Basic Principles of Organic Chemistry — Practice

Question 1

Classify each as aromatic, antiaromatic or non-aromatic: cyclobutadiene, cycloheptatrienyl cation, cyclooctatetraene, pyrrole.

Accepted Answer

**Cyclobutadiene:** cyclic, planar, conjugated, $4$ $\pi$ electrons — a $4n$ count, so **antiaromatic**.

**Cycloheptatrienyl cation:** all seven carbons conjugated with $6$ $\pi$ electrons — **aromatic**, which is why the tropylium salt is stable.

**Cyclooctatetraene:** $8$ $\pi$ electrons would make it antiaromatic if planar, so it puckers into a tub and becomes **non-aromatic**.

**Pyrrole:** the nitrogen lone pair joins the ring system, giving $6$ $\pi$ electrons — **aromatic**, and correspondingly a very weak base.

Question 2

Explain why pyrrole is a much weaker base than pyridine, although both are nitrogen heterocycles.

Accepted Answer

In pyrrole the nitrogen lone pair occupies a $p$ orbital and is part of the six-electron aromatic sextet.

Protonating that nitrogen would remove the pair from the ring and **destroy the aromaticity**, which is prohibitively costly.

In pyridine the lone pair lies in an $sp^{2}$ orbital in the ring plane, perpendicular to the $\pi$ system and not part of it.

Protonating pyridine therefore leaves the aromatic sextet untouched, and pyridine behaves as an ordinary weak base.

Question 3

Arrange in order of increasing acidity: acetic acid, chloroacetic acid, dichloroacetic acid, trichloroacetic acid, and explain the spacing.

Accepted Answer

Each added chlorine stabilises the carboxylate further by inductive withdrawal.

$pK_a$ values: acetic $4.76$, chloroacetic $2.86$, dichloroacetic $1.29$, trichloroacetic $0.65$.

Order of acidity: acetic $<$ chloro $<$ dichloro $<$ trichloro.

The increments shrink as more chlorines are added, since each additional one operates on an already partly stabilised anion and the effects are not simply additive.

Question 4

Explain why $2$-chlorobutanoic acid is much stronger than $4$-chlorobutanoic acid.

Accepted Answer

The inductive effect is transmitted through sigma bonds and falls off sharply with distance, roughly by a factor of three for each additional bond.

In $2$-chlorobutanoic acid the chlorine is adjacent to the carboxyl group and stabilises the anion strongly.

In the $4$-substituted isomer it is three bonds away, so only about a twenty-seventh of the effect survives.

$pK_a$ values are about $2.9$ and $4.5$ respectively, against $4.8$ for butanoic acid itself.

Question 5

Arrange in order of increasing basicity in aqueous solution: ammonia, methylamine, dimethylamine, trimethylamine, aniline.

Accepted Answer

Aniline is weakest, since its lone pair is delocalised into the ring and protonation destroys that conjugation.

Ammonia has no alkyl groups and so no inductive assistance.

Among the methylamines, induction favours more substitution while solvation of the cation favours less, and the balance puts the secondary amine highest.

Order: aniline $<$ ammonia $<$ trimethylamine $<$ methylamine $<$ dimethylamine.

Question 6

Explain why phenol is far more acidic than cyclohexanol but far less acidic than acetic acid.

Accepted Answer

Cyclohexoxide has its negative charge localised entirely on one oxygen with no delocalisation at all: $pK_a$ about $16$.

Phenoxide delocalises that charge into the ring, onto the ortho and para carbons. Carbon bears negative charge poorly, so the stabilisation is real but modest: $pK_a$ $10$.

Carboxylate shares its charge between **two equivalent oxygens**, both highly electronegative, giving far greater stabilisation: $pK_a$ $4.8$.

The ranking follows the quality of the atoms carrying the charge, not merely whether delocalisation occurs.

Question 7

Predict the major product when neopentyl bromide is treated with a weak nucleophile in a polar solvent, and explain.

Accepted Answer

Ionisation gives a **primary** neopentyl cation, which is very unstable.

A methyl group migrates from the adjacent quaternary carbon, converting it into a **tertiary** cation.

The nucleophile then attacks that rearranged cation, so the product has the $2$-methylbutane skeleton rather than the neopentyl one.

Alternatively the cation loses a proton to give $2$-methylbut-2-ene. In either case the carbon skeleton has changed, which is the signature of a carbocation route.

Question 8

Rank these carbanions by stability and explain: $	ext{CH}_3^-$, $\left(	ext{CH}_3ight)_3	ext{C}^-$, $	ext{C}_6	ext{H}_5	ext{CH}_2^-$, $	ext{HC}\equiv	ext{C}^-$.

Accepted Answer

The acetylide is most stable: its charge sits in an $sp$ orbital with fifty per cent $s$ character, holding the pair close to the nucleus.

Benzylic is next, its charge delocalised into the aromatic ring.

Methyl follows, with no destabilising alkyl groups.

Tertiary butyl is least stable, since three electron-donating alkyl groups intensify the negative charge.

Order: $\left(	ext{CH}_3ight)_3	ext{C}^-<	ext{CH}_3^-<	ext{C}_6	ext{H}_5	ext{CH}_2^-<	ext{HC}\equiv	ext{C}^-$

Question 9

Explain why $	ext{SN}1$ reactions of allyl chloride can give two different products from a single starting material.

Accepted Answer

Ionisation gives the allyl cation, whose positive charge is **delocalised** over both terminal carbons through resonance.

The nucleophile can therefore attack at either end.

With an unsymmetrical allyl system the two attacks give constitutionally different products, one retaining the original position of the double bond and one with it shifted.

This allylic rearrangement is direct evidence that the intermediate is genuinely delocalised rather than a rapidly equilibrating pair of localised cations.

Question 10

How many stereoisomers exist for $2,3$-dibromobutane and for $2,3$-dibromopentane? Explain the difference.

Accepted Answer

$2,3$-Dibromobutane has two stereocentres, each bearing methyl, bromine, hydrogen and the rest of the chain. The two are **equivalent**, so a meso form exists.

Total: **three** stereoisomers, a pair of enantiomers plus one meso compound.

$2,3$-Dibromopentane has two stereocentres carrying different substituent sets, so no internal mirror plane is possible.

Total: **four** stereoisomers, two pairs of enantiomers. The upper bound of $2^{n}$ is reached only in the second case.

Question 11

Explain why the anti conformation of butane is lower in energy than the gauche, and why neither can be isolated.

Accepted Answer

In the anti arrangement the two methyl groups are one hundred and eighty degrees apart, minimising steric repulsion between them.

In the gauche arrangement they are sixty degrees apart and crowd one another, raising the energy by about $3.8$ kJ mol$^{-1}$.

Neither can be isolated because they interconvert by rotation about a single bond, with a barrier of only about $20$ kJ mol$^{-1}$.

At room temperature that barrier is crossed billions of times a second, so the two forms exist only as a rapidly equilibrating mixture.

Question 12

Explain why phenol exists entirely as the enol tautomer while cyclohexanone exists entirely as the keto form.

Accepted Answer

The keto tautomer of phenol would be cyclohexa-2,4-dien-1-one, in which one ring carbon is $sp^{3}$ and the aromatic sextet is destroyed.

The aromatic stabilisation lost is far larger than anything the carbonyl group could provide, so the enol form is overwhelmingly favoured.

Cyclohexanone has no aromatic ring to preserve, so the ordinary preference applies: a carbon-oxygen double bond is stronger than a carbon-carbon double bond plus an oxygen-hydrogen bond.

The keto form therefore dominates completely, with an enol content around $10^{-4}$ per cent.

Practice — Basic Principles of Organic Chemistry

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