CHEM 219 | CHEM219 Module 4: Principles of
Organic Chemistry with Lab - Portage Learning
Updated and Latest Questions and Correct
Answers with Rationale
1. What is the hybridization and bond angle of the carbon atoms involved in a double bond in
an alkene?
A. sp2, 120 degrees
B. sp, 180 degrees
C. sp3, 109.5 degrees
D. sp2, 109.5 degrees
Correct Answer: A
Expert Explanation: Alkene carbons involved in a double bond are sp2 hybridized
because they are bonded to three other atoms. The geometry associated with sp2
hybridization is trigonal planar. This arrangement results in bond angles of approximately
120 degrees. This specific geometry allows for the sideways overlap of p-orbitals to form
the pi bond. Understanding this geometry is fundamental to predicting the reactivity and
physical properties of alkenes.
2. Which of the following molecules can exhibit geometric (cis/trans) isomerism?
A. 1-Butene
B. 2-Methyl-2-butene
C. Propene
D. 2-Butene
Correct Answer: D
Expert Explanation: Geometric isomerism occurs when each carbon of the double bond is
attached to two different groups. In 2-butene, each carbon is bonded to a hydrogen and a
methyl group, allowing for cis and trans arrangements. 1-butene fails this because the first
carbon has two hydrogen atoms. Similarly, 2-methyl-2-butene has two methyl groups on
one carbon, preventing isomerism. Recognizing these requirements is essential for
determining the structural diversity of unsaturated hydrocarbons.
3. In the E/Z system of nomenclature, how is the priority of groups determined?
A. By the length of the carbon chain
B. By the number of hydrogen atoms attached to the group
,C. By the total molecular weight of the group
D. By the atomic number of the atoms directly attached to the double bond
Correct Answer: D
Expert Explanation: The E/Z system uses the Cahn-Ingold-Prelog (CIP) priority rules
based on atomic numbers. Atoms with higher atomic numbers receive higher priority when
comparing groups on a single carbon. If the atoms directly attached are the same, the
system looks at the next set of atoms along the chain. ‘Z’ stands for ‘zusammen’ (together)
while ‘E’ stands for ‘entgegen’ (opposite). This system provides a more comprehensive way
to name complex alkenes than the simple cis/trans method.
4. What is the expected observation when an alkene reacts with a solution of bromine in
carbon tetrachloride (Br2/CCl4)?
A. The reddish-brown color persists
B. The solution turns bright blue
C. A white precipitate forms immediately
D. The reddish-brown color disappears
Correct Answer: D
Expert Explanation: The bromine test is a classic laboratory procedure used to detect
unsaturation in organic compounds. Bromine is a reddish-brown liquid that reacts with the
double bond of an alkene via electrophilic addition. As the bromine is consumed to form a
colorless dibromoalkane, the original color fades. This rapid decolorization indicates the
presence of pi bonds in the sample. Alkanes do not react in this manner under normal
laboratory conditions without light.
5. Which reagent is used in the Baeyer test to identify the presence of double or triple bonds?
A. Silver nitrate in ammonia
B. Dilute aqueous potassium permanganate
C. Concentrated sulfuric acid
D. Sodium hydroxide and iodine
Correct Answer: B
Expert Explanation: The Baeyer test utilizes a cold, dilute solution of potassium
permanganate (KMnO4). When an alkene is added, the purple color of the permanganate
ions disappears. In its place, a dark brown precipitate of manganese dioxide (MnO2) forms.
This reaction involves the syn-hydroxylation of the alkene to produce a vicinal diol. It
serves as a qualitative test for unsaturation similar to the bromine test.
, 6. According to Markovnikov’s rule, in the addition of HBr to an asymmetric alkene, where
does the hydrogen atom attach?
A. To the carbon with the most alkyl groups
B. To either carbon with equal probability
C. To the carbon with the least hydrogen atoms
D. To the carbon with the most hydrogen atoms
Correct Answer: D
Expert Explanation: Markovnikov’s rule predicts the regioselectivity of electrophilic
addition reactions. It states that the hydrogen atom adds to the carbon atom of the double
bond that already has more hydrogens. This pathway is preferred because it leads to the
formation of the more stable carbocation intermediate. For example, in propene, the
hydrogen adds to C1, creating a secondary carbocation at C2. Understanding this rule is
crucial for predicting the major product of addition reactions.
7. What is the primary product formed when propene reacts with water in the presence of an
acid catalyst?
A. 1-Propanol
B. Cyclopropanol
C. Propanal
D. 2-Propanol
Correct Answer: D
Expert Explanation: The acid-catalyzed hydration of propene follows Markovnikov’s rule.
The reaction begins with the protonation of the double bond to form the more stable
secondary carbocation. Water then acts as a nucleophile, attacking the secondary
carbocation at the second carbon. Subsequent deprotonation results in the formation of 2-
propanol (isopropanol). This reaction demonstrates how hydration can convert alkenes
into functional alcohols.
8. Which of the following carbocations is the most stable?
A. Methyl carbocation
B. Primary carbocation
C. Tertiary carbocation
D. Secondary carbocation
Correct Answer: C
Organic Chemistry with Lab - Portage Learning
Updated and Latest Questions and Correct
Answers with Rationale
1. What is the hybridization and bond angle of the carbon atoms involved in a double bond in
an alkene?
A. sp2, 120 degrees
B. sp, 180 degrees
C. sp3, 109.5 degrees
D. sp2, 109.5 degrees
Correct Answer: A
Expert Explanation: Alkene carbons involved in a double bond are sp2 hybridized
because they are bonded to three other atoms. The geometry associated with sp2
hybridization is trigonal planar. This arrangement results in bond angles of approximately
120 degrees. This specific geometry allows for the sideways overlap of p-orbitals to form
the pi bond. Understanding this geometry is fundamental to predicting the reactivity and
physical properties of alkenes.
2. Which of the following molecules can exhibit geometric (cis/trans) isomerism?
A. 1-Butene
B. 2-Methyl-2-butene
C. Propene
D. 2-Butene
Correct Answer: D
Expert Explanation: Geometric isomerism occurs when each carbon of the double bond is
attached to two different groups. In 2-butene, each carbon is bonded to a hydrogen and a
methyl group, allowing for cis and trans arrangements. 1-butene fails this because the first
carbon has two hydrogen atoms. Similarly, 2-methyl-2-butene has two methyl groups on
one carbon, preventing isomerism. Recognizing these requirements is essential for
determining the structural diversity of unsaturated hydrocarbons.
3. In the E/Z system of nomenclature, how is the priority of groups determined?
A. By the length of the carbon chain
B. By the number of hydrogen atoms attached to the group
,C. By the total molecular weight of the group
D. By the atomic number of the atoms directly attached to the double bond
Correct Answer: D
Expert Explanation: The E/Z system uses the Cahn-Ingold-Prelog (CIP) priority rules
based on atomic numbers. Atoms with higher atomic numbers receive higher priority when
comparing groups on a single carbon. If the atoms directly attached are the same, the
system looks at the next set of atoms along the chain. ‘Z’ stands for ‘zusammen’ (together)
while ‘E’ stands for ‘entgegen’ (opposite). This system provides a more comprehensive way
to name complex alkenes than the simple cis/trans method.
4. What is the expected observation when an alkene reacts with a solution of bromine in
carbon tetrachloride (Br2/CCl4)?
A. The reddish-brown color persists
B. The solution turns bright blue
C. A white precipitate forms immediately
D. The reddish-brown color disappears
Correct Answer: D
Expert Explanation: The bromine test is a classic laboratory procedure used to detect
unsaturation in organic compounds. Bromine is a reddish-brown liquid that reacts with the
double bond of an alkene via electrophilic addition. As the bromine is consumed to form a
colorless dibromoalkane, the original color fades. This rapid decolorization indicates the
presence of pi bonds in the sample. Alkanes do not react in this manner under normal
laboratory conditions without light.
5. Which reagent is used in the Baeyer test to identify the presence of double or triple bonds?
A. Silver nitrate in ammonia
B. Dilute aqueous potassium permanganate
C. Concentrated sulfuric acid
D. Sodium hydroxide and iodine
Correct Answer: B
Expert Explanation: The Baeyer test utilizes a cold, dilute solution of potassium
permanganate (KMnO4). When an alkene is added, the purple color of the permanganate
ions disappears. In its place, a dark brown precipitate of manganese dioxide (MnO2) forms.
This reaction involves the syn-hydroxylation of the alkene to produce a vicinal diol. It
serves as a qualitative test for unsaturation similar to the bromine test.
, 6. According to Markovnikov’s rule, in the addition of HBr to an asymmetric alkene, where
does the hydrogen atom attach?
A. To the carbon with the most alkyl groups
B. To either carbon with equal probability
C. To the carbon with the least hydrogen atoms
D. To the carbon with the most hydrogen atoms
Correct Answer: D
Expert Explanation: Markovnikov’s rule predicts the regioselectivity of electrophilic
addition reactions. It states that the hydrogen atom adds to the carbon atom of the double
bond that already has more hydrogens. This pathway is preferred because it leads to the
formation of the more stable carbocation intermediate. For example, in propene, the
hydrogen adds to C1, creating a secondary carbocation at C2. Understanding this rule is
crucial for predicting the major product of addition reactions.
7. What is the primary product formed when propene reacts with water in the presence of an
acid catalyst?
A. 1-Propanol
B. Cyclopropanol
C. Propanal
D. 2-Propanol
Correct Answer: D
Expert Explanation: The acid-catalyzed hydration of propene follows Markovnikov’s rule.
The reaction begins with the protonation of the double bond to form the more stable
secondary carbocation. Water then acts as a nucleophile, attacking the secondary
carbocation at the second carbon. Subsequent deprotonation results in the formation of 2-
propanol (isopropanol). This reaction demonstrates how hydration can convert alkenes
into functional alcohols.
8. Which of the following carbocations is the most stable?
A. Methyl carbocation
B. Primary carbocation
C. Tertiary carbocation
D. Secondary carbocation
Correct Answer: C
