IB DP Chemistry Structure 3.2 Functional groups: Classification of organic compounds IB Style Question Bank HL Paper 1

IB DP PhysicsIB DP Maths IB DP ChemistryIB DP Biology

Question

Which formula represents an ether?
A. \(C_6H_5OH\)
B. \(CH_3CHO\)
C. \(CH_3COCH_3\)
D. \(CH_3OCH_3\)

▶️Answer/Explanation

Markscheme: D

An ether is a compound that has an oxygen atom bonded to two alkyl or aryl groups. Looking at the given formulas:

A. \(C_6H_5OH\) – This is phenol, not an ether.

B. \(CH_3CHO\) – This is acetaldehyde, not an ether.

C. \(CH_3COCH_3\) – This is acetone, not an ether.

D. \(CH_3OCH_3\) – This formula represents dimethyl ether, which is indeed an ether.

So, the correct answer is:

D. \(CH_3OCH_3\)

Question

Why does benzene undergo substitution more readily than addition?
A. Benzene is unsaturated.
B. Addition could produce an alkane.
C. Resonance makes carbon–carbon bonds too strong to break.
D. A benzene molecule is planar.

▶️Answer/Explanation

Markscheme: C

The correct answer is C. Resonance makes carbon–carbon bonds too strong to break.

Benzene is a stable aromatic compound with a special stability due to resonance. In benzene, the pi electrons are delocalized over the entire ring, leading to a more stable system. The resonance in benzene results in a uniform distribution of electron density, and this stability makes carbon-carbon bonds in benzene stronger than typical carbon-carbon single or double bonds.

This increased bond strength makes benzene less reactive toward addition reactions (which would break the strong carbon-carbon bonds) and more prone to substitution reactions. In a substitution reaction, one atom or group is replaced by another without breaking the carbon-carbon bonds.

So, the correct explanation is that resonance makes carbon–carbon bonds too strong to break, leading to benzene’s preference for substitution reactions over addition reactions.

Question

What is the product of the reaction of but-2-ene with bromine?
A. 1,2-dibromobutane
B. 2,2-dibromobutane
C. 2,3-dibromobutane
D. 3,3-dibromobutane

▶️Answer/Explanation

Markscheme: C

The reaction of but-2-ene with bromine typically involves the addition of bromine across the double bond. This is a halogenation reaction. The product is formed by the addition of one bromine atom to each carbon of the double bond.

For but-2-ene (CH₃CH=CHCH₃) and bromine (Br₂), the product is 2,3-dibromobutane. Each carbon of the double bond reacts with one bromine atom, resulting in a product with bromine atoms attached at the 2nd and 3rd carbon positions.

Question

Which molecule is optically active?
A. 2,2-dichloropropane
B. 1,2-dichloropropane
C. 1,3-dichloropropane
D. 1,2,3-trichloropropane

▶️Answer/Explanation

Markscheme: B

To determine if a molecule is optically active, you need to examine its molecular structure and identify the presence of a chiral center (asymmetric carbon) or a chiral axis. A molecule is optically active if it contains chiral elements that do not have a superimposable mirror image.

A chiral center is a carbon atom bonded to four different substituents. A molecule with one or more chiral centers is generally optically active.

To determine if a molecule is optically active, we need to check if it has a chiral center (asymmetric carbon). A chiral center is a carbon atom bonded to four different substituents.

Let’s evaluate each option:

A. 2,2-dichloropropane:

This molecule does not have a chiral center. Both carbon atoms have two chlorine atoms and two hydrogen atoms as substituents, so it is not optically active.

B. 1,2-dichloropropane:

This molecule has a chiral center at the second carbon atom (C2) because it is bonded to two different chlorine atoms, a hydrogen atom, and a methyl group (CH₃). Therefore, it is optically active.

C. 1,3-dichloropropane: