Chapter 1 Classifying Organic Compounds

CHEMISTRY 12

Note to Teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the full set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act on Your Strategy, and Check Your Solution. Where a shorter solution is appropriate, the Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check Your Solution step is also included in the shorter Sample Problems.

Solutions for Practice Problems Student Textbook page 10

1. Problem Predict and sketch the three-dimensional shape of each single-bonded atom. (a) C and O in CH3OH (b) CinCH4 Solution (a) The carbon atom has four single bonds, so it will have a tetrahedral shape. The oxygen atom has two single bonds, so it will have a bent shape. H

C • • • • HH O OH H CH3 around the carbon atom around the oxygen atom

(b) the carbon atom has four single bonds, so it will have a tetrahedral shape. H

C HH H

2. Problem Predict and sketch the three-dimensional shape of each multiple-bonded molecule. (a) HCCH (b) H2CO Solution (a) Each carbon atom has one triple bond and one single bond. The shape around each carbon atom is linear, so the shape of this molecule is linear. H C C H (b) The carbon atom has one double bond and two single bonds, so the shape around the carbon atom (and the shape of the molecule) will be trigonal planar. O

C H H Chapter 1 Classifying Organic Compounds • MHR 1 CHEMISTRY 12

3. Problem Identify any polar bonds that are present in each molecule in questions 1 and 2. Solution The molecule in question 1(a) has two polar bonds: CO and OH. The molecule in question 1(b) has only CH bonds, which are usually considered to be non-polar. The molecule in question 2(a) has only CC and CH bonds, which are usually considered to be non-polar. The molecule in question 2(b) has one polar bond: CO.

4. Problem For each molecule in questions 1 and 2, predict whether the molecule as a whole is polar or non-polar. Solution For question 1(a): Step 1 The molecule has polar bonds. Step 2 There are two polar bonds: CO and OH Step 3 Because there is a bent shape around the oxygen atom, the polar bonds do not counteract each other. The molecule is polar. For question 1(b): Step 1 There are no polar bonds, so the molecule is non-polar. For question 2(a): Step 1 There are no polar bonds, so the molecule is non-polar. For question 2(b): Step 1 The molecule has a polar bond. Step 2 Since there is only one polar bond, the molecule is polar.

Solutions for Practice Problems Student Textbook pages 16–17

5. Problem Name each hydrocarbon.

(a)H3C CH3 (e)

(b)H2C CH (f) CH3

CH3 H2C CH

(c)CH3 (g)

CH3 C CH CH2 CH3

(d)

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Solution (a) Step 1 There are only two carbon atoms, so the root is -eth-. Step 2 The name of this compound is ethane. (b) Step 1 There are four carbon atoms in a ring. The root is -cyclobut-. Step 2 This is an alkene, so the name of the compound is cyclobutene. (c) Step 1 There are five carbon atoms in the main chain, so the root is -pent-. Step 2 There is a double carbon-carbon bond, so the suffix is -ene. Step 3 Number the compound from the left to give the lowest position number to the double bond. Step 4 There is a methyl group on the second carbon, so the prefix is 2-methyl-. Step 5 The full name is 2-methyl-2-pentene. (d) Step 1 There are six carbon atoms in the longest chain, so the root is -hex-. Step 2 The compound is an alkane, so the suffix is -ane. Step 3 Number from the left to give the branch the lowest possible position number. Step 4 There is a methyl group on the third carbon atom, so the prefix is 3-methyl-. Step 5 The full name is 3-methylhexane. (e) Step 1 There are seven carbon atoms in the longest chain that contains the double bond. The root is -hept-. Step 2 The compound is an alkene, so the suffix is -ene. Step 3 Number from the left to give the lowest possible position number to the double bond. Step 4 There is a methyl group on the second carbon atom, and an ethyl group on the third carbon atom. In alphabetical order, the prefix is 3-ethyl-2-methyl-. Step 5 The full name is 3-ethyl-2-methyl-2-heptene. (f) Step 1 There are six carbon atoms in the main ring, so the root is -cyclohex-. Step 2 The compound is an alkane, so the suffix is -ane. Step 3 Start numbering at a branch, so that the two branches have the lowest possible position numbers. Step 4 There are two methyl groups, so the prefix is 1,2-dimethyl-. Step 5 The full name is 1,2-dimethylcyclohexane. (g) Step 1 There are five carbon atoms in the main chain, so the root is -pent-. Step 2 There is a triple bond, so the suffix is -yne. Step 3 Number from the left to give the lowest possible position number to the triple bond. Step 4 There are no branches. Step 5 The full name is 2-pentyne.

6. Problem Draw a condensed structural diagram for each hydrocarbon. (a) propane (b) 4-ethyl-3-methylheptane (c) 3-methyl-2,4,6-octatriene Solution (a) Step 1 The root is -prop-, so there are three carbon atoms in the main chain. Step 2 The compound is an alkane, so all the bonds are single. Step 3 There are no branches. Step 4 CH3CH2CH3

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(b) Step 1 The root is -hept- so there are seven carbon atoms in the main chain. Step 2 There are no double or triple bonds. Step 3 The ethyl group is attached to carbon 4. The methyl group is attached to carbon 3. Step 4 CH3

CH3 CH2 CH CH CH2 CH2 CH3

CH2 CH3 (c) Step 1 The root is -oct- so there are eight carbon atoms in the main chain. Step 2 There are three double bonds, between carbons 2 and 3, 4 and 5, and 6 and 7. Step 3 There is a methyl branch at carbon 3.

Step 4 CH3

CH3 CH C CH CH CH CH CH3 7. Problem Identify any errors in the name of each hydrocarbon. (a) 2,2,3-dimethylbutane (b) 2,4-diethyloctane (c) 3-methyl-4,5-diethyl-2-nonyne Solution (a) There are three methyl groups, so the name should be 2,2,3-trimethylbutane. (b) If you draw this compound, you can see that the main chain has more than eight carbon atoms (the ethyl group on carbon 2 should be counted as part of the main chain). The correct name is 5-ethyl-3-methylnonane. (c) If you draw this compound, you will see that the third carbon atom forms more than four covalent bonds. This is not possible. One solution would be to change the name to 3-methyl-4,5-diethyl-2-nonene.

8. Problem Correct any errors so that each name matches the structure beside it.

(a) 4-hexyne CH3 CH2 CH CH CH2 CH3

(b) 2,5-hexene CH3 C C C C CH3 Solution (a) This compound has a double bond, not a triple bond. Also, the bond is located at carbon 3. The correct name is 3-hexene. (b) This compound has triple bonds, not double bonds. Also, since there are two triple bonds, the preﬁx -di- should be used. Finally, the triple bonds are located at carbons 2 and 4. The correct name is 2,4-hexadiyne.

9. Problem Use each incorrect name to draw the corresponding hydrocarbon. Examine your draw- ing, and rename the hydrocarbon correctly. (a) 3-propyl-2-butene (b) 1,3-dimethyl-4-hexene (c) 4-methylpentane

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Solution (a) CH3 CH C CH3 ➀➁ ➂

CH2 CH2 CH3 ➃➄➅ The propyl group should be part of the main chain. The correct name is 3-methyl-2-hexene. ➆ (b) CH3 CH3

CH2 CH2 CH CH CH CH3 ➅ ➄ ➃ ➂ ➁ ➀ The ﬁrst methyl group should be part of the main chain. Also, you should number in the opposite direction to give a lower position number to the double bond. The correct name is 4-methyl-2-heptene. (c) CH3

CH3 CH2 CH2 CH CH3 ➄ ➃ ➂ ➁ ➀ You should number in the opposite direction to give the lowest possible position number for the methyl group. The correct name is 2-methylpentane.

Solutions for Practice Problems Student Textbook page 19

10. Problem Name the following aromatic compound.

CH3

H3C CH3

Solution Step 1 Start numbering at one of the branches. Since they are identical and spaced evenly, it doesn’t matter which one. Step 2 There are methyl groups at carbons 1, 3, and 5. Step 3 The name is 1,3,5-trimethylbenzene.

11. Problem Draw a structural diagram for each aromatic compound. (a) 1-ethyl-3-methylbenzene (b) 2-ethyl-1,4-dimethylbenzene (c) para-dichlorobenzene (Hint: Chloro refers to the chlorine atom, Cl.) Solution (a) (b) (c) CH2 CH3 CH3 Cl

CH2CH3

CH3 CH3 Cl 12. Problem Give another name for the compound in question 11(a). Solution The compound can also be called meta-ethylmethylbenzene.

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13. Problem Draw and name three aromatic isomers with the molecular formula C10H14. (Isomers are compounds that have the same molecular formula, but different struc- tures. See the Concepts and Skills Review for a review of structural isomers.) Solution Aside from the six carbons in the benzene ring, there are an additional four carbon atoms that exist as branches. Three possible isomers are 1,2,3,4-tetramethylbenzene, 1,2,3,5-tetramethylbenzene, and 1-methyl-4-propylbenzene, shown below. There are many other possible isomers.

CH3 CH3 CH2CH2CH3

CH3 CH3 CH3

CH3

CH3 CH3 CH3

Solutions for Practice Problems Student Textbook pages 26–27

14. Problem Name each alcohol. Identify it as primary, secondary, or tertiary. OH

(a)CH3 CH2 CH2 OH (d) CH3 CH CH CH2 CH3

(b)OH (e) OH

Solution (a) Step 1 The main chain has three carbon atoms. The name of the parent alkane is propane. Step 2 Replacing -e with -ol gives propanol. Step 3 The OH group is at carbon 1, giving 1-propanol. This is a primary alcohol. (b) Step 1 There are four carbon atoms in the main chain, so the parent alkane is butane. Step 2 The base name is butanol. Step 3 A position number is needed, so the compound is 2-butanol. This is a secondary alcohol. (c) Step 1 There are four carbons in the main ring, so the parent alkane is cyclobutane. Step 2 The name of the compound is cyclobutanol. No position number is needed. This is a secondary alcohol. (d) Step 1 There are ﬁve carbons in the main chain, so the parent alkane is pentane. Step 2 There are two OH groups, so the base name is pentanediol.

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Step 3 The OH groups are located on carbon 2 and carbon 3 of the main chain. The compound name is 2,3-pentanediol. Both OH groups are secondary. (e) Step 1 There are seven carbons in the main chain, so the parent alkane is heptane. Step 2 The base name is heptanol. Step 3 The OH group is located on carbon 1, giving 1-heptanol. (Note that an OH group is always given priority over alkyl groups such as a methyl group.) Step 4 There are methyl groups at carbon 2 and carbon 4. The preﬁx is 2,4-dimethyl-. Step 5 The full name is 2,4-dimethyl-1-heptanol. This is a primary alcohol.

15. Problem Draw each alcohol. (a) methanol (d) 3-ethyl-4-methyl-1-octanol (b) 2-propanol (e) 2,4-dimethyl-1-cyclopentanol (c) 2,2-butanediol Solution (a) There is only one carbon atom in this molecule. CH3 OH (b) There are three carbon atoms in the main chain. The OH group is at carbon 2. OH

CH3 CH CH3 (c) There are four carbon atoms in the main chain. There are two OH groups, both located at carbon 2. OH

CH3 C CH2 CH3

OH (d) There are eight carbons in the main chain. The OH group is at carbon 1, and there is an ethyl group at carbon 3 and a methyl group at carbon 4. CH2 CH3

HO CH2 CH2 CH CH CH2 CH2 CH2 CH3

CH3 (e) There are ﬁve carbons, and the compound is ring-shaped. The OH group is at carbon 1, and there are two methyl groups at carbons 2 and 4. OH

CH3

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16. Problem Identify any errors in each name. Give the correct name for the alcohol. (a) 1,3-heptanol OH

HO CH2 CH2 CH CH2 CH3 (b) 3-ethyl-4-ethyl-1-decanol

CH3

CH2 CH CH CH3

CH3 OH Solution (a) There are ﬁve carbon atoms, so the root should be -pent-. Also, there are two OH groups. The correct name is 1,3-pentanediol. (b) The two ethyl groups should be grouped together, so the correct name is 3,4-diethyl-1-decanol. (c) There are ﬁve carbons in the main chain so the root should be -pent-, not -but-. The OH group should be given the lowest possible position number, so the correct name is 3-methyl-2-pentanol.

17. Problem Sketch a three-dimensional diagram of methanol. Hint: Recall that the shape around an oxygen atom is bent. Solution Note to teacher: If students complete a three-dimensional diagram of methanol successfully, have them attempt a similar diagram of ethanol, as shown below. H H C H H H C CH H • • • • O H •O • •• H H methanol ethanol

Solutions for Practice Problems Student Textbook page 28

18. Problem Draw a condensed structural diagram for each alkyl halide. (a) bromoethane (b) 2,3,4-triiodo-3-methylheptane

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Solution (a) There are two carbon atoms in the main chain. No position number is needed because the two carbon atoms are indistinguishable. CH3CH2Br (b) There are seven carbon atoms in the main chain. There is one methyl group, at carbon 3, and three iodine atoms, at carbons 2,3, and 4.

I CH3 I

CH3 CH C CH CH2 CH2 CH3

19. Problem Name the alkyl halide at the right. F Then draw a condensed structural diagram to represent it. F Solution There are six carbon atoms in the main ring, and the compound is a cyclic alkane. The parent alkane is cyclohexane. There are two ﬂuorine atoms at positions 1 and 2. The name is 1,2-diﬂuorocyclohexane. CH2

CH2 CH F

CH2

20. Problem Draw and name an alkyl halide that has three carbon atoms and one iodine atom. Solution 1-iodopropane CH3CH2CH2I

21. Problem Draw and name a second, different alkyl halide that matches the description in the previous question. Solution 2-iodopropane CH3 CH CH3

Solutions for Practice Problems Student Textbook page 30

22. Problem Use the IUPAC system to name each ether.

(a)H3C O CH3 (c) CH3 CH2 CH2 CH2 O CH3

CH3

(b) H3C O CH

CH3

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Solution (a) Step 1 The longest alkyl group is based on methane. Step 2 The alkoxy group is also based on methane. The prefix is 1-methoxy, or just methoxy. (No 2-methoxy position is possible.) Step 3 The full name is methoxymethane. (b) Step 1 The longest alkyl group is based on propane. Step 2 The prefix is 2-methoxy. Step 3 The full name is 2-methoxypropane. (c) Step 1 The longest alkyl group is based on butane. Step 2 The prefix is 1-methoxy. Step 3 The full name is 1-methoxybutane.

23. Problem Give the common name for each ether.

(a)H3C O CH2CH3 (b) H3C O CH3 Solution (a) Step 1 The two alkyl groups are methyl and ethyl. Step 2 The full name is ethyl methyl ether. (b) Step 1 The two alkyl groups are both methyl. Step 2 The full name is dimethyl ether.

24. Problem Draw each ether. (a) 1-methoxypropane (b) 3-ethoxy-4-methylheptane (c) tert-butyl methyl ether Solution (a) The main alkyl group has three carbon atoms. The alkoxy group has one carbon atom, and is attached to the ﬁrst carbon of the larger group. CH3OCH2CH2CH3 (b) The parent alkane is 4-methylheptane. Draw the parent alkane, then add an ethoxy group at the third carbon atom of the main chain.

CH2 CH3 O

CH3 CH2 CH CH CH2 CH2 CH3

CH3 (c) There are two alkyl groups: the tert-butyl group, and the methyl group. These two groups are connected by an oxygen atom.

CH3

CH3 C O CH3

CH3

25. Problem Sketch diagrams of an ether and an alcohol with the same number of carbon atoms. Generally speaking, would you expect an ether or an alcohol to be more soluble in water? Explain your reasoning.

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Solution The alcohol will be more soluble in water because it has an OH bond, and can form hydrogen bonds with water molecules. The ether can accept hydrogen bonds from water molecules, but cannot reciprocate them, as shown below. O H H CH CH CH O 3 2 2 O H H H O H CH3CH2 O H CH3 alcohol ether

Solutions for Practice Problems Student Textbook page 32

26. Problem Name each amine.

(a)CH3 NH2 (c) CH3 CH2 CH CH3

NH2

CH3

(b)C(CH3)3CH2 N CH2CH3 (d) N CH3

H Solution (a) Step 1 The parent alkane is methane. Step 2 The name of the compound is methanamine. No position number is needed. (b) Step 1 The largest group has three carbon atoms in the main chain, and two methyl groups on the second carbon atom. (This may be easier to see if you draw a full structural diagram of the alkyl branch.) The parent alkane is 2,2-dimethylpropane. Step 2 The nitrogen atom is attached at carbon 1, so the base name is 2,2-dimethyl-1-propanamine. Step 3 An ethyl group is attached to the nitrogen atom. The corresponding preﬁx is N-ethyl. Step 4 The full name is N-ethyl-2,2-dimethyl-1-propanamine. (c) Step 1 The parent alkane is butane. Step 2 The nitrogen atom is attached at carbon 2, numbering from the right, so the full name is 2-butanamine. (d) Step 1 The parent alkane is cyclopentane. Step 2 No position number is needed, since there are no other branches on the ring. The base name is cyclopentanamine. Step 3 The preﬁx is N,N-dimethyl-. Step 4 The full name is N,N-dimethylcyclopentanamine.

27. Problem Draw a condensed structural diagram for each amine. (a) 2-pentanamine

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(b) cyclohexanamine (c) N-methyl-1-butanamine (d) N,N-diethyl-3-heptanamine Solution (a) NH2

CH3 CH CH2 CH2 CH3

(b) CH2

CH2 CH NH2

CH2 CH2

CH2

(d) CH3 CH2 CH CH2 CH2 CH2 CH3

CH3 CH2 CH2 CH3

28. Problem Classify each amine in the previous question as primary, secondary, or tertiary. Solution (a) primary (b) primary (c) secondary (d) tertiary

29. Problem Draw and name all the isomers with the molecular formula C4H11N. Solution NH NH2

NH2 1-butanamine N-methyl-2-propanamine 2-butanamine

NH NH2 N

N-ethylethanamine 2-methyl-1-propanamine N,N-dimethylethanamine

NH NH2 N-methyl-1-propanamine 2-methyl-2-propanamine

Solutions for Practice Problems Student Textbook page 36

30. Problem Name each aldehyde or ketone. O O

(a)HC CH2 CH2 CH3 (b)

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O O

(c)CH3 CH C CH2 CH2 CH3 (d) CH3 CH2 CH CH

CH3 CH2CH3

(e)

Solution (a) Step 1 The parent alkane is butane. Step 2 The compound’s name is butanal. (b) Step 1 The parent alkane is octane. Step 2 Changing the suffix gives octanone. Step 3 The carbonyl group is on the third carbon from the right, so the full name is 3-octanone. (c) Step 1 The parent alkane is 2-methylhexane. Step 2 Changing the suffix gives 2-methylhexanone. Step 3 The carbonyl group is on carbon 3, so the full name is 2-methyl-3-hexanone. (d) Step 1 The parent alkane is 2-ethylbutane. (Note that the main chain must contain the carbonyl group.) Step 2 The full name is 2-ethylbutanal. (e) Step 1 The parent alkane is 4-ethyl-3-methylheptane. Step 2 Replacing the suffix gives 4-ethyl-3-methylheptanone. Step 3 The carbonyl group is at carbon 2, so the full name is 4-ethyl-3-methyl-2-heptanone.

31. Problem Draw a condensed structural diagram for each aldehyde or ketone. (a) 2-propylpentanal (b) cyclohexanone (c) 4-ethyl-3,5-dimethyloctanal Solution O

(a) HC CH CH2 CH2 CH3

CH2 CH2 CH3 (b) O

CH2 CH2

CH2

O CH2 CH3

CH3 CH3

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32. Problem Is a compound with a CO bond and the molecular formula C2H4O an aldehyde or a ketone? Explain. Solution This compound must be an aldehyde. There are only two carbon atoms, so it is not possible for the carbonyl group to have an alkyl group on each side.

33. Problem Draw and name ﬁve ketones and aldehydes with the molecular formula C6H12O. Solution O

O O 2-hexanone 3-hexanone hexanal

O 2-methylpentanal 4-methyl-2-pentanone

Solutions for Practice Problems Student Textbook page 40

34. Problem Name each carboxylic acid.

O CH3 O

(a)HO C CH2 CH3 (b) CH3 CH2 C C OH

CH3 O

Solution (a) Step 1 The parent alkane is propane. Step 2 Replacing the suffix gives propanoic acid. (b) Step 1 The parent alkane is butane. Step 2 Replacing the suffix gives butanoic acid. Step 3 The two methyl branches are both at carbon 2, since the carboxyl group is carbon 1. The full name is 2,2-dimethylbutanoic acid. (c) Step 1 The parent alkane is hexane. (Note: Even though a longer carbon chain is possible, the main chain must include the carboxyl group.) Step 2 Replacing the suffix gives hexanoic acid. Step 3 There is an ethyl group at carbon 2; and two methyl groups at carbon 4 and carbon 5. The full name is 2-ethyl-4,5-dimethylhexanoic acid.

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35. Problem Draw a condensed structural diagram for each carboxylic acid. (a) hexanoic acid (b) 3-propyloctanoic acid (c) 3,4-diethyl-2,3,5-trimethylheptanoic acid Solution O

(a) CH3 CH2 CH2 CH2 CH2 C OH O

(b) CH3 CH2 CH2 CH2 CH2 CH CH2 C OH

CH2 CH2 CH3

O CH3 CH3 CH2 CH3

CH2 CH3 CH3 36. Problem Draw a line structural diagram for each compound in question 35. Solution O O

(a) (b) OH OH

HO (c)

O 37. Problem Draw and name two carboxylic acids with the molecular formula C4H8O2. Solution O O

OH OH butanoic acid 2-methylpropanoic acid

Solutions for Practice Problems Student Textbook page 45

38. Problem Name each ester. O O

(a)CH3CH2 O CH (b) CH3CH2CH2C O CH3

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Solution (a) Step 1 The parent acid is methanoic acid. (Note that this is not the largest group; rather, it is the group that contains the CO bond.) Step 2 Replacing the suffix gives methanoate. Step 3 An ethyl group is attached to the oxygen atom. Step 4 The full name is ethyl methanoate. (b) Step 1 The parent acid is butanoic acid. Step 2 Replacing the suffix gives butanoate. Step 3 A methyl group is attached to the oxygen atom. Step 4 The full name is methyl butanoate. (c) Step 1 The parent acid is pentanoic acid. Step 2 Replacing the suffix gives pentanoate. Step 3 A pentyl group is attached to the oxygen atom. Step 4 The full name is pentyl pentanoate, or n-pentyl pentanoate.

39. Problem For each ester in the previous question, name the carboxylic acid and the alcohol that are needed to synthesize it. Solution (a) methanoic acid/ethanol (b) butanoic acid/methanol (c) pentanoic acid/1-pentanol

40. Problem Draw each ester. (a) methyl pentanoate (d) propyl octanoate (b) heptyl methanoate (e) ethyl 3,3-dimethylbutanoate (c) butyl ethanoate Solution O O (a) (d) O O O O

(b) (e) O O

(c) O 41. Problem Write the molecular formula of each ester in the previous question. Which esters are isomers of each other? Solution (a) C6H12O2 (d) C11H22O2 (b) C8H16O2 (e) C8H16O2 (c) C6H12O2 Isomers: (a) and (c); (b) and (e)

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42. Problem Draw and name ﬁve ester isomers that have the molecular formula C5H10O2. Solution O O O O O O methyl butanoate ethyl propanoate methyl 2-methylpropanoate O O O O butyl methanoate isopropyl ethanoate (n-butyl methanoate)

Solutions for Practice Problems Student Textbook page 48

43. Problem Name each amide. O N (a)CH3CH2CH2C NH2 (c)

O O

(b) H3C NH CCH2CH2CH2CH2CH3 Solution (a) Step 1 The parent acid is butanoic acid. Step 2 Replacing the suffix gives butanamide. Step 3 The compound is a primary amide, so no prefix is needed. The full name is butanamide. (b) Step 1 The parent acid is hexanoic acid. Step 2 Replacing the suffix gives hexanamide. Step 3 This is a secondary amide. The prefix is N-methyl-. Step 4 The full name is N-methylhexanamide. (c) Step 1 The parent acid is 3-methylheptanoic acid. Step 2 Replacing the suffix gives 3-methyl heptanamide. Step 3 The compound is a tertiary amide. The prefix is N,N-diethyl-. Step 4 The full name is N,N-diethyl-3-methylheptanamide.

44. Problem Draw each amide. (a) nonanamide (b) N-methyloctanamide (c) N-ethyl-N-propylpropanamide (d) N-ethyl-2,4,6-trimethyldecanamide

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Solution

NH2 N (a) (c)

O O O O

(b) (d) NH NH

45. Problem Name each amide. (a) CH3CONH2 (b) CH3CH2CH2CH2CH2CH2CONHCH3 (c) (CH3)2CHCON(CH3)2 Solution (a) Step 1 The parent acid is ethanoic acid. Step 2 Replacing the suffix gives ethanamide. This is the name of the compound. (b) Step 1 The parent acid is heptanoic acid. Step 2 Replacing the suffix gives heptanamide. Step 3 This is a secondary amide. The prefix is N-methyl-. Step 4 The full name is N-methylheptanamide. (c) Step 1 The parent acid is 2-methylpropanoic acid. Step 2 Replacing the suffix gives 2-methylpropanamide. Step 3 This is a tertiary amide. The prefix is N,N-dimethyl-. Step 4 The full name is N,N-dimethyl-2-methylpropanamide.

46. Problem Draw a line structural diagram for each amide in the previous question. Solution O

(a) (c) N NH2 O O

(b) NH

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