Halogenoalkanes: Question 2

Worked solution

Part (a)(i): Product with aqueous NaOH

Hydroxide ion, $\text{OH}^-$, is a nucleophile that substitutes for the chlorine atom: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{Cl}^-$

The organic product is pentan-1-ol.

Part (a)(ii): The SN2 mechanism

The carbon bearing the chlorine (C1) is a primary carbon: it is attached to only one other carbon, so it is not very crowded and there is plenty of room for a nucleophile to approach.

The hydroxide ion, using a lone pair of electrons on its oxygen atom, approaches and attacks this carbon from the side directly opposite to the C–Cl bond (“backside attack”). As the new C–O bond begins to form, the C–Cl bond simultaneously begins to break, so bond-forming and bond-breaking happen together in a single step, through one transition state, rather than through a separate carbocation intermediate. The chloride ion leaves as the C–O bond finishes forming.

Because this is a single concerted step involving two species (the halogenoalkane and the hydroxide ion) in the rate-determining step, it is called SN2 (substitution, nucleophilic, bimolecular): the rate depends on the concentration of both 1-chloropentane and $\text{OH}^-$. This mechanism dominates for primary halogenoalkanes because there is too little steric hindrance to prevent backside attack, and a primary carbocation (which SN1 would require) is not stable enough to form readily.

Part (b): Product with ethanolic KCN

Cyanide ion, $\text{CN}^-$, acts as the nucleophile, attacking through the carbon atom of the $\text{CN}^-$ ion (not the nitrogen) and displacing $\text{Cl}^-$: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{CN}^- \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CN} + \text{Cl}^-$

The organic product is hexanenitrile. Because the carbon of the $-\text{CN}$ group itself becomes part of the new carbon skeleton, the product chain has one more carbon atom than the starting halogenoalkane: the 5-carbon pentyl chain plus the 1 carbon from $\text{CN}^-$ gives a 6-carbon nitrile.

Part (c): Product with excess ethanolic ammonia

Ammonia, $\text{NH}_3$, acts as the nucleophile via the lone pair on its nitrogen atom, displacing $\text{Cl}^-$ to give a protonated intermediate, which then loses a proton (to another $\text{NH}_3$ molecule) to give the neutral amine: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2 + \text{NH}_4\text{Cl}$

The organic product is pentan-1-amine (pentylamine).

An excess of ammonia is used because the primary amine product is itself a nucleophile and could go on to react with more unreacted 1-chloropentane, giving secondary and tertiary amines and eventually a quaternary ammonium salt. Using a large excess of $\text{NH}_3$ makes it statistically far more likely that a molecule of 1-chloropentane collides with an $\text{NH}_3$ molecule than with an amine molecule, which minimises this further substitution and maximises the yield of the primary amine.

Final answers

• (a)(i) Pentan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$
• (a)(ii) SN2: concerted, one-step backside attack by $\text{OH}^-$ with simultaneous loss of $\text{Cl}^-$
• (b) Hexanenitrile, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CN}$ (chain extended by the carbon of $\text{CN}^-$)
• (c) Pentan-1-amine, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2$ (excess $\text{NH}_3$ minimises further substitution to secondary/tertiary amines)