University Chemistry: Module 2 Scenario-Based Problem Assignment

This assignment addresses a scenario-based problem in organic chemistry, focusing on fixing errors in a chemical database. The student first identifies and draws the structural isomers of C5H10, providing their IUPAC names, and identifies those with geometric isomers, drawing and labeling the Z and E isomers. The assignment then analyzes different chemical reactions, identifying the reaction type, reactants, and products for various functional groups. Furthermore, the student names a complex chemical structure according to IUPAC nomenclature, identifies four functional groups present in the structure, determines the chemical formula and calculates the molecular weight of the compound. The solution demonstrates a comprehensive understanding of organic chemistry principles, including isomerism, nomenclature, reaction mechanisms, and functional group identification, providing a detailed breakdown of each question and its solution.

Running head: MODULE 2 SCENARIO-BASED PROBLEM 1
Module 2 Scenario-Based Problem: Fixing dis-functional organic chemical data
Firstname Lastname
Name of Institution

MODULE 2 SCENARIO-BASED PROBLEM 2
Question 1: Drawing the structural isomers of C5H10 with their IUPAC name
The molecular formula C5H10 has over 13 isomers where the alkenes ar upto 6 isomers having
both cis and trans (E/Z) of pent-2-ene. We also have cycloalkanes which include
methylcyclobutane, cis-1, 2-dimethylcyclopropane, 1,1-dimethylcyclopropane, Trans-1,2-
dimethylcyclopropane, ethylcyclopropane etc.
The isomers of C5H10 are as shown below, their IUPAC are also given in column 3, the IUPAC
name varies for example 1-pentene is same as pentene
Isomer
Number
Structural Isomer of C5H10 IUPAC name
1 2-Methylbut-2-ene
2 3- methylbutene
3 2-methylbutene
4 1-Pentene
5 2-pentene
6 ethyl cyclopropane

MODULE 2 SCENARIO-BASED PROBLEM 3
7 cyclopentane
8 methylcyclobutane
9 1,1-dimethylcyclopropane
10 1,2-dimethylcyclopropane
Question 1(b): Isomers of C5H10 having geometric isomers
Two of the above isomers have geometrical isomers. These include:
2-pentene
1,2-dimethylcyclopropane
The geometric isomers occurs where restriction has e placed particularly between double bond
2-pentene can have geometrical isomers as shown
1, 2-dimethylcyclopropane can form geometrical isomers as illustrated below. The geometrical
isomer is reached since C-C bond is restricted at cyclopropane ring. Having methyl groups on
opposite side, Tran’s isomer is formed
Question 1(c). Drawing and labels of Z and E isomers from part (b)
2-pentene

MODULE 2 SCENARIO-BASED PROBLEM 4
1,2-dimethylcyclopropane
Question 2: Naming the reaction type, reactant and product of functional groups
Reaction Reaction type Reactant
(functional group)
Product
(functional group)
1 Addition Ethene Ethane
2 Addition Ethene 1, 2 dibromoethane
3 Addition Ethene Bromoethane
4 Elimination Bromoethane Ethene
5 Acid/base Bromoethane ethanol
6 Substitution ethanol Chloroethane
7 Acid / base Acetic Acid Methyl acetate
Question 3: Giving the IUPAC name of the chemical structure given below
The name of the stricture is derived from the longest chain where we have the longest carbon
chain to be having seven carbon hence it belongs to heptane group while and it has two butyl
group at the 4th carbon and methyl group at the 2nd carbon. Hence, its name can be derived as
follows

MODULE 2 SCENARIO-BASED PROBLEM 5
IUPAC name: 4-dibutyl-2-methylheptane
Question 4. Four functional Groups present in the structure
Entry Names of Functional
groups
1 Alcohol
2 Aromatic Ring
3 Ketone
4 Ester
Question 4 (b). The chemical formulae of the organic compound is
The chemical formula of the organic compound is determined by considering the number of
carbon element, hydrogen element making the compound and lastly oxygen atom making the
element
C27H24O20
X = 27
Y = 24
Z =20
Question 4 (c). The molecular weight of the compound is computed as follows:

MODULE 2 SCENARIO-BASED PROBLEM 6
Considering that the relative atomic mass of the element making the compound i.e carbon ,
hydrogen and oxygen is as follows, Hydrogen H = 1.01 g/mol, Carbon C = 12.01 g/mol and
Oxygen O = 16.00 g/mol. The relative atomic mass given for elements making the compound
are replaced in the formula to give the total value of molecular weight. The formula shows that
we have 27 carbon atoms, 24 hydrogen atoms and 20 oxygen atom making the element. Thus
during calculation, these values are multiplied by the relative atomic mass. The weight value has
no units since it is a ratio or relative mass
Therefore molecular weight of the organic compound is equal to MW = C27H24O20 = (27 x 12.01)
+ (24 x 1.01) + (20 x 16.00) = 646.67

MODULE 2 SCENARIO-BASED PROBLEM 7
References
Ingold, C. K. (2006). Structure and mechanism in organic chemistry. Cornell University Press;
Ithaca; New York.
Smith, M. B., & March, J. (2007). March's advanced organic chemistry: reactions, mechanisms,
and structure. John Wiley & Sons.
Carey, F. A., & Sundberg, R. J. (2007). Advanced organic chemistry: part A: structure and
mechanisms. Springer Science & Business Media.
Greenberg, A., & Liebman, J. F. (2013). Strained Organic Molecules: Organic Chemistry: A
Series of Monographs (Vol. 38). Academic Press.
Roberts, J. D., & Caserio, M. C. (1977). Basic principles of organic chemistry. WA Benjamin,
Inc..