IB Chemistry HL R3.4 Electron-pair sharing reactions questions

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Question 1

Multiple choiceEasy1 mark

In a reaction mechanism for bromomethane, a curly arrow starts at the C–Br bond and points to Br. Which equation represents the products of this bond cleavage?

Show answer and explanation

Answer: C. The curly arrow from the bond to Br shows the bonding electron pair moving to Br. Carbon loses its share of the pair, giving CH₃⁺; bromine gains both electrons, giving Br⁻.
A: splits the pair between the fragments, producing radicals by homolytic fission.
B: reverses the electron-pair movement, assigning both electrons to carbon.
D: represents cleavage of a C–H bond rather than the specified C–Br bond.
Remember: a curly arrow from a bond to an atom shows that the bonding pair moves to that atom.

Question 2

Paper 2Medium6 marks

Boron trifluoride forms a 1:1 adduct with ammonia:

Reaction I: BF₃ + NH₃ → BF₃·NH₃

A proton reacts with water to form a hydronium ion:

Reaction II: H⁺ + H₂O → H₃O⁺

(a)

Identify the electrophile in reaction I and state which reactant supplies both electrons in the new B–N bond.

[2]

(b)

Identify the electrophile in reaction II and explain the origin of both electrons in the new O–H bond.

[2]

(c)

A student claims that an electrophile must be positively charged. Assess this claim using both reactions.

[2]

Question 3

Paper 2Hard12 marks

Boron trifluoride reacts with ammonia to form a boron-containing adduct. A sample initially contains $1.00\ \mathrm{mmol}$ of $\mathrm{BF_3}$ Ammonia is added in measured amounts, and the amount of adduct formed is determined.

The following two stages occur when propanone reacts with cyanide ions in the presence of hydrogen cyanide:

\mathrm{(CH_3)_2C{=}O + CN^- \longrightarrow (CH_3)_2C(O^-)(C{\equiv}N)}

\mathrm{(CH_3)_2C(O^-)(C{\equiv}N) + H{-}C{\equiv}N \longrightarrow (CH_3)_2C(OH)(C{\equiv}N) + CN^-}

In the product of the first stage, the carbon atom of $\mathrm{CN^-}$ is bonded to the carbon atom that was part of the carbonyl group.

(a)

Use Figure 1 to estimate the amount of $\mathrm{NH_3}$ at which the amount of adduct formed first becomes approximately constant. Hence, deduce the stoichiometric ratio of $\mathrm{BF_3}$ to $\mathrm{NH_3}$ in the adduct.

[2]

(b)

Write an equation for the formation of the adduct, showing the coordinate bond, and identify the Lewis acid and Lewis base.

[2]

(c)

State why the formation of the adduct in (b) is not a Brønsted–Lowry acid–base reaction.

[1]

(d)

A student claims, “Since each stage in the propanone reaction involves an electron-pair donor and an electron-pair acceptor, both stages must also be Brønsted–Lowry acid–base reactions.”

Annotate the two stages with curly arrows to show electron-pair movement. Identify the Lewis acid and Lewis base in each stage, and evaluate the student’s claim.

[7]

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