Chapter 1 Classifying Organic Compounds - Quia 1 Classifying Organic Compounds Note to Teacher:You will notice that there are two different formats for the Sample Problems in the student

Chapter 1

ClassifyingOrganic Compounds

Note to Teacher: You will notice that there are two different formats for the SampleProblems in the student textbook. Where appropriate, the Sample Problem contains thefull set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act onYour Strategy, and Check Your Solution. Where a shorter solution is appropriate, theSample 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 ProblemsStudent Textbook page 10

1. ProblemPredict and sketch the three-dimensional shape of each single-bonded atom.(a) C and O in CH3OH (b) C in CH4

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.

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

2. ProblemPredict and sketch the three-dimensional shape of each multiple-bonded molecule.(a) HC�CH (b) H2C�O

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.

(b) The carbon atom has one double bond and two single bonds, so the shape aroundthe carbon atom (and the shape of the molecule) will be trigonal planar.

C

O

HH

H C C H

C

H

HH H

CH3HO

••

around the oxygen atom

••

around the carbon atom

C

H

OHH H

1Chapter 1 Classifying Organic Compounds • MHR

CHEMISTRY 12

3. ProblemIdentify any polar bonds that are present in each molecule in questions 1 and 2.

SolutionThe molecule in question 1(a) has two polar bonds: C�O and O�H. The mole-cule in question 1(b) has only C�H bonds, which are usually considered to be non-polar.

The molecule in question 2(a) has only C�C and C�H bonds, which are usuallyconsidered to be non-polar.

The molecule in question 2(b) has one polar bond: C�O.

4. ProblemFor each molecule in questions 1 and 2, predict whether the molecule as a whole ispolar or non-polar.

SolutionFor question 1(a):Step 1 The molecule has polar bonds.Step 2 There are two polar bonds: C�O and O�HStep 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 ProblemsStudent Textbook pages 16–17

5. ProblemName each hydrocarbon.

(a) (e)

(b) (f)

(c) (g)

(d)

CH3

C CH3CH3 CH2CH

CH3

CH3

H2C CH

H2C CH

H3C CH3


CHEMISTRY 12

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. ProblemDraw 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 CH3�CH2�CH3


CHEMISTRY 12

(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

(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

7. ProblemIdentify 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 mainchain). The correct name is 5-ethyl-3-methylnonane.

(c) If you draw this compound, you will see that the third carbon atom forms morethan four covalent bonds. This is not possible. One solution would be to changethe name to 3-methyl-4,5-diethyl-2-nonene.

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

(a)

(b)

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 atcarbons 2 and 4. The correct name is 2,4-hexadiyne.

9. ProblemUse 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

2,5-hexene C C C C CH3CH3

CH24-hexyne CH CH3CH3 CH2CH

CH3 CH C

CH3

CH CH CH CH CH3

CH2

CH2

CH2

CH3

CHCHCH3

CH3

CH2 CH3


CHEMISTRY 12

Solution(a)

The propyl group should be part of the main chain. The correct name is 3-methyl-2-hexene.

(b)

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 doublebond. The correct name is 4-methyl-2-heptene.

(c)

You should number in the opposite direction to give the lowest possible positionnumber for the methyl group. The correct name is 2-methylpentane.


10. ProblemName the following aromatic compound.

SolutionStep 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. ProblemDraw 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)

12. ProblemGive another name for the compound in question 11(a).

SolutionThe compound can also be called meta-ethylmethylbenzene.

Cl

Cl

CH3

CH2CH3

CH3

CH2 CH3

CH3

CH3

CH3H3C

CH3

CH3

CH2➀➁➂➃➄

CH2 CH CH3

CH3

CH2 CH2 CH➀➁➂➃➄➅

CH3

CH CH CH3

➆

CH3

CH2

CH C CH3

CH2 CH3

➀ ➁ ➂

➃ ➄ ➅


CHEMISTRY 12

13. ProblemDraw 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.)

SolutionAside from the six carbons in the benzene ring, there are an additional four carbonatoms 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 aremany other possible isomers.

Solutions for Practice ProblemsStudent Textbook pages 26–27

14. ProblemName each alcohol. Identify it as primary, secondary, or tertiary.

(a) (d)

(b) (e)

(c)

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 isneeded. 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.

OH

OHOH

CH CH2 CH3CH3 CH

OH

OH

OHCH2CH3 CH2

CH2CH2CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3

CH3


CHEMISTRY 12

Step 3 The �OH groups are located on carbon 2 and carbon 3 of the mainchain. 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 amethyl 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. ProblemDraw 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.

(b) There are three carbon atoms in the main chain. The �OH group is at carbon 2.

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

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

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

OH

CH3

CH3

CH2 CH2CHCH

CH3

HO

CH2

CH2

CH3

CH2 CH2 CH3

CH2

OH

OH

CH3 CH3C

CH3

OH

CH3CH

CH3 OH


CHEMISTRY 12

16. ProblemIdentify any errors in each name. Give the correct name for the alcohol.

(a) 1,3-heptanol

(b) 3-ethyl-4-ethyl-1-decanol

(c) 1,2-dimethyl-3-butanol

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. ProblemSketch a three-dimensional diagram of methanol. Hint: Recall that the shape aroundan oxygen atom is bent.

SolutionNote to teacher: If students complete a three-dimensional diagram of methanol successfully, have them attempt a similar diagram of ethanol, as shown below.


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

C

H

HH

O

ethanol

C

• •

••

HH

H

C H

H

HH O •

••

•

methanol

CH CH3CH2 CH

CH3

CH3 OH

OH

CH CH2 CH3HO CH2 CH2

OH


CHEMISTRY 12

Solution(a) There are two carbon atoms in the main chain. No position number is needed

because the two carbon atoms are indistinguishable.CH3�CH2�Br

(b) There are seven carbon atoms in the main chain. There is one methyl group, atcarbon 3, and three iodine atoms, at carbons 2,3, and 4.

19. ProblemName the alkyl halide at the right. Then draw a condensed structural diagram to represent it.

SolutionThere 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.

20. ProblemDraw and name an alkyl halide that has three carbon atoms and one iodine atom.

Solution1-iodopropaneCH3�CH2�CH2�I

21. ProblemDraw and name a second, different alkyl halide that matches the description in theprevious question.

Solution2-iodopropane


22. ProblemUse the IUPAC system to name each ether.(a) (c)

(b) H3C

CH3

CH3

CHO

CH3OCH2CH2 CH2CH3OH3C CH3

CH3

I

CH3CH

CH2

CH2

CH

CH2

CH2

F

CH F

F

F

CH

I I

CH3 C

CH3

I

CH CH2CH2 CH3


CHEMISTRY 12

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, orjust 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. ProblemGive the common name for each ether.(a) (b)

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. ProblemDraw 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.CH3�O�CH2�CH2�CH3

(b) The parent alkane is 4-methylheptane. Draw the parent alkane, then add anethoxy group at the third carbon atom of the main chain.

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

25. ProblemSketch 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 inwater? Explain your reasoning.

CH3OC

CH3

CH3

CH3

CH CH

O

CH2CH3

CH3

CH2 CH2 CH3

CH3CH2

OH3C CH3OH3C CH2CH3


CHEMISTRY 12

SolutionThe alcohol will be more soluble in water because it has an O�H bond, and canform hydrogen bonds with water molecules. The ether can accept hydrogen bondsfrom water molecules, but cannot reciprocate them, as shown below.


26. ProblemName each amine.(a) (c)

(b) (d)

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 twomethyl groups on the second carbon atom. (This may be easier to see if you draw a full structural diagram of the alkyl branch.) The parentalkane 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. ProblemDraw a condensed structural diagram for each amine.(a) 2-pentanamine

CH3

N CH3C(CH3)3CH2 N

H

CH2CH3

CHCH2 CH3CH3

NH2

NH2CH3

CH3CH2 OCH3

HO

H

ether

CH3CH2CH2 OH

HO

H

O H

H

alcohol


CHEMISTRY 12

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

Solution(a)

(b)

(c) CH3�CH2�CH2�CH2�NH�CH3

(d)

28. ProblemClassify each amine in the previous question as primary, secondary, or tertiary.

Solution(a) primary (b) primary (c) secondary (d) tertiary

29. ProblemDraw and name all the isomers with the molecular formula C4H11N.

Solution


30. ProblemName each aldehyde or ketone.

(a) (b)

O

CH3CH2 CH2HC

O

2-methyl-2-propanamine

NH2

N-methyl-1-propanamine

NH

N,N-dimethylethanamine

N

2-methyl-1-propanamine

NH2

N-ethylethanamine

NH

2-butanamine

NH2

N-methyl-2-propanamine

NH

1-butanamine

NH2

CH3 CH2

N

CH2CH2CH CH2 CH3

CH3 CH2 CH2 CH3

CH2

CH2

CH

CH2

CH2

NH2

CH2

CH3

NH2

CH2CH2 CH3CH


CHEMISTRY 12

(c) (d)

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

Solution

(a)

(b)

(c) HC CH

O

CH3

CH

CH3

CH2 CH

CH3CH2

CH2CH2 CH3

C

O

CH2

CH2

CH2

CH2

CH2

HC CH

O

CH2

CH2CH2 CH3

CH2 CH3

O

CH3 CH

O

CH

CH2CH3

CH2CH2 CH3CH2CCH

CH3

O

CH3


CHEMISTRY 12

32. ProblemIs a compound with a C�O bond and the molecular formula C2H4O an aldehydeor a ketone? Explain.

SolutionThis compound must be an aldehyde. There are only two carbon atoms, so it is notpossible for the carbonyl group to have an alkyl group on each side.

33. ProblemDraw and name five ketones and aldehydes with the molecular formula C6H12O.

Solution


34. ProblemName each carboxylic acid.

(a) (b)

(c)

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 ispossible, 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.

O

OH

OHC

CH3

C

CH3

O

CH2CH3HO

O

C CH2 CH3

4-methyl-2-pentanone

O

2-methylpentanal

O

hexanal

O3-hexanone

O

2-hexanone

O


CHEMISTRY 12

35. ProblemDraw 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

(a)

(b)

(c)

36. ProblemDraw a line structural diagram for each compound in question 35.

Solution

(a) (b)

(c)

37. ProblemDraw and name two carboxylic acids with the molecular formula C4H8O2.

Solution


38. ProblemName each ester.

(a) (b)

(c)

O

CH3CH2CH2CH2C O CH2CH2CH2CH2CH3

O

CH3CH2CH2C O CH3

O

CH3CH2 O CH

OH

O

2-methylpropanoic acid

OH

O

butanoic acid

HO

O

OH

O

OH

O

CHCHO

O CH3 CH3

CHC

CH3CH2

CH

CH3CH3CH2

CH3CH2

CH2 CH2 CHCH2CH3 CH2 CH2

O

OHC

CH2 CH2 CH3

CH3 CH2 CH2CH2 CH2

O

OHC


CHEMISTRY 12

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 C�O 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. ProblemFor each ester in the previous question, name the carboxylic acid and the alcohol thatare needed to synthesize it.

Solution(a) methanoic acid/ethanol(b) butanoic acid/methanol(c) pentanoic acid/1-pentanol

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

Solution

(a) (d)

(b) (e)

(c)

41. ProblemWrite the molecular formula of each ester in the previous question. Which esters areisomers of each other?

Solution(a) C6H12O2 (d) C11H22O2

(b) C8H16O2 (e) C8H16O2

(c) C6H12O2

Isomers: (a) and (c); (b) and (e)

O

O

O

O

O

O

O

O

O

O


CHEMISTRY 12

42. ProblemDraw and name five ester isomers that have the molecular formula C5H10O2.

Solution


43. ProblemName each amide.

(a) (c)

(b)

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

NH CCH2CH2CH2CH2CH3

O

H3C

O

NCH3CH2CH2C

O

NH2

O

O

butyl methanoate(n-butyl methanoate)

O

O

isopropyl ethanoate

O

O

methyl 2-methylpropanoate

O

O

ethyl propanoate

O

O

methyl butanoate


CHEMISTRY 12

Solution

(a) (c)

(b) (d)

45. ProblemName 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. ProblemDraw a line structural diagram for each amide in the previous question.

Solution

(a) (c)

(b)NH

O

N

O

NH2

O

NH

O

NH

O

N

O

NH2

O


CHEMISTRY 12

Documents

Chapter 1 Classifying Organic Compounds - Quia 1 Classifying Organic Compounds Note to Teacher:You will notice that there are two different formats for the Sample Problems in the student