Below is a sample revision test covering key topics typically addressed from
Chapters 1 through 29 in Organic Chemistry. Each question is followed by its
answer and a brief rationale explaining the underlying concept.
Revision Test for Organic Chemistry (Chapters 1–29)
Question 1: Structure & Bonding (Ch. 1)
Question:
Determine the hybridization of the carbon atom in methane (CH₄).
Answer:
sp³
Rationale:
In CH₄, carbon forms four sigma bonds with hydrogen atoms. The tetrahedral arrangement of these
bonds requires sp³ hybridization, where one s orbital mixes with three p orbitals.
Question 2: Nomenclature (Ch. 2)
Question:
Provide the IUPAC name for the compound with the formula CH₃CH₂COOH.
Answer:
Propanoic acid
Rationale:
The carboxylic acid group (–COOH) is the highest priority functional group. When numbering, the
carboxyl carbon is assigned as carbon 1. With a three-carbon chain, the correct name is propanoic acid.
Question 3: Stereochemistry (Ch. 3)
Question:
Explain what makes a molecule chiral and why chirality is significant in organic chemistry.
Answer:
A molecule is chiral if it cannot be superimposed on its mirror image, often due to the presence of an
asymmetric carbon (a stereocenter). Chirality is significant because enantiomers can exhibit different
biological activities and interactions with other chiral systems.
Rationale:
The concept of chirality underpins many aspects of organic reactivity and drug design, where one
enantiomer may be therapeutically active and the other inactive or harmful.
, Question 4: Reaction Mechanisms (Ch. 4)
Question:
Compare the SN1 and SN2 reaction mechanisms, highlighting key differences in their processes and
conditions.
Answer:
SN1:
o Two-step mechanism (formation of a carbocation intermediate, then nucleophilic
attack)
o Favored by tertiary substrates, polar protic solvents, and weak nucleophiles
o Reaction rate depends only on the substrate concentration
SN2:
o One-step, concerted mechanism with simultaneous bond-making and bond-breaking
o Favored by primary (or sometimes secondary) substrates, polar aprotic solvents, and
strong nucleophiles
o Reaction rate depends on both substrate and nucleophile concentrations
Rationale:
Understanding the distinctions helps predict reaction outcomes based on substrate structure and
reaction conditions.
Question 5: Acidity & Inductive Effects (Ch. 5)
Question:
How does the presence of an electron-withdrawing group affect the acidity of a carboxylic acid?
Answer:
Electron-withdrawing groups increase the acidity of a carboxylic acid by stabilizing the negative charge
on its conjugate base via the inductive effect.
Rationale:
By delocalizing the negative charge, the conjugate base becomes more stable, making proton loss (acid
behavior) more favorable.
Question 6: Electrophilic Addition (Ch. 10)
Question:
When propene reacts with HBr, which product is formed as the major product and why?
Chapters 1 through 29 in Organic Chemistry. Each question is followed by its
answer and a brief rationale explaining the underlying concept.
Revision Test for Organic Chemistry (Chapters 1–29)
Question 1: Structure & Bonding (Ch. 1)
Question:
Determine the hybridization of the carbon atom in methane (CH₄).
Answer:
sp³
Rationale:
In CH₄, carbon forms four sigma bonds with hydrogen atoms. The tetrahedral arrangement of these
bonds requires sp³ hybridization, where one s orbital mixes with three p orbitals.
Question 2: Nomenclature (Ch. 2)
Question:
Provide the IUPAC name for the compound with the formula CH₃CH₂COOH.
Answer:
Propanoic acid
Rationale:
The carboxylic acid group (–COOH) is the highest priority functional group. When numbering, the
carboxyl carbon is assigned as carbon 1. With a three-carbon chain, the correct name is propanoic acid.
Question 3: Stereochemistry (Ch. 3)
Question:
Explain what makes a molecule chiral and why chirality is significant in organic chemistry.
Answer:
A molecule is chiral if it cannot be superimposed on its mirror image, often due to the presence of an
asymmetric carbon (a stereocenter). Chirality is significant because enantiomers can exhibit different
biological activities and interactions with other chiral systems.
Rationale:
The concept of chirality underpins many aspects of organic reactivity and drug design, where one
enantiomer may be therapeutically active and the other inactive or harmful.
, Question 4: Reaction Mechanisms (Ch. 4)
Question:
Compare the SN1 and SN2 reaction mechanisms, highlighting key differences in their processes and
conditions.
Answer:
SN1:
o Two-step mechanism (formation of a carbocation intermediate, then nucleophilic
attack)
o Favored by tertiary substrates, polar protic solvents, and weak nucleophiles
o Reaction rate depends only on the substrate concentration
SN2:
o One-step, concerted mechanism with simultaneous bond-making and bond-breaking
o Favored by primary (or sometimes secondary) substrates, polar aprotic solvents, and
strong nucleophiles
o Reaction rate depends on both substrate and nucleophile concentrations
Rationale:
Understanding the distinctions helps predict reaction outcomes based on substrate structure and
reaction conditions.
Question 5: Acidity & Inductive Effects (Ch. 5)
Question:
How does the presence of an electron-withdrawing group affect the acidity of a carboxylic acid?
Answer:
Electron-withdrawing groups increase the acidity of a carboxylic acid by stabilizing the negative charge
on its conjugate base via the inductive effect.
Rationale:
By delocalizing the negative charge, the conjugate base becomes more stable, making proton loss (acid
behavior) more favorable.
Question 6: Electrophilic Addition (Ch. 10)
Question:
When propene reacts with HBr, which product is formed as the major product and why?
