Pharmaceutical and Medicinal Organic Chemistry Lab - Week 1 PDF

PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO Excess reactant WEEK 1 - The reactant in a chemical reaction that STOICHIOMET...

PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO Excess reactant WEEK 1 - The reactant in a chemical reaction that STOICHIOMETRY remains when a reaction stops when the limiting reactant is completely consumed. STOICHIOMETRY. Involves relationship between reactants and/or products in chemical reaction to determine quantitative data 8 car bodies + 48 tires = 8 cars and 16 tires excess side notes: - its the field of chemistry that is concerned Side note: with a relative quantities of a reactant and products in a chemical reaction - in this picture you will see 8 cars bodies plus - it is founded in a law of conservation of mass 48 cars. in every car you only need 4 tires for where the total mass of reactant equals the the 8 cars you only need 32 tires to complete total mass of the products leading to the the 8 cars and there will be a remaining 16 insight that the relations among the tires in excess. no matter how many tires are quantities of the reactants and products there if there are only 8 cars bodies then only typically formed a ratio of positive integer. 8 cars can be made this means that if the amount of the separate IN CHEMISTRY reactants are known the amount of the products can be calculated conversely if the - If there is a certain only amount of one one reactant has known quantity and the reactant is available for reaction the reaction quantity of the product can be empirically must stop when that reactant is consumed determined then the amount of other whether or not the other reactant has been reactant can also be calculated used up. STEPS 1. Balance the chemical equation 2. Find the limiting reagent Chemical reaction in real the word don’t go exactly as 3. Find the theoretical yield planned on paper in a course of experiment many 4. Find the actual yield things will contribute to the formation of less product 5. Find the percentage yield that with be the predicted. Besides pills and experimental errors there are usually losses due to incomplete reaction, undesirable side reactions and Limiting reactant etc.. Therefore, we need a measurement that - The reactant in a chemical reaction that limits indicates how successful are reaction has been, this the amount of product that can be formed. measurement is called the PERCENTAGE YIELD The reaction will stop when all of the limiting reactants is consumed. Percentage yield U.S. SIDDIQUE & M.K. JOSE | PH3YB-1 Page | 1 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO PERCENTAGE YIELD FORMULA = 1. Balance the chemical equation 𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑥100 C= 2 C= 1 x2 = 2 - Ratio of actual yield to theoretical yield H= 4 H= 2 x 2 = 4 expressed in percentage O= 2 x 3 = 6 O= 3 = 5 = 6 Theoretical yield C2H4 + 3O2 🡪 2CO2 + 2H2O After balancing, second step we need to look for their - Amounts of products calculated from the complete reaction of the limiting reagent molecular weight because our goal is to convert grams → moles → grams, because based on the problem the given is in grams and then your products, Actual yield or the required unit for product is also in grams so we - Amount actually produced of a product need to convert grams → moles → grams SIde notes: - To compute for the percentage yield it is necessary to determine how much of the C2- 12g x 2= O- 16g x 2= C- 12 x 1 = 12g product should be formed based on 246g 16g O2 - 16 x 2 = stoicheometry and that is called the H4- 1g x 4= 32g THEORETICAL YIELD - The maximum amount 4g of products that could be formed from the given amounts of the reactants Total= 28g/n Total= 32g/n Total = 44g/n - The ACTUAL YIELD - is the amount of the product that is actually formed when the reaction is carried out in laboratory. so lets start with ethylene: - 4.95𝑔 𝐶2𝐻4 1𝑚𝑜𝑙 𝐶2𝐻4 2𝑚𝑜𝑙 𝐶𝑂2 44𝑔 𝐶𝑂2 x x 1𝑚𝑜𝑙 𝐶2𝐻4 x 1𝑚𝑜𝑙 𝐶𝑂2 =15g CO2 Balance chemical equation 28𝑔 𝐶2𝐻4 Al + Cl2 → AlCl3 So next we need to find how much CO2 are formed in BaCl2 + Al2(SO4)+3 → BaSO4 + AlCl3 O2 C3H8 + O2 🡪 CO2 + H2O 3.25𝑔 𝑂2 1𝑚𝑜𝑙 𝑂2 2𝑚𝑜𝑙 𝐶𝑂2 44𝑔 𝐶𝑂2 x 32𝑔 𝑂2 x 3𝑚𝑜𝑙 𝑜2 x 1𝑚𝑜𝑙 𝐶𝑂2 = 2g CO2 CH3 NH2 + O2 🡪 CO2 + H2O + N2 2. Find the limiting reagent C6H6 + O2 🡪 CO2 + H2O As mentioned earlier the limiting reactant —--------------------------------------------------------------------- will be the reactant with the lowest 1. If 4.95 g of ethylene (C2H4) are combusted number of product that is formed with 3.25 g of oxygen. - OXYGEN IS THE LIMITING REACTANT a. What is the limiting reagent? C2H4 + O2 🡪 CO2 + H2O U.S. SIDDIQUE & M.K. JOSE | PH3YB-1 Page | 2 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO 3. Find the theoretical yield: How many grams of CO2 3. Determine the grams of water from are formed? propane - So whatever the value of limiting reagent is that will be the value of theoretical yield so Note: Convert from grams to moles to grams our theoretical yield is ( g→ moles → g ) = 2 GRAMS OF CARBON DIOXIDE 12𝑔 𝐶3𝐻8 1𝑚𝑜𝑙 𝐶3𝐻8 4𝑚𝑜𝑙 𝐻2𝑂 18𝑔 𝐻2𝑂 x 44𝑔 𝐶3𝐻8 x 1𝑚𝑜𝑙 𝐶3𝐻8 x 1𝑚𝑜𝑙 𝐻2𝑂 =19g H2O SIDE NOTES: here in this problem the reactants are Final Answer: 19g of H2O ethylene and oxygen and the products are —------—------------------------------------------------------------ carbon dioxide and water 3. If 5 g of salicylic acid reacts with 10 mL of acetic —--------------------------------------------------------------------- anhydride, how grams of ASA are formed? 2. If 12 g of propane is burned in excess oxygen, how many grams of water are formed? C3H8 + O2 🡪 CO2 + H2O 1. Check if balance ang chemical Salicylic acid + Acetic anhydride→ Aspirin (ASA) + Acetic ā equation C7H6O3 + C4H603 → C9H8O4 + C2H4O2 Write first the number of atoms 1. Check if balance ang chemical equation present in both sides C= 3 C= 1 C= 7+4= 11 C= 9+2= 11 H= 8 H= 2 H= 6+6= 12 H= 8+4= 12 O= 2 O= 2 + 1 = 3 O= 3+3= 6 O= 4+2= 6 2. Pag nabalance na, Compute for the Then balance the equation Molecular weight ng bawat molecule na C3H8 + 5O2 🡪 3CO2 + 4H2O kailangan 2. Pag nabalance na, Compute for the Salicylic Acid Molecular weight ng bawat molecule C7H6O3 na kailangan. C7= 12g x 7= 84g H6= 1g x 6= 6g O3= 16g x3= 48g C3- 12g x 3= 36g H2- 1g x 2= 2g Total= 138g/n H8- 1g x 8= 8g O- 16g x 1= 16g Acetic anhydride Total= 44g/n Total= 18g/n C4H6O3 C4= 12g x 4= 48g n= “mol” ang pagbasa ni maam H6= 1g x 6= 6g O3= 16g x 3= 48g Total=102g/n U.S. SIDDIQUE & M.K. JOSE | PH3YB-1 Page | 3 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO Aspirin —--------------------------------------------------------------------- C9H8O4 4. In an experiment, 378 g of Aspartame (C14 H18N2O5) C9= 12g x 9= 108g react with water and produced aspartic acid H8= 1g x 8= 8g (C4H7NO4), methanol (CH3OH) and phenylalanine O4= 16g x 4= 64g (C9H11NO2) Total= 180g/n C14 H18N2O5 + H2O🡪 C4H7NO4 + CH3OH + C9H11NO2 n= “mol” ang pagbasa ni maam 1. Check if balance ang chemical equation C= 14 C= 4+1+9= 14 3. Determine the grams of ASA that will be H= 18+(2x2)= 22 H= 7+4+11= 22 produced from salicylic acid and acetic N= 2 N= 2 anhydride O= 5+(2x1)= 7 O= 7 C14 H18N2O5 + 2H2O🡪 C4H7NO4 + CH3OH + C9H11NO2 Note: Convert from grams to moles to grams ( g→ moles → g ) 2. Pag nabalance na, Compute for the Molecular weight ng bawat molecule na kailangan Salicylic Acid (SA): 5𝑔 𝑆𝐴 1𝑚𝑜𝑙 𝑆𝐴 1𝑚𝑜𝑙 𝐴𝑆𝐴 180𝑔 𝐴𝑆𝐴 x 138𝑔 𝑆𝐴 x 1𝑚𝑜𝑙 𝑆𝐴 x 1𝑚𝑜𝑙 𝐴𝑆𝐴 =6g ASA Aspartame (A) C14H18N2O5 C14= 12g x 14= 168g Note: If mL ang given sa mga reactants, we have to H18= 1g x 18= 18g know its density. N2= 14g x 2= 28g Acid Anhydride(AA): O5= 16g x 5= 80g Since given in mL si Acid anhydride, we have Total= 294g/n to know its density. Density of Acid Anhydride: 1.08g/mL Phenylalanine (P) C9H11NO2 C9= 12g x 9= 108g 10𝑚𝐿 1.08𝑔 𝐴𝐴 1𝑚𝑜𝑙 𝐴𝑆𝐴 180𝑔 𝐴𝑆𝐴 x 𝑚𝐿 x 1𝑚𝑜𝑙𝐴𝐴 X 1𝑚𝑜𝑙𝐴𝑆𝐴 = 19g ASA H11 1g x 11= 11g N= 14g x 1= 14g O2= 16g x 2= 32g Total= 165g/n To compute for the theoretical yield, we need to identify between the Salicylic acid and Acid 1. How many grams of Phenylalanine are anhydride will form the least amount. produced? 378𝑔 𝐴 1𝑚𝑜𝑙 𝐴 1𝑚𝑜𝑙 𝑃 165𝑔 𝑃 Therefore that is the Salicylic acid. x 294𝑔 𝐴 X 1𝑚𝑜𝑙 𝐴 x 1𝑚𝑜𝑙 𝑝 = 212g Phenylalanine To answer the question, how grams of ASA are formed? Answer: 6g of Aspirin U.S. SIDDIQUE & M.K. JOSE | PH3YB-1 Page | 4 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO 2. If the actual yield is 150 g, compute for the percent yield? Formula: 𝐴𝑐𝑡𝑢𝑎𝑙 𝑌𝑖𝑒𝑙𝑑 %yield= 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑 𝑥100 150𝑔 %yield= 212𝑔 𝑥100= 70.75% U.S. SIDDIQUE & M.K. JOSE | PH3YB-1 Page | 5 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO - Trihydric alcohol EXERCISE 1 o 3 hydroxyl groups are present Chemical Reactions of Organic Compounds EXERCISE 1A – Alcohols, Aldehydes, and Ketones ALCOHOLS. Compounds that have -OH (Hydroxyl) group o 1,2,3 – Propanetriol/ Glycerol bonded to a saturated, alkane-like carbon atom. R-OH 2. According to number of carbon atoms are An -OH can be attached to alkane, alkene or bound to the carbon atom to which the hydroxyl alkyne. group is bound. An -OH can be aromatic if it is attached to a (Primary, Secondary, or Tertiary) benzene ring, when -OH is bonded to Benzene - Primary alcohol ring it is called Phenol. o The carbon where (-OH) hydroxyl is attached has 2 available hydrogen. Phenol is an enol- double bond with an -OH o Primary alcohol can be oxidized to group attached to it. Aldehyde and then further oxidized to form Carboxylic acid. 1°Alcohol🡪 Aldehyde🡪 Carboxylic acid CLASSIFICATIONS OF ALCOHOL 1. According to number of hydroxyl group attached to their molecule - Primary- Yung carbon kung saan nakaattach (Monohydric, Dihydric, Trihydric) yung -OH group ay nakabond lang sa isang carbon. - Monohydric alcohol o 1 hydroxyl group present - Secondary alcohol o The carbon where (-OH) hydroxyl is attached has 1 available hydrogen. o Secondary alcohol can be oxidized o Example: propanol to form ketone. 2°Alcohol🡪 Ketone - Dihydric alcohol o 2 hydroxyl groups are present - Secondary- Yung carbon kung saan nakaattach yung -OH group ay nakabond sa dalawang carbon. o Example: 1,2-Ethanediol Siddique, U.S. & Noveno, K.M. |BSPHARM 3-YA-1 PAGE|6 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO Unlike the primary alcohol where -OH is always located at the terminal part of the chain - Tertiary alcohol o The carbon where (-OH) hydroxyl is Secondary Alcohol (R2CHOH) attached has NO available - converted into ketones (RC=OR) on hydrogen. treatment with oxidizing agents. - Secondary alcohol will always have one available hydrogen on the carbon where -OH is attached - 2 ° Alcohol 🡪 KETONE - Tertiary- Yung carbon kung saan nakaattach yung -OH group ay nakabond sa tatlong carbon. - There will be a removal of Hydrogen which will be replaced by a formation of carbonyl ALCOHOLS group forming Ketone. Different kinds of carbonyl-containing products - The carbonyl part of the ketone can be are formed depending on the structure of the found at any part of the chain- either in a starting alcohol and on the reaction conditions. ring or in a chain but not in number 1 position. Because the carbonyl of a ketone Primary alcohol (RCH2OH) must be in between two carbons - 1 ° to ALDEHYDE to CARBOXYLIC ACID - The lowest number of carbon that can produce ketone is 3 = Propanone/ Propan-2-one or Acetone Tertiary alcohol (R3COH) - Tertiary alcohols don’t normally react with - The -OH of the primary alcohol is always on oxidizing agents because they don’t have the last part. When oxidized, it is converted hydrogen on the carbon atom to which the to a carbonyl group and the carbonyl group -OH group is bonded. of an aldehyde is always at position 1. - If tertiary alcohol is tested in terms of rate - Aldehyde is further converted to an acid, of oxidation, it should not change since it the carbonyl group of an acid is always at cannot be oxidized. position 1. - NO REACTION 1°Alcohol🡪 Aldehyde🡪 Carboxylic acid - ACID no longer undergo oxidation - YOU CANNOT FORM AN ALDEHYDE FROM A SECONDARY ALCOHOL since the -OH of a secondary alcohol is found within a chain, Siddique, U.S. & Noveno, K.M. |BSPHARM 3-YA-1 PAGE|7 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO CHEMICAL C. Oxidation of Alcohols: TERTIARY ALCOHOL PROPERTIES…………………………….. - No Available Hydrogen therefore there is no oxidation. 1. OXIDATION - NO REACTION A. Oxidation of Alcohols: PRIMARY ALCOHOL Oxidizing agents: Periodinane, CrO3(Chromate) Na2Cr2O7 (Sodium dichromate), KMNO4 (Potassium permanganate) 2. ELIMINATION - Removal of water ✔ 2 hydrogen atoms are removed from - The product formed is either alkene or alcohol ether and the product formed depends o 1 Hydrogen from -OH group on the temperature that is used. o 1 Hydrogen from the carbon A. Dehydration of alcohol → Alkene atom bonded to the -OH ✔ New C=O bond is formed and a C-H bond is broken - In oxidation, there is addition of -OH and - when you remove water from alcohol, removal of Hydrogen the -OH from the carbon in where it is - IF THERE IS NO AVAILABLE HYDROGEN, attached, and hydrogen sa tabing OXIDATION WILL NOT TAKES PLACE. carbon. - The removal of the hydrogen will cause the - Dehydration of Alcohol at 170°celcius formation of carbonyl moiety, forming now with the use of Concentrated Sulfuric an ALDEHYDE-where the carbonyl group is acid (H2SO4) as the catalyst will form an at position 1. alkene. - This process is reversible if water is B. Oxidation of Alcohols: SECONDARY ALCOHOL added to alkene, you will form alcohol. Oxidizing agents: Periodinane, CrO3(Chromate) Na2Cr2O7 (Sodium dichromate) - Dehydration of alcohol forms alkene at - Hydration of alkene forms alcohol - isang uri lang ng alcohol ang gagamitin pero it depends parin sa temperature. ✔ 2 hydrogen atoms are removed from alcohol B. Dehydration of excess alcohol → Ether o 1 Hydrogen from -OH group o 1 Hydrogen from the carbon atom bonded to the -OH - 2 alcohols ang gagamitin, you remove - The removal of the hydrogen will cause the the water portion/ water content at formation of carbonyl moiety, forming RCOR- KETONE Siddique, U.S. & Noveno, K.M. |BSPHARM 3-YA-1 PAGE|8 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO 145°C using Concentrated Sulfuric acid , → Ferric Chloride (FeCl3)— traditional colorimetric test you form ether (ROR). for phenol - meaning if you combine two alcohols, - Ferric Chloride is yellow (may pagkayellow, pwedeng primary at secondary at may pagka orange) niremove mo yung water you will form ether. - When a liquid with enol structure is added with ferric chloride, it will turn to color purple or violet. - Ferric chloride is used in colometric test, it means it produces a distinct color para malaman mo if may presence ba ng enol structure. - Enols interconvert with carbon compounds 3. SUBSTITUTION that have an alpha-hydrogen, like ketones - The -OH group is changed by a halogen and aldehydes. - Halogen (Chlorine, Bromine, Iodine or - The compound is deprotonated on one side Fluorine) and protonated on another side, whereas a - Cl, Br, I, or F but never Astatine single bond and a double bond are - Astatine- synthetic halogen exchanged. This is called keto-enol - You add heat and remove the water portion, tautomerism. you form an alkyl halide. - Positive test (+) = Purple iron complex Test for Alcohols…………………………………… … 1. TEST FOR ENOL STRUCTURE - Enols (AKA: Alkenols, Alkene alcohol) - alkENe alcohOL - A molecule that has a hydroxyl group (OH) group directly bonded to an alkene (C=C) - N-butyl/Butan-1-ol/Butanol. Siddique, U.S. & Noveno, K.M. |BSPHARM 3-YA-1 PAGE|9 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO - Upon the addition of ferric - chloride, it turned violet. (CH3-CH2-CH2-CH2-OH) - Colorimetric test for phenol can - There is no double bond, but it also be color blue, red or dark has an -OH group colored. - When ferric chloride is added, it - In one of the experiment, there will turn yellow are color red, dark red, dark - Ethanol. orange, However, mas distinct - (CH3-CH2-OH) kung mas fresh or di pa sira - There is an -OH pero walang yung phenol. double bond - Fresh na phenol= color violet - When ferric chloride is added, it ang distinct na color. will turn yellow - Glycerol. 2. RATE OF OXIDATION Permanganate Oxidation - - Potassium permanganate is used - Trihydric alcohol - oxidizes both primary and secondary - you have 3 carbon present, and alcohols in either basic or acidic solutions. all the 3 carbon have attached - Basic solution- Sodium Hydroxide (NaOH)- -OH. to make the environment basic - It contains -OH group but no - Acidic solution- 6 Normal Sulfuric Acid double bond is present (6.0N H2SO4) for the acidic environment - Benzyl alcohol. - In permanganate oxidation, it will only - oxidize primary and secondary alcohol. - It has a carbon before the -OH - ONLY the primary and secondary alcohol are group. it means that the -OH oxidized. does not directly bonded to a double bond. - In the experiment, the tertiary alcohol must - Benzyl alcohol will turn yellow not change. If the tertiary alcohol changes, in the presence of ferric it means there has been contamination that chloride happened during the conduct of the - Phenol. experiment. Usually there is no change in tertiary alcohol since it is resistant to - oxidation. - Benzene ring and the -OH is attached to any part of the benzene ring. - Kahit saan dumikit ang -OH, nakadikit parin ung -OH sa double bond. Siddique, U.S. & Noveno, K.M. |BSPHARM 3-YA-1 PAGE|10 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO - The example in the experiment are N-butyl, Sec-butyl, Tert-butyl, control. - N-butyl is the primary alcohol - Sec-butyl is the secondary alcohol - Butan-2-ol - Tert-butyl is the tertiary alcohol - No available hydrogen= no oxidation - The precipitate upon the addition of N-butyl and Sec-butyl, the discoloration is fast, or the change in color of permanganate, from violet to color green. - ideally walang magbabago sa tertiary alcohol, - Usually color green, which will take kasi siya ay resistant to oxidation. Pero baka around 5 seconds since oxidation is fast, mali yung nagamit na reagent sa picture o kaya the fastest is the primary alcohol. nagamit yung reagent ng pipette NG Sec-butyl - There is also a conversion or a reduction at pinang kuha ng Tert-butyl, thats why we must of ketone or of permanganate, make sure to use the right pipette. Because Potassium permanganate is the contamination will prevent you from getting the oxidizing agent correct result. - Oxidizing agents are the one being - Control, meaning wala ka talagang result na reduced. makukuha. - The precipitate is the Manganese dioxide (MNO2)- - So, in KMnO4, the oxidation state of Potassium permanganate (KMNO4) that has been Manganese (Mn) is 7, it will be reduced reduced to Manganese dioxide (MNO2) to 4 - And if nareduced siya sa 4, dahil siya ay oxidizing agent ay siya ay merereduced, magkakaroon ka ng formation of brown precipitate. ALDEHYDES AND KETONES. - The brown precipitate is Manganese Dioxide (MnO2)- has an Oxidation of 5 Aldehydes and Ketones are the 2 simplest families of carbonyl compounds. ALDEHYDES - Aroma of cinnamon, vanilla, or almond are natural flavors and aromas from. KETONES Siddique, U.S. & Noveno, K.M. |BSPHARM 3-YA-1 PAGE|11 PHARMACEUTICAL AND MEDICINAL ORGANIC CHEMISTRY LABORATORY- MS. JOBELLE ABRIO - Jasmone from jasmine flower - become more alkane-like and less - Muscone from the male musk deer water-soluble, or they become more - Vital to the complex formulas of expensive lipophilic and more nonpolar perfumes Flammability - lower-boiling aldehydes and ketones - PROPERTIES OF ALDEHYDES AND KETONES flammable - Moderately polar compound Toxicity - since aldehydes & ketones do not have - Simple ketone - generally low toxicity an OH group, they are moderately polar - Simple aldehydes - toxic (esp. - They have higher boiling points than formaldehyde) alkanes with similar Molecular Weight (MW). Biochemistry - No hydrogen atoms bonded to oxygen or - simplest sugars called monosaccharides nitrogen, their molecules don’t form contain either an aldehyde group (aldose) or hydrogen bond with each other ketone group (ketose) Compounds with similar MW CHEMICAL PROPERTIES OF ALDEHYDES AND KETONES - Alkane - lowest boiling - lowest bp because it does not contain an 1. Oxidation of Primary Alcohols to form OH group Aldehydes - usually, alkanes are nonpolar - Alcohol - the highest boiling - alcohol is polar since it has the ability to form hydrogen bond and has an OH group - Aldehydes and Ketones fall in between State - Primary alcohol can be oxidized to form an - Simple aldehydes (RCH) and ketones (RCOR) aldehyde are LIQUID, however, as the number of - Oxidation of ethanol will form ethanal carbon increases, they become more solid (acetaldehyde). The IUPAC name of - >12 C— SOLID acetaldehyde is ethanal. Solubility 2. Oxidation of Secondary Alcohols to form - RCHO and RCOR soluble in common organic Ketones solvents -