edited by
exam que~tionexchange Using Sequencing Questionsin Chemistry Examinations
JOHN J.
ALEXANDER University of Cincinnali Cincinnati. OH 45221
(1) Arrange the following compounds in order of decreasing boiling point by placing the letter representing the compound in the appropriate space on the right. a. pentane - (highest b.p.) h. 2-methylbutane c. 22-dimethylpropane d. heaane (lowest b.p.) ~~
Wllllam M. Hemmerlln Pacific Union College Angwin. CA 94508
~
Quesllon
Many teachers are searching for examination questions that are quick and easy to grade hut which require careful thoueht on the art of the student to answer. I have found "seq;encing" q;estions to be very useful. They are relatively easv to write, require considerable knowledge and thinking to Hnswer correctly, and are easy to Three typical examples follow:
(2) Arrange the following compounds in the order of decreasing
acidity. a. p-methoxybenzoic acid b. p-cyanobenzoic acid c. henzoic acid d. 1-propanol
(strongest acid) -
(weakest acid)
Partlal Credit Asignments in Four-Response Sequence Questions A B C D
A B D C
A C B D
A C D B
A D B C
A D C B
B A C D
B A D C
B C A D
B C D A
B D A C
B D C A
C A B D
C A D B
C B A D
C B D A
C D A B
C D B A
D A B C
D A C B
D B A C
D B C A
D C A B
D C B A
6 5 5 4 4 3
5 6 4 3 5 4
5 4 6 5 3 4
4 3 5 6 4 5
4 5 3 4 6 5
3 4 4 5 5 6
5 4 4 3 3 2
4 5 3 2 4 3
4 3 3 2 2 1
3 2 2 1 1 0
3 4 2 1 3 2
2 3 1 0 2 1
4 3 5 4 2 3
3 2 4 5 3 4
3 2 4 3 1 2
2 1 3 2 0 1
2 1 3 4 2 3
1 0 2 3 1 2
3 4 2 3 5 4
2 3 3 4 4 5
2 3 1 2 4 3
1 2 0 1 3 2
1 2 2 3 3 4
0 1 1 2 2 3
CABD CADB CBAD CBDA CDAB CDBA
4 3 3 2 2 1
3 2 2 1 1 0
5 1 4 3 3 2
4 2 3 5 3 4 3 1 2 2 0 1 4 2 3 3 1 2
3 2 4 3 1 2
2 1 3 2 0 1
4 3 5 4 2 3
3 2 4 5 3 4
1 0 2 3 1 2
2 1 3 4 2 3
DABC DACB DBAC OBCA DCAB DCBA
3 2 2 1 1 0
4 3 3 2 2 1
2 3 1 0 2 1
3 4 2 1 3 2
2 1 3 2 0 1
3 2 4 3 1 2
1 0 2 3 1 2
2 1 3 4 2 3
4 3 5 4 2 3
3 2 4 5 3 4
1 2 0 1 3 2
2 3 1 2 4 3
0 1 1 2 2 3
1 2 2 3 3 4
3 4 2 3 5 4
2 3 3 4 4 5
6 5 5 4 4 3
5 6 4 3 5 4
5 4 6 5 3 4
4 3 5 6 4 5
4 5 3 4 6 5
3 4 4 5 5 6
ABCD ABDC ACBD ACDB ADBC ADCB BAC D - BADC BCAD BCDA BDAC BDCA ~
532
5 4 4 3 3 2
4 5 3 2 4 3
Journal of Chemical Education
4) Arrange the following compounds in order of decreasing mart i v i r ~t ~ ~ v a electruphil~ rd anmntrc rul,rtitutim. a. benzoic acid - (fastest) h. toluene e. phenol d. benzene - (slowest)
Acceptable Solutlon Although this type of question can be graded "all or nothing", a partial credit system is easy to derive and is perceived as more fair by the student. Asan example, we lookat question # I for which the following sequence was given by a student: Student's Sequence
Correct Sequence
Part A The compound, 3-methylpentane, undergoes free radical chlorination to produce eight total monochloro compounds of formula C6HI3C1. Gas chromatography separates the product mixture into five components. Identify the components completely giving both their correct structure and name. Stereochemistry and correct usage of the R, S system of nomenclature is necessary. Part B Ootional. Calculate the oercentaee " comoosition of the five components of the mixture assuming the relative rates of hydrogen atom abstraction at constant reaction temperature is in the ratio of 30:2*:1° = 5:4:1.
Acceptable Solutlon
To answer this question correctly, the student must be familiar with two factors that influence boiling points in alkanes: (1) boiling point increases with increasing molecular weight, and (2) boiling point decreases with branching. When assigning partial credit for the response, it must he noted that the student correctly deduced that both 22-dimethylpropane and 2-methylbutane had lower boiling points than either pentane or hexane. Further, 2-methylbutane was assigned a higher boiling point than 2,2-dimethylpropane since it is less branched. The only error was in placing pentane higher than hexane. In grading this type of question, one point will be given for each correct relationship. There are a total of six relationships when four compounds are used: d should he higher than a, h, or e (three relationships); a should be higher than h and c (two relationships); and h should be higher than e (one relationship). The only relationship that thestudent didnot correctly identify was hetweend and a. Therefore, the answer will be assigned a seore of 5 out of apossihle 6. Since there are 24 possible combinations of sequences, it is helpful to create a table that lists all possible combinations and the credit for eachsequence. In the table the correct sequence can he found in either the horizontal or vertical set, and the credit can he obtained hy locating the number where the correct sequence column meets the column of the incorrect sequence. For question #1 the correct sequence is found in the horizontal set DABC while the incorrect sequence is found in the vertical set ADBC. A score of 5 is found at the intersection of the two columns. It is common to use a multiple-choice format for sequencing questions, hut the approach described above forces the student to come up with the entire sequence rather than picking from a set of sequences given by the instructor. Also, it provides for more flenihility inscoring the response in that partial credit isallowed in proportion to the number of relationships the student correctly deduced. Acknowledgment T h e suthur greatly apprecistes the assistance provided by Kentun Abel in designing the -partial credit table and helpful ~discussion regarding assignment of partial credit.
Part A Component I 50% (3R)-l-ehloro-3-methylpentane
FE CHcHCH,&4mm,,CI
Component 3 50% (2R,3S)-2-chloro-3-methyl~entanemcHCH, ,,,,,,,F--F""'cB
r f
Component 4 3-chloro-3-methylpentane
Component 5 3-chloromethylpentane
Stereoisomerlsmas a Result of Free Radical Chlorination of a Nonstereoisomeric Substance John E. Fulkroad University of Minnesota Technical College. Wareca Wasem. MN 56093 T h e r e has been a number of articles recently devoted t o stereochemistry and techniques t o aid students i n applying t h e Cahn, Ingold, a n d Prelog rules (1-6). Of course, i t is hoped t h a t after studvine stereochemistw a n d different teihniques of assigning th;: R or S configuration, students will be able t o recoanize enantiomers. disatereomers.. etc... and draw structures a n d name substances correctly. T h e followina question is useful anytime after t h e topics of free radical hacogenation, isomeris& a n d stereoche&try have been discussed.
Part S The five components must totsl 100%. The simplest solution involves letting component 4 be equal to x% (since there is only one 3O component) and then calculating the ratio of all other components compared to component 4. The probability factor multiplied by the ratio of the abstraction of 3' versus 2' versus 1' hydrogen atoms will give the component ratio. The relative probabilities of components 1through 5 based on the number of hydrogen atoms is 6:2:2:1:3. The ratios are found as follows
Component I _--no. of loH reactivity of I>, X Cunnponrnt 4 no. ot 3OH rracriviry of YH
Volume 84
Number 8
June 1987
533
reactivity of 2'H reactivity of 39H
Component 2 - no. of 2'H Component 4 no. of 3'H
Component 3 Component 4
-- (same as) Component 2
Component 2 = 1.6%= 26.67%
Component 4
- 1.6x
Component 3 = 1 . 6 = ~ 26.67% = 16.67% Component4 = x Component 5 = 0.61: = 10%
I:
Component 5 - no. of 1'H Component 4 no. of 3OH
X
Llterature Clted
reactivity of l 0 H reactivity of 3OH 3 =-Y
534
Journal of Chemical Education
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Huhoey, J.E., J . Chem.Edue. 1986.63.598. Aelund, M.P.: Pinmck, J. A,, J. Chrm. Educ. i986,63,6W Mstfern,D. L..J.ChemEduc. 1985.62.191, Beauchamp.P. S . J . Chem.Educ. 1384.61.686. Eneiand. D..J . Chrm. Ed-
