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CHEM 210 MODULE 2 ACTUAL PAPER 2026 QUESTIONS WITH SOLUTIONS GRADED A PLUS COMPLETE REVIEW SET.pdf

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CHEM 210 MODULE 2 ACTUAL PAPER 2026 QUESTIONS WITH SOLUTIONS GRADED A PLUS COMPLETE REVIEW SET.pdf

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CHEM 210 MODULE 2 ACTUAL PAPER 2026
QUESTIONS WITH SOLUTIONS GRADED A
PLUS COMPLETE REVIEW SET

◉ What are the two major classes of hydrocarbons?
Answer: Aromatic (benzene) and aliphatic (alkanes, alkenes,
alkynes)


◉ What is the main source of hydrocarbons?
Answer: Fossil fuels like coal, petroleum, and natural gas obtained
by mining and drilling.


◉ What are Alkanes?
Answer: They are saturated hydrocarbons because their skeleton is
only composed of C-C single bonds.


◉ How can alkanes exist?
Answer: They can be linear, branched or exist in ring formations
called cycloalkanes.


◉ Describe a higher/more complex alkane structure:

,Answer: They will have more carbons and/or increased structural
diversity.


◉ What is the general formula of Alkanes?
Answer: CnH2n+2


◉ The names of the alkanes reflect?
Answer: The number of carbons present.


◉ How many isomers does each alkane have from 1-10
Answer: 1,1,1,2,3,5,9,18,35,75


◉ Name the first ten prefix of alkanes:
Answer: Methane, ethane, propane, butane, pentane, hexane,
heptane, octane, nonane, decane.


◉ As the molecular weight increases:
Answer: The number of isomers increase.


◉ What does the suffix -ane mean?
Answer: Denotes a saturated hydrocarbon. It is used for all acyclic
saturated alkanes

, ◉ What is nomenclature?
Answer: The naming of a molecular structure. Historically they were
named after their source, now they are systemic.


◉ Why is nomenclature important?
Answer: Because we cannot remember all the prefixes since a MF
composed of carbons can have several different patterns of
connectivity.


◉ What is it meant by the systemic approach?
Answer: There are now methods of naming molecules so that they
all have unique names. You can then determine the structure from
only the name and vice-versa.


◉ IUPAC rules for naming branced alkanes:
Answer: 1. Find the parent name by identifying the longest
continuous chain of carbons.
2. Number the parent chain. (begin at the end of the chain nearest to
the first branch.)
3. If there are two equally long continuous chains, select the one
with most branches.
4. If the first branch point is the same distance, choose the one that
gives the lowest substituents.

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