CHM 1025C Homework Packet    Name: __________________


Module 16  Chapter 19: Introduction to Organic Chemistry

A. ______ (02) Alkane Series Section 19.3 Table 19.1 p 445 Answers

B. ______ (02) Alkyl Radicals Section 19.3 Table 19.2 p 449 Answers
C. ______ (08) Structural Isomer Problem Section 19.3 446-450

D. ______ (02) Recognition of 1o, 2o, 3o, 4o carbons ( 0o, 1o, 2o, 3o , 4o) & hydrogen  Answers
E. ______ (05) Nomenclature of Alkanes and Cycloalkanes Section 19.3-5 p445-455 Answers
F. ______ (03) Functional Group Recognition Section 19.6 Table 19.4 p460 Answers
    _______(22) = ______% Total Chapter 19


Take-Home Lab Part C Structural Isomer Number Problems

    ______ (01) Structural Isomer Take-Home Problem #1 C6H14  (makeup-See #7)

    ______ (04) Structural Isomer Take-Home  Problem #2 C7H16  (makeup C8H18)

    ______ (04) Structural Isomer Take-Home Problem #3 C5H11Br (makeup C6H13Br)

    ______ (04) Structural Isomer Take-Home Problem #4 C4H8Br2 (makeup C5H10Br2)     

    ______ (04) Structural Isomer Take-Home Problem #5 C6H12 Cycloalkanes only (makeup C7H14)

    ______ (04) Structural Isomer Take-Home Problem #6 C6H12 Alkenes only (makeup C7H14)

    ______ (00)  Makeup Structural Isomer Take-Home Problem C4H10O   Alcohols & Ethers

________(21) Module 16 Take-Home Lab Total


Foundations of College Chemistry, 14th Edition

Table of Contents


Chapter 19 Introduction to Organic Chemistry 441

19.1 The Beginnings of Organic Chemistry 442

19.2 Why Carbon? 442

19.3 Alkanes 445

19.4 Alkenes and Alkynes 452

19.5 Aromatic Hydrocarbons 456

19.6 Hydrocarbon Derivatives 459

19.7 Alcohols 461

19.8 Ethers 465

19.9 Aldehydes and Ketones 467

19.10 Carboxylic Acids 469

19.11 Esters 471

19.12 Polymers—Macromolecules 473

Review 474

Review Questions, Paired Exercises 477

Additional Exercises 481

Answers to Practice Exercises 482










Module 16 Alkane Series   2 points


Name the first ten members of the alkane series and give its chemical, semi-structural, or structural  formula:

                             Name                                          Chemical Formula






























See Section 19.3  Table 19.1 Page 445
















CHM 1025C                  


Module 16 (Chapter 19) Part B: Alkyl Radicals   2 points


Draw the structural or semi-structural formulas for all the alkyl radicals of the first four members of the alkane series, insert a X in place of the hydrogen which is removed to make the radical:


Methyl                                                                                                Ethyl










n-propyl                                                                                          isopropyl











n-butyl                                                                                               sec-butyl










Isobutyl                                                                                 t-butyl or tert-butyl










See Table 19.2  Section 19.2  Pages 554-555



Module 16  Part B1: Additional Alkyl and Aryl Radicals   1 bonus point

Draw the structural or semi-structural formulas for all the alkyl/aryl radicals of the following. Place a X in place of the hydrogen which is removed to make the radical: (Not in Textbook)

n-pentyl (amyl)                             isopentyl or isoamyl




neopentyl                                    sec-amyl or sec-pentyl




 Allyl                                             vinyl



 Benzyl                                        phenyl





o-tolyl                               m-tolyl                                 p-tolyl





 CHM 1025C M-16 Homework Packet

Module 16 (former M-4III) Part C: Structural Isomer Number Problem:

Alkanes, Alkyl halides, and Cycloalkanes   8 points


Draw the structural or semi-structural formulas for all the isomers of the following chemical formulas, then give the IUPAC name for each: (Use additional paper)


Test Item #1 C6H14   5 isomers;   1 point  (no Makeup)

Download Real Player and then Download Hexane Isomer Video:




















Test Item #2 C5H12 and C4H10   5 total isomers between the two alkanes 1 point

















See Section 19.2 Guidelines for Drawing Hydrocarbon Isomers Table p553
See Section 19.2 Guidelines for Naming Alkanes p355-6  See
Exercise 19.1 p556

Web Page: Structure and Nomenclature of Hydrocarbons:

Naming Organic Compounds:

IUPAC Rules for Alkane Nomenclature

 1.   Find and name the longest continuous carbon chain.
 2.   Identify and name groups attached to this chain.
 3.   Number the chain consecutively, starting at the end nearest a substituent group.
 4.   Designate the location of each substituent group by an appropriate number and name.
 5.   Assemble the name, listing groups in alphabetical order using the full name (e.g. cyclopropyl before isobutyl).
    The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.

Test Item#3  C4H9Br  4 isomers; 1 point  (no makeup)

Organic Halides: Reference Section 19.6 Pages 564-565

Directions take four butyl radicals in Part B and replace the X with Bromine and name the alkyl bromides using the IUPAC

















Test Item#4 C3H6Br2   4 isomers;   1 Point (No Makeup)




















      Cycloalkanes have one or more rings of carbon atoms. The simplest examples of this class consist of a single, unsubstituted carbon ring, and these form a homologous series similar to the unbranched alkanes. The IUPAC names of the first five members of this series are given in the following table. The last (yellow shaded) column gives the general formula for a cycloalkane of any size. If a simple unbranched alkane is converted to a cycloalkane two hydrogen atoms, one from each end of the chain, must be lost. Hence the general formula for a cycloalkane composed of n carbons is CnH2n. Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so such compounds are still considered to be saturated with hydrogen.

Examples of Simple Cycloalkanes


















Substituted cycloalkanes are named in a fashion very similar to that used for naming branched alkanes. The chief difference in the rules and procedures occurs in the numbering system. Since all the carbons of a ring are equivalent (a ring has no ends like a chain does), the numbering starts at a substituted ring atom.

IUPAC Rules for Cycloalkane Nomenclature

 1.   For a monosubstituted cycloalkane the ring supplies the root name (table above) and the substituent group is named as usual. A location number is unnecessary.
 2.   If the alkyl substituent is large and/or complex, the ring may be named as a substituent group on an alkane.
 3.   If two different substituents are present on the ring, they are listed in alphabetical order, and the first cited substituent is assigned to carbon #1. The numbering of ring carbons then continues in a direction (clockwise or counter-clockwise) that affords the second substituent the lower possible location number.
 4.   If several substituents are present on the ring, they are listed in alphabetical order. Location numbers are assigned to the substituents so that one of them is at carbon #1 and the other locations have the lowest possible numbers, counting in either a clockwise or counter-clockwise direction.
 5.   The name is assembled, listing groups in alphabetical order and giving each group (if there are two or more) a location number. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.




CHM 1025C M-16 Homework Packet

Test Item #5 C4H8  5 total isomers (cycloalkanes plus alkenes) 1 Point No Makeup  





















Test Item #6 C5H10 5 isomers (cycloalkanes only); 1 Point (No Makeup)






























Test Item #7: C5H10  5 isomers (alkenes only) 1 Point (No Makeup)

Reference Section 19.3 pages 557-558

Read Guidelines for Naming Alkenes p557-558 Work Exercise 19.2

IUPAC Rules for Alkene and Cycloalkene Nomenclature

 1.   The ene suffix (ending) indicates an alkene or cycloalkene.
 2.   The longest chain chosen for the root name must include both carbon atoms of the double bond.
 3.   The root chain must be numbered from the end nearest a double bond carbon atom. If the double bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts.
 4.   The smaller of the two numbers designating the carbon atoms of the double bond is used as the double bond locator. If more than one double bond is present the compound is named as a diene, triene or equivalent prefix indicating the number of double bonds, and each double bond is assigned a locator number.
 5.   In cycloalkenes the double bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule.
 6.   Substituent groups containing double bonds are:
2C=CH–   Vinyl group
2C=CH–CH2   Allyl group



















CHM 1025C M-16 Homework Packet

Test Item #8  C6H4Br2  3 aromatic isomers 1 Point (No Makeup)

Reference: Section 19.4 Arenes Pages 561-562










CHM 1025C M-16 Homework Packet

Problem #2 C7H16  9 isomers (makeup C8H18)  4 points

Hint: Parent: one heptane; two hexanes; five pentanes; and one butane


















































CHM 1025C M-16 Homework Packet

Problem #3 C5H11Br  8 isomers (makeup C6H13Br)  4 Points


















































CHM 1025C M-16 Homework Packet

Problem #4: C4H8Br2   9 isomers   (makeup C5H10Br2) 4 Points




























































CHM 1025C M-16 Homework Packet

Problem#5 C6H1213 isomers (Cycloalkanes only (makeup C7H14) 4 Points












































































 CHM 1025C M-16  Homework Packet  

Problem #6 C6H1211isomers (alkenes only) (makeup C7H14) 4 Points













































CHM 1025C M-16 Homework Packet

Module 16 Chapter 19  Part D:                          2 points

Recognition of 1o, 2o, 3o, 4o carbons & 1o, 2o, 3o hydrogen

Classify the carbon or hydrogen atom in the below structure as 1o, 2o, 3o, 4o

or primary, secondary, tertiary, or neo (or quaternary)  carbon or hydrogen atoms: The numbers below on each image refer to the parent hydrocarbon of the chain::

Image #2





1. C1:_____

2. C2: _____

3. H2: _____

4. C4: _____

5. C5 :_____

6. C6: _____



7. C5: _____

8.  C3:_____

9.  C2:_____

10. C1:_____



11. C5:_____

12. C4:_____

13. H3_____

14. C2:_____

15.C1: _____



16. C5:_____

17. H4:_____

18. C3:_____

19. C2:_____

20. C1:_____


No Reading Reference: See Notes from Lecture/See Web Site Links
0o; 1o; 2o; 3o; 4o Carbon Atoms :Methyl/Primary Carbon Atoms    Secondary Carbon Atom     Tertiary Carbon Atoms     Neo Carbon Atoms



CHM 1025C Module 16 Homework Packet

Module 16   Part E: Nomenclature of Alkanes and Cycloalkanes   5 points


Give the IUPAC Name for the following compounds:





1. ___________________________






2. ___________________________








3. ____________________________







4. ___________________________






5. ___________________________








Classification of Organic Compounds Via Functional Groups








CHM 1025C  Module-16 Homework Packet           

Module 16 Part F: Functional Group Recognition   3 points


Classify the following compounds according to their Functional Group:

A. Alkane                  F. Aromatic Hydrocarbon   K. Ketone               P. Amide

B. Alkene                  G. Alkyl/Aryl Halide               L. Carboxylic Acid

C. Alkyne                  H. Alcohol                                M. Ester

D. Cycloalkane        I.  Ether                                    N. Amine

E. Cycloalkene        J.  Aldehyde                          O. Amino Acid

1. ___

2. ___

3. ___

4. ___

5. ___

6. ___

7. ___

8. ___

9. ___

10. __

11. __

12. __

13. __

14. __

15. __

16. __

1.   2.    3.     4.

5.        6.    7.  

8. 9.

10.     11.

12.    13.

14.   15. 16.