IB DP Chemistry - Structure 3.2 Functional groups- IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question 

Which compounds could exist as stereoisomers?

I. \( \mathrm{CH_2CClBr} \)
II. \( \mathrm{CHBrCl} \)
III. \( \mathrm{CHBrClCClBr} \)

(A) I and II only
(B) I and III only
(C) II and III only
(D) I, II and III
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

Compound I: \( \mathrm{CH_2CClBr} \) contains a terminal \( \mathrm{CH_2} \) carbon, so the carbon-carbon double bond does not have two different groups attached to each carbon. It therefore cannot show geometrical stereoisomerism.

Compound II: \( \mathrm{CHBrCl} \) represents a carbon atom attached to different groups and can have a stereogenic centre when the complete structure is considered, allowing stereoisomerism.

Compound III: \( \mathrm{CHBrClCClBr} \) contains carbon atoms with different groups attached and can therefore have stereoisomers.

Therefore, compounds II and III could exist as stereoisomers, so the correct answer is (C).

Question 

What is the IUPAC name of this compound?

(A) 3-methylpent-2-enoic acid
(B) 3-ethylbut-2-enoic acid
(C) 2-ethylbut-2-en-4-oic acid
(D) 2-methylpent-3-en-5-oic acid
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{A}} \)

The carboxylic acid group has the highest priority, so numbering begins from the carbon of the \( \mathrm{-COOH} \) group.

The longest chain containing the carboxylic acid carbon has five carbon atoms, giving the parent name pentenoic acid.

The carbon-carbon double bond is between carbon \(2\) and carbon \(3\), giving:

\( \mathrm{pent\text{-}2\text{-}enoic\ acid} \)

There is a methyl substituent on carbon \(3\).

Therefore, the IUPAC name is 3-methylpent-2-enoic acid.

Therefore, the correct answer is (A).

Question

Which statement best explains the first ionization energy of sulfur being lower than that of phosphorus?
A. Sulfur has more protons than phosphorus.
B. Phosphorus does not have paired electrons in the outer \(p\) sub-level.
C. Sulfur has an unpaired electron in the outer \(p\) sub-level.
D. Phosphorus is more reactive than sulfur.
▶️ Answer/Explanation
Detailed solution

✅ Correct answer: (B)

The first ionization energy is the energy needed to remove one mole of electrons from one mole of gaseous atoms. Across a period, ionization energy generally increases because nuclear charge increases and electrons are held more strongly.

However, there is an exception between phosphorus and sulfur due to their electron configurations in the outer \(3p\) sub-level:

Phosphorus: \( [\mathrm{Ne}]\,3s^2\,3p^3 \)  → three unpaired electrons in \(3p\).
Sulfur: \( [\mathrm{Ne}]\,3s^2\,3p^4 \)  → one of the \(3p\) orbitals now contains a pair of electrons.

In sulfur, the pairing of two electrons in the same \(3p\) orbital causes extra electron–electron repulsion, making it easier to remove one of these electrons. This lowers the first ionization energy of sulfur compared with phosphorus.

Statement B correctly highlights that phosphorus does not have paired electrons in the outer \(p\) sub-level, while sulfur does. This difference explains why sulfur has a lower first ionization energy than phosphorus.

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