CAIE A-Level Chemistry 14. Hydrocarbons Question Bank

Practise alkane and alkene reactions, from cracking and combustion to radical substitution, addition and oxidative cleavage.

Covered syllabus points

14.1 Alkanes 14.2 Alkenes

[Maximum number: 1]

Different hydrocarbon mixtures produced from fractional distillation of crude oil have different uses.

State the compound that is heated with long-chain hydrocarbons to produce more useful smaller alkanes and alkenes.

$\mathrm{Al}_{2} \mathrm{O}_{3}$ / aluminium oxide

[Maximum number: 1]

Species such as $\mathrm{NH}_{4}{ }^{+}, \mathrm{CO}_{3}{ }^{2-}$ and $\mathrm{PO}_{4}{ }^{3-}$ are examples of molecular ions.

$\mathrm{OH}^{-}(\mathrm{aq})$ reacts with 2-bromo-2-methylpropane in an $\mathrm{S}_{\mathrm{N}} 1$ reaction. The molecular ion $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$ forms as the intermediate in this reaction.

Organic compound M forms when 2-bromo-2-methylpropane is heated with ethanolic $\mathrm{OH}^{-}$ .

Draw the structure of M.

Figure for Question 14. Hydrocarbons question 2 — CAIE A-Level Chemistry AS

[Maximum number: 5]

Allyl chloride is an important chemical used in the manufacture of plastics, pharmaceuticals and pesticides.

allyl chloride

(a)

Allyl chloride can be produced by many different methods. The most common method is chlorination of propene which proceeds via a free-radical substitution mechanism.

Figure for Question (a) — CAIE A-Level Chemistry AS

[ 3 ]

(i)

The initiation step in this reaction is the formation of chlorine radicals ( $\mathrm{C} l^{\bullet}$ ) from $\mathrm{Cl}_{2}$ molecules.

State the conditions required to initiate this reaction.

[ 1 ]

(ii)

The propenyl radical, $\mathrm{CH}_{2}=\mathrm{CHCH}_{2}$ , is formed in the first propagation step of the reaction. Write an equation to show the formation of $\mathrm{CH}_{2}=\mathrm{CHCH}_{2}{ }^{\bullet}$ in this propagation step.

[ 1 ]

$\begin{aligned} \mathrm{CH}_{2}=\mathrm{CHCH}_{3}+\mathrm{C} l^{\bullet} \rightarrow \mathrm{CH}_{2}=\mathrm{CHCH}_{2}^{\bullet}+\mathrm{HCl} \mathrm{OR} \mathrm{C}_{3} \mathrm{H}_{6}+\mathrm{C} l^{\bullet} \rightarrow \rightarrow \mathrm{C}_{3} \mathrm{H}_{5}{ }^{\bullet}+\mathrm{HCl} \end{aligned}$

(iii)

Explain why the free-radical substitution reaction gives a low yield of allyl chloride.

[ 1 ]

(b)

2-bromo-1-chloropropane, $\mathrm{CH}_{3} \mathrm{CHBrCH}_{2} \mathrm{Cl}$ , is the major product of the reaction of allyl chloride with HBr .

Figure for Question (b) — CAIE A-Level Chemistry AS

Explain why 2-bromo-1-chloropropane is the major product of this reaction.

[ 2 ]

M1 major product formed from more stable intermediate / carbocation

OR

(intermediate has) $2^{\circ}$ carbocation which is (more) stable

M2 (positive) inductive effect / (+)I of alkyl groups (on the intermediate)

[Maximum number: 1]

Organic compounds can be distinguished using chemical tests and analytical techniques.

Organic compound E contains three carbon atoms. E reacts with cold dilute acidified $\mathrm{KMnO}_{4}(\mathrm{aq})$ to form a single compound F with $M_{\mathrm{r}}=154.9$ .
Fig. 3.1 shows the infrared spectrum of E.
Fig. 3.2 shows the infrared spectrum of F.

Fig. 3.1

Fig. 3.2

Table 3.2

Both spectra show absorptions between 2850 and $2950 \mathrm{~cm}^{-1}$ owing to C-H bonds in each molecule.

Complete the equation for the reaction of E with cold dilute acidified $\mathrm{KMnO}_{4}(\mathrm{aq})$ to form F. In the equation, $[\mathrm{O}]$ represents cold dilute acidified $\mathrm{KMnO}_{4}(\mathrm{aq})$ .

\mathrm{H}_{2} \mathrm{O}+[\mathrm{O}]+

→

(1) $\mathrm{H}_{2} \mathrm{O}+(1)[\mathrm{O}]+\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{Br} \rightarrow \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{BrO}_{2}$

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Covered syllabus points

14. Hydrocarbons question 1

14. Hydrocarbons question 2

14. Hydrocarbons question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

14. Hydrocarbons question 4