนเรียน 5% 2. สมุดการวางแผนและบันทึกผลการทดลอง 5% 3. รายงานผลการทดลอง 55% 4. สอบอัตนัย 35% เกณฑ์การประเมินผล (สามารถปรับเปลี ยนได้ตามความเหมาะสม) 45 + 3.5SD % ขึ้นไป ระดับคะแนน A 45 + 1.5SD % ระดับคะแนน C 45 + 3.0SD % ระดับคะแนน B+ 45 + 1.0SD % ระดับคะแนน D+ 45 + 2.5SD % ระดับคะแนน B 45+ 0.5SD % ระดับคะแนน D 45 + 2.0SD % ระดับคะแนน C+ ต ากว่า 45 % ระดับคะแนน F

8 การเขียนรายงาน เรื่อง...... วันที่ทดลอง...... กลุ่มที่...... ผู้เขียนรายงาน ชื อ) ...... รหัส...... สาขา...... ผู้ร่วมงาน ชื อ)...... รหัส...... สาขา...... ชื อ)...... รหัส...... สาขา...... วัตถุประสงค์ ให้บอกวัตถุประสงค์ของการทดลองให้ชัดเจนว่า เพื อต้องการหาค่าอะไร ของสารอะไร โดยวิธีไหน (ตามวิธีท าการทดลองจริง)

ทฤษฎี (หลักการ) ไม่เกิน 1 หน้ากระดาษ ให้อธิบายทฤษฎีหรือหลักการโดยย่อ พร้อมทั้งสูตรและสัญลักษณ์ ในการทดลองหรือการค านวณ เพื อให้ได้ผลการทดลองตรงตามวัตถุประสงค์ที ต้องการ

วิธีการทดลอง สมการเคมี (ถ้ามี) อธิบายวิธีท าการทดลองโดยย่อ ตามล าดับขั้นที ท าการทดลองในห้องปฏิบัติการ ตั้งแต่เริ มต้นจนกระทั ง ถึงขั้นสุดท้ายของการทดลอง แสดงรูปประกอบ Flow Chart

การค านวณ ให้แสดงวิธีการค านวณที ส าคัญ ให้แสดงผลลัพธ์หรือค่าเฉลี ยของผลลัพธ์ที ได้จากการค านวณสาร ก าหนดปริมาณ, %yield, %conversion และ %recovery

ผลการทดลอง ได้สารอะไร ลักษณะทางกายภาย สี ได้สารกี กรัม

สรุปและวิจารณ์ผลการทดลอง สรุปตามจุดประสงค์ ว่าได้สาร(ชื อ) ได้สารกี กรัม คิดเป็น ร้อยละ (%yield) การวิจารณ์ผลการทดลอง วิจารณ์ผลว่าท าไมถึงเป็นเช่นนั้น และ ท าไมถึงไม่เป็นเช่นนั้น อาจอิงตามจุดประสงค์การทดลอง

เอกสารอ้างอิง เอกสารอ้างอิง อย่างน้อย 3 เอกสาร (หนังสือหรือเวปไซด์) 1 Laboratory Safety and MSDS Safety! Organic chemistry laboratory can provide social benefit, basic scientific discoveries, and intellectual satisfaction. Chemical experiments are not just fun, But can also be very hazardous, some experiments inherently so. Complacency is often observed by experience. One often forgets that chemistry is a potentially dangerous enterprise; a cavalier attitude often results in disastrous consequences. Therefore, extreme caution should be exercised at all time, especially when one handles large-scale reactions that are exothermic or when dealing with toxic chemicals.

Personal Protection Equipment 1. Safety Glasses If a chemical splashes into your eyes, it could do serious and sometimes permanent damage to your vision. The most common forms of eye protection include safety glasses (with side shields), goggles, and face shields. Prescription eye glasses are acceptable provided that the lenses are impact resistant and they are equipped with side shields. 2. Gloves Laboratory gloves are an essential part of safe laboratory practice and must be worn while handling chemicals. Despite practicing good safety techniques, tragedy may still strike. The types of gloves available and their recommended use. The manufacturer’s description of the gloves should be consulted. 3. Laboratory Coats Laboratory coats provide an important barrier for your clothes and, more important, your skin from chemicals. The laboratory coat should fit comfortably, have long sleeves and should be clean. 4. Material Safety Data Sheets (MSDS) When dealing with chemicals, caution is warranted, especially with reactive chemicals, carcinogens, and toxic reagents. A useful resource is the Material Safety Data Sheets (MSDS). There are a variety of resources that can be accessed online, including the following: MSDS Solution [www. msds.com (accessed June. 10, 2013), MSDS online [www.msdsonline.com (accessed June. 10, 2013), Cornell University [www.msds.ehs.cornell.edu (accessed June. 10, 2013), Chemical suppliers/manufactures such as Sigma-Aldrich (www.sigmaaldrich.com), (accessed June. 10, 2013), 2 www.msds.ped.go.th (accessed June. 10, 2013) 5. Useful Preparations Setting up the reaction is probably the most import job for an organic chemist. Once a reaction is initiated, there is little left that needs to be done to change the outcome.

3 Experiment 1 Melting Point and Boiling Point Determination

The physical properties of organic compounds are those properties that are intrinsic to a given compound when it is pure. A compound may often be identified simply by determining its physical properties. The most commonly recognized physical properties of a compound include its color, melting point, boiling point, density, refractive index, molecular weight and optical rotation. In modern chemistry, organic compounds would include the various types of spectra (infrared, nuclear magnetic resonance, mass, and ultraviolet-visible) among the physical properties of a compound. A compound's spectra does not vary from one pure sample to another. Many reference books list the physical properties of substances. The most useful works for finding list of values for the non-spectroscopic physical properties include: Sigma-Aldrich index The Merck index

1.1 Melting Point (M.P.) Determination The M.P. of a substance is the temperature at which a material changes from a solid state to a liquid state. The pure crystalline substances have a clear, sharply defined melting point. During the melting process, all of the energy added to a substance is consumed as heat of fusion, and the temperature remains constant. A pure substance melts at a precisely defined temperature, characteristic of every crystalline substance and dependent only on pressure (though the pressure dependency is generally considered insignificant). Determining the M.P. is a simple and fast method used in many diverse areas of chemistry to obtain a first impression of the purity of a substance. This is because even small quantities of impurities change the melting point. The melting point is easy to measure and classify. Extensive collections of tables give the exact values of many pure organic compounds. The M.P. determination is a fast and cost-effective technique which remains a strong link to the vast pre-instrumental chemistry literature.

Sample Preparation: Careless preparation of a sample is the leading cause of inaccurate and irreproducible results in melting point determinations. Any substance being loaded into a melting point capillary must be: 4 1. Fully dry (oven-dried) 2. Homogeneous 3. In powdered form

Packing Sample tube: Capillary: Place the sample in a piece of thin-walled capillary tubing (1 mm X 100 mm) that has been sealed at one end. To pack the tube: Fill a capillary tube with crystals about 0.5-1.0 cm high. To transfer the crystals to the closed end of the tube, drop the capillary tube, closed end first, down all length of glass tubing, which is held upright on the desk top. "When the capillary tube hits the desk top, the crystals will pack down into the bottom of the tube. Capillary tube attached to the thermometer: It is held by a rubber band or a thin slice of rubber tubing. It is important that this rubber band be above the level of the oil (allowing for expansion of the oil on heating) so that the oil does not soften the rubber and allow the capillary tubing to fall into the oil. If a cork or a rubber stopper is used to hold the thermometer, a triangular wedge should be sliced in it to allow pressure equalization.

powdered

Figure 1.1: Packing sample and determination melting point with Thiele tube 5 Determining with Thiele tube There are two principal types of melting point apparatus available: the Thiele tube and electrically heated instruments. The Thiele tube (shown in Figure 1.1) is the simple device and was once widely used. It is a glass tube designed to contain a heating oil (paraffin oil) and a thermometer to which a capillary tube containing the sample is attached. Set melting point apparatus as Figure 1.1. Place the capillary tube in the Thiele tube that contains paraffin oil. The Thiele tube is usually heated by a micro-burner. During the heating, the rate of temperature increase should be regulated. For melting point determination, follow these four basic steps:

1. The heating stand is rapidly preheated, to a user-specified start temperature, selected just a few degrees below the expected melting point of the samples. 2. Once the temperature is stable, the sample capillaries are inserted into the oven and after the temperature stabilizes a heating ramp is immediately launched. 3. The temperature of the samples continues to rise at the user-specified ramping rate until a user-specified stop temperature is reached. Automated and/or visual observations of the melting point, melting range, Melting range = T1 – T2 T1: sample start to melt T2: sample complete melting (transparent)

4. At the end of the heating ramp, the capillaries are discarded and the heater stand is rapidly cooled down using a low flame, move the burner slowly back and forth along the bottom of the arm of the Thiele tube.

Table 1.1: The Melting point of common organic compounds Compounds M.P. (◦C) Water 0 p-Dichlorobenzene 53 m-Dinitrobenzene 90 Acetanilide 114 Benzoic acid 122 6 Benzamide 128 Urea 133 Adipic acid 152 Salicylic acid 157 Benzanilide 161 Succinic acid 189 3,5-Dinitrobenzoic acid 205 p-Nitrobenzoic acid 239 Anthraquinone 289 N,N’-Diacetylbenzidine 332

1.2 Boiling Point (B.P.) Determination If a sample of a liquid is placed in an otherwise empty space, some of it will vapourise. The pressure in the space above the liquid will rise and will finally reach some constant value. The pressure under these conditions is due entirely to the vapour of the liquid, and is called the equilibrium vapour pressure. The phenomenon of vapour pressure is interpreted in terms of molecules of liquid escaping into the empty space above the liquid. In order for the molecules to escape from the liquid phase into the vapour phase, the intermolecular forces (in order of increasing strength: 1. Van der Waals, 2. Dipole-Dipole, 3. Hydrogen bonding These forces have to be overcome which requires energy. Since the nature of the intermolecular forces is determined by the molecular structure, then the amount of energy required to vapourise the sample also depends on the molecular structure.

The vapour pressure-temperature curve for water is shown on the Figure 1.2. When the vapour pressure of a liquid is equal to the atmospheric (or applied) pressure then boiling occurs. The temperature at which this occurs, for a given pressure, is the boiling point. It should be noted, therefore, that the boiling point of a liquid decreases

7

Figure 1.2: The vapor pressure-temperature curve for water As the atmospheric (or applied) pressure decreases. This is illustrated by the Vapor Pressure- Temperature Curve above in Figure 1.2 For example, The atmospheric pressure is 760 mm Hg and pure water boils at 100 oC. For example if we are on a mountain 3500 feet above sea level, the atmospheric pressure would be approximately 660 mm Hg, and water boils at about 96.5 oC.

As a rule of thumb, the boiling point of many liquids will drop about 0.5 oC for a 10 mm decrease in pressure in the vicinity of 760 mm Hg. At lower pressures, a 10 drop in boiling point is observed for each halving of the pressure. A rough measure that seems to work relatively well for calculating the boiling point of a liquid in on our mountain is to subtract 1 oC for every 14-16 oC of temperature above 50 oC, So, in order to convert an experimental measurement taken at a top of a mountain (higher altitude and lower pressure) to that reported at sea level (higher pressure) one needs to ADD a correction factor since the boiling point at sea level is higher than that at higher altitudes. The correction factor can be obtained by algebraically rearranging the equation given above. There are more accurate methods available for this calculation. At lower pressures, a boiling point monograph or temperature-pressure alignment chart (shown below) can be used. 8

Figure 1.3: The pressure-temperature alignment monograph The basic principle is that a line through two known points on any two different scales (A, B and C) can be used to read off the value on the third scale. There are two ways in which a pressure monograph can be used (1) to determine the boiling point at atmospheric pressure (760 mm Hg) given the boiling point at a lower pressure and (2) to determine the boiling point at a lower pressure given the boiling point at atmospheric pressure. Firstly, let's say we have a compound with a boiling point of 100 oC at 1 mm Hg pressure. What is the boiling point at 760 mm Hg To do this we need to draw a line from 100 oC on scale A (left side, observed boiling point) to 1.0 mm Hg on scale C (right side, pressure "P" mm). We can then read off the boiling point at 760 mm on line B, it is about 280 oC.

9

Figure 1.4: Using a pressure nomograph: Determining the boiling point at 760mm Hg (scale B) for a sample that boils at 100 oC (scale A) at 1mm Hg (scale C)

Now what temperature would that same compound boil at 10 mm Hg pressure? Now we draw a line that passes through 280 oC on scale B (middle scale, the boiling point at 760 mm Hg) and to 10mm Hg on scale C. By extending that line to scale A, we can read off the new boiling point on scale A (left side) as being about 140 oC. Of course, you don't really have to "draw" the line, it can be done using the edge of a ruler or something else straight (Figure 1.5)

Figure 1.5: Using a pressure monograph: Determining the boiling point (scale A) at 10 mm Hg (scale C) for a sample that boils at 280 oC at 760mm Hg (scale B) 10 Sample Preparation: Careless preparation of a sample is the leading cause of inaccurate and irreproducible results in boiling point determinations. The substance must be: 1. Fully dry (anhydrous) 2. Homogeneous 3. In liquid form Packing Sample tube: Put the sample liquid 1.0 – 2.0 mL in a small test tube and a short piece of melting point capillary tubing (sealed at one end) is dropped in with the open end down. The small test tube assembly is attached to a thermometer with a rubber band or a thin slice of rubber tubing. (Figure 1.6). Sample tube attached to the thermometer: It is held by a rubber band or a thin slice of rubber tubing. It is important that this rubber band be above the level of the oil (allowing for expansion of the oil on heating) so that the oil does not soften the rubber and allow the capillary tubing to fall into the oil. If a cork or a rubber stopper is used to hold the thermometer, a triangular wedge should be sliced in it to allow pressure equalization.

Determining:

Sample Capillary (1. 0-2.0 ml) Figure 1.6: Packing sample and determination boiling point

Set boiling point apparatus as Figure 1.6. The boiling point determination procedure, followed by virtually every modern instrument, involves three basic steps: 11 1. Initial stream of bubbles as dissolved air is expelled and then, a little later, a rapid and continuous stream of bubbles emerges from the inverted capillary tube. 2. At this point stop heating. Soon the stream of bubbles will slow down and stop. When they stop, the liquid will enter the capillary tube. The moment when the liquid enters the capillary corresponds to the boiling point of the liquid, and the temperature is recorded (T1) 3. The slowly heating. The bubbles will come out the capillary tube. The moment when the bubbles come out the capillary corresponds to the boiling point of the liquid, and the temperature is recorded (T2) Boiling range = T1 – T2 T1: liquid enters the capillary T2: bubbles come out the capillary During the initial heating, the air trapped in the capillary tube expands and leaves the tube and vapour from the liquid also enters the tube. There is always vapour in equilibrium with a heated liquid. This gives rise to the initial stream of bubbles. When the temperature reaches the boiling point, the vapour pressure inside the capillary tube equals the atmospheric pressure. As the temperature rises just above the boiling point then the vapour will start to escape: the second set of bubbles. Once the heating is stopped, the only vapour left in the capillary comes from the heated liquid which seals its open end. As the liquid cools, its vapour pressure will decrease and when the vapour pressure drops just below atmospheric pressure, the liquid will be drawn into the capillary tube (forced there by the higher atmospheric pressure. The common organic compounds are shown in Table 1.2.

Table 1.2: The Boiling point of common organic compounds

Compounds B.P. (◦C) Water 100.0 Methanol 64.6 Ethanol 78.5 Acetone 56.4 n-Butanol 117.6 s-Butanol 98.0 t-Butanol 82.2 12 Tetrahydrofuran (THF) 66.0 Hexane 69.0 Heptane 98.0 Ethyl acetate 77.0 Cyclohexane 80.7 Toluene 110.6 o-Xylene 144.0 m-Xylene 139.1 p-Xylene 138.4

13 Procedures Part I: Melting point Determination of Pure Compounds A 1) Set the apparatus in Figure 1.1 2) Determine the M.P. with the Thiele tube 3) Compare the M.P. range with standard on Table 1.1 4) Conclusion; what is Compound A

Part II: Boiling point Determination of Pure Compounds B 1) Set the apparatus in Figure 1.6 2) Determine the B.P. 3) Compare the B.P. range with standard on Table 1.2 4) Conclusion; what is Compound B

Part III: Determination of Unknown..X… Liquid or Solid Sample 1) Set the apparatus 2) Determine M.P. or B.P. 3) Compare the M.P. or B.P. range with standard on Tables 1.1 and 1.2. 4) Conclusion; what is the unknown liquid or solid?

14 Experiment 2 Chromatographic Adsorption

The most common method of separation mixtures available to the organic chemist all involve chromatography. Chromatography is defined as the separation of a mixture compounds or ions by distribution between two phases, one of which is Stationary phase and the other is Mobile phase. Many types of chromatography are possible, depending on the nature of the two phases involved: Solid-liquid (column, thin-layer and paper) Liquid-liquid (high-performance liquid) Gas-liquid (vapor-phase) All chromatography works on much the same principle. The methods depend on the differential solubility or absorptivity of the substances to be separated relative to the two phases between which they are to be partitioned. In this experiment, two types of Solid- liquid chromatography are presented: column chromatography and Thin- layer chromatography (TLC). Stationary phase (Adsorbents): Column chromatography is a technique based on both absorptivity and solubility. It is a solid-liquid phase-partitioning technique. The solid may be almost any material that does not dissolve in the associated liquid phase; the most commonly used solids are silica gel

SiO2.xH2O, also called silicic acid, and alumina Al2O3.xH2O. These compounds are used in their powdered or finely ground forms. Interactions: The most important interactions are those typical of polar organic compounds. Either these forces are of the dipole- dipole type or they involve some direct interaction (coordination, hydrogen bonding, or salt formation). These types of interactions are illustrated in Figure 2.1, which for convenience shows only a portion of the alumina structure. Similar interactions occur with silica gel. The strengths of such interactions vary in the following approximate order:

Salt formation > coordination > hydrogen bonding> dipole-dipole > van der Waals

15 Strength of interaction varies among compounds. For instance, a strongly basic amine would bind more strongly than a weakly basic one (by coordination). In fact, strong bases and strong acids often interact so strongly that they dissolve alumina to some extent.

Figure 2.1: Interaction between silica gel and polar organic compound Parameters Affecting Separation The versatility of column chromatography results from the many factors that can be adjusted. These include 1. Adsorbent chosen 2. Polarity of the solvents chosen 3. Size of the column (both length and diameter) relative to the amount of material to be chromatographed 4. Rate of elution (or flow)

2.1 Thin-layer Chromatography (TLC) TLC is a simple, quick, and inexpensive procedure that gives the chemist a quick answer as to how many components (qualitative analysis) are in a mixture. TLC is also used to support the identity of a compound in a mixture when the R of a compound is compared with the R of a known compound (preferably both run on the same TLC plate) TLC is a very important technique for the rapid separation and qualitative analysis of small amounts of organic compounds. It is ideally suited for the analysis of mixtures and reaction products in experiments. This technique is closely related to column chromatography. In fact, TLC can be considered to be a type of column chromatography. TLC plates are coated with silica gel. In order to visualize UV-active organic compounds. The 16 silica gel is coated with a layer of fluorescent dye. Therefore, UV-active compounds can be detected under UV light and non-UV-active compounds are also seen.

Retention Factor (R) is a measurement of how far up a plate the compound travels.

R value is calculated by the distance of the center of the spot from the baseline (P or SM) divided by the distance of the solvent front from the baseline (S).

푡ℎ푒 푑𝑖푠푡푎푛푐푒 표푓 푡ℎ푒 푐푒푛푡푒푟 표푓 푡ℎ푒 푠푝표푡 푓푟표푚 푡ℎ푒 푏푎푠푒푙𝑖푛푒 (푃) R = 푡ℎ푒 푑𝑖푠푡푎푛푐푒 표푓 푡ℎ푒 푠표푙푣푒푛푡 푓푟표푛푡 푓푟표푚 푡ℎ푒 푏푎푠푒푙𝑖푛푒 (푆) In general, the eluting strength of commonly used solvents for normal phase chromatography where the stationary phase is silica gel neutral alumina (SiO2/Al3O3) and the mobile phase is increasing order proceeds as follows: hexane < cyclohexane < toluene < diethyl ether < dichloromethane < ethyl acetate < acetone < ethanol < methanol < acetic acid. The TLC developing procedure, followed by virtually every modern instrument, involves three basic steps:

1. Prepare the developing tank The developing tank for TLC can be a specially designed chamber or a beaker with a watch glass on the top. Pour the mobile phase into the chamber to a depth of just less than 0.5 cm. Cover the beaker with a watch glass, swirl it gently, and allow it to stand while you prepare your TLC plate.

2. Prepare the TLC plate TLC plates used are 2.5 cm x 5 cm sheets. Each large sheet is cut horizontally into plates which are 2.5 cm tall by various widths; the more samples you plan to run on a plate, the wider it needs to be. Handle the plates carefully so that you do not disturb the coating of adsorbent or get them dirty

3. Spot the TLC plate The sample should be soluble in a solvent. Usually dissolve about 1 mg in 1 mL of a volatile solvent such as ethyl acetate or methylene chloride works well for TLC. Take a micro capillary, Dip the micro capillary into the solution and then gently touch the end of it onto the proper location on the TLC plate. 17

4. Develop the plate Place the prepared TLC plate in the developing tank, cover the tank with the watch glass, and leave it undisturbed on your bench top. The solvent will rise up the TLC plate by capillary action. Make sure the solvent does not cover the spot. Allow the plate to develop until the solvent is about half a centimeter below the top of the plate. Remove the plate from the beaker and immediately mark the solvent front with a pencil. Allow the plate to dry.

5. Visualize the spots If there are any colored spots, circle them lightly with a pencil. Most samples are not colored and need to be visualized with a UV lamp. Hold a UV lamp over the plate and circle any spots you see. If samples are not visualized in UV light select one reagent for detect it. Beware! UV light is damaging both to your eyes and to your skin! Make sure you are wearing your goggles and do not look directly into the lamp. Protect your skin by wearing gloves. If the TLC plate runs samples which are too concentrated, the spots will be streaked and/or run together. If this happens, you will have to start over with a more dilute sample to spot and run on a TLC plate

a b c

18

d e Figure 2.2: TLC preparation (a-b), Development (c) and Location (d-e) If there are not visualize in UV light. The reagent that can be used to detect organic compound is as follows

A: Iodine (I2) Everything stains yellow. Add Solid I2 to a developing chamber. The spots will fade, but the plate can be redeveloped. B: p-Anisaldehyde Stain The greatest advantage is that stains have different colors and are manifested on TLC on heating for different molecules. C: Hanessian stain One of the most sensitive stains which detects most functional groups. The disadvantage is that everything stains blue.

D: KMnO4 It can detect molecules with an oxidizable . Everything stains yellow and the spots will be indistinguishable from the background. E: Ninhydrin Stain Especially sensitive to amino acids including amines and amilines 19

Scheme 2.1: The formation with Ninhydrin

2.2 Column Chromatography (CC) Column chromatography is a method used to purify individual chemical compounds from mixtures of compounds. It is often used for preparative applications on scales from micrograms up to kilograms. The main advantage of column chromatography is the relatively low cost and disposability of the stationary phase used in the process. The latter prevents cross-contamination and stationary phase degradation due to recycling. The classical preparative chromatography column, is a glass tube with a diameter from 5 mm to 50 mm and a height of 5 cm to 1 m with a tap and some kind of a filter at the bottom. Two methods are generally used to prepare a column: the dry method, and the wet method. For the dry method, the column is first filled with dry stationary phase powder, followed by the addition of mobile phase, which is flushed through the column until it is completely wet, and from this point is never allowed to run dry. For the wet method, a slurry is prepared of the eluent with the stationary phase powder and then carefully poured into the column. Care must be taken to avoid air bubbles. A solution of the organic material is pipetted on top of the stationary phase. This layer is usually topped with a small layer of sand or with cotton or glass wool to protect the shape of the organic layer from the velocity of newly added eluent. Eluent is slowly passed through the column to advance the organic material. Often a spherical eluent reservoir or an eluent- filled and stoppered separating funnel is put on top of the column. The individual components are retained by the stationary phase differently and separate from each other while they are running at different speeds through the column with the eluent. At the end of the column they elute one at a time. During the entire chromatography process the eluent is collected in a series of fractions. Fractions can be collected automatically by means of fraction collectors. The productivity of chromatography can be increased by running several columns at a time. In this case multi stream collectors are used. The composition of the eluent flow can be monitored and each fraction is analyzed 20 for dissolved compounds, e.g. by analytical chromatography, UV absorption, or fluorescence. Colored compounds (or fluorescent compounds with the aid of an UV lamp) can be seen through the glass wall as moving bands.

Figure 2.3: Chromatographic column As the components of the mixture are separated, they begin to form moving bands (or zones), with each band containing a single component. If the column is long enough and the other parameters (column diameter, adsorbent, solvent, and flow rate) are correctly chosen, the bands separate from one another, leaving gaps of pure solvent in between. As each band (solvent and solute) passes out from the bottom of the column, it can be collected before the next band arrives. If the parameters mentioned are poorly chosen, the various bands either overlap or coincide, in which case either a poor separation or no separation is the result. A successful chromatographic separation is illustrated in Figure 1.3.

21

Compound A Compound B Figure 2.4: Sequences of steps in a chromatographic separation

22 Procedures 1 The following procedures can be used to identify an unknown organic mixture and separation of plant pigments. 1. Preparation of the developing tank - 10 mL of mobile phase is a 20:80 mixture of ethyl acetate and hexane. Pour “mobile phase” into a tank. 2. Preparation of TLC - With a lead pencil, draw a base line and solvent font line. - Along this line, mark off tick marks 0.5 cm apart. 4 tick marks. - Using a glass capillary tube, dip a small volume of: 1) -naphthol solution 2) Diphenylamine solution 3) Unknown No. ….X…. 4) Pigment plant - Gentle touching the capillary to one of the tick marks. - Allow the spots to dry before developing the chromatogram. 3. Development of the chromatogram - Put the TLC at the development tank. Cover it. - Let the solvent run up the chromatogram until get the solvent front (top of the chromatogram). At this point remove the TLC. 4. Location of the organic compound of chromatogram - Put the chromatogram into UV box. The spots are visible. - Circle spots with a pencil. Mark the center of each spot, as best you can judge and record color of the spot.

For your report

1. Draw TLC plate with located spot and R of organic compound (each spot) 2. Based on your results, how many components are in unknown No….and pigment plant. 3. Compare the unknown No…X…. and pigment plant spot with standard -naphthol solution and Diphenylamine. Explain why.

23 Procedures 2 1. Packing the Column: Slurry Method 1. Combine the solid stationary phase (Silica gel) with a small amount of solvent (……………….) in a Erlenmeyer flask 2. Position a plug of cotton with solvent in the bottom . 3. Pour this homogeneous mixture into the column as carefully as possible using a spatula to scrape out the solid as you pour the liquid. 4. Once the column is loaded, open the stopcock and allow the solvent level to drop to the top of the packing, but do not allow the solvent layer to go below this point. Allowing this solvent level to go below the stationary phase should always be avoided.

2. Adding the Sample The sample can be loaded directly to the top of the column. 1. Use a minimum amount of a polar solvent, 5-10 drops to dissolve the mixture. 2. The solution is then carefully added to the top of the column using a dropper without disrupting the flat top surface of the column (A thin horizontal band of sample is best for an optimal separation). 3. Continuously add the solvent eluent while collecting small fractions at the bottom of the column. Using a dropper to add the first bit of solvent on top of the packing, and sample will minimize disturbance of the column and diluting the sample. 4. Collecting small fractions (1-3 mL) or as your observe (color band). 3. TLC Identification of a fractions band using TLC plate. Using a glass capillary tube, dip a small volume of the collecting fractions 1. Developing the chromatogram in development tank using mobile phase system, visualize with UV light.

2. Draw TLC plate with located spot and R of organic compound (each spot). Compare fractions band spot. Explain why

24 Experiment 3 Acid-Base Extractions, Separations and Drying Agents

A commonly used method for separation a mixture of organic compounds is known as liquid-liquid extraction. Most reactions of organic compounds require extraction at some stage of product purification. In this experiment you will use extraction techniques to separate a mixture of an organic acid, a base, and a neutral compound. Organic acids and bases can be separated from each other and from neutral compounds by extraction using aqueous solutions of different pH values.

Organic acids: Most organic carboxylic acids are insoluble or slightly soluble in water, but these compounds are highly soluble in dilute aqueous sodium hydroxide because the acid is deprotonated by the base producing the sodium carboxylate salt.

Scheme 3.1: Benzoic acid in NaOH solution The carboxylic acid can be selectively isolated by dissolving the mixture in an organic solvent that is immiscible with water, and then extracting the solution with sodium hydroxide. The basic aqueous solution containing the carboxylate salt is acidified, causing the sodium carboxylate salt to convert back to the carboxylic acid, which is not water soluble. The acid will precipitate from the solution, as shown here.

Scheme 3.2: Sodium benzoate in HCl solution Organic bases (e.g., amines): that are insoluble in water can be separated by extraction with hydrochloric acid. Addition of HCl to the amine produces the corresponding ammonium salt, which is soluble in water but not in organic solvents.

Scheme 3.3: p-nitroaniline in HCl solution 25 The amine can be recovered from the aqueous solution by treatment with a base, converting the ammonium salt back to the amine. The amine is not water-soluble and will precipitate, as shown here.

Scheme 3.4: p-nitroaniline ammonium chloride in NaOH solution Using your understanding of these properties, separation of a mixture containing a carboxylic acid, an amine, and a neutral compound can be carried out via sequential acid and base extractions. The precipitates will be collected and characterized by melting temperature analysis.

26 Procedures Part 1: Extraction 1. Weigh 1.50 g of the mixture sample. Transfer the mixture to a 100 mL separation funnel and dissolve it in 30 mL of dichloromethane (DCM). 2. Clamp the support ring onto a ring stand and place the separation funnel into the ring. Add 10 mL of 6.0 M hydrochloric acid (HCl). 3. Gently shake several times, open the stopcock to release the pressure. Place the funnel on a support ring and allow the solvent and aqueous layer to separate. 4. Open valve of separator funnel; keep the lower organic layer (DCM layer) into a 50 mL Erlenmeyer flask. And drain the upper aqueous layer A into the Erlenmeyer flask 5. Repeat the extraction with another 10 mL of 6.0 M hydrochloric acid (HCl), draining the second aqueous layer into the same Erlenmeyer flask (Aqueous layer A). And save the organic layer (DCM layer) in the separator funnel for later use. - Cool the flask containing the acidic aqueous extracts (Aqueous layer A) into an ice water bath. Slowly add 6.0 M sodium hydroxide (NaOH) until the aqueous layer is basic. Use pH paper to test. (Solid appear) - Collect the Solid A using vacuum filtration and save the solid for TLC analysis in Part II. 6. Extract the Organic layer (DCM layer) in the separator funnel with 10 mL of 3.0 M sodium hydroxide (NaOH). Drain the aqueous layer into a 50 mL Erlenmeyer flask (Aqueous layer B). Repeat the extraction with another 10 mL of 3.0 M sodium hydroxide (NaOH). Save the Organic layer (DCM layer) in the separator funnel. - Cool the flask containing the basic aqueous extract (Aqueous layer B) in an ice water bath. Slowly add 6.0 M hydrochloric acid (HCl) until the aqueous layer is acidic. Use pH paper to test. - Collect the Solid B using vacuum filtration and save the solid for TLC analysis in Part II. 7. Add 20 mL of saturated aqueous sodium chloride solution to the organic layer remaining in the separator funnel and shake gently. Allow the layers to separate. 8. Open valve; keep the organic layer into an Erlenmeyer flask containing approximately

1 g of anhydrous sodium sulfate (NaSO4) and allow it to stand for about 10 minutes, swirling occasionally (Organic layer C). 9. Gravity filter into a clean 100 mL round-bottom flask. Evaporate the DCM in the Rotary evaporator (Solid C). 27

10. Weigh the recovered solid and record the Mass, Calculate % recovery and Save the solid for the TLC analysis in Part II.

Part 2: Qualitative with TLC 1. Obtain a small amount of the isolated solid (A-C) dissolve in a small amount of DCM solvent. And spot on TLC plate with standard benzoic acid, p-nitroaniline and . 2. Developing the chromatogram in development tank using mobile phase system EtOAc : Hexane (30:70), visualize with UV light. 3. Report sheet. R of organic compounds 4. Complete the box in Scheme 3.1

For your report 1. Calculate the percent yield of your product. 2. Complete the boxes in Scheme 3.5. Explain why. 3. Based on your results, Draw TLC plate and identify the product as standard.

28 Scheme 3.5: Acid-base Extractions

29 Experiment 4 Alkane, Alkene and Alkyne

Alkane and are saturated hydrocarbons which contain only hydrogen and carbon connected together by only single bonds. They are generally chemically inert, however, the reaction of alkanes can occur at appropriate conditions. In halogenation reaction, light or elevated temperature is needed as a catalyst to facilitate the reaction. Hydrogen atom will be substituted by halogen atom such as bromine (Br) or chlorine (Cl).

Order of reactivity of halogen are as following; X2; F2 > Cl2 > Br2, however, I2 is not reacting. This reaction is purposed to take place through free-radical mechanism. There are various products from this halogenation reaction depended on stability of radical intermediate and possibility of each hydrogen atom on deprotonating step. As an example;

This reaction is called “Substitution reaction” when a hydrogen atom of alkane is replaced by halogen atom to give a mix of halogenated compounds and an acid. This acid can be tested by placing wetted Litmus paper over reaction tube which turns from blue to red in the presence of acids.

Alkene Alkenes and cycloalkenes are classified as unsaturated hydrocarbons which contain double bonds (1-bond+1-bond). This group of hydrocarbons is generally reactive in chemical reactions especially addition reactions of electrophilic reagents to the double bonds as following;

30

-electron of alkene will act as a nucleophile reacting with an electrophile (A-B) to give a saturated product. Mechanism of this reaction are generally founded in two ways, ionic addition reaction and free radical addition reaction depending on reaction conditions and reagents.

Ionic mechanism,

Free radical mechanism

Unlike substitution reactions, addition reactions do not require catalyst. Various types of reagents can react with alkene such as H2, Br2, Cl2, HX, H2O and KMnO4, etc.

a) Oxidation by potassitum permanganese (KMnO4)

Oxidation of unsaturated compounds by a strong oxidant KMnO4 causes the decoloration of KMnO4 and formation of brown precipitate of manganese dioxide (MnO2). This method is used as a qualitative test for unsaturation called “Baeyer’s Test for unsaturation”. Saturated compounds and aromatic compounds will be negative to this test.

b) Addition reaction with bromine

31

Decoloration of bromine by unsaturated compounds is an addition reaction forming 1, 2-dibromo compound. Unlike the substitution reaction of alkanes with bromine, HBr will not be produced from this reaction.

c) Addition reaction with sulfuric acid

Alkene reacts with concentrated H2SO4 to produce alkyl hydrogen sulfate and heat which unlike alkane. This reaction can be used to distinguish alkane from alkene.

d) Hydrohalogenation of alkene

Hydrohalogenation is an addition reaction of HX to alkene which generates alkyl halide. When adding HX to unsymmetrical alkene, the reaction follows Markovnikov’s rule which hydrogen will first add to the least substituted carbon generated the most stable carbocation then halide anion will add to the carbocation.

Alkyne 32 Alkyne contains triple bonds which composes of one -bond and two -bonds. They are very reactive to the reactions including addition reactions same as alkenes but the reaction can occur twice (have two -bonds to add). The reaction mechanism is the same as what happen in alkene.

For terminal alkyne which has a hydrogen attach to the sp-carbon, this hydrogen is acidic and reactive with metals such as Na, K, Ag, etc., generating flammable hydrogen gas and precipitate.

Acetylene gas can be freshly prepared in laboratory from the reaction of calcium carbide (CaC2) with water. This reaction gives pungent smell of hydrides of P, As and S while pure acetylene gas does not smell. Acetylene gas will be purged into the reaction vessel containing other reagents (HX = HCl, HBr and HI) to pursue the reaction.

The generation of acetylene gas

Other alkynes can be prepared from various reactions such as following;

a) Alkylation of acetylene or monoalkylated acetylene using alkylating agent such as;

b) Elimination of two moles of HX from dihalides using alcohol, KOH or NaNH2

33

c) Elimination of four halogen atoms from tetrahalides using Zinc/alcohol

34 Procedures Reagents and equipment • Equipment 1. test tube 2. dropper 3. beaker 4. gas collecting kit • Reagents

1. An alkane 8. 2% NH4OH 2. An alkene 9. CaC2 3. concentrated H2SO4 10. diluted HNO3 4. 0.3% KMnO4 11. H2O 5. 1% Br2/CCl4 12. Ethanol 6. 5% AgNO3 13. Benzene 7. 5%NaOH 14. Na metal

Procedure Part 1 Alkane vs. Alkene A. Solubility test Put 1.0 mL of water, ethanol and benzene in three separated test tubes then add 3 drops of an alkane (drop-wise) into each tube, observe and record the result. Repeat the whole experiment with an alkene. B. Decoloration of bromine

Put 0.5 mL of an alkane into two test tubes. Put 5 drops of 1% Br2/CCl4 into each tube, cap each tube with a wooden cork and gently check the tubes. Put one tube in dark cupboard and put another tube under light (or sunlight) for 15 minutes. Compare color of each tube and record the results. Put wetted blue litmus paper on top of the tubes to test for HBr, record your observation. Repeat the whole experiment with an alkene.

C. Decoloration of KMnO4 (neutral) Put 0.5 mL of an alkane into a test tube together with 0.5 mL of 0.3% KMnO4. Check the tube and record your observation. Repeat the whole experiment with an alkene.

35

D. Addition reaction of H2SO4 Put 1.0 mL of an alkane into a test tube, slowly add 1.0 mL of conc. H2SO4. Check the tube and record your observation. Is there any heat generated? Repeat the whole experiment with an alkene.

Part 2 Reactions of Alkyne A. Reaction with bromine In hood, purge acetylene gas generated from gas collecting kit into 0.5 mL of bromine solution in test tube. Shake the tube and record your observation. B. Baeyer’s Test for unsaturation In hood, purge acetylene gas generated from gas collecting kit into 0.5 mL of 0.3%

KMnO4 solution in test tube. Shake the tube and record your observation. C. Acidity test Put 0.5 mL of benzene into test tube. Purge acetylene gas into the solution then put a small piece of Na metal. Pour the mixture onto a watch glass (put undissolved Na metal in ethanol), leave benzene evaporates to dryness. Record the observation then add 4-5 drops of water, test acidity by litmus paper then record the result.

CAUTION! Put remaining undissolved Na metal in ethanol. DO NOT dispose it in the sink since Na vigorously reacts with water which can catch fire or cause skin and eye damage.

D. Reaction with ammoniacal silver nitrate Put 3.0 mL of ammoniacal silver nitrate solution into test tube then purge acetylene gas into the solution. Record your observation. Filter out acetylide precipitates then dissolve them back in diluted nitric acid (CAUTION! these precipitates can explode when they dry).

Note Ammoniacal silver nitrate solution can be prepared from adding 2.0 mL of 5%

AgNO3 and 2 drops of 5% NaOH into a test tube. Shake and then slowly add 2% NH4OH drop- wise while shaking. Keep adding 2% NH4OH until Ag2O precipitate formed is completely dissolved. Do not add excess NH4OH.

36 Experiment 5 Extractions and Crystallization of Caffeine Part 1

The primary method for purify organic compounds is recrystallization. The reaction mixtures or natural product mixtures always contain impurities. The impurities may include some combination of insoluble, soluble, and colored impurities. The solubility of organic compounds is an important thing to consider

Figure 5.1: Structure of Purine and Caffeine Caffeine is an alkaloid compound. Alkaloids are containing basic compounds that are found in plants. They usually taste bitter and often are physiologically active in humans. The names of some of these compounds are familiar to you even if the structures aren’t: nicotine, morphine, strychnine, and cocaine. The role or roles these compounds play in the life of the plants in which they are found is not well understood. In some cases they may act as pesticides; nicotine is found in tobacco and has been sprayed onto other plants, in which it is not normally found, to function as an insecticide. The structure of caffeine is shown to the right of Figure 4.1. It can be considered to be constructed from the purine ring system, which is important biologically, being found in nucleic acids and elsewhere. Caffeine is found in a number of foods ingested by people. Caffeine acts as a stimulant. It stimulates the heart, respiration, the central nervous system, and is a diuretic. Its use can cause nervousness, insomnia and headaches. It is physically addictive. A person who drinks as few as 4 cups of coffee a day and who attempts to stop “cold turkey” may experience headache, insomnia, and possibly nausea as the result of withdrawal.

Table 4.1: The amount of caffeine in coffee tea Type of coffee and tea Quantity (mg/30 mL) Coffee, espresso 40-75 Mcdonald’s brewed 6 37 Mcdonald’s Mocha 7 Starbucks Latte 9 Starbucks brewed 20 Black tee 2-8 Green tea 0-2 Lipton brisk lemon 1 Coca-Cola 2-3 7UP 0 Pepsi 3-4

Adapted from Journal of Food Science, 2010; Pediatrics, 2011; USDA National Nutrient Database for Standard Reference, Release 23, 2010; Journal of Analytical Toxicology, 2006; Starbucks, 2011; McDonald's, 2011 Tea has been consumed as a beverage for almost 2,000 years, starting in China. It is a beverage produced by steeping in freshly boiled water the young leaves and leaf buds of the tea plant, Camellia sinensis. Today, two principal varieties are used, the small-leaved China plant (C. sinensis sinensis) and the large-leaved Assam plant (C. sinensis assamica). Hybrids of these two varieties are also grown. The leaves may be fermented or left unfermented. Fermented teas are referred to as black tea, unfermented teas as green tea, and partially fermented teas as oolong. As trade routes opened to Asia in the 17th century, tea was imported to Europe. Today, you are going to make a small but strong cup of tea and extract the caffeine from it.

Scheme 5.1: Extraction of tea leaves

38 Procedures Part 1: Dissolution of Caffeine in Water

1. Weigh about 15.00 g of tea and 4.00 g of calcium carbonate (CaCO3) in the 250 mL beaker. Record actual weight. 2. Add 100.00 mL of distilled water to the beaker. 3. Boil on hot plate for 15-20 minutes while stirring occasionally. 4. After the boiling period is over, filter the hot solution with cotton to remove any solid particles. Leave it to room temperature.

Part 2: Transfer of Caffeine from water to dichloromethane (DCM) (Extraction) 1. Extraction with 40.0 mL of DCM 2 time. 2. Save the Organic layer (DCM layer) in the Erlenmeyer flask.

3. Add anhydrous MgSO4 to remove water. 4. Carefully drain the DCM solution (about 40 mL) in Erlenmeyer flask A and B. 5. Remove the DCM from Solution flask A and B using hot plate in the fume Hood (BP. of DCM is 39.6 OC). Reduce the solution down to about 2 mL and then remove from the hot plate. “For Flask B, transfer to culture dish and Keep for Experiment 5”

Part: 3 Crystallization of Caffeine Flask A (dissolve at hot but non-dissolve at cool)

1. Dissolve the crude Flask A with hot acetone (CH3COCH3) on hot plate. Until, The solid completely dissolves. 2. Remove from the hot plate. Allowing the hot mixture to slowly cool to room temperature and cooling further in an ice bath. Crystals will appear. If not, remove the acetone with hot plate. 3. Collect the caffeine crystal with vacuum filtration. Record actual weight. What was your percentage yield of caffeine in tea leaves?

39 Experiment 6 Sublimation: Caffeine Part 2 amd Simple Distillation of Mixed Compounds

6.1 Sublimation: Caffeine Part 2 The sublimation process is the transition of a substance from a solid phase to an immediate gas phase, where it does not pass through a transitional liquid phase. Normally the sublimation process is used to purify the crude substance without decomposition and the impurities cannot sublimate. Sublimation is much easier than evaporation from the melt, because the pressure of their triple point is very high, and it is difficult to obtain them as liquids. Sublimation is a faster method of purification than recrystallization but not so selective. It is most effective in removing a volatile substance from a nonvolatile substance, particularly a salt or other inorganic material. Sublimation is also effective in removing highly volatile bicyclic or other symmetrical molecules from less volatile reaction products such as , isoborneol and camphor. In this experiment, Caffeine can be purified by sublimation process because of the properties of caffeine, M.P. 235-238 oC and B.P. 178 oC that mean caffeine does not pass through a transitional liquid phase.

Figure 6.1: Sublimation of caffeine with Culture dish

40 Procedures From Experiment 5 caffeine extracted Flask B 1. Close the culture dish with the lid of the culture dish (weigh it), put on the hot plate. 2. Take the ice bath (beaker) on the top culture dish. 3. Gently warming, white crystal will appear on top lid of the culture dish. Remove from the heat plate (Caution!!!) 4. Record actual weight of lid culture dish. Collect the caffeine crystal, What was your %yield of caffeine in tea leaves? 5. Comparisons the result with experiment 5.. What is a good process for purify caffeine. Explain why?

41 6.2 Simple Distillation of Mixed Compounds Distillation is the process of vaporizing a liquid, condensing the vapor, and collecting the condensate in another container. It is very useful for separating, identification and purification a liquid mixture. When the components have different boiling points, or when one of the components will not distill. It is one of the principal methods of purifying a liquid. Four basic distillation methods are available to the chemist: 1. Simple distillation 2. Steam distillation (experiment 7) 3. Fractional distillation (experiment 9) 4. Vacuum distillation (distillation at reduced pressure) Separation and Purification: It is a physical process used to separate from a mixture by the difference in how easily they vaporize. As the mixture is heated, the temperature rises until it reaches the temperature of the lowest boiling substance in the mixture, while the other components of the mixture remain in their original phase in the mixture. The resultant hot vapor passes into a condenser and is converted to the liquid, which is then collected in a receiver flask. The other components of the mixture remain in their original phase until the most volatile substance has all boiled off. Only then does the temperature of the gas phase rises again until it reaches the boiling point of a second component in the mixture, and so on. Identification: The boiling point of a substance is a useful physical property for the characterization of pure compounds.

Figure 6.2: Simple distillation or Fractional distillation apparatus

42

Procedures Purification of mixtures liquid A and B (1:1) 1. Obtain 30.0 mL of liquid mixtures in 100 mL round-bottom flask. Add a few of boiling chips 2. Assemble a simple distillation apparatus as shown in Figure 6.2 (the bulb of the thermometer must be below the arm of the distillation head) 3. Slowly heat the heating mantle. - The lower boiling component is distilled, the boiling point of the mixture in the distillation flask will increase. “the temperature remains constant” - Record the temperature after the first drop is collected. Collect it until the temperature changes. Keep Liquid A and record the volume of Liquid A. 4. Increase heating, Record the temperature after the first drop is collected. Keep Liquid B and record the volume of Liquid B. !!!DO NOT DISTILL TO DRYNESS!!!.

For your report - What is the %yield of Liquid A and Liquid B? - Conclude what is organic compounds of the liquid A and B? (Table 1.2 page No. 4). Explain why?

43 Experiment 7 Stream Distillation of an Essential Oil: Eugenol oil

Eugenol oil can be extracted from Clove trees (กานพลู or Eugenia caryophyllata tree) which is made up of the extract of dried flower buds and leaves. Normally, Clove contains 85-90% Eugenol and 9-10% Eugenol acetate. It is used for the purposes of food flavoring, cosmetic, dermal drug delivery, dental, aromatic, soaps detergents.

Figure 7.1: Clove, Eugenol and Eugenol acetate structure. Distillation can also be performed on mixtures in which the two compounds are not miscible. This process is called co-distillation. When one of the compounds is water, the process is called steam distillation. The apparatus like simple distillation as show in Figure 6.1. Simple and fractional distillations are carried out on miscible mixtures. Ideal mixtures follow Raoult’s law: the total vapor pressure of the system is determined by adding together the products of the vapor pressure and the respective mole fraction of each compound. Characterization of Eugenol, it contains double bond (alkene) and an aromatic hydroxyl group (phenol). These functional groups provide the basis for simple chemical tests used to characterize the clove oil.

Characterize double bone with Br2 reaction; 44

Scheme 7.1: Br2 addition reaction

Characterize double bone with KMnO4 oxidation reaction;

Scheme 7.2: KMnO4 oxidation reaction 3+ Characterize phenol complex with FeCl3 reaction; Phenols (ArOH) react with the Fe ion in iron(III) chloride (FeCl3) to give complexes that are blue, green, red, or purple.

Scheme 7.3: Phenol complex with Fe3+ In this experiment, you will carry out steam distillation of eugenol oil from dried ground cloves and characterization of eugenol oil.

45 Procedures Part I: Steam Distillation 1. Weight 30 g of dried cloves and transfer to 500 mL round‐bottom flask. Add 250 mL of distilled water and a few of boiling chips. 2. Assemble the distillation apparatus (Figure 6.1). 3. Heat the mixture. Adjust the heat to maintain a distillation rate of approximately 1 drop per second. 4. Stop the distillation when approximately 150 – 200 mL of distillate has been collected. !!!DO NOT DISTILL TO DRYNESS!!!. 5. Use a dropper to remove water (lower layer) as much as you can.

6. Add anhydrous MgSO4 for remove water. 7. Drain the oil to bottle (weigh it). 8. Report the weight and percent yield of clove oil.

Part II: Characterization 1. Dissolve 3 drops the clove oil with 5 drops of methanol into three test tubes. 2. Add 5 drops of bromine in chloroform to tube 1. Gently swirl and record your observation.

3. Add 5 drops of KMnO4 solution to test tube 2. Gently swirl and record your observation.

4. Add a 5 drops of FeCl3 solution to test tube 3. Gently swirl and record your observation. 5. Identification the functional group of clove oil. Explain why?

46 Experiment 8 The Oxidation of 1o-, 2 o –Alcohol and Identification of Aldehyde and Ketone with 2,4-DNP

Identifications of simple chemical reactions that produce color changes or form precipitates, can be used to differentiate alcohols, aldehydes, and ketones and also to provide further structural information.

The oxidation [O] of primary and secondary alcohols is oxidized by K2Cr2O7 to carboxylic acids and ketones respectively. Primary alcohols can be oxidized to aldehydes and to carboxylic acids. The equation for the oxidation of ethanol to the ethanal is shown below.

Scheme 8.1: The oxidation of primary alcohol (ethanol) When preparing aldehydes in the laboratory, you will need to distil the aldehyde from the reaction mixture as it is formed. This prevents the aldehyde from being oxidized further to a carboxylic acid. When making the carboxylic acid, the reaction mixture is usually heated under reflux before distilling the product off. Secondary alcohols can be oxidized to ketones. Unlike aldehydes, ketones are not oxidized further. Tertiary alcohols can’t be oxidized

Scheme 8.2: The oxidation of secondary alcohol (2-propanol) Aldehydes react with Schiff's reagent to produce a color change (purple). The sulfur in the bisulfite ion acts as a nucleophile and adds to the carbonyl carbon. Because this is such a bulky nucleophile, it will add only to a relatively steric unhindered carbonyl. This requires that the carbonyl be part of an aldehyde in which one of the R groups is the very small 47 hydrogen, or a ketone having small R and R’ groups. A ketone having large groups attached to the carbonyl will not react with bisulfite.

Scheme 8.3: The Schiff’s reagent of aldehyde In the same way, the Schiff’s reagent acts as a nucleophile that adds to the carbonyl group of an aldehyde. Because this nucleophile is extremely bulky, a ketone, which is more steric crowded than an aldehyde at the carbonyl carbon, does not react with Schiff's reagent, and thus does not produce the purple color. Production of the purple color therefore indicates that the unknown is an aldehyde

Aldehydes and ketones react with 2,4-dinitrophenylhydrazine (DNP) to form yellow, orange or reddish- orange precipitates, whereas alcohols do not react. Formation of a precipitate therefore indicates the presence of an aldehyde or ketone. The precipitate from this test also serves as a solid derivative.

Scheme 8.4: The combination of DNP and aldehyde or ketone

Table 8.1 The Melting point (M.P.) of 2,4-dinitrophenylhydrazone (DNP derivative) Aldehyde M.P. (oC) Ketone M.P. (oC) Pentanal 98 2-Butanone 115 Hexanal 107 Acetone 128 48 Butanal 123 2-pentanone 144 Propanal 155 3-pentanone 156 2-methylpropanal 187 cyclohexanone 162

In this experiment, the primary and secondary alcohol are oxidize to aldehyde or ketone then complex with 2,4-dinitrophenylhydrazine (DNP) and identify 2,4-dinitrophenylhydrazone derivatives (DNP derivatives) by melting point.

49 Procedures Part I: Oxidation reaction of 1o - or 2o-alcohol

1. Place 2.0 mL of conc. H2SO4 and saturated K2Cr2O7 into a 50 mL round- bottom flask. 2. Stir the mixture for 3 minutes in an ice‐water bath. 3. Slowly add 4.0 mL of Unknown alcohol No… and stir for 5 minutes. 4. Add 10.0 mL of distilled water. 5. Assemble the distillation apparatus (Figure 6.1). Distillation the mixture untill organic layer appear and Keep Liquid A (6 mL) (ketone or aldehyde)

Part II: Functional group classification with Schiff’s reagent 1. Add 3 drops of Liquid A and 2.0 mL of distilled water in test tube 2. Slowly add 10 drops of Schiff’s reagent into test tube. Gently swirl and record the color of the mixture solution. 3. Is it a positive test? Why? 4. Conclusion. Liquid A is? (aldehyde or ketone) and type of alcohol (1o or 2o)

Part III: Synthesis of 2,4-dinitrophenylhydrazone (DNP derivative) 1. Place 10.0 mL of 2,4-dinitrophenylhydrazine in a 50 mL Erlenmeyer flask 2. Slowly drop a Liquid A. Stir gently with a glass stirring rod 3. Collect the precipitate by vacuum filtration. 4. Recrystallize the product using ethyl alcohol. “dissolve at hot but non-dissolve at cool” 5. Collect the precipitate by vacuum filtration (weight). 6. Determine the M.P. of DNP Derivative 7. What is actually of aldehyde or ketone of Liquid A. and 1o or 2o of alcohol. Compare with Table 10.1.

50 Experiment 9 Stereochemistry, Addition Reaction: Camphor Reduction

Reactions and interconversions of compounds in the monoterpene series have been of great importance in studying the mechanism of carbocation rearrangements and the stereoselectivity of various reagents and synthetic reactions. Typical monoterpenes are the ketone camphor and the corresponding epimeric alcohols borneol and isoborneol. In this experiment, you will reduce camphor, a naturally occurring ketone, using sodium borohydride (NaBH4). Camphor is an example of a bridged : a molecule with multiple rings that share non-adjacent atoms called bridgeheads. Camphor is a substituted bicyclo[ 2. 2. 1] heptane. The enclosed numbers represent the size of the “bridges” that are attached to the bridgeheads.

Figure 9.1: The Bridges bicyclic molecule Bridged bicyclic systems of camphor have rigid conformations. The six- membered ring highlighted in the figure is in a boat conformation ( recall the chair conformation of cyclohexane). When there is a planar functional group like an alkene or a carbonyl, the faces are defined as exo and endo, referring to the “outside” and “inside” of the boat-shaped ring. Reacting partners can approach from either of these faces, resulting in different products. There are many reagents that are useful for the reduction of various carbonyl compounds. The complex hydrides NaBH4 and LiAlH4 are among the most useful reagents available for the conversion of carbonyl compounds to alcohols. Sodium borohydride is the less reactive of the two; for example, esters and acids are not affected. Sodium borohydride is very convenient to use, since reactions can be carried out in aqueous or alcoholic solutions. The reduction of bicyclic ketones such as camphor and norcamphor with these hydrides is quite stereoselective, with one of the two diastereomeric alcohols being formed in over nine times the amount of the other. 51

Scheme 9.1: The reduction reaction of camphor Procedure 1. Weigh 1.00 g of camphor and dissolve in 25 mL anhydrous MeOH in 100 mL of round-bottom flask.

2. Add 0.62 g of NaBH4 to the camphor solution, stirring the reaction mixture during the addition. 3. Attach an air condenser and gently reflux the mixture for 10 min. 4. After cooling the reaction mixture to room temperature, carefully add 10 mL of ice-cold water. 5. Extracted with 15 mL of dichloromethane (two times) and dry the solution over

Na2SO4. 6. Remove the solvent by rotary evaporator or gently evaporate the solvent on a hot plate under the hood. 7. Recrystallize with MeOH and collect the solid with vacuum filtration. 8. Determine the weight of the dry solid and calculated %yield. 9. Identify the product with TLC use mobile phase as EtOAc : Hexane (20:80) and visualize with p-Anisoldehyde solution.

For your report 1. Calculate the percent yield of your product. 2. Draw the mechanism of Major and Minor products. Explain why. 3. Based on your results, draw TLC plate and identify the product as either the endo or exo product. Explain why?

52

Experiment 10 The Aldol Condensation

Aldol condensation combines both Nucleophilic addition and Condensation reaction. It is a very useful synthetic reaction for generating new carbon-carbon bond and double bond. In an Aldol reaction, it connects between alpha carbon of one molecule and the carbon of carbonyl of the second molecule. This forms a beta hydroxyl carbonyl compound. Sometime it is followed by dehydration to get alpha-beta unsaturated carbonyl compound (Aldol products).

Scheme 10.1: The aldol condensation reaction of acetaldehyde

Mechanistically, the aldol condensation reaction is a simple nucleophilic addition reaction between an enolate ion and a carbonyl group. The enolate is formed when the base (NaOH) present abstracts an -Hydrogen of a carbonyl one. This Hydrogen is particularly acidic photon because of the resonance stabilization of the resulting enolate. This enolate nucleophilically attacks carbonyl groupings in the typical fashion, forming the aldol product.

Scheme 10.2: The resonance form of enolate anion

In this experiment, we will synthesize the substitute dibenzalacetone from crossed Aldol condensation of acetone and two molecule of a substituted benzaldehyde.

53

Scheme 10.3: The aldol condensation reaction of substituted benzaldehyde Procedure Synthesis of Substituted dibenzalacetone 1. A: Place1.4 mL (1.0 eq., 1.458 g) of Benzaldehyde and 0.5 mL (0.5 eq., 0.4 g) of acetone and 10 mL of ethyl alcohol in a 50 mL Erlenmeyer flask B: Place1.4 mL (1.0 eq., 1.567 g) of anisaldehyde and 0.4 mL (0.5 eq., 0.34 g) of acetone and 10 mL of ethyl alcohol in a 50 mL Erlenmeyer flask 2. Stir the mixture for 3 minutes in an ice‐water bath. 3. Slowly add 0.5 g of sodium hydroxide (NaOH) in 2 mL of water. Stir for 5 minutes. 4. Remove the ice‐water bath. Stir the mixture at room temperature for another 10 minutes. The reaction mixture should at first be clear and homogeneous, but should shortly become milky with solid particles of the product. 5. Add the ice and stir until the precipitation occurs. 6. Collect the precipitate by vacuum filtration. 7. Wash the filtrate with 150 mL of distilled water to remove the basicity. 8. Transfer the precipitate into an Erlenmeyer flask. 9. Recrystallize the product using ethyl alcohol. “dissolve at hot but non-dissolve at cool” 10. Collect the crystal with vacuum filtration. Allow the solid to dry completely 11. Determine weight and %yield of the product.

Explanations! 1. What is the structure and configuration (cis-cis or trans –tran or cis-trans) of the product? “The trans-tran-form melts at 110-111 oC; The cis-trans-form melts at 60 oC The cis-cis-form is an oil” 2. What is the mechanism of the product? 54 3. Draw the structure of your product in stable-form.

Organic Laboratory 2 คม254: ปฏิบัติการเคมีอินทรีย์ 2

สาขาวิชาเคมี คณะวิทยาศาสตร์ มหาวิทยาลัยแม่โจ้ ประจำปีการศึกษา 2/2562

Contents

1. Electrophilic Aromatic Substitution (Nitration reaction): 3 2. Nucleophilic Aromatic Substitution with KI 7 3. Diazonium salt Coupling reaction with -naphthol 10 4. Lipid : Saponification Number 12 5. Carbohydrate: Isolation of lactose from Milk 18 6. The Aldol condensation 21 7. The Claisen condensation 23 8. Synthesis of carboxylic derivative; Aspirin 25 9. Cannizzaro Reaction of benzaldehyde 28 10. Identification of Cannizzaro Reaction Product by FT-IR technique 31

หลักสูตรวิทยาศาสตรบัณฑิต สาขาวิชาเคมี คณะวิทยาศาสตร์ มหาวิทยาลัยแม่โจ้ ประมวลการสอนรายวิชา (Course Syllabus) ประจำปีการศึกษาที่ 2562 ภาคเรียนที่ 2

ชื่อรายวิชา ปฏิบัติการเคมีอินทรีย์ 2 รหัสวิชา คม254 จำนวนหน่วยกิตรวม 3 หน่วยกิต [ปฏิบัติการ 3 ชั่วโมง/สัปดาห์] ผู้สอน อ.ดร. ปิยธิดา กล่ำภู่ และ อ.ดร. วชิระ ชุ่มมงคล ผู้ประสานงาน อ.ดร. วชิระ ชุ่มมงคล

คำอธิบายรายวิชา การศึกษาเทคนิคปฏิบัติการ ปฏิกิริยาเคมีและการสังเคราะห์สมัยใหม่ กลไก การแยก และการระบุเอกลักษณ์ ของสารอินทรีย์ ของสารประกอบคาร์บอนิล อัลดีไฮด์ คีโตน เอมีน แอโรมาติก กรดคาร์บอกซิลิก อนุพันธ์ ฟีนอล คาร์โบไฮเดรต กรดอะมิโน และไขมัน วัตถุประสงค์รายวิชา 1. เพื่อให้นักศึกษานำความรู้จากภาคทฤษฎีมาใช้ในการอภิปรายผลจากภาคปฏิบัติได้อย่างถูกต้อง 2. เพื่อให้นักศึกษาเข้าใจเนื้อหาวิชาเคมีอินทรีย์ที่เรียนมากขึ้น โดยการเรียนรู้จากการปฏิบัติจริง 3. เพื่อให้นักศึกษารู้จักคิดวิเคราะห์ และอภิปรายผลการปฏิบัติ 4. เพื่อให้นักศึกษาเรียนรู้ปฏิกิริยาของอะโรมาติก กรดคาร์บอกซิลิกและอนุพันธ์ 5. เพื่อให้นักศึกษามีความรู้ทางปฏิกิริยาเคมีอินทรีย์ที่ทันสมัย อาจารย์ผู้สอน วันอังคาร 15.00-18.00 น. section 1 ผู้สอน อ.ดร. ปิยธิดา กล่ำภู่ วันศุกร์ 15.00-18.00 น. section 3 ผู้สอน อ.ดร. วชิระ ชุ่มมงคล แผนที่แสดงการกระจายความรับผิดชอบมาตรฐานผลการเรียนรู้สู่รายวิชา (Curriculum Mapping) วิทยาศาสตร์และคณิตศาสตร์ ทักษะ ทักษะวิเคราะห์เชิง ระหว่าง ตัวเลข การสื่อการ ความรู้ ทักษะทาง คุณธรรมและจริยธรรม บุคคลและ และการใช้ รายวิชา ปัญญา ความ เทคโนโลยี รับผิดชอบ สารสนเทศ 1 2 3 4 5 1 2 3 4 1 2 3 1 2 3 1 2 3 4 คม254 ปฏิบัติ การเคมีอินทรีย์ 2                รายละเอียดการสอนของเนื้อหาวิชา สัปดาห์ที่ หัวข้อ/รายละเอียด จำนวน กิจกรรมการ สื่อการเรียนรู้ที่ ชั่วโมง เรียนการสอน ใช้ 1 แนะนำห้องปฏิบัติการ อธิบายข้อตกลงและข้อควร 3 การบรรยาย คู่มือปฏิบัติการ (18-22 พ.ย. 62) ปฏิบัติในห้องปฏิบัติการตรวจสอบรายชื่อและ แบ่งกลุ่มปฏิบัติการ ตรวจเช็คอุปกรณ์และเครื่อง แก้ว 2 แนะนำและอธิบายวิธีการทดลอง การเก็บข้อมูล 3 การบรรยาย PowerPoint (25-29 พ.ย. 62) การคำนวณ การสรุปผล และการอภิปรายผลการ คู่มือปฏิบัติการ ทดลอง ของทุกปฏิบัติการ 3 ปฏิบัติการที่ 1 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ (2-6 ธ.ค. 62) Electrophilic Aromatic Substitution (Nitration reaction): 4 ปฏิบัติการที่ 2 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ (9-13 ธ.ค. 62) Nucleophilic Aromatic Substitution with KI อังคาร ที่ 10 ธันวาคม วันรัฐธรรมนูญ (งด) ทำแลป วันพุธ ที่ 11 ธันวาคม 11.00 น 5 ปฏิบัติการที่ 3 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ (16-20 ธ.ค. 62) Diazonium salt Coupling reaction with - naphthol 6 ปฏิบัติการที่ 4 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ (23-27 ธ.ค. 62) Carbohydrate: Isolation of lactose from Milk 7 (30 ธ.ค. 62-3 หยุดปฏิบัติการ ม.ค. 63) 8 ปฏิบัติการที่ 5 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ (6-10 ม.ค. 63) Lipid: Saponification Number สอบกลางภาค 1 13 -19 ม.ค. 2563 9 ปฏิบัติการที่ 6 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ 20-24 ม.ค. 63 The Aldol condensation 10 ปฏิบัติการที่ 7 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ 27-31 ม.ค. 63 The Claisen condensation 11 ปฏิบัติการที่ 8 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ 3-7 ก.พ. 63 Synthesis of carboxylic derivative; Aspirin 12 ปฏิบัติการที่ 9 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ 10-14 ก.พ. 63 Cannizzaro Reaction of benzaldehyde 13 ปฏิบัติการที่ 10 3 การฝึกปฏิบัติ คู่มือปฏิบัติการ 17-21 ก.พ. 63 Identification of Cannizzaro Reaction Product by FT-IR technique 14 ตรวจเช็คอุปกรณ์และเครื่องแก้ว ทบทวน 3 การบรรยาย คู่มือปฏิบัติการ 24-28 ก.พ. 63 ปฏิบัติการ 15 ทบทวนปฏิบัติการ 3 การบรรยาย คู่มือปฏิบัติการ 2-6 มี.ค. 63 สอบปลายภาค ตามประกาศมหาวิทยาลัย 9-22 มีนาคม 2563

เกณฑ์การให้คะแนน 1. สังเกตพฤติกรรมการเข้าปฏิบัติการ แบบทดสอบก่อนเรียน 5% 2. สมุดการวางแผนและบันทึกผลการทดลอง 5% 3. รายงานผลการทดลองและการเสนอรายงานกลุ่ม 55% 4. สอบปลายภาค 35%

เกณฑ์การตัดเกรด 45 + 3.5SD % ขึ้นไป ระดับคะแนน A 45 + 1.5SD % ระดับคะแนน C 45 + 3.0SD % ระดับคะแนน B+ 45 + 1.0SD % ระดับคะแนน D+ 45 + 2.5SD % ระดับคะแนน B 45+ 0.5SD % ระดับคะแนน D 45 + 2.0SD % ระดับคะแนน C+ ต่ำกว่า 45 % ระดับคะแนน F

ตำราและเอกสารประกอบการสอน 1. VOGEL’s Textbook of practical organic chemistry 5th ed. Brian S. Furniss; Antony J. Hannaford; Peter W. G. Smith; Austin R. Tatchell. 1989 2. Microscale and Macroscale Techniques in the Organic Laboratory. Domald L. Pavia; Gary M. Lampman; George S. Kriz; Randall G. Engel. 2001 3. Modern Organic Synthesis in the Laboratory: A Collection of Standard Experimental Procedures. Jie Jack Li; Chris Limberakis; Derek A. Pflum. 2007 รายงานการออกแบบรายวิชาด้วย Course Learning Outcome (CLO) วิชา คม254 ปฏิบัติการเคมีอินทรีย์ 2 หลักสูตร วิทยาศาสตร์บัณฑิต (หลักสูตรปรับปรุง พ.ศ. 2560) ตารางที่ 1 ผลการเรียนรู้ระดับรายวิชาและสอดคล้องกับ Bloom Taxonomy CLO Outcome statement Specif Gener Level ic LO ic LO 1. สามารถเข้าใจการสังเคราะห์และกลไกการเกิดปฏิกิริยาของ  U Electrophilic Aromatic Substitution ( Nitration reaction): Nucleophilic Aromatic Substitution with KI Diazonium salt Coupling reaction with -naphthol ของสารอะโรมาติกได้ 2. สามารถเข้าใจการแยกสารและปฏิกิริยาของ  U Carbohydrate: Isolation of lactose from Milk Lipid: Saponification Number 3. สามารถสังเคราะห์สารและกลไกการเกิดปฏิกิริยาของ  U The Aldol condensation และ The Cliasen condensation ได้ 4 สามารถสังเคราะห์สารและกลไกการเกิดปฏิกิริยาของ  Synthesis of carboxylic derivative; Aspirin Cannizzaro Reaction of benzaldehyde 5 สามารถยืนยันโครงสร้างสารด้วยเทคนิค FT-IR ของผลิตภัณฑ์จากปฏิกิริยา  Cannizzaro ได้ Bloom’s taxonomy: U= Remembering/Understanding, A= Applying/Analyzing, E= Evaluating/Creating ตารางที่ 2 ความสัมพันธ์ระหว่างผลการเรียนรู้กับหลักสูตรกับความต้องการของผู้มีส่วนได้ส่วนเสีย CLO รายละเอียด สกอ ผู้ใช้งาน วิชาชีพ ภาค ศิษย์ บัณฑิต สังคม เก่า 1. สามารถเข้าใจการสังเคราะห์และกลไกการเกิดปฏิกิริยาของ F F F P M Electrophilic Aromatic Substitution Nitration Nucleophilic Aromatic Substitution with KI Diazonium salt Coupling reaction with -naphthol ของสารอะโรมาติกได้ 2. สามารถเข้าใจการแยกสารและปฏิกิริยาของ F F F M M Carbohydrate: Isolation of lactose from Milk Lipid: Saponification Number 3. สามารถสังเคราะห์สารและกลไกการเกิดปฏิกิริยาของ F F F P M The Aldol condensation แ ล ะ The Cliasen condensation ได้ 4 สามารถสังเคราะห์สารและกลไกการเกิดปฏิกิริยาของ F F F P M Synthesis of carboxylic derivative; Aspirin Cannizzaro Reaction of benzaldehyde 5 สามารถยืนยันโครงสร้างสารด้วยเทคนิค FT-IR ของ F F F P M ผลิตภัณฑ์จากปฏิกิริยา Cannizzaro ได้ F-Fully fulfilled M-Moderately fulfilled P-Partially fulfilled วิทยาศาสตร์และคณิตศาสตร์ CLO ทักษะวิเคราะห์เชิง ทักษะระหว่าง ตัวเลข การสื่อการ ความรู้ ทักษะทาง บุคคลและ คุณธรรมและจริยธรรม และการใช้ ปัญญา ความ เทคโนโลยี รับผิดชอบ สารสนเทศ 1 2 3 4 5 1 2 3 4 1 2 3 1 2 3 1 2 3 4 CLO1 • •  • • • • • •   CLO2 • •  • • • • • •   CLO3 • •  • • • • • •   CLO4 • •  • • • • • •   CLO5 • •  • • • • • •  

ทำเครื่องหมาย •  ลงในช่องที่มีความสัมพันธ์กัน

การเขียนรายงาน เรื่อง...... วันที่ทดลอง...... กลุ่มที่...... ผู้เขียนรายงาน ชื่อ) ...... รหัส...... สาขา...... ผู้ร่วมงาน ชื่อ)...... รหัส...... สาขา...... ชื่อ)...... รหัส...... สาขา...... วัตถุประสงค์ ให้บอกวัตถุประสงค์ของการทดลองให้ชัดเจนว่า เพื่อต้องการหาค่าอะไร ของสารอะไร โดยวิธีไหน (ตามวิธีทำการ ทดลองจริง) ทฤษฎี (หลักการ) ให้อธิบายทฤษฎีหรือหลักการโดยย่อ พร้อมทั้งสูตรและสัญลักษณ์ ในการทดลองหรือการคำนวณ เพื่อให้ได้ผลการ ทดลองตรงตามวัตถุประสงค์ที่ต้องการ วิธีการทดลอง สมการเคมี (ถ้ามี) อธิบายวิธีทำการทดลองโดยย่อ ตามลำดับขั้นที่ทำการทดลองในห้องปฏิบัติการ ตั้งแต่เริ่มต้นจนกระทั่งถึงขั้น สุดท้ายของการทดลอง แสดงรูปประกอบ Flow Chart การคำนวณ ให้แสดงวิธีการคำนวณที่สำคัญ ให้แสดงผลลัพธ์หรือค่าเฉลี่ยของผลลัพธ์ที่ได้จากการคำนวณสารกำหนดปริมาณ, %yield, %conversion และ %recovery ผลการทดลอง ได้สารอะไร ลักษณะทางกายภาย สี ได้สารกี่กรัม สรุปและวิจารณ์ผลการทดลอง สรุปตามจุดประสงค์ ว่าได้สาร(ชื่อ) ได้สารกี่กรัม คิดเป็น กี่% การวิจารณ์ผลการทดลอง วิจารณ์ผลว่าทำไมถึงเป็นเช่นนั้น และ ทำไมถึงไม่เป็นเช่นนั้น อาจอิงตามจุดประสงค์ การทดลอง เอกสารอ้างอิง เอกสารอ้างอิง อย่างน้อย 3 เอกสาร (หนังสือหรือเวปไซด์)

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Laboratory Safety and MSDS Safety! Organic chemistry laboratory can provide social benefit, basic scientific discoveries, and intellectual satisfaction. Chemical experiments are not just fun, But can also be very hazardous, some experiments inherently so. Complacency is often observed by experience. One often forgets that chemistry is a potentially dangerous enterprise; a cavalier attitude often results in disastrous consequences. Therefore, extreme caution should be exercised at all time, especially when one handles large-scale reactions that are exothermic or when dealing with toxic chemicals.

Personal Protection Equipment 1. Safety Glasses If a chemical splashes into your eyes, it could do serious and sometimes permanent damage to your vision. The most common forms of eye protection include safety glasses (with side shields), goggles, and face shields. Prescription eye glasses are acceptable provided that the lenses are impact resistant and they are equipped with side shields. 2. Gloves Laboratory gloves are an essential part of safe laboratory practice and must be worn while handling chemicals. Despite practicing good safety techniques, tragedy may still strike. The types of gloves available and their recommended use. The manufacturer’s description of the gloves should be consulted. 3. Laboratory Coats Laboratory coats provide an important barrier for your clothes and, more important, your skin from chemicals. The laboratory coat should fit comfortably, have long sleeves and should be clean. 4. Material Safety Data Sheets (MSDS) When dealing with chemicals, caution is warranted, especially with reactive chemicals, carcinogens, and toxic reagents. A useful resource is the Material Safety Data Sheets (MSDS). There are a variety of resources that can be accessed online, including the following: MSDS Solution [www. msds.com (accessed June. 10, 2013), 2

MSDS online [www.msdsonline.com (accessed June. 10, 2013), Cornell University [www.msds.ehs.cornell.edu (accessed June. 10, 2013), Chemical suppliers/manufactures such as Sigma-Aldrich (www.sigmaaldrich.com), (accessed June. 10, 2013), www.msds.ped.go.th (accessed June. 10, 2013) 5. Useful Preparations Setting up the reaction is probably the most import job for an organic chemist. Once a reaction is initiated, there is little left that needs to be done to change the outcome.

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Experiment 1: Electrophilic Aromatic Substitution (Nitration reaction):

Aromatic substitution reactions are generally electrophilic, due to the high electron density of the benzene ring. The species reacting with the aromatic ring is usually a positive ion or the positive end of a dipole. This electron deficient species, or electrophile, may be produced in various ways, but the reaction between the electrophile and the aromatic ring is essentially the same in all cases. The overall directing and rate effects of a can be classified into three groups: 1. ortho‐para-directing activators 2. ortho-para-directing deactivators 3. meta-directing deactivators. Any substituent that activates the aromatic ring is an ortho-para director. Figure 1.1 shows the resonance forms of the arenium ion associated with a monosubstituted aromatic system. Electrophilic attack at either the ortho- or para-position placed a positive charge on the carbon. This stabilization is not possible when electrophile attacks at the meta-position

Figure 1.1 The classification of substituent aromatic

Normally, electron donating group (EDG) substituent of the aromatic ring was substituted electrophile at ortho- and para- positions. The stabilization of the positive charge can occur. In another hand, electron withdrawing group substituent of the aromatic ring was substituted electrophile at meta- positions. 4

Scheme 1.1: Electrophilic aromatic substitution and their mechanism The most common electrophilic aromatic substitution mechanism is the arenium ion mechanism, shown in above. In the first step of the reaction, benzene donates an electron pair to the electrophilic species, designated E+. A carbocation intermediate is formed, called an arenium ion or a sigma complex. This arenium ion can be written in three resonance forms. Although the arenium ion is stabilized by these resonance forms, it is destabilized by the loss of aromatic stability (~36 - 62 kcal/mole). This aromatic stability is regained in the second step of the reaction, consisting of elimination of a proton from the arenium ion, forming a substituted benzene. This process is called "re-aromatization".

Nitration is one of the most important examples of electrophilic aromatic substitution. Aromatic nitro compounds are used in products ranging from explosives to pharmaceutical + synthetic intermediates. The electrophile in nitration is the nitronium ion ( NO2 ) . The 5 nitronium ion is generated from nitric acid by protonation and loss of water, using sulfuric acid as the dehydrating agent as shown in Scheme 1.2.

Scheme 1.2: Preparation of nitronium ion

Nitration of aniline was give ortho-nitroaniline as major product and para-nitroaniline as minor product. In this experiment, you will perform a nitration of acetanilide to ortho- nitroacetanilide and para-nitroacetanilide products. Then, hydrolysis with acid to give ortho- nitroaniline and para-nitroaniline product.

Scheme 1.3: Nitration reaction of aniline

Scheme 1.4: Nitration reaction of acetanilide and hydrolysis reaction

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Procedures Part 1: Syntheses of Nitroacetanilide

1. Weigh 1.50 g of acetanilide and 3.0 ml of conc. H2SO4 in 250 ml beaker, stir on an ice bath (< 5 oC)

2. Slowly drop nitrating solution (1.5 ml of H2SO4 and 1.5 ml of HNO3) “carefully the solution temperature must below < 5 oC” and stir for 10 minute. 3. Add 100 ml of cool water into the solution. The particle of Nitroacetanilide will appear. 4. Collect the Nitroacetanilide with vacuum filtration

Part 2: Hydrolysis of Nitroacetanilide 1. Obtain all of Nitroacetanilide in 100 ml of round bottom flask. 2. Add 10 ml of HCl and 20 ml of water. 3. Reflux the solution for 15 minute. 4. Decant the solution into 100 ml of cool water (beaker) 5. Basicity with 50% NaOH (Slowly drop). The particle of Nitroaniline will appear. 6. Collect the Nitroaniline with vacuum filtration and dried with oven. Record actual weight. What was your percentage yield? 7. Explain and complete Scheme 3.4, what the major and minor of your products. Why?

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Experiment 2: Nucleophilic Aromatic Substitution with KI; Diazotization reaction

Nucleophilic aromatic substitution (SNAr) reactions offer a useful way to functionalize an aromatic ring. The high –electron density of an aromatic ring results in predominant reactivity towards electrophiles; however, if the aromatic ring is activated with electron withdrawing groups ( EWG) ortho and/ or para to a good leaving group, a nucleophilic substitution reaction is possible. Halogens are the most common leaving groups for SNAr reactions and functional groups such as - NO2, - SO2R, - NR2, - CF3 and - CN are electron withdrawing enough to render the aromatic ring susceptible to reaction with an electron-rich nucleophile, such as an amine.

Scheme 2.1: Nucleophilic aromatic substitution and their mechanism

The reaction follows an addition-elimination two-step reaction sequence. It is generally accepted that the first step, in which a tetrahedral cyclohexadienyl anion called a Meisenheimer complex is formed, is the rate-determining step (rds). This is generated by the addition of the nucleophile to the carbon bearing the leaving group. Subsequent elimination of the halogen substituent (leaving group) leads to regeneration of the aromaticity in the ring.

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Diazonium compounds or diazonium salts are a group of organic compounds sharing a common functional group R-N2+ X-. Diazonium salts, especially those where R is an aryl group, are important intermediates in the organic synthesis. The process of forming diazonium compounds is called "diazotization". The most important method for the preparation of diazonium salts is treatment of aromatic amines such as aniline with nitrous acid (HNO2) or Sodium nitrite (NaNO2). Usually the nitrous acid is generated in situ (in the same flask) from sodium nitrite and mineral acid. In aqueous solution diazonium salts are unstable at temperatures above +5 oC; the –N2+group tends to be lost as N2 (nitrogen gas). Often, diazonium compounds are not isolated and once prepared, used immediately in further reactions. This approach is illustrated in the preparation of an aryl halide compound.

Scheme 2.2: Nucleophilic aromatic substitution of diazonium salt with small nucleophile

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Procedures Part 1: Preparation of Diazonium salt 1. Weigh about 0.50 g of p-nitroaniline, 5 ml of conc. HCl and 10 ml of water in 250 ml beaker, stir on an ice bath (< 5 oC)

2. Add solution of 0.5 g NaNO2/ 2 ml water. Carefully the solution temperature must below 5 oC

Part 2: Synthesis of p-iodonitrobenzene 1. Slowly drop solution of 1.0 g KI in 4 ml water in to Diazonium salt. continuous stirring the solution until no bubble appear. Carefully the solution temperature must below 5 oC 2. Collect the crude of p-iodonitrobenzene with vacuum filtration.

Part: 3 Crystallization of p-iodonitrobenzene 1. Dissolve the crude of p-iodonitrobenzene with hot Ethanol (EtOH) on hot plate. until the solid completely dissolves. 2. Remove from the hot plate. Allowing the hot mixture to slowly cool to room temperature and cooling further in an ice bath. Crystals will appear. If not, remove the solvent with hot plate. 3. Collect the caffeine crystal with vacuum filtration. Record actual weight. What was your percentage yield of p-iodonitrobenzene? 4. Explain, what the major of your products. Why?

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Experiment 3: Diazonium Salt coupling with -naphthol: Diazo coupling

An azo coupling is an organic reaction between a diazonium compound and another aromatic compound that produces an azo compound. In this electrophilic aromatic substitution reaction, the aryl diazonium cation is the electrophile and the activated aromatic is a nucleophile. The coupling product results from reaction of the terminal diazonium nitrogen atom at the para- or ortho-position to the activated aromatic.

Scheme 3.1: Diazo coupling with bulky nucleophile Products of diazonium coupling reactions are normally highly conjugated, colored compounds and dye. Important azo dyes include methyl red and pigment red. As shown in scheme 3.2.

Scheme 3.2: Diazo coupling with bulky nucleophile 11

One example is the synthesis of the dye. Para red can produce from aniline and - naphthol couples with phenyl diazonium electrophile to produce an intense orange-red dye.

Scheme 3.3: Diazo coupling between Diazonium ion and -naphthol

Procedures Part 1: Preparation of Diazonium salt 1. Weigh about 0.50 g of p-nitroaniline, 5 ml of conc. HCl and 10 ml of water in 250 ml beaker, stir on an ice bath (< 5 oC)

2. Add solution of 0.5 g NaNO2/ 2 ml water. Carefully the solution temperature must below 10 oC

Part 2: Syntheses of para red 1. Slowly drop solution of 0.2 g of -naphthol in 4 ml of 10% NaOH in to Diazonium salt. Continuous stirring the solution until no bubble appear. Carefully the solution temperature must below 10 oC and stir for 10 minute. 2. Add dilute HCl till the solution turn to acidity. 3. Collect the para red with vacuum filtration and dried with oven. Record actual weight. What was your percentage yield? 4. Explain and complete Scheme 3.4, what the major and minor of your products. Why?

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Experiment 4: Carbohydrate: Isolation of lactose from Milk

Milk is a food of exceptional interest. Not only is milk an excellent food for the very young, but humans have also adapted milk, specifically cow’s milk, as a food substance for persons of all ages. Many specialized milk products like cheese, yogurt, butter, and ice cream are staples of our diet. Milk is probably the most nutritionally-complete food that can be found in nature. This property is important for milk, since it is the only food young mammals consume in the nutritionally significant weeks following birth. Whole milk contains vitamins (principally thiamine, riboflavin, pantothenic acid, and vitamins A, D, and K), minerals (calcium, potassium, sodium, phosphorus, and trace metals), proteins (which include all the essential amino acids), carbohydrates (chiefly lactose), and lipids (fats).

The only important elements in which milk is seriously deficient are iron and Vitamin C. Infants are usually born with a storage supply of iron large enough to meet their needs for several weeks. Vitamin C is easily secured through an orange juice supplement. The average composition of the milk of each of several mammals is summarized in the accompanying table. Table 6.1 Average percentage composition of Milk from Various Mammals Cow Human Goat Sheep Horse Water 87.1 87.4 87.0 82.6 90.6 Protein 3.4 1.4 3.3 5.5 2.0 Fats 3.9 4.0 4.2 6.5 1.1 Carbohydrates 4.9 7.0 4.8 4.5 5.9 Minerals 0.7 0.2 0.7 0.9 0.4

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Fats Whole milk is an oil-water type of emulsion, containing about 4% fat dispersed as very small (5-10 microns in diameter) globules. The globules are so small that a drop of milk contains about a million of them. Because the fat in milk is so finely dispersed, it is digested more easily than fat from any other source. The fat emulsion is stabilized to some extent by complex phospholipids and proteins that are adsorbed on the surfaces of the globules. Skimmed milk, except for lacking the fats and vitamins A and D, has approximately the same composition as whole milk. If milk is homogenized, its fatty content will not separate. Milk is homogenized by forcing it through a small hole. This breaks up the fat globules and reduces their size to about 1 to 2 microns in diameter.

Proteins Proteins may be classified broadly in two general categories: fibrous and globular. Globular proteins are those that tend to fold back on themselves into compact units that approach nearly spheroidal shapes. These types of proteins do not form intermolecular interactions between protein units (H bonds and so on) as fibrous proteins do, and they are more easily solubilized as colloidal suspensions. There are three kinds of proteins in milk: caseins, lactalbumins, and lactoglobulins. All are globular.

Casein is a phosphoprotein, which has phosphate groups are attached to some of the amino acid side- chains. These are attached mainly to the hydroxyl groups of the serine and threonine moieties. Actually, casein is a mixture of at least three similar proteins, which differ primarily in molecular weight and amount of phosphorus they contain (number of phosphate groups). Casein exists in milk as the calcium salt, calcium caseinate. This salt has a complex structure. It is composed of α,  and κ-caseins which form a micelle, or a solubilized unit. Neither the α nor the  casein is soluble in milk, singly or in combination. If κ casein is added to either one, or to a combination of the two, however, the result is a casein complex that is soluble owing to the formation of the micelle.

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A structure proposed for the casein micelle is shown on the following page. The κ casein is thought to stabilize the micelle. Since both α and  casein are phosphoproteins, they are precipitated by calcium ions.

Calcium caseinate has its isoelectric (neutrality) point at pH 4.6. Therefore, it is insoluble in solutions of pH less than 4.6. The pH of milk is about 6.6; therefore casein has a negative charge at this pH and is solubilized as a salt. If acid is added to milk, the negative charges on the outer surface of the micelle are neutralized (the phosphate groups are protonated) and the neutral protein precipitates:

2+ Ca Caseinate + 2HCl Casein + CaCl2 The calcium ions remain in solution. When milk sours, lactic acid is produced by bacterial action (see below), and the consequent lowering of the pH causes the same clotting reaction. The isolation of casein from milk will be carried out in this experiment.

The casein in milk can also be clotted by the action of an enzyme called rennin. Rennin is found in the fourth stomach of young calves. However, both the nature of the clot and the mechanism of clotting differ when rennin is used. The clot formed using rennin, calcium paracaseinate, contains calcium.

Ca2+Caseinate + rennin Ca2+Paracaseinate + a small peptide Rennin is a hydrolytic enzyme (peptidase) and acts specifically to cleave peptide bonds between phenylalanine and methionine residues. It attacks the κ casein, breaking the peptide chain so as to release a small segment of it. This destroys the water-solubilizing surface of the κ casein, which protects the inner α and  caseins and causes the entire micelle to precipitate as calcium paracaseinate. Milk can be decalcified by treatment with oxalate ion, 15 which forms an insoluble calcium salt. If the calcium ions are removed from milk, a clot will not be formed when the milk is treated with rennin.

The clot, or curd, formed by the action of rennin is sold commercially as cottage cheese. The liquid remaining is called the whey. The curd can also be used in producing various types of cheese. It is washed, pressed to remove any excess whey, and chopped. After this treatment, it is melted, hardened and ground. The ground curd is then salted, pressed into molds, and set aside to age. CARBOHYDRATES When the fats and the proteins have been removed from milk, the carbohydrates remain, as they are soluble in aqueous solution. The main carbohydrate in milk is lactose. Lactose, a disaccharide, is the only carbohydrate that mammals synthesize. It is synthesized in the mammary glands. Hydrolyzed, it yields one molecule of D-glucose and one of D- galactose.

Lactose is an example of a disaccharide. It is made up of two sugar molecules: galactose and glucose. In the following structures, the galactose portion is on the left and glucose is on the right. Galactose is bonded through an acetal linkage to glucose

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Notice that the glucose portion can exist in one of two isomeric hemiacetal structures: α- lactose and - lactose. Glucose can also exist in a free aldehyde form. This aldehyde form (open form) is an intermediate in the equilibration (interconversion) of α and  -lactose. Very little of this free aldehyde form exists in the equilibrium mixture. The isomeric α and - lactose are diastereomers because they differ in the configuration at one carbon atom, called the anomeric carbon atom.

The sugar α -lactose is easily obtainable by crystallization from a water-ethanol mixture at room temperature. On the other hand, -lactose must be obtained by a more difficult process, which involves crystallization from a concentrated solution of lactose at temperatures about 93.5°C . In the present experiment, α-lactose is isolated by the simpler experimental procedure indicated above. α- Lactose undergoes numerous interesting reactions. First, α-lactose interconverts, via the free aldehyde form, to a large extent, to the -isomer in aqueous solution. This causes a change in the rotation of polarized light from +92.60o to +52.30o with increasing time. The process that causes the change in optical rotation with time is called mutarotation.

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Procedures Part I: Isolation of Lactose 1. Weigh out 5.00 g of non-fat powdered milk. Dissolve the milk in 100 mL of warm water in a 250mL beaker. 2. Heat the milk solution to 40 oC. 3. Slowly drop 6 mL of 10% Acetic Acid to precipitate out the Casein, and stir the mixture slowly and briefly. Avoid adding excess Acetic Acid. Work the Casein into a mass and remove it with a stirring rod or spoon. 4. Decant the solution and Filter with cotton wool. 5. Add 1.25 g of powdered Calcium Carbonate. Stir the mixture thoroughly . 6. Heat the mixture almost to boiling for about 10 minutes; stirring continuously. This should precipitate the remaining proteins. 7. Filter the hot mixture with cotton wool, collecting the filtrate in a 250mL beaker. 8. Stir the filtrate continuously while boiling it down to about 7.5 mL, and then add 50 mL of 95% Ethanol. 9. Filter the warm Ethanol solution with vacuum, collecting the filtrate in a 125mL Erlenmeyer flask. Stopper the flask and place it in your lab drawer until the next lab period. 10. Filter off the crystals of Lactose. Allow them to dry. Weigh the crystals. 11. Compare your percentage recovery of Lactose with the percentage carbohydrate as reported on the nutritional label of the powdered milk you used.

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Experiment 5: Lipid; Biodiesel; Saponification Fats and oils are the principle stored forms of energy in many organisms. They are highly reduced compounds and are derivatives of fatty acids. Fatty acids are carboxylic acids with hydrocarbon chains of 4 to 36 carbons, they can be saturated or unsaturated. The simplest lipids constructed from fatty acids are triacylglycerols or triglycerides. Triacylglycerols are composed of three fatty acids each in ester linkage with a single glycerol. Since the polar hydroxyls of glycerol and the polar carboxylates of the fatty acids are bound in ester linkages, triacyl glycerols are non-polar, hydrophobic molecules, which are insoluble in water.

Saponification is the hydrolysis of fats or oils under basic conditions to afford glycerol and the salt of the corresponding fatty acid. Saponification literally means "soap making". It is important to the industrial user to know the amount of free fatty acid present, since this determines in large measure the refining loss. The amount of free fatty acid is estimated by determining the quantity of alkali that must be added to the fat to render it neutral. This is done by warming a known amount of the fat with strong aqueous caustic soda solution, which converts the free fatty acid into soap. This soap is then removed and the amount of fat remaining is then determined. The loss is estimated by subtracting this amount from the amount of fat originally taken for the test. 19

The saponification number is the number of milligrams of potassium hydroxide required to neutralize the fatty acids resulting from the complete hydrolysis of 1g of fat. Saponification number = mg of KOH consumes by 1g of fat

It gives information concerning the character of the fatty acids of the fat- the longer the carbon chain, the less acid is liberated per gram of fat hydrolyzed. It is also considered as a measure of the average molecular weight (or chain length) of all the fatty acids present. The long chain fatty acids found in fats have low saponification value because they have a relatively fewer number of carboxylic functional groups per unit mass of the fat and therefore high molecular weight. The average molecular weight (MW) of fat can calculate from the saponification number.

The average molecular weight of fat = 3 x Saponification number

Table 10.1: the saponification number of lipids Lipids Saponification number Coconut oil 248-285 Palm oil 190-209 Olive oil 190-195 Soy oil 189-194

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Procedures Saponification and titration 1. Weigh 4.0 g of fat (Sample X) in 250 mL of round-bottom flask A. And blank (not thing) in 250 mL of round-bottom flask B. 2. Slowly add 50.0 mL of 0.5N ethanolic KOH to both the flasks

(Weigh 1.4 g of KOH dissolve in EtOH : H2O/ 3 : 1) and stir for 5 minutes. 3. Reflux and stir the solution for 30 minute, until the solution no longer has two separate layers. 4. Cool the flasks to room temperature 5. Add phenolphthalein indicator to both the flasks and titrate with 0.5N HCl. 6. Note down the endpoint of test (flask A) and blank (flask B). 7. Calculate the saponification value using the formula

8. Compare your result with table 10.1. Explain why? 9. Show calculation of the average molecular weight (MW) of your fat.

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Experiment 6: The Aldol Condensation

Aldol condensation combines both Nucleophilic addition and Condensation reaction. It is a very useful synthetic reaction for generating new carbon-carbon bond and double bond. In an Aldol reaction, it connects between alpha carbon of one molecule and the carbon of carbonyl of the second molecule. This forms a beta hydroxyl carbonyl compound. Sometime it is followed by dehydration to get alpha-beta unsaturated carbonyl compound (Aldol products).

Scheme 4.1: The aldol condensation reaction of acetaldehyde Mechanistically, the aldol condensation reaction is a simple nucleophilic addition reaction between an enolate ion and a carbonyl group. The enolate is formed when the base (NaOH) present abstracts an α-Hydrogen of a carbonyl one. This Hydrogen is particularly acidic photon because of the resonance stabilization of the resulting enolate. This enolate nucleophilically attacks carbonyl groupings in the typical fashion, forming the aldol product.

Scheme 4.2: The resonance form of enolate anion In this experiment, we will synthesize the substitute dibenzalacetone from crossed Aldol condensation of acetone and two molecule of a substituted benzaldehyde.

Scheme 4.3: The aldol condensation reaction of substituted benzaldehyde 22

Procedure Synthesis of Substituted dibenzalacetone 1. A: Place1.4 mL (1.0 eq., 1.458 g) of Benzaldehyde and 0.5 mL (0.5 eq., 0.4 g) of acetone and 10 mL of ethyl alcohol in a 50 mL Erlenmeyer flask B: Place1.4 mL (1.0 eq., 1.567 g) of anisaldehyde and 0.4 mL (0.5 eq., 0.34 g) of acetone and 10 mL of ethyl alcohol in a 50 mL Erlenmeyer flask 2. Stir the mixture for 3 minutes in an ice water bath. 3. Slowly add 0.5 g of sodium hydroxide (NaOH) in 2 mL of water. Stir for 5 minutes. 4. Remove the ice water bath. Stir the mixture at room temperature for another 10 minutes. The reaction mixture should at first be clear and homogeneous, but should shortly become milky with solid particles of the product. 5. Add the ice and stir until the precipitation occurs. 6. Collect the precipitate by vacuum filtration. 7. Wash the filtrate with 150 mL of distilled water to remove the basicity. 8. Transfer the precipitate into an Erlenmeyer flask. 9. Recrystallize the product using ethyl alcohol. “dissolve at hot but non-dissolve at cool” 10. Collect the crystal with vacuum filtration. Allow the solid to dry completely 11. Determine weight and %yield of the product. 12. Determine the melting point. Explanations! 1. What is the configuration (cis-cis or trans –tran or cis-trans) of the product? “The trans-tran-form melts at 110-111 oC; The cis-trans-form melts at 60 oC The cis-cis-form is an oil” 2. What is the mechanism of the product? 3. Draw the structure of your product in stable-form.

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Experiment 7: The Claisen Condensation

One of the important enolate condensation reactions is the Claisen condensation, which involves reaction of an enolate with the carbonyl compound of an ester. The enolate is generated by removal of a slightly acidic hydrogen from the a-carbon of a ketone, nitrile or ester using a relatively strong base. The enolate then attacks the carbonyl carbon of an ester to yield an intermediate which undergoes rapid loss of the alkoxide leaving group to yield a b-dicarbonyl compound. As usual, all steps in the sequence are completely reversible; the reaction is driven to the product side by deprotonation of the relatively acidic product by the alkoxide ion produced in the condensation. The final product is therefore a new enolate, which can be extracted from the reaction mixture by washing with water, in which it is soluble because it is a salt. It is finally liberated by acidifying the aqueous extract.

Scheme 5.1: The Claisen condensation reaction of ester Mechanistically, Claisen condensation reaction is a simple nucleophilic addition reaction between an enolate ion and a carbonyl group. The enolate is formed when the base (NaH) present abstracts an α-Hydrogen of a carbonyl one. This Hydrogen is particularly acidic photon because of the resonance stabilization of the resulting enolate. This enolate nucleophilically attacks carbonyl groupings in the typical fashion, forming the claisen product.

Scheme 5.2: The resonance form of enolate anion In this experiment, we will synthesize the substitute benzoyl ester from crossed Claisen condensation of methyl acetate and two molecule of a substituted benzoyl ester. 24

Scheme 5.3: The Claisen condensation reaction

Procedure Synthesis of Substituted dibenzalacetone 1. Place1.4 mL (1.0 eq., 1.458 g) of Benzoyl ester and 0.5 mL (1.0 eq.,0.4 g) of methyl acetate and 10 mL of ethyl alcohol in a 50 mL Erlenmeyer flask 2. Stir the mixture for 3 minutes in an ice water bath. 3. Slowly add 0.5 g of sodium hydroxide (NaH) in 2 mL of water. Stir for 5 minutes. 4. Remove the ice water bath. Stir the mixture at room temperature for another 10 minutes. The reaction mixture should at first be clear and homogeneous, but should shortly become milky with solid particles of the product. 5. Add the ice and stir until the precipitation occurs. 6. Collect the precipitate by vacuum filtration. 7. Wash the filtrate with 150 mL of distilled water to remove the basicity. 8. Transfer the precipitate into an Erlenmeyer flask. 9. Recrystallize the product using ethyl alcohol. “dissolve at hot but non-dissolve at cool” 10. Collect the crystal with vacuum filtration. Allow the solid to dry completely 11. Determine weight and %yield of the product. 12. Determine the melting point. Explanations! 1. What is the mechanism of the product? 2. Draw the structure of your product in stable-form.

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Experiment 8: Synthesis of carboxylic derivative; Aspirin

Aspirin (acetylsalicylic acid) is a synthetic organic derived from salicylic acid. Salicylic acid is a natural product found in the bark of the willow tree and was used by the ancient Greeks and Native Americans, among others, to counter fever and pain. However, salicylic acid is bitter and irritates the stomach.

Scheme 6.1: aspirin structure and pills

A German chemist named Felix Hoffman is credited with being the first to synthesize aspirin in 1897. Hoffman's father had severe arthritis but could not tolerate salicylic acid he was taking for pain relief. The name given for Hoffman's new compound was A-spirin. Apparently this comes from acetylation (A-), together with Spirin, part of the name for Meadow-sweet (Spiraea ulmaria), a plant rich in salicylates.

Friedrich Bayer, the employer of Hoffman, patented the name and began marketing the product in 1899. It was a huge success and sales grew rapidly. Bayer's company set up by himself, is generally reckoned to have been the first pharmaceutical company, and the production of aspirin is generally accepted to have laid the foundation of the modern pharmaceutical industry.

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In this experiment you will synthesize aspirin (acetylsalicylic acid, C9H8O4), purify it, and determine the percent yield. The purity of the product will be confirmed by qualitative analysis and by measuring its melting point range.

Scheme 6.2: synthesis of acetyl salicylic acid The reaction that is used for the synthesis is shown below. In this reaction, an excess of acetic anhydride (C4H6O3) is added to a measured mass of salicylic acid (C7H6O3) in the presence of a catalyst, sulfuric acid (H2SO4). The mixture is heated to form the acetylsalicylic acid (C9H8O4) and acetic acid (C2H4O2). After the reaction takes place, water is added to destroy the excess acetic anhydride and cause the product to crystallize. The aspirin is then collected, purified by recrystallization, and its melting temperature measured.

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Procedure Synthesis of Aspirin 1. Staring Materials (1.0 eq., 5.00 g) of salicylic acid and 7.0 mL of acetic acid from graduated cylinder in 125 mL Erlenmeyer flask. And 5 drops of sulfuric acid (catalyst) to the flask. ……Be sure to do this in the hood and wearing your goggles…… ……Don't let the acetic acid contact your skin and don't get the vapors in your eyes… 2. Reflux the mixture for 15 minutes. 3. Slowly add 20 mL of distilled water to the flask. Swirl the flask to mix the contents. 4. Place it in an ice bath and cool until the crystallization of the aspirin appears complete (approx. ? min.). If crystals do not appear, you can scratch the walls of the flask with a stirring rod to induce crystallization. 5. Collect the precipitate by Buchner funnel on a vacuum filtration. Wash the filtrate with cool distilled water. 6. Transfer the precipitate into an Erlenmeyer flask. Recrystallizing the product using ethanol. “Dissolve at hot but non-dissolve at cool” 7. Collect the crystal with vacuum filtration. Allow the solid to dry completely 8. Determine weight and %yield of the product.

Explanations! 1. Propose the mechanism of the products? 2. What is the possible products? (3 structures) 3. What is the acetyl derivative can use to prepare aspirin?

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Experiment 9: Cannizzaro Reaction of benzaldehyde

As a general rule, nucleophilic addition reactions are characteristic only of aldehydes and ketones, not of carboxylic acid derivatives. The reason for the difference of is structural; the tetrahedral intermediate produced by addition of a nucleophile to a carboxylic acid derivative can eliminate a leaving group, leading to a net nucleophilic acyl substitution reaction. The tetrahedral intermediate produced by addition of a nucleophile to an aldehyde or ketone, however, has only alkyl or hydrogen and thus can't usually expel a leaving group. One exception to this rule, however, is the Cannizaro reaction, discovered in 1853. The Canizzaro reaction takes place by nucleophilic addition of OH- to an unenolizable aldehyde (bearing no −H) to give a tetrahedral intermediate, which expels hydride ion as a leaving group and is thereby oxidized. A second aldehyde molecule accepts the hydride ion in another nucleophilic addition step and is thereby reduced

Scheme 7.1: Cannizzaro Reaction of benzaldehyde The Cannizzaro reaction is that of aldehydes that do not contain alpha hydrogens to give carboxylic acids and alcohols (alpha hydrogens cause an Aldol reaction to take place). This occurs in the presence of a strong base. Benzaldehyde, which does not contain alpha hydrogens, was used for this reaction.

The reflux technique will be used in order to allow for a complete reaction between all the components. On completion of the reaction, the extraction technique will enable the separation of benzoic acid and benzyl alcohol. This will be achieved by the ionization of 29

Benzoic acid which will separate from the organic benzyl alcohol. Reacidification will reform the Benzoic acid.

Procedure 1. Place about 3.0 ml of benzaldehyde, 15 ml of methanol and 2 ml of aqueous 10 M KOH in a 100 ml round bottom flask; add a couple of boiling chips finally. 2. Reflux the reaction mixture and heat the reaction mixture to the boiling point of methanol for 30 minute. Allow the mixture to cool down and place in an ice bath.

3. Transfer the mixture carefully to a separator funnel. Addition 30 ml of CH2Cl2 and Extraction two time with 30 ml of water 4. The aqueous phase was collected in Flask A. and the aqueous phase was work up with HCl to appear crystal. 5. Collect the crystal with vacuum filtration. Allow the solid to dry completely, Determine weight and %yield of the product A.

Organic Phase Work-up:

1. The CH2Cl2 phase was collected in Flask B. it was dries over by anhydrous MgSO4. Add enough drying agent so, the solution appears clear as opposed as cloudy.

2. Rinse the CH2Cl2 to beaker (weight).

3. In the hood, evaporate the CH2Cl2 by using a hot plate. 4. Determine the yield of the oily residue of the product B. ------Then take the product A and B ot IR spectroscopy------

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Scheme 7.2 the synthesis and extraction of cannizzaro reaction 31

Experiment 10: Identification of Cannizzaro Reaction Product by FT-IR technique

Infrared spectroscopy is certainly one of the most important analytical technique available to today’s scientists. Infrared spectroscopy is a technique based on the vibrations of the atoms of a molecule. An infrared spectrum is commonly obtained by passing infrared radiation through a sample and determining what fraction of the incident radiation is absorbed at a particular energy. The energy at which any peak in an absorption spectrum appears corresponds to the frequency of a vibration of a part of a sample molecule. In this introductory chapter, the basic ideas and definitions associated with infrared spectroscopy will be described. The vibrations of molecules will be looked at here, as these are crucial to the interpretation of infrared spectra. This equation may be modified so that direct use of the wavenumber values for bond vibrational frequencies can be made, namely:

The two primary modes of vibration are stretching and bending Stretching modes are typically of higher energy than bending modes Stretching modes are often divided into two a symmetric and asymmetric stretch; the asymmetric stretch is usually of higher energy

Scheme 8.1 the mode of vibration in organic compounds

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A second and more important use of the infrared spectrum is Functional Groups of a molecule. The absorptions of each type of bond (N−H, C−H, O−H, C−X, C−O, C=O, C−C, C=C, CC, CN, and so on) are regularly found only in certain small portions of the vibrational infrared region. A small range of absorption can be defined for each type of bond. The same type of range applies to each type of bond. Scheme 8.1 illustrates schematically how these are spread out over the vibrational infrared. Try to fix this general scheme in your mind for future convenience.

Scheme 8. 2 The approximate regions where various common types of bonds absorb (stretching vibrations only; bending, twisting, and other types of bond vibrations have been omitted for clarity).

In general, the frequencies range of vibration stretching of single, double and triple bonds between two atoms are following bonds C−H 2900-3300 cm-1 -1 C−C 1200 cm-1 C=C 1650 cm-1 CC 2150 cm-1 C−O 1100 cm-1 C=O 1800 cm-1 CN 2250 cm-1 O−H 3200-3600 cm-1 N−H 3300 cm-1

The Canizzaro reaction for previous experiment benzoic acid and phenyl alcohol were produced. The infrared (IR) absorption occurs from the stretching and bending of the covalent bonds in molecules to be accompanied by IR absorption a stretch or bend must change the dipole moment of the molecule 33

Sample Preparation for IR Work Three major methods of sample preparation 1. Sample is mixed with a mulling agent such mineral oil and pressed between plates made of Sodium chloride is used because it has no IR absorptions; glass or plastic plates would have IR absorptions of their own Sodium chloride plates are good from 4000 to 650 cm-1; below 650 cm-1 they begin to absorb Potassium bromide plates can be used in place of sodium chloride and are transparent to 400 cm-1, but they are more expensive. Downside of this method is absorptions due to the mineral oil 2. Sample is mixed with solid potassium bromide and pressed into a pellet under high pressureNo absorptions from mulling agent. Only works for solids 3. Sample is dissolved in carbon tetrachloride and pressed between salt plates

Downside of this method is absorptions due to the CCl4

Procedure Sample Preparation 1. The product A and B were prepared for FT-IR spectrometer. 2. Try to assign at least three of the most prominent bands in the spectrum. Are they consistent with the Functional Group you believe is present in your sample? 3. The laboratory work involves identification of an unknown by recording its infrared spectrum, investigating the major absorption bands, and comparing the spectrum product A and B with spectra of benzoic acid and phenyl alcohol.

Explanations! 1. Give the mechanism of the reaction and complete Scheme 7.1 2. How would you isolate the alcohol from the organic layer that contains also the portion of unreacted aldehyde? 3. Calculate the mass of unreacted benzaldehyde.