Chemistry 2050 Introduction to Organic Chemistry Fall Semester 2011 Dr. Rainer Glaser

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1 Chemistry 2050 Introduction to Organic Chemistry Fall Semester 2011 Dr. Rainer Glaser Examination #4 Practice Edition Carbonyl Compounds and Amines. Wednesday, November 16, 2011, 10 10:50 am Name: Question 1. Aldehydes and Ketones 20 Question 2. Carboxylic Acids and Anhydrides 20 Question 3. Carboxylic Acid Esters 20 Question 4. Carboxylic Acid Halides 20 Question 5. Amines and Diazonium Ions 20 Total 100 1

2 Question 1. Aldehydes and Ketones. (20 points) (a) Provide abbreviated structural formulas of the substrate and the product of the reaction of acetophenone with (1) lithium aluminum hydride and (2) dilute aqueous acid. (4 points) (b) Provide abbreviated structural formulas of the substrate, the reagent and the product of the reaction of cyclopentanone with (1) sodium acetylide and (2) dilute aqueous acid. (4 points) (c) Provide abbreviated structural formulas of the substrate, the reagent and the product of the reaction of benzaldehyde with aniline. The product is an (name the functional group). (4 points) (d) Cinnamaldehyde H 5 C 6 CH=CH CHO can be prepared by a mixed aldol condensation. The mixed aldol is formed first by addition of the enolate anion of (name of the enolizable aldehyde) and the carbonyl component (name of the carbonyl component). The aldol reaction requires (base, acid) catalysis to form the enolate anion. Provide abbreviated structural formulas of both substrate, of the intermediate aldol, and of the final product. (8 points) 2

3 Question 2. Carboxylic Acids and Anhydrides. (20 points) (a) The boiling points of propane (C 3 H 8, MM = 44 g/mol), ethanol (C 2 H 5 OH, MM = 46 g/mol), and formic acid (HCO 2 H, MM = 46 g/mol), respectively are -42 C, 78 C, and 101 C, respectively. Note that the molecular masses (MM) of the three molecules are very similar. Briefly explain (1) why the boiling points of ethanol and formic acid are so much higher than the one of propane and (2) why the boiling point of formic acid is higher than the one of ethanol. (8 points) (b) The trivialname capric acid comes from the latin word for goats and the IUPAC name of capric acid is decanoic acid. Draw the abbreviated structural formula of capric acid. Decanoic acid is a (gas, liquid, solid) at room temperature. Considering its structure, argue whether capric is soluable in water. (4 points) (c) 2-Butendioic acid HOOC CH=CH COOH forms two stable isomers: cis-2-butendioic acid is called and trans-2-butendioic acid is called. Heating one of these isomers leads to water elimination and the formation of an acid anhydride (a condensation reaction). Draw the structure of the acid anhydride formed in the appropriate box. What product do you think might form with water elimination by heating the other isomer? (8 points) cis-2-butendioic acid trans-2-butendioic acid 3

4 Question 3. Carboxylic Acid Esters. (20 points) (a) The ester methyl benzoate H 5 C 6 CO OCH 3 can be made by Fischer esterification of the carboxylic acid (formic acid, acetic acid, benzoic acid) and the alcohol (methanol, ethanol, phenol). Fischer esterification means that the ester formation is catalyzed by a strong mineral acid. If we wanted to make the 18 O-labeled ester H 5 C 6 CO 18 OCH 3, then we would employ the 18 O- labeled (alcohol, carboxylic acid). (4 points) (b) Consider the formation of methyl benzoate H 5 C 6 CO OCH 3 using catalysis by a strong mineral acid. Abbreviate the phenyl ring as H 5 C 6 or as Ph and draw the complete structural formulas of the species indicated (draw all lone pairs, formal charges). The neutral gem tri-functional intermediate contains (1, 2, or 3 hydroxyl group(s)) and (1, 2, or 3 ether group(s)). (10 points) Protonated carboxylic acid, i.e., the actual substrate for alcohol addition: Product formed by addition of alcohol to the protonated carboxylic acid: Neutral gem tri-functional intermediate: Protonated gem tri-functional intermediate that will lead to water elimination: (c) Consider the reaction of γ-butyrolactone with ammonia. The lactone γ-butyrolactone is the cyclic ester of -hydroxybutanoic acid because the α-, β- and γ-positions of butanoic acid are C, C and C. Provide the structural formulas of the substrate lactone and of the reaction product formed by ammonolysis. (6 points) 4

5 Question 4. Carboxylic Acid Halides. (20 points) (a) Consider the preparation of benzoyl chloride using thionyl chloride. Draw structural formulas of the substrate, of the reagent, of the organic product (the benzoyl chloride), and also of the inorganic products. (5 points) (b) Considering the specific case of acetyl chloride, explain why acyl halides may be irritating to your nose. [If a reaction happens in the nose, then provide the reaction equation for this reaction.] (5 points) (d) Devise a synthesis of 4-methylphenyl propyl ketone from toluene and butanoic acid as starting materials. For each of two reaction steps, specify substrate, reagent(s), catalyst(s) and product formed. [Hint: Friedel-Crafts Acylation!] (10 points) 5

three molecules of methyl iodide, and provide the names of these products.

6 Question 5. Amines and Diazonium Ions. (20 points) (a) Draw the structures of the products obtained by the reaction of aniline with one, two and three molecules of methyl iodide, and provide the names of these products. (8 points) (b) Devise a synthesis of benzoic acid from aniline. For each of three reaction steps, specify substrate, reagent(s), catalyst(s) and product formed. [Hint: via benzenediazonium ion!] (12 points) 6

Chemistry 2050 Introduction to Organic Chemistry Fall Semester 2011 Dr. Rainer Glaser

Chemistry 2050 Introduction to Organic Chemistry Fall Semester 2011 Dr. Rainer Glaser Examination #4 Carbonyl Compounds and Amines. Wednesday, November 16, 2011, 10 10:50 am Name: Answer Key Question 1.