IB DP Chemistry - Reactivity 3.3 Electron sharing reactions - IB Style Questions For HL Paper 1A -FA 2025

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

Question 

Which product may be obtained by the reduction of \( \mathrm{CH_3CH_2COOH} \)?

(A) \( \mathrm{CH_3CH(OH)CH_3} \)
(B) \( \mathrm{CH_3CH_2CH_2OH} \)
(C) \( \mathrm{CH_3CH_2OCH_3} \)
(D) \( \mathrm{CH_3COOCH_3} \)
▶️ Answer/Explanation

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

\( \mathrm{CH_3CH_2COOH} \) is propanoic acid. Reduction of a carboxylic acid produces a primary alcohol.

The reaction can be represented as:

\( \mathrm{CH_3CH_2COOH+4[H]\rightarrow CH_3CH_2CH_2OH+H_2O} \)

Therefore, propanoic acid is reduced to propan-1-ol, \( \mathrm{CH_3CH_2CH_2OH} \).

Therefore, the correct answer is (B).

Question 

What is the main product of the reaction between but-1-ene, \( \mathrm{CH_2CHCH_2CH_3} \), and hydrogen bromide, HBr?

(A) \( \mathrm{CH_3CHBrCH_2CH_3} \)
(B) \( \mathrm{CH_2BrCH_2CH_2CH_3} \)
(C) \( \mathrm{CH_2BrCHBrCH_2CH_3} \)
(D) \( \mathrm{CH_2BrCH(CH_3)_2} \)
▶️ Answer/Explanation

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

But-1-ene undergoes an electrophilic addition reaction with HBr. The \( \mathrm{H-Br} \) bond breaks and H and Br add across the carbon-carbon double bond.

According to Markovnikov’s rule, hydrogen adds to the carbon that already has more hydrogen atoms, while bromine attaches to the more substituted carbon.

The main product is therefore 2-bromobutane:

\( \mathrm{CH_2=CHCH_2CH_3+HBr\rightarrow CH_3CHBrCH_2CH_3} \)

Therefore, the correct answer is (A).

Question

Which of these reactions proceeds by a free radical mechanism in the presence of UV light?
A.  \( \text{C}_6\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{HCl} \)
B.  \( \text{C}_6\text{H}_6 + 3\text{H}_2 \rightarrow \text{C}_6\text{H}_{12} \)
C.  \( \text{CH}_2\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br} \)
D.  \( \text{CH}_3\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{HCl} \)
▶️ Answer/Explanation
Detailed solution

Free radical substitution occurs when alkanes react with halogens in the presence of UV light. UV light causes homolytic fission of the halogen molecule, forming radicals.

  • Option A: Substitution on benzene occurs by electrophilic substitution, not radical substitution.
  • Option B: This is hydrogenation of benzene, not a radical reaction.
  • Option C: This is electrophilic addition of HBr to an alkene, not a radical mechanism.
  • Option D: Reaction of ethane with chlorine under UV light is a classic free radical substitution.

Thus, the reaction that proceeds by a free radical mechanism under UV light is:

\(\boxed{\text{CH}_3\text{CH}_3 + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{HCl}}\)

✅ Answer: (D)

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