(1) Recall the classification system for substituted alkenes. (2) Look at the alkene indicated. Count the number of bonds to non-hydrogen groups.

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1 Organic Chemistry - Problem Drill 10: Alkenes, Alkynes, and Dienes No. 1 of What is the substitution pattern for alkene indicated below? (A) mono (B) di (C) tri (D) tetra (E) unsubstituted Mono is the classification for alkenes with one non-hydrogen group attached to it. How many groups are attached to the indicated alkene? Di is the classification for alkenes with two non-hydrogen groups attached to the double bond. How many groups are attached to the indicated alkene? C. Correct! The alkene indicated has three non-hydrogen groups attached to the carbons of the double bond. It is classified as a tri-substituted alkene. Tetra is the classification for alkenes with four non-hydrogen groups attached to the double bond. How many groups are attached to the indicated alkene? Unsubstituted is the classification for an alkene with four hydrogens attached to the double bond. How many groups are attached to the indicated alkene? (1) Recall the classification system for substituted alkenes. Unsubstituted is the classification for an alkene with four hydrogens attached to the double bond. Mono is the classification for alkenes with one non-hydrogen group attached to double bond. Di is for alkenes with two non-hydrogen groups attached, tri for alkenes with three non-hydrogen groups attached, and tetra is for alkenes with four non-hydrogen groups attached. (2) Look at the alkene indicated. Count the number of bonds to non-hydrogen groups. H It often helps students to draw in any hydrogens not indicated by the skeletal structure and then to circle all non-hydrogen groups. There are three circles/groups so the alkene is tri-substituted. Therefore, the correct answer is (C).

2 No. 2 of What is the correct IUPAC name for the compound below? H (A) E-4-chloro-3-methyl-3-hepten-6-yne (B) E-4-chloro-5-methyl-4-hepten-1-yne (C) Z-4-chloro-3-methyl-3-hepten-6-yne (D) Z-4-chloro-5-methyl-4-hepten-1-yne (E) E-3-chloro-2-ethyl-2-hexen-5-yne When both an alkene and alkyne are present in a molecule, the chain should be numbered so that the multiple bonds have the lowest possible numbers. Go back and see if you can number the chain so that the multiple bonds receive even lower numbers. B. Correct! The tetrasubstituted alkene is E. The parent chain, location numbers and substituents are listed correctly. The alkene configuration and the location numbers are incorrect in this choice. Go back and check your numbering of the parent chain and the assignment of group priority. While the substituents, numbering and the parent chain is correct, the configuration of the alkene is incorrect. Go back and check your determination of the alkene configuration again. There is a continuous chain longer than 6 carbons present above. Go back and determine the correct parent chain for the molecule. (1) Find the longest carbon chain and the main functional group in the molecule to determine the parent. 7 1 H 5 4 The longest carbon chain is comprised of seven carbons. The molecule is an enyne (the enyne suffix is enyne) so the parent s name is heptenyne. When both an alkene and alkyne are present, the chain should be numbered so that the multiple bonds have the lowest possible numbers. For this molecule, the lowest possible numbers are obtained by numbering from right to left. The parent with location numbers for the multiple bonds is 4-hepten-1-yne. (2) Determine the substituents and their correct location numbers. There are two substituents: a chlorine atom and a methyl group. Since the multiple bonds determined the numbering of the chain, you already know the location number of the substituents: 4-chloro and 5- methyl (3) Determine the configuration of the alkene using E/Z. Since this is a tetrasubstituted alkene, the E/Z system must be used. Split the alkene in half and look at each end. Determine the higher priority group on each end using the E/Z rules. Here, the ethyl group and the chlorine atom are the higher priority groups. Since they are on opposite sides of the double bond, the alkene has an E configuration H (4) Put the substituents in alphabetical order (ignoring any numerical prefixes) and place in front of the parent name. Place the alkene configuration in front of the substituents. E-4-chloro-5-methyl-4-hepten-1-yne. Therefore, the correct answer is (B).

3 No. 3 of Predict the product formed in highest yield in the following reaction: H 2 SO 4 heat O OSO 3 H I II III IV (A) I (B) II (C) III (D) IV (E) V V Sulfuric acid will not oxidize an alcohol. Review the reactions in the tutorial and try to determine what the acid does to the alcohol. While sulfuric acid does dehydrate the alcohol, this alkene will not be the one formed in highest yield. Review Zaitsev s rule and determine why this alkene is not the correct answer. C. Correct! According to Zaitsev s rule, this alkene should be the one formed in the higher yield during the dehydration reaction because it is the more substituted alkene. Sulfuric acid will not form an enol from an alcohol. Review your alkene reactions and try to determine what the acid does to the alcohol. A sulfuric acid would not be obtained under these conditions. Go back and review the reactions alkenes undergo. (2) Determine what kind of reaction is taking place. The starting material is an alcohol. The other reactant is sulfuric acid which is a very strong acid. You may have already recognized this reaction as a dehydration of the alcohol to form an alkene. If you did not, go back and review the reactions in this tutorial. It is not enough to just recognize the type of reaction taking place. You must also understand the mechanism in order to accurately predict the product. Recall that the dehydration of an alcohol using a strong acid proceeds via a carbocation intermediate that forms on the carbon to which the alcohol was attached. A base removes a proton from a carbon adjacent to the carbocation to form the double bond. (3) Predict the product(s). This molecule has two carbons that can be deprotonated in the last step of the mechanism to form the double bond. Is there a preference for which one is deprotonated? In this case, yes. Recall Zaitsev s rule: the product that contains the more substituted alkene will be formed in higher yield. If a proton is removed from the carbon to the left of the alcohol, then a trisubstituted alkene will be formed. If a proton is removed from the carbon to the right of the alcohol, then a monosubstituted alkene will be formed. Therefore, according to Zaitsev s rule, the trisubstituted alkene will be formed in higher yield. Therefore, the correct answer is (C).

4 No. 4 of Predict the product formed in highest yield in the following reaction: excess K I II III IV (A) I (B) II (C) III (D) IV (E) V HO V A. Correct! The excess base deprotonates the carbon adjacent to the geminal dihalide to form a triple bond between carbons 1 and 2. While this is a possible intermediate of the reaction, it is not the final product. Remember, excess base is used in this reaction so both chlorine atoms should be displaced. If both chlorine atoms are displaced by deprotonation of an adjacent carbon, what will the final product be? Careful! Review the mechanism for this reaction. Which carbon will the base deprotonate twice? While this is a possible intermediate of the reaction, it is not the final product. Remember, excess base is used in this reaction so both chlorine atoms should be displaced. If both chlorine atoms are displaced by deprotonation of an adjacent carbon, what will the final product be? An acetal would not be obtained under these conditions. Go back and review the reactions covered in the tutorial. (2) Determine what kind of reaction is taking place. The starting material is a geminal dihalide. The other reactant is potassium hydroxide which is a very strong base. You may have already recognized this reaction as a dehydrohalogenation of the dihalide to form an alkyne. If you did not, go back and review the reactions in this tutorial. It is not enough to just recognize the type of reaction taking place. You must also understand the mechanism in order to accurately predict the product. Recall that dehydrohalogenation of a geminal dihalide occurs when a strong base deprotonates an adjacent carbon twice, displacing each chlorine and forming a triple bond. (3) Predict the product(s). This molecule has only one adjacent carbon to deprotonate so predicting the product will not be difficult. A triple bond will form between carbons 1 and 2 on the chain. Therefore, the correct answer is (A).

5 No. 5 of Predict the product formed in highest yield in the following reaction: H 2 O H 2 SO 4 OSO 3 H I II III IV (A) I (B) II (C) III (D) IV (E) V V Recall that this reaction proceeds with Markovnikov addition. According to Markovnikov s rule, where will the water add? In order to obtain an alkane product, the double bond would have to be reduced entirely. The stated conditions will not reduce the alkene. Go back and review the types of reactions alkenes undergo. These conditions will not affect a dihydroxylation of the alkene. Go back and review the mechanism for this reaction. D. Correct! These reaction conditions will essentially add water over the double bond yielding an alcohol as the final product. Since the reaction proceeds with Markovnikov addition, the alcohol attaches to the more substituted carbon of the double bond. These conditions will not add sulfuric acid to the double bond. Go back and review the reactions of alkenes covered in this tutorial. (2) Determine what kind of reaction is taking place. The starting material is an alkene. The other reactants are sulfuric acid (which is a very strong acid) and water. You may have already recognized this reaction as the hydration of an alkene (using an acid catalyst) to form an alcohol. If you did not, go back and review the reactions in this tutorial. It is not enough, however, to just recognize the type of reaction taking place. You must also understand the mechanism in order to accurately predict the product. Recall that acid catalyzed hydration of an alkene proceeds via a carbocation intermediate that forms on the more substituted carbon of the double bond (Markovnikov addition). Water then adds to the carbocation to form the alcohol. (3) Predict the product(s). Recall that hydration of an alkene follows Markovnikov s rule. The carbocation will form on the more substituted carbon of the double bond. In this example, the left end of the double bond has one substituent while the right end of the double bond has two substituents. The carbocation will form on the right end of the double bond. Once the carbocation forms on this carbon, water will attack it leading to the formation of the alcohol product. Therefore, the correct answer is (D).

6 No. 6 of Predict the product formed in highest yield in the following reaction: excess 2 I II III IV V (A) I (B) II (C) III (D) IV (E) V A. Correct! The bromine adds over the triple bond twice because the alkyne is composed of two pi bonds that can react with the halogen. The result is the formation of a tetrahaloalkane. This product could not be formed under these conditions. With each addition of halogen, two bromines are added over one of the pi bonds of the alkyne. Go back and review the mechanism for this reaction. Careful! The location of the triple bond will determine the placement of the halides in the final product. Go back and number your parent chain in the alkyne. Then number each parent chain in the possible answers to determine which has the correct placement of the halides. This product could not be formed under these conditions. With each addition of halogen, two bromines are added over one of the pi bonds of the alkyne. Go back and review the mechanism for this reaction. This dibromide would not be formed under these conditions. With each addition of halogen, two bromines are added over one of the pi bonds of the alkyne. Go back and review the mechanism of this reaction. (2) Determine what kind of reaction is taking place. The starting material is an alkyne. The other reactant is bromine. You may have already recognized this reaction as the halogenation of the alkyne to form a tetrahaloalkane. If you did not, go back and review the reactions in this tutorial. It is not enough, however, to just recognize the type of reaction taking place. You must also understand the mechanism in order to accurately predict the product. Recall that the halogenation of an alkyne can proceed via syn or anti addition of the halogen across the triple bond. However, since an excess of halogen is used, we do not have to worry about the stereochemistry of the reaction since any cis or trans haloalkenes formed will be consumed when the excess halogen adds over their double bonds. (3) Predict the product(s). Recall that the halogen can add to an alkyne twice because there are two pi bonds. Since there is an excess of the halogen present, the reaction will not stop at the haloalkene formed after the addition of the first equivalent of halogen. The addition will occur a second time to form a tetrahaloalkane. Therefore, the correct answer is (A).

7 No. 7 of What is the product of the reaction of an alkene with potassium permanganate (KMnO 4 )? (A) An anti diol (B) An alcohol (C) A syn diol (D) A cis dihalide (E) A geminal diol ose! Go back and review what kind of addition (syn or anti) is seen with the reaction of potassium permanganate and alkenes. An alcohol would not be the product obtained from reacting an alkene with potassium permanganate. Go back and review the oxidation reactions of alkenes. C. Correct! Potassium permanganate oxidizes a double bond by adding two hydroxyl groups to it. Both hydroxyl groups will add to the same side of the double bond (syn addition) to give a syn diol. Go back and review the structure of potassium permanganate. There are no halogens in its structure so it can not add any over a double bond. Go back and review the oxidation reactions of alkenes. A geminal diol, a compound with two hydroxyl groups on the same carbon, would not be obtained under these conditions. Go back and review the oxidation reactions of alkenes. (2) Determine what kind of reaction is taking place and then predict the product(s). Potassium permanganate is an oxidant. If combined with an alkene, one would expect that the alkene would be oxidized. Indeed, an oxidation of the alkene to a diol does take place. The tutorial states that the oxidation proceeds with syn addition giving a syn diol as the final product. Therefore, the correct answer is (C).

8 No. 8 of What is the product(s) of the following reaction? 1. O 3 2. H 2 O (A) Two ketones (B) Two aldehydes (C) A ketone and an aldehyde (D) An aldehyde and a carboxylic acid (E) Two carboxylic acids Two ketones would not be formed as products under these conditions. Go back and review the oxidation reactions of alkynes. Two aldehydes would not be formed as products under these conditions. Go back and review the oxidation reactions of alkynes. An aldehyde and a ketone would not be formed as products under these conditions. Go back and review the oxidation reactions of alkynes. ose! An aldehyde would not be formed as a product under these conditions. Go back and review the oxidation reactions of alkynes. E. Correct! Two carboxylic acids are formed from the ozonolysis of an alkyne. (2) Determine what kind of reaction is taking place. The starting material is an alkyne. The other reactant is ozone. You may have already recognized this reaction as the ozonolysis of an alkyne (oxidation reaction). If you did not, go back and review the reactions in this tutorial. Recall that ozone adds to the triple bond of an alkyne. The carbons that participated in the triple bond are split apart and are converted into carboxylic acid functional groups. (3) Predict the product(s). Two carboxylic acids are formed from the ozonolysis of an alkyne. In this case, the alkyne is converted into one molecule of propanoic acid and one molecule of butanoic acid. Therefore, the correct answer is (E).

9 No. 9 of What is the trade name of the following polymer? F C F F C F n (A) Polyethylene (B) Polypropylene (C) Orlon (D) Polystyrene (E) Teflon Polyethylene consists of monomers of ethylene. Go back and review the polymerization of alkenes. Polypropylene consists of monomers of propene. Go back and review the polymerization of alkenes. Orlon is the trademarked name of polyacrylonitrile which is made up of monomers of acrylonitrile. Go back and review the polymerization of alkenes. Polystryene consists of monomers of styrene. Go back and review the polymerization of alkenes. E. Correct! Teflon is the trade name of the polymer polytetrafluoroethylene which consists of monomers of tetrafluoroethylene. (1) Recall the different types of polymers covered in the tutorial. A monomer is the simplest unit that makes up a polymer and a polymer is named by the monomer it is derived from. For an example, when ethylene (also known as ethene) polymerizes, it forms polyethylene. However, some polymers become better known as their trade name (the name given to the polymer by the company that develops it) rather than a generic name like polyethylene. An example of this would be Orlon. In those instances, a student must memorize the trade name and its corresponding monomer as the trade name often does not give a hint to the monomer used like a generic name would. (2) Determine the type of monomer used in a polymer. The molecular structure of a polymer is often pictured by showing the monomer, minus its double bond as it has already undergone reaction, in brackets. Look at the structure above and see if you can determine the monomer used to make the polymer. The structure shows two carbons bonded together as the backbone of the polymer. Each carbon is bonded to two fluorine atoms. One could surmise that this polymer is derived from tetrafluoroethylene: F F C C F F (3) Determine the name of the polymer from its monomer. The polymer derived from tetrafluoroethylene is known by its trade name: Teflon. Therefore, the correct answer is (E).

10 No. 10 of Which monomer below constitutes the polymer polyvinyl chloride? (A) Styrene (B) Ethylene (C) Propylene (D) Vinyl chloride (E) Acrylonitrile Styrene, once polymerized, would give rise to polystyrene. Go back and review the polymerization of alkenes. Ethylene, once polymerized, would give rise to polyethylene. Go back and review the polymerization of alkenes. Propylene, once polymerized, would give rise to polypropylene. Go back and review the polymerization of alkenes. D. Correct! When it undergoes polymerization, vinyl chloride gives rise to polyvinyl chloride. Acrylonitrile, once polymerized, would give rise to polyacrylonitrile (also known as Orlon). Go back and review the polymerization of alkenes. (1) Recall the different types of polymers covered in the tutorial. A monomer is the simplest unit that makes up a polymer and a polymer is named by the monomer it is derived from. For an example, when ethylene (also known as ethene) polymerizes, it forms polyethylene. However, some polymers become better known as their trade name (the name given to the polymer by the company that develops it) rather than a generic name like polyethylene. An example of this would be Orlon. In those instances, a student must memorize the trade name and its corresponding monomer as the trade name often does not give a hint to the monomer used like a generic name would. (2) Can the monomer be determined from the name? Often, if the polymer s name begins with poly-, the monomer can be guessed by the portion of the name following the word poly. In this case, polyvinyl chloride is derived from vinyl chloride. Therefore, the correct answer is (D).