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A-Level Chemistry – Organic Mechanisms

Substitution, addition, elimination, aromatic and carbonyl mechanisms.

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12 questions  ·  53 marks  ·  answers included  ·  A-Level Chemistry (Edexcel/AQA/OCR style, Year 1 and Year 2 organic mechanisms)
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This worksheet covers:

  • Nucleophilic substitution (S_N1/S_N2)
  • Electrophilic addition
  • Elimination
  • Electrophilic aromatic substitution
  • Carbonyl (nucleophilic addition and addition–elimination) mechanisms

Since no diagrams are used, describe curly arrows in words (e.g. ‘a curly arrow moves from the lone pair on the nucleophile to the carbon atom’). No calculator needed. Allow about 50 minutes.

Questions

Section A – Substitution and addition (skills)

1

Bromoethane, $\text{CH}_3\text{CH}_2\text{Br}$, reacts with aqueous sodium hydroxide by nucleophilic substitution ($S_N2$).(a) Identify the nucleophile and explain why it can act as a nucleophile.(b) Describe, step by step, the mechanism for this reaction.

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(a) The nucleophile is the hydroxide ion, $\text{OH}^-$. It can act as a nucleophile because the oxygen atom carries a lone pair of electrons (and a negative charge), which can be donated to an electron-deficient carbon atom.(b) The $\text{C}$–$\text{Br}$ bond is polarised ($\text{C}^{\delta+}$–$\text{Br}^{\delta-}$) because bromine is more electronegative than carbon. A curly arrow moves from a lone pair on the oxygen of $\text{OH}^-$ to the electron-deficient carbon atom, forming a new $\text{C}$–$\text{O}$ bond; simultaneously a curly arrow moves from the $\text{C}$–$\text{Br}$ bond to the bromine atom, breaking the bond heterolytically and releasing $\text{Br}^-$. This occurs in a single step (both bonds forming/breaking simultaneously) via a five-coordinate transition state, giving ethanol, $\text{CH}_3\text{CH}_2\text{OH}$, and a bromide ion.
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2

Explain why tertiary halogenoalkanes typically react by an $S_N1$ mechanism rather than $S_N2$, and outline the two steps of the $S_N1$ mechanism for 2-bromo-2-methylpropane reacting with water.

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Tertiary halogenoalkanes have three bulky alkyl groups around the carbon bearing the halogen; this steric hindrance makes it very difficult for a nucleophile to approach and attack that carbon directly ($S_N2$), and the three alkyl groups also stabilise the intermediate carbocation by their inductive (electron-donating) effect, favouring a two-step ($S_N1$) pathway.Step 1: the $\text{C}$–$\text{Br}$ bond breaks heterolytically (a curly arrow moves from the bond to the bromine atom), forming a tertiary carbocation intermediate and a bromide ion — this is the slow, rate-determining step.Step 2: a lone pair on the oxygen of a water molecule forms a new bond to the carbocation (curly arrow from a lone pair on oxygen to the positive carbon), giving a protonated alcohol, which then rapidly loses a proton (to another water molecule) to form the neutral alcohol, 2-methylpropan-2-ol.
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3

But-1-ene reacts with hydrogen bromide, $\text{HBr}$, by electrophilic addition.(a) Explain why but-1-ene can undergo electrophilic addition.(b) Describe the mechanism, and state which product (1-bromobutane or 2-bromobutane) is the major product, explaining your answer using Markovnikov’s rule / carbocation stability.

Show answer
(a) But-1-ene contains a $\text{C=C}$ double bond with a region of high electron density (the $\pi$ bond), which can act as a nucleophile and attack electrophiles.(b) The slightly positive hydrogen atom of $\text{HBr}$ ($\text{H}^{\delta+}$–$\text{Br}^{\delta-}$) is attacked by the $\pi$ electrons of the double bond: a curly arrow moves from the $\text{C=C}$ bond to the hydrogen atom, forming a new $\text{C}$–$\text{H}$ bond, while a curly arrow moves from the $\text{H}$–$\text{Br}$ bond to the bromine atom, breaking it heterolytically to release $\text{Br}^-$ and forming a carbocation intermediate. The bromide ion then donates a lone pair to the positively charged carbon (curly arrow from lone pair on $\text{Br}^-$ to the carbocation), forming the $\text{C}$–$\text{Br}$ bond. The major product is 2-bromobutane, because addition proceeds via the more stable secondary carbocation (rather than the less stable primary carbocation that would lead to 1-bromobutane), since alkyl groups stabilise positive charge by electron donation (positive inductive effect).
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4

Ethene reacts with bromine, $\text{Br}_2$, by electrophilic addition even though bromine is a non-polar molecule. Explain how bromine becomes polarised as it approaches the double bond, and name the type of mechanism involved.

Show answer
As the $\text{Br}_2$ molecule approaches the electron-rich $\text{C=C}$ double bond, the double bond’s electrons repel the electrons in the $\text{Br}$–$\text{Br}$ bond, inducing a temporary dipole in the bromine molecule ($\text{Br}^{\delta+}$–$\text{Br}^{\delta-}$), so the nearer bromine atom becomes electrophilic. This is an electrophilic addition mechanism.
[3]

Section B – Elimination and aromatic substitution (exam-style)

5

2-bromopropane is heated under reflux with potassium hydroxide dissolved in ethanol, giving propene by elimination.(a) State the role of the hydroxide ion in this mechanism.(b) Describe the mechanism, and explain why using ethanolic (rather than aqueous) $\text{KOH}$ favours elimination over substitution.

Show answer
(a) The hydroxide ion acts as a base (proton acceptor), not as a nucleophile, in this mechanism.(b) A curly arrow moves from a lone pair on the oxygen of $\text{OH}^-$ to a hydrogen atom on the carbon adjacent to the one bearing the bromine, removing that proton and forming a new $\text{O}$–$\text{H}$ bond in water; simultaneously a curly arrow moves from the $\text{C}$–$\text{H}$ bond to form a new $\text{C=C}$ $\pi$ bond, and a further curly arrow moves from the $\text{C}$–$\text{Br}$ bond onto bromine, releasing $\text{Br}^-$; these occur together, giving propene, water and bromide ion. Ethanolic conditions favour elimination because the ethanolic solvent does not solvate/stabilise the hydroxide ion as strongly as water does, keeping it more strongly basic (rather than acting as a nucleophile), which favours proton removal (elimination) over nucleophilic attack on carbon (substitution).
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6

Benzene reacts with bromine in the presence of a halogen carrier catalyst, iron(III) bromide, $\text{FeBr}_3$, by electrophilic substitution to form bromobenzene.(a) Explain the role of the catalyst in generating the electrophile.(b) Describe the mechanism, referring to the delocalised ring of electrons in benzene.

Show answer
(a) $\text{FeBr}_3$ accepts a lone pair from a bromine atom of $\text{Br}_2$, polarising the $\text{Br}$–$\text{Br}$ bond heterolytically; this generates a more strongly electrophilic $\text{Br}^+$ species (and $\text{FeBr}_4^-$).(b) The delocalised ring of $\pi$ electrons in benzene acts as a nucleophile: a curly arrow moves from the delocalised ring to the electrophilic bromine ($\text{Br}^+$), forming a new $\text{C}$–$\text{Br}$ bond and breaking the delocalisation over the ring at that carbon, giving a positively charged (non-aromatic) intermediate in which the remaining electrons are delocalised over the other five carbons. A curly arrow then moves from the $\text{C}$–$\text{H}$ bond on the carbon bearing both the $\text{H}$ and the new $\text{Br}$ substituent to reform the delocalised ring, releasing $\text{H}^+$ (which reacts with $\text{FeBr}_4^-$ to regenerate the $\text{FeBr}_3$ catalyst and form $\text{HBr}$).
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7

Explain why benzene reacts with bromine only in the presence of a halogen carrier catalyst by substitution, whereas alkenes such as ethene react readily with bromine at room temperature by addition, with reference to the stability of the delocalised ring of electrons in benzene.

Show answer
The delocalised ring of $\pi$ electrons in benzene is spread over all six carbon atoms, giving benzene extra thermodynamic stability (lower energy) compared with a hypothetical structure with three localised double bonds. This delocalisation makes benzene’s electron density less concentrated/available than the localised $\pi$ bond in an alkene, so benzene cannot polarise bromine sufficiently on its own and needs a catalyst to generate a stronger electrophile; addition would also destroy the stable delocalised ring, so substitution (which regenerates the aromatic ring) is energetically favoured over addition.
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8

Predict the major organic product formed when 2-bromo-2-methylpropane is heated with excess ethanolic sodium hydroxide, and explain, in terms of mechanism, why substitution and elimination products can both form from tertiary halogenoalkanes under these conditions.

Show answer
The major product is 2-methylprop-1-ene (elimination product); some 2-methylpropan-2-ol (substitution, $S_N1$) may also form. Tertiary substrates readily form a stable tertiary carbocation intermediate. Hydroxide ions reacting with this carbocation can either donate a lone pair to the positive carbon (nucleophilic attack, giving substitution/$S_N1$) or remove a proton from an adjacent carbon (acting as a base, giving elimination); since hydroxide is a reasonably strong base as well as a nucleophile, and ethanolic conditions favour its basic character, elimination tends to dominate, but both pathways are mechanistically possible from the same carbocation intermediate.
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Section C – Carbonyl mechanisms (challenge)

9

Propanal, $\text{CH}_3\text{CH}_2\text{CHO}$, reacts with hydrogen cyanide (generated in situ from $\text{KCN}$ and dilute acid) to form a hydroxynitrile by nucleophilic addition.(a) Explain why the carbonyl carbon is susceptible to nucleophilic attack.(b) Describe the mechanism, and explain why the product is formed as a racemic mixture.

Show answer
(a) The $\text{C=O}$ bond is polarised ($\text{C}^{\delta+}$=$\text{O}^{\delta-}$) because oxygen is more electronegative than carbon, making the carbonyl carbon electron-deficient and susceptible to attack by nucleophiles.(b) The cyanide ion, $\text{CN}^-$, attacks the electrophilic carbonyl carbon: a curly arrow moves from the lone pair on the carbon of $\text{CN}^-$ to the carbonyl carbon, forming a new $\text{C}$–$\text{C}$ bond; simultaneously a curly arrow moves from the $\text{C=O}$ $\pi$ bond onto the oxygen atom, forming an alkoxide intermediate (negatively charged oxygen). This alkoxide then accepts a proton, typically from $\text{HCN}$ or $\text{H}^+$ in solution (curly arrow from the lone pair on the negative oxygen to a hydrogen atom, or a curly arrow from an $\text{H}$–$X$ bond onto oxygen), giving the hydroxynitrile product. The carbonyl carbon is planar (trigonal), so the cyanide nucleophile can attack from either face with equal probability, generating equal amounts of both enantiomers — a racemic mixture.
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10

Ethanoyl chloride, $\text{CH}_3\text{COCl}$, reacts vigorously with water to form ethanoic acid and hydrogen chloride, by nucleophilic addition–elimination.(a) Identify the nucleophile.(b) Describe the addition–elimination mechanism.

Show answer
(a) Water, $\text{H}_2\text{O}$ (specifically the lone pair on its oxygen atom).(b) A lone pair on the oxygen of water attacks the electrophilic carbonyl carbon (curly arrow from the lone pair on oxygen to the carbon), forming a new $\text{C}$–$\text{O}$ bond while a curly arrow moves from the $\text{C=O}$ $\pi$ bond onto the oxygen, giving a tetrahedral intermediate bearing a positively charged oxygen (from water) and a negatively charged carbonyl oxygen, plus the chlorine substituent. A proton is then lost from the positively charged oxygen (curly arrow from the $\text{O}$–$\text{H}$ bond to the oxygen, released as $\text{H}^+$), and in the elimination step a curly arrow moves from a lone pair on the carbonyl oxygen to reform the $\text{C=O}$ $\pi$ bond, while a curly arrow moves from the $\text{C}$–$\text{Cl}$ bond onto chlorine, expelling $\text{Cl}^-$ (which combines with $\text{H}^+$ to give $\text{HCl}$). This regenerates the carbonyl group, giving ethanoic acid, $\text{CH}_3\text{COOH}$.
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11

Explain why acyl chlorides such as ethanoyl chloride react much more readily with nucleophiles than carboxylic acids such as ethanoic acid do, with reference to the leaving group and the electronegativity of the substituent attached to the carbonyl carbon.

Show answer
In an acyl chloride, chlorine is a good leaving group (as $\text{Cl}^-$ is a stable, weakly basic ion) and is highly electronegative, which further increases the partial positive charge on the carbonyl carbon, making it more strongly electrophilic; this makes both nucleophilic attack and loss of the leaving group easier. In a carboxylic acid, the leaving group would be a hydroxide/alkoxide-type ion ($\text{OH}^-$), which is a much poorer (more strongly basic, less stable) leaving group, and the $\text{OH}$ group’s lone pair can also donate electron density into the carbonyl carbon by resonance, reducing its electrophilicity; both effects make carboxylic acids far less reactive towards nucleophiles than acyl chlorides.
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12

Propanone, $\text{CH}_3\text{COCH}_3$, undergoes nucleophilic addition with $\text{NaBH}_4$ (a source of hydride ion, $\text{H}^-$) to form propan-2-ol. Describe this mechanism, and explain why aldehydes generally react faster than ketones in nucleophilic addition reactions.

Show answer
A curly arrow moves from the hydride ion, $\text{H}^-$, to the electrophilic carbonyl carbon of propanone, forming a new $\text{C}$–$\text{H}$ bond, while a curly arrow moves from the $\text{C=O}$ $\pi$ bond onto the oxygen, forming an alkoxide intermediate; this alkoxide then gains a proton (from water/dilute acid on work-up) to give propan-2-ol. Aldehydes react faster than ketones because ketones have two alkyl groups attached to the carbonyl carbon, which are electron-donating (positive inductive effect) and reduce the partial positive charge on that carbon, and are also bulkier, sterically hindering nucleophilic attack; aldehydes have only one alkyl group (and a hydrogen), so the carbonyl carbon is both more electrophilic and more accessible to the incoming nucleophile.
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