SCH4U GRADE 12 CHEMISTRY FINAL
EXAM COMPREHENSIVE 150 MCQ
EVALUATION WITH CORRECT
VERIFIED ANSWERS AND DETAILED
RATIONALES GRADE A+ PREMIUM
RESOURCE | INSTANT DOWNLOAD
1. Which of the following statements best describes the primary
reason why primary alcohols have significantly higher boiling
points than their corresponding isomeric ethers?
A) Primary alcohols exhibit strong intermolecular hydrogen
bonding due to the highly polar \(-OH\) group, whereas ethers
only experience weaker dipole-dipole and London dispersion
forces.
B) Ethers contain a bent molecular geometry around the oxygen
atom which prevents close packing in the liquid state, drastically
lowering their boiling points.
C) The carbon-oxygen bonds in ethers are non-polar, preventing
them from interacting with adjacent molecules through any
electrostatic attractions.
D) Primary alcohols possess a greater molecular mass than their
corresponding isomeric ethers, which naturally increases their
London dispersion forces.
Correct Answer: A
Rationale: The presence of the hydroxyl group (\(-OH\)) in
primary alcohols allows for the formation of strong
intermolecular hydrogen bonds between molecules. Isomeric
ethers contain a carbon-oxygen-carbon linkage which, while
polar, lacks a hydrogen atom directly bonded to an
electronegative oxygen atom, meaning they cannot form
intermolecular hydrogen bonds with themselves. This difference
in intermolecular force strength requires significantly more
thermal energy to overcome in alcohols.
2. An organic compound with the molecular formula
\(C_{4}H_{8}O\) does not undergo oxidation when treated with
, mild oxidizing agents such as Tollens' reagent, but it can be
reduced back to a secondary alcohol using lithium aluminum
hydride (\(LiAlH_{4}\)). Which of the following is the IUPAC
name for this compound?
A) Butanal
B) Butan-1-ol
C) Butanone
D) Cyclobutanone
Correct Answer: C
Rationale: The molecular formula \(C_{4}H_{8}O\) fits the
general formula for an aliphatic aldehyde or ketone
(\(C_nH_{2n}O\)). Because the compound does not undergo
oxidation with a mild oxidizing agent like Tollens' reagent, it
cannot be an aldehyde (such as butanal). Ketones resist mild
oxidation but are readily reduced by strong reducing agents like
\(LiAlH_{4}\) to yield secondary alcohols. Therefore, the
compound must be the four-carbon ketone, which is butanone.
3. Consider the elimination reaction of 2-bromobutane in the
presence of a strong base like sodium ethoxide. According to
Zaitsev's rule, what is the major organic product formed and why?
A) 1-butene, because the elimination of a primary hydrogen atom
is kinetically favored due to less steric hindrance.
B) but-2-ene, because the more highly substituted alkene is
thermodynamically more stable due to hyperconjugation.
C) 2-ethoxybutane, because substitution reactions always
outcompete elimination reactions when using alkoxide bases.
D) butadiene, because the removal of multiple halogen atoms
increases the overall entropy of the system.
Correct Answer: B
Rationale: Zaitsev's rule states that in an elimination reaction,
the major alkene product is the one that is more highly substituted
(the double bond is attached to more carbon substituents). For 2-
bromobutane, elimination can yield either 1-butene or but-2-ene.
But-2-ene is the major product because the internal double bond
is stabilized by a greater degree of alkyl group substitution and
hyperconjugation, making it thermodynamically more stable.
4. Which of the following sequences correctly ranks the functional
groups in order of decreasing priority according to IUPAC
, nomenclature rules for multi-functional organic compounds?
A) Carboxylic acid > Ester > Aldehyde > Ketone > Alcohol > Amine
B) Alcohol > Amine > Ketone > Aldehyde > Ester > Carboxylic acid
C) Ester > Carboxylic acid > Alcohol > Ketone > Aldehyde > Amine
D) Carboxylic acid > Alcohol > Aldehyde > Ester > Amine > Ketone
Correct Answer: A
Rationale: IUPAC nomenclature rules establish a strict
hierarchy of functional group priority for numbering and naming
multi-functional molecules. Carboxylic acids hold the highest
priority, followed by derivatives like esters, then carbonyl groups
where aldehydes outperform ketones. Hydroxyl groups (alcohols)
come next, followed by amino groups (amines), alkenes, alkynes,
and finally alkanes or halogens.
5. What type of reaction mechanism is primarily responsible for the
polymerization of ethene into polyethene under high pressure and
in the presence of an organic peroxide initiator?
A) Electrophilic addition polymerization
B) Free-radical addition polymerization
C) Nucleophilic substitution polymerization
D) Condensation elimination polymerization
Correct Answer: B
Rationale: The polymerization of ethene to polyethene using an
organic peroxide initiator follows a free-radical mechanism. The
peroxide undergoes homolytic cleavage to generate highly
reactive free radicals, which attack the \(\pi \)-bond of an ethene
molecule, generating a new carbon-centered radical. This radical
attacks another ethene monomer, propagating the chain. Because
no small molecules are eliminated during this process, it is an
addition polymerization driven by free radicals.
6. According to the quantum mechanical model of the atom, which of
the following sets of quantum numbers \((n, l, m_l, m_s)\) is
physically permissible for an electron occupying a valence orbital
in a ground-state sulfur atom?
A) \((3, 2, -1, +\frac{1}{2})\)
B) \((3, 1, 0, -\frac{1}{2})\)
C) \((2, 1, +1, +\frac{1}{2})\)
D) \((3, 0, +1, -\frac{1}{2})\)
Correct Answer: B
, Rationale: Sulfur has an atomic number of 16, with a ground-
state electron configuration of \([Ne] 3s^2 3p^4\). Its valence
electrons reside in the \(n = 3\) shell. For the \(3s\) orbital, \(l =
0\), meaning \(m_{l}\) must be 0. For the \(3p\) orbitals, \(l =
1\), which allows \(m_{l}\) values of \(-1, 0, +1\). Choice B has
\(n = 3\), \(l = 1\), \(m_l = 0\), and \(m_s = -\frac{1}{2}\), which
is valid and represents an electron in a \(3p\) orbital. Choice A
describes a \(3d\) orbital (\(l = 2\)), which is empty in ground-
state sulfur. Choice C represents \(n = 2\), which is core, not
valence. Choice D is invalid because when \(l = 0\), \(m_{l}\)
cannot be \(+1\).
7. Why does the first ionization energy of oxygen deviate from the
general periodic trend by being lower than the first ionization
energy of nitrogen, despite oxygen having a higher nuclear charge?
A) Oxygen has a filled valence shell which shields the incoming
electrons more effectively from the nucleus.
B) The electron being removed from nitrogen resides in a more
stable \(2s\) orbital rather than a \(2p\) orbital.
C) Oxygen experiences electron-electron repulsion within its
doubly occupied \(2p\) orbital, making it easier to remove one
electron.
D) Nitrogen has a higher electronegativity value than oxygen,
which causes it to hold onto its valence electrons tightly.
Correct Answer: C
Rationale: Nitrogen has a valence electron configuration of
\(2s^{2}2p^{3}\), where the three \(2p\) electrons each occupy a
separate orbital singly, according to Hund's rule, maximizing
stability. Oxygen has a configuration of \(2s^{2}2p^{4}\), where
one of the \(2p\) orbitals contains a pair of electrons. The
electrostatic repulsion between these two electrons sharing the
same spatial orbital raises the energy level of that electron pair,
making it easier to remove the first electron from oxygen
compared to the spin-stabilized, half-filled \(2p\) shell of
nitrogen.
8. Which of the following molecules features a central atom that is
\(sp^{3}d\) hybridized and displays a T-shaped molecular
geometry?
A) \(SF_{4}\)
B) \(BF_{3}\)
EXAM COMPREHENSIVE 150 MCQ
EVALUATION WITH CORRECT
VERIFIED ANSWERS AND DETAILED
RATIONALES GRADE A+ PREMIUM
RESOURCE | INSTANT DOWNLOAD
1. Which of the following statements best describes the primary
reason why primary alcohols have significantly higher boiling
points than their corresponding isomeric ethers?
A) Primary alcohols exhibit strong intermolecular hydrogen
bonding due to the highly polar \(-OH\) group, whereas ethers
only experience weaker dipole-dipole and London dispersion
forces.
B) Ethers contain a bent molecular geometry around the oxygen
atom which prevents close packing in the liquid state, drastically
lowering their boiling points.
C) The carbon-oxygen bonds in ethers are non-polar, preventing
them from interacting with adjacent molecules through any
electrostatic attractions.
D) Primary alcohols possess a greater molecular mass than their
corresponding isomeric ethers, which naturally increases their
London dispersion forces.
Correct Answer: A
Rationale: The presence of the hydroxyl group (\(-OH\)) in
primary alcohols allows for the formation of strong
intermolecular hydrogen bonds between molecules. Isomeric
ethers contain a carbon-oxygen-carbon linkage which, while
polar, lacks a hydrogen atom directly bonded to an
electronegative oxygen atom, meaning they cannot form
intermolecular hydrogen bonds with themselves. This difference
in intermolecular force strength requires significantly more
thermal energy to overcome in alcohols.
2. An organic compound with the molecular formula
\(C_{4}H_{8}O\) does not undergo oxidation when treated with
, mild oxidizing agents such as Tollens' reagent, but it can be
reduced back to a secondary alcohol using lithium aluminum
hydride (\(LiAlH_{4}\)). Which of the following is the IUPAC
name for this compound?
A) Butanal
B) Butan-1-ol
C) Butanone
D) Cyclobutanone
Correct Answer: C
Rationale: The molecular formula \(C_{4}H_{8}O\) fits the
general formula for an aliphatic aldehyde or ketone
(\(C_nH_{2n}O\)). Because the compound does not undergo
oxidation with a mild oxidizing agent like Tollens' reagent, it
cannot be an aldehyde (such as butanal). Ketones resist mild
oxidation but are readily reduced by strong reducing agents like
\(LiAlH_{4}\) to yield secondary alcohols. Therefore, the
compound must be the four-carbon ketone, which is butanone.
3. Consider the elimination reaction of 2-bromobutane in the
presence of a strong base like sodium ethoxide. According to
Zaitsev's rule, what is the major organic product formed and why?
A) 1-butene, because the elimination of a primary hydrogen atom
is kinetically favored due to less steric hindrance.
B) but-2-ene, because the more highly substituted alkene is
thermodynamically more stable due to hyperconjugation.
C) 2-ethoxybutane, because substitution reactions always
outcompete elimination reactions when using alkoxide bases.
D) butadiene, because the removal of multiple halogen atoms
increases the overall entropy of the system.
Correct Answer: B
Rationale: Zaitsev's rule states that in an elimination reaction,
the major alkene product is the one that is more highly substituted
(the double bond is attached to more carbon substituents). For 2-
bromobutane, elimination can yield either 1-butene or but-2-ene.
But-2-ene is the major product because the internal double bond
is stabilized by a greater degree of alkyl group substitution and
hyperconjugation, making it thermodynamically more stable.
4. Which of the following sequences correctly ranks the functional
groups in order of decreasing priority according to IUPAC
, nomenclature rules for multi-functional organic compounds?
A) Carboxylic acid > Ester > Aldehyde > Ketone > Alcohol > Amine
B) Alcohol > Amine > Ketone > Aldehyde > Ester > Carboxylic acid
C) Ester > Carboxylic acid > Alcohol > Ketone > Aldehyde > Amine
D) Carboxylic acid > Alcohol > Aldehyde > Ester > Amine > Ketone
Correct Answer: A
Rationale: IUPAC nomenclature rules establish a strict
hierarchy of functional group priority for numbering and naming
multi-functional molecules. Carboxylic acids hold the highest
priority, followed by derivatives like esters, then carbonyl groups
where aldehydes outperform ketones. Hydroxyl groups (alcohols)
come next, followed by amino groups (amines), alkenes, alkynes,
and finally alkanes or halogens.
5. What type of reaction mechanism is primarily responsible for the
polymerization of ethene into polyethene under high pressure and
in the presence of an organic peroxide initiator?
A) Electrophilic addition polymerization
B) Free-radical addition polymerization
C) Nucleophilic substitution polymerization
D) Condensation elimination polymerization
Correct Answer: B
Rationale: The polymerization of ethene to polyethene using an
organic peroxide initiator follows a free-radical mechanism. The
peroxide undergoes homolytic cleavage to generate highly
reactive free radicals, which attack the \(\pi \)-bond of an ethene
molecule, generating a new carbon-centered radical. This radical
attacks another ethene monomer, propagating the chain. Because
no small molecules are eliminated during this process, it is an
addition polymerization driven by free radicals.
6. According to the quantum mechanical model of the atom, which of
the following sets of quantum numbers \((n, l, m_l, m_s)\) is
physically permissible for an electron occupying a valence orbital
in a ground-state sulfur atom?
A) \((3, 2, -1, +\frac{1}{2})\)
B) \((3, 1, 0, -\frac{1}{2})\)
C) \((2, 1, +1, +\frac{1}{2})\)
D) \((3, 0, +1, -\frac{1}{2})\)
Correct Answer: B
, Rationale: Sulfur has an atomic number of 16, with a ground-
state electron configuration of \([Ne] 3s^2 3p^4\). Its valence
electrons reside in the \(n = 3\) shell. For the \(3s\) orbital, \(l =
0\), meaning \(m_{l}\) must be 0. For the \(3p\) orbitals, \(l =
1\), which allows \(m_{l}\) values of \(-1, 0, +1\). Choice B has
\(n = 3\), \(l = 1\), \(m_l = 0\), and \(m_s = -\frac{1}{2}\), which
is valid and represents an electron in a \(3p\) orbital. Choice A
describes a \(3d\) orbital (\(l = 2\)), which is empty in ground-
state sulfur. Choice C represents \(n = 2\), which is core, not
valence. Choice D is invalid because when \(l = 0\), \(m_{l}\)
cannot be \(+1\).
7. Why does the first ionization energy of oxygen deviate from the
general periodic trend by being lower than the first ionization
energy of nitrogen, despite oxygen having a higher nuclear charge?
A) Oxygen has a filled valence shell which shields the incoming
electrons more effectively from the nucleus.
B) The electron being removed from nitrogen resides in a more
stable \(2s\) orbital rather than a \(2p\) orbital.
C) Oxygen experiences electron-electron repulsion within its
doubly occupied \(2p\) orbital, making it easier to remove one
electron.
D) Nitrogen has a higher electronegativity value than oxygen,
which causes it to hold onto its valence electrons tightly.
Correct Answer: C
Rationale: Nitrogen has a valence electron configuration of
\(2s^{2}2p^{3}\), where the three \(2p\) electrons each occupy a
separate orbital singly, according to Hund's rule, maximizing
stability. Oxygen has a configuration of \(2s^{2}2p^{4}\), where
one of the \(2p\) orbitals contains a pair of electrons. The
electrostatic repulsion between these two electrons sharing the
same spatial orbital raises the energy level of that electron pair,
making it easier to remove the first electron from oxygen
compared to the spin-stabilized, half-filled \(2p\) shell of
nitrogen.
8. Which of the following molecules features a central atom that is
\(sp^{3}d\) hybridized and displays a T-shaped molecular
geometry?
A) \(SF_{4}\)
B) \(BF_{3}\)
