Hydrocarbons — Practice Questions with Answers

Markovnikov's rule states that when an unsymmetrical reagent (like HX) adds across an unsymmetrical alkene's double bond, the hydrogen atom preferentially attaches to the carbon already bearing the greater number of hydrogen atoms, while the halogen attaches to the more substituted carbon.

Mechanistically, protonation of the alkene occurs first, generating a carbocation intermediate — the reaction proceeds through whichever pathway generates the more stable (more substituted) carbocation, since this represents the lower-energy, kinetically favoured pathway; the halide ion then attacks this more stable, more substituted carbocation, giving the Markovnikov product.

Benzene's actual structure is not simply three isolated, alternating double and single bonds — instead, all six p-orbitals (one from each carbon) overlap continuously around the ring, creating a fully delocalised system of six pi electrons spread evenly over all six carbons.

This extensive delocalisation results in significant additional stabilisation energy (resonance/delocalisation energy) compared to a hypothetical, non-delocalised structure with genuinely alternating single and double bonds — this substantial extra stability is precisely why benzene behaves quite differently from typical alkenes, showing far greater resistance to typical addition reactions and strongly favouring substitution reactions instead, which preserve this valuable aromatic stabilisation.

Initiation: $Cl_2 \xrightarrow{h\nu} 2Cl^\bullet$ (homolytic cleavage of chlorine, generating two chlorine radicals under UV light)

Propagation (a repeating cycle of two steps): $CH_4 + Cl^\bullet \rightarrow CH_3^\bullet + HCl$, followed by $CH_3^\bullet + Cl_2 \rightarrow CH_3Cl + Cl^\bullet$ (the regenerated chlorine radical continues the chain).

Termination: Two radicals combine to form a stable, non-radical product, ending the chain — for example, $CH_3^\bullet + Cl^\bullet \rightarrow CH_3Cl$, or $2Cl^\bullet \rightarrow Cl_2$, or $2CH_3^\bullet \rightarrow C_2H_6$.

Aromaticity refers to the special, additional stability exhibited by certain cyclic, planar, fully conjugated molecules, arising from the complete delocalisation of pi electrons around the ring.

Huckel's rule: For a molecule to be aromatic, it must be cyclic, planar, fully conjugated (every ring atom must have a p-orbital available for overlap), and possess exactly $(4n+2)$ pi electrons in the delocalised system, where $n$ is a non-negative whole number (0, 1, 2, ...) — benzene, with 6 pi electrons ($n=1$), satisfies this rule.

Although alkynes have two pi bonds (compared to an alkene's single pi bond), the carbon atoms in a triple bond are $sp$ hybridised, making them more electronegative (holding electrons somewhat more tightly/closer to the nucleus) compared to the $sp^2$ hybridised carbons of an alkene.

This increased effective electronegativity of $sp$ carbons means the pi electrons in an alkyne's triple bond are held somewhat more tightly (less available/accessible) compared to an alkene's more loosely held, more polarisable pi electron cloud — this reduced availability of the alkyne's pi electrons for initial electrophilic attack is why alkynes generally react somewhat more slowly than comparable alkenes in typical electrophilic addition reactions.

Wurtz reaction involves treating an alkyl halide with sodium metal in dry ether, producing a higher alkane through coupling of two alkyl groups: $2R-X + 2Na \rightarrow R-R + 2NaX$.

Limitation: This method is primarily useful for preparing symmetrical alkanes (where both alkyl groups are identical) with an even number of carbon atoms — using two different alkyl halides typically produces an unwanted, difficult-to-separate mixture of three different possible products (each possible R-R combination), making Wurtz reaction impractical for the targeted synthesis of unsymmetrical alkanes.

Syn addition occurs when both new atoms/groups add to the same face (side) of the double bond — catalytic hydrogenation (using a metal catalyst like Pd or Pt) is a classic example, where both hydrogen atoms are delivered from the metal surface to the same face of the alkene simultaneously.

Anti addition occurs when the two new atoms/groups add to opposite faces of the double bond — the addition of bromine ($Br_2$) to an alkene is a classic example, proceeding through a cyclic bromonium ion intermediate, which is subsequently attacked by the bromide ion from the opposite (back) face, resulting in anti addition of the two bromine atoms.

In a terminal alkyne (like $HC\equiv CH$), the carbon bonded to the acidic hydrogen is $sp$ hybridised — $sp$ orbitals have greater "s-character" (50% s-character) than $sp^2$ (33%) or $sp^3$ (25%) orbitals, meaning the bonding electrons (and consequently the negative charge on the resulting carbanion after losing $H^+$) are held closer to, and more stabilised by, the more electronegative $sp$ carbon nucleus.

This greater s-character-associated electronegativity of $sp$ carbons makes the resulting acetylide carbanion comparatively more stable than the corresponding vinyl ($sp^2$) or alkyl ($sp^3$) carbanions that would form from alkenes or alkanes — this relative stabilisation is precisely why terminal alkynes show measurable (though still weak) acidity, while alkanes and alkenes do not show comparable acidic behaviour under similar conditions.

Friedel-Crafts alkylation involves treating benzene with an alkyl halide in the presence of a Lewis acid catalyst (typically anhydrous $AlCl_3$), which generates a carbocation electrophile that then undergoes electrophilic aromatic substitution with benzene, producing an alkylbenzene.

Limitation: This reaction is prone to overalkylation — since the alkyl group introduced is electron-donating (activating the ring further towards subsequent substitution), the initially formed alkylbenzene product is actually more reactive towards further alkylation than benzene itself, often leading to an undesired mixture of poly-alkylated products rather than the clean, single-substitution product typically desired.

Cracking is the process of breaking down large, higher-molecular-weight alkane molecules (obtained from crude petroleum fractions, which have relatively lower demand/value) into smaller, lower-molecular-weight alkanes and alkenes (which are typically in much higher demand, especially as fuels and as feedstock for the petrochemical industry).

This process is industrially crucial because it allows refineries to better match the actual supply of various petroleum fractions with fluctuating market demand — converting excess, lower-value heavier fractions into more valuable, more useful smaller hydrocarbon products (like petrol/gasoline components and alkenes for plastics manufacturing), significantly increasing the overall economic value extracted from each barrel of crude oil processed.

Step 1 (protonation): The pi electrons of propene ($CH_3-CH=CH_2$) attack a proton from HBr. Two possible carbocations could form: a secondary carbocation ($CH_3-CH^+-CH_3$) if $H^+$ adds to the terminal carbon, or a primary carbocation ($CH_3-CH_2-CH_2^+$) if $H^+$ adds to the internal carbon.

Step 2: The secondary carbocation is considerably more stable than the primary carbocation (due to greater hyperconjugative/inductive stabilisation from two adjacent alkyl groups versus one) — this more stable secondary carbocation therefore forms preferentially and more rapidly (lower activation energy pathway).

Step 3 (nucleophilic attack): The bromide ion ($Br^-$) then attacks this more abundant secondary carbocation, giving 2-bromopropane as the major product — consistent with Markovnikov's rule, since hydrogen added to the carbon (terminal $CH_2$) already bearing more hydrogens.

Balanced equation: $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$

By Avogadro's law (equal volumes contain equal moles at the same T and P), volume ratios directly match the mole ratios in the balanced equation: 2 volumes of $C_2H_6$ require 7 volumes of $O_2$.

Volume of $O_2$ required for 5 L of $C_2H_6$: $\dfrac{7}{2}\times5 = \mathbf{17.5\text{ L}}$

2-methylbut-2-ene has structure $(CH_3)_2C=CH-CH_3$. Ozonolysis cleaves the double bond, with each doubly-bonded carbon becoming a carbonyl (C=O) carbon in the products.

Mechanism: Ozone adds across the double bond to form an unstable ozonide intermediate, which is then typically hydrolysed (in the presence of zinc dust, to prevent oxidation of any aldehyde products to carboxylic acids) — this cleaves the C-C bond that was originally the double bond, with each fragment gaining an oxygen atom.

Products: $(CH_3)_2C=O$ (acetone) and $CH_3-CHO$ (acetaldehyde) — since the more substituted carbon ($(CH_3)_2C=$) becomes a ketone, and the less substituted carbon ($=CH-CH_3$) becomes an aldehyde.

Step 1 (partial reduction to alkene, using Lindlar's catalyst): $HC\equiv CH + H_2 \xrightarrow{\text{Lindlar's catalyst}} CH_2=CH_2$

(Lindlar's catalyst, a poisoned/deactivated palladium catalyst, stops the reduction cleanly at the alkene stage, preventing further reduction — and produces cis-alkene specifically for internal alkynes, though ethyne gives simply ethene here.)

Step 2 (complete reduction to alkane, using a standard, active catalyst): $CH_2=CH_2 + H_2 \xrightarrow{Pd/Pt \text{ or } Ni} CH_3-CH_3$

Overall: $HC\equiv CH + 2H_2 \rightarrow CH_3-CH_3$ (ethane)

Generation of the electrophile: Concentrated nitric acid reacts with concentrated sulfuric acid (acting as a catalyst/proton source): $HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-$, generating the nitronium ion ($NO_2^+$) electrophile.

Step 1 (electrophilic attack): The benzene ring's delocalised pi electrons attack the electrophilic $NO_2^+$, forming a resonance-stabilised carbocation intermediate (an "arenium ion" or "sigma complex"), temporarily disrupting the ring's aromaticity.

Step 2 (deprotonation/rearomatisation): A base (typically $HSO_4^-$) removes a proton from the carbon bearing the new $NO_2$ group, restoring the aromatic ring's full delocalisation and yielding nitrobenzene as the final product, with regeneration of $H_2SO_4$.

Since complete hydrogenation of an alkyne adds 2 moles of $H_2$ (converting the triple bond fully to a single bond), and produces n-hexane ($C_6H_{14}$, a straight 6-carbon chain), the starting alkyne must have the same 6-carbon straight-chain skeleton, but with 4 fewer hydrogen atoms (2 $H_2$ molecules, 4 H atoms, removed in reverse — i.e., added during hydrogenation).

Molecular formula: $C_6H_{14} - 4H = C_6H_{10}$

This corresponds to hex-1-yne ($HC\equiv C-CH_2-CH_2-CH_2-CH_3$), assuming the triple bond is at the terminal position (though hex-2-yne or hex-3-yne, with internal triple bonds, would also satisfy the same molecular formula and give n-hexane upon complete hydrogenation).

In 1,3-butadiene, the two double bonds are conjugated (separated by only a single bond), allowing continuous overlap of all four p-orbitals across the four-carbon chain — this delocalisation provides additional stabilisation energy compared to a hypothetical, non-conjugated diene with isolated double bonds.

This delocalisation also explains the characteristic 1,4-addition pattern often observed with conjugated dienes — when an electrophile adds to one terminal carbon (C1), the resulting allylic-type carbocation intermediate is itself stabilised by resonance, with positive charge delocalised onto C2 and C4; a nucleophile can then attack either the C2 position (giving the simple 1,2-addition product) or the C4 position (giving the 1,4-addition product, where the remaining double bond has shifted to the C2-C3 position) — the availability of this delocalised, resonance-stabilised intermediate is precisely why conjugated dienes characteristically undergo this distinctive dual addition pattern.

An alkyne's triple bond consists of one sigma and two pi bonds. Complete addition (saturation) requires adding across both pi bonds, each requiring one molecule of $Br_2$: $R-C\equiv C-R' + 2Br_2 \rightarrow R-CBr_2-CBr_2-R'$

So, each mole of alkyne requires 2 moles of $Br_2$ for complete saturation.

For 2 moles of alkyne: moles of $Br_2$ required $= 2\times2 = \mathbf{4\text{ moles}}$

In the presence of peroxides, HBr addition proceeds via a free radical mechanism, rather than the usual ionic (carbocation) mechanism.

Initiation: Peroxides decompose to generate radicals, which abstract a hydrogen from HBr, generating a bromine radical ($Br^\bullet$).

Propagation: The bromine radical adds to the terminal carbon of propene (forming the more stable secondary free radical, $CH_3-CH^\bullet-CH_2Br$, rather than a less stable primary radical) — this secondary radical then abstracts a hydrogen atom from another HBr molecule, giving the final product and regenerating a bromine radical to continue the chain.

Major product: 1-bromopropane ($CH_3-CH_2-CH_2Br$) — this is the "anti-Markovnikov" product, since bromine ends up on the terminal carbon (opposite to what simple Markovnikov ionic addition would predict), a direct consequence of the different (radical, rather than ionic) mechanism operating specifically in the presence of peroxides.

Degree of unsaturation (DoU) formula for a hydrocarbon $C_nH_m$: $DoU = \dfrac{2n+2-m}{2}$

$DoU = \dfrac{2(5)+2-8}{2} = \dfrac{10+2-8}{2} = \dfrac{4}{2} = \mathbf{2}$

A degree of unsaturation of 2 could correspond to: (a) one triple bond (as in pent-1-yne, $HC\equiv C-CH_2-CH_2-CH_3$, since a triple bond counts as 2 degrees of unsaturation), or (b) two double bonds, or (c) one ring plus one double bond, or (d) two separate rings — pent-1-yne is one valid, simple example structure consistent with $C_5H_8$ and 2 degrees of unsaturation.