IB Chemistry HL Reactivity 3: What are the mechanisms of chemical change? questions

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Question 1

Multiple choiceHard1 mark

The alkane shown reacts with $\mathrm{Cl_2}$ under UV light. Which sequence shows initiation followed by the two propagation steps in which one H atom at C2 is replaced by Cl? A dot denotes one unpaired electron.

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Answer: C. UV light causes homolytic fission of $\mathrm{Cl_2}$ producing two chlorine radicals. A chlorine radical abstracts H from C2, forming $\mathrm{HCl}$ and $\mathrm{CH_3CH\boldsymbol{\cdot}CH_3}$ This radical reacts with $\mathrm{Cl_2}$ to form $\mathrm{CH_3CHClCH_3}$ and regenerate $\mathrm{Cl\boldsymbol{\cdot}}$ sustaining the chain. A: abstracts H at a terminal carbon, so the product is 1-chloropropane. B: uses radical–radical combination, a termination step that consumes the radicals rather than regenerating one. D: shows heterolytic fission, producing ions rather than the radicals needed to initiate the chain. Remember: a propagation step consumes one radical and produces another.

Question 2

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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