"Basic Principles of Organic Chemistry (GOC)" — Practice

Question 1

State the hybridisation of each carbon in $\mathrm{CH_3{-}CH{=}CH{-}C \equiv N}$ and describe the overall shape.

Accepted Answer

The first carbon has four sigma bonds and no lone pairs, so it is $sp^3$ and tetrahedral. The second and third are joined by a double bond, so each has three sigma bonds and is $sp^2$, trigonal planar at 120 degrees. The fourth is triple bonded to nitrogen, so it has two sigma bonds and is $sp$, linear. The molecule therefore has a tetrahedral end, a planar central section and a linear tail. Pi bonds are never counted when assigning hybridisation, only sigma bonds plus lone pairs.

Question 2

Explain why ethyne is measurably acidic while ethane is not.

Accepted Answer

The carbon in ethyne is $sp$ hybridised with 50 per cent s character against 25 per cent for the $sp^3$ carbon in ethane. An s orbital holds electrons closer to the nucleus, so an $sp$ carbon is significantly more electronegative and grips the electrons of its C-H bond more tightly. What decides the acidity, though, is the anion left behind: the acetylide lone pair sits in an orbital of high s character, close to the nucleus and comparatively comfortable. Ethene sits between, and $\mathrm{p}K_a$ runs 25, 44 and 50 in exactly the order of s character.

Question 3

Name $\mathrm{HOCH_2CH_2CHO}$ and explain the choice of suffix.

Accepted Answer

The compound contains both an alcohol and an aldehyde. The functional group seniority order places aldehyde above alcohol, so the aldehyde becomes the suffix and the alcohol is expressed as a hydroxy prefix. The longest chain has three carbons and the aldehyde carbon must be C1, since an aldehyde group is necessarily terminal. The name is therefore **3-hydroxypropanal**. Naming it as a propanol with an aldehyde prefix would be wrong, and this seniority decision appears in almost every nomenclature question.

Question 4

Distinguish metamerism from position isomerism, with an example of each.

Accepted Answer

Position isomerism arises when the same functional group occupies a different position on the same carbon skeleton, as in propan-1-ol and propan-2-ol. Metamerism arises when the carbon atoms are distributed differently on either side of a divalent functional group such as an ether oxygen or a ketone carbonyl, as in diethyl ether and methyl propyl ether, both $\mathrm{C_4H_{10}O}$. The distinction matters because metamers necessarily have different alkyl groups attached to the same functional group, while position isomers keep the same groups and merely move the functionality.

Question 5

Tartaric acid has two stereocentres. How many stereoisomers does it actually have, and why is this fewer than the maximum?

Accepted Answer

The maximum is $2^2 = 4$. Tartaric acid has only three, because its two stereocentres are identical, so one of the four possibilities possesses an internal plane of symmetry making one half the mirror image of the other. That structure, the $RS$ form, is superimposable on its own mirror image and is a meso compound, optically inactive despite containing stereocentres. The remaining two, $RR$ and $SS$, are a pair of enantiomers. A racemic mixture of those two is also inactive, but by external compensation and is separable, whereas a meso compound is a single substance and cannot be resolved.

Question 6

What conditions must be met for a compound to show geometrical isomerism?

Accepted Answer

Two conditions. First, there must be restricted rotation about a bond, supplied either by a pi bond, since rotating about a double bond would break the sideways overlap, or by a ring. Second, each of the two atoms involved must carry two **different** groups, because if either carries two identical groups then swapping them produces the same molecule and there is no second arrangement. Where each carbon has one identical pair, cis and trans are used; where all four groups differ, the E and Z system based on Cahn-Ingold-Prelog priorities is required.

Question 7

Arrange in decreasing order of stability: methyl cation, ethyl cation, tert-butyl cation, benzyl cation, and justify.

Accepted Answer

The order is benzyl, then tert-butyl, then ethyl, then methyl. Benzyl is most stable because the positive charge is delocalised into the aromatic ring by resonance, spreading it over four positions rather than one. Tert-butyl comes next, stabilised by the $+I$ effect of three methyl groups and by nine alpha hydrogens available for hyperconjugation. Ethyl has one methyl group and three alpha hydrogens, and methyl has neither, so each step down removes one stabilising mechanism. The underlying principle throughout is that a charge is stabilised by anything that spreads it out.

Question 8

Why is the stability order of carbanions the reverse of that of carbocations?

Accepted Answer

Because the same electronic effect acts on charges of opposite sign. An alkyl group donates electron density by the inductive effect and also participates in hyperconjugation. Donating into a carbocation relieves the positive charge and stabilises it; donating into a carbanion, which already carries a full negative charge on carbon, intensifies a charge it is already struggling to accommodate. The alkyl group did the same thing both times and only the sign of the charge changed. Carbanion stability therefore runs methyl above primary above secondary above tertiary, and carbanions are stabilised instead by electron-withdrawing groups.

Question 9

Distinguish the inductive effect from the resonance effect on four points.

Accepted Answer

The inductive effect operates through sigma bonds while resonance operates through pi systems and lone pairs. The inductive effect involves only a partial shift of electron density, while resonance involves genuine delocalisation over three or more atoms. The inductive effect weakens rapidly with distance and is negligible beyond about three carbons, whereas resonance does not weaken with distance, which is why a para substituent on a benzene ring still exerts its full effect. Both are permanent, unlike the electromeric effect, which appears only when an attacking reagent is present. Resonance additionally requires the system to be planar.

Question 10

Explain why hyperconjugation makes the tert-butyl cation more stable than the ethyl cation, and quantify the comparison.

Accepted Answer

Hyperconjugation delocalises the electrons of a sigma C-H bond into the adjacent empty p orbital of the cationic carbon, spreading the positive charge and lowering the energy. Only hydrogens on the carbon immediately next to the charged centre, the alpha hydrogens, can do this. The tert-butyl cation has three methyl groups and therefore nine alpha hydrogens, while the ethyl cation has one methyl group and three. More alpha hydrogens means more contributing structures and greater stabilisation, which is why the observed order tracks the alpha hydrogen count exactly: nine, six, three, none.

Question 11

Arrange acetic, chloroacetic, dichloroacetic and trichloroacetic acid in increasing order of acid strength and explain the trend.

Accepted Answer

The order is acetic, chloroacetic, dichloroacetic, trichloroacetic, with $\mathrm{p}K_a$ falling from 4.76 to about 0.7. Acid strength depends on the stability of the conjugate base, since a more stable anion releases the proton more readily. Each chlorine exerts a $-I$ effect that pulls electron density away from the carboxylate, spreading the negative charge over a larger region and stabilising it, so more chlorines means more withdrawal. The effect also weakens with distance, so 3-chlorobutanoic acid is considerably weaker than 2-chlorobutanoic acid.

Question 12

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

Accepted Answer

All three lose a proton from oxygen, so the comparison is between the stabilities of the three anions. The ethoxide ion has nowhere to put its negative charge, which stays localised on one oxygen and is made worse by the $+I$ effect of the ethyl group, so ethanol is the weakest at $\mathrm{p}K_a$ 16. The phenoxide ion delocalises its charge into the aromatic ring by resonance, spreading it over the oxygen and three ring carbons, giving $\mathrm{p}K_a$ 10. The acetate ion spreads its charge over two **equivalent** oxygens, far more electronegative than carbon, giving 4.76 and making acetic acid the strongest of the three.

Practice — "Basic Principles of Organic Chemistry (GOC)"

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities