RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Chemistry · 10 questions · 24 marks
Hydrocarbons are the raw material of the whole petrochemical industry and the simplest place to watch organic mechanisms in action. You will move from the sluggish alkanes, which react only by free radical substitution, through the electron-rich alkenes and alkynes that welcome electrophiles, to benzene, whose delocalised ring makes it prefer substitution over addition.
The addition of HBr to propene in the absence of peroxides gives mainly:
Answer
2-bromopropane is the main product. By Markovnikov's rule the hydrogen adds to the doubly bonded carbon that already carries more hydrogens, so the bromine ends up on the middle carbon. The mechanistic reason is that the proton can add in two ways. Adding to carbon 1 gives the secondary cation CH₃−CH⁺−CH₃, which is stabilised by hyperconjugation from six alpha hydrogens; adding to carbon 2 would give the far less stable primary cation CH₃−CH₂−CH₂⁺. The reaction runs through the lower-energy secondary cation, which the bromide ion then attacks at the central carbon. 1,2-dibromopropane would need Br₂ rather than HBr, and the products are not formed equally because the two possible intermediates differ greatly in stability.
Which of the following species is aromatic according to Huckel's rule?
Answer
The cyclopentadienyl anion, with 6 π electrons in a planar ring, is aromatic. Huckel's rule needs a system that is cyclic, planar, fully conjugated and holds (4n + 2) π electrons. The cyclopentadienyl anion has 6 π electrons, which fits n = 1, and its five carbons all lie in one plane with a continuous overlapping p orbital system, so it is aromatic and unusually stable for a carbanion. Cyclobutadiene has 4 π electrons, which fits 4n rather than 4n + 2, making it antiaromatic and highly unstable. Cyclohexane has no π electrons at all, since every carbon is sp³, so conjugation is impossible. Cyclooctatetraene has 8 π electrons and escapes antiaromaticity by folding into a non-planar tub shape, which breaks the conjugation and leaves it behaving like an ordinary polyene.
The number of different monochlorinated products (ignoring stereoisomers) obtained on free radical chlorination of 2-methylbutane is:
Answer
4 different monochloro products are obtained. The structure is CH₃−CH(CH₃)−CH₂−CH₃. Free radical chlorination can replace a hydrogen at any distinct type of carbon, so count the sets of equivalent hydrogens. The two methyl groups attached to carbon 2 are equivalent to each other and give 1-chloro-2-methylbutane. The tertiary hydrogen on carbon 2 gives 2-chloro-2-methylbutane. The CH₂ hydrogens give 2-chloro-3-methylbutane. The terminal CH₃ of the ethyl part gives 1-chloro-3-methylbutane. That makes four distinct products, and it is exactly why free radical halogenation is a poor way to prepare a single pure haloalkane in the laboratory.
An alkene of formula C₄H₈ on ozonolysis followed by treatment with zinc and water gives ethanal (CH₃CHO) as the only carbonyl product. The alkene is:
Answer
But-2-ene is the alkene. Ozonolysis with a reductive work-up cuts the molecule cleanly at the double bond, and each doubly bonded carbon becomes the carbon of a carbonyl group, keeping whatever groups it carried. Getting only one product means the two halves must be identical, so the double bond sat exactly in the middle of a symmetrical molecule. CH₃−CH=CH−CH₃ splits into two molecules of CH₃CHO, which matches. But-1-ene would give methanal and propanal, two different products. 2-methylpropene would give methanal and propanone. Cyclobutane has no double bond at all and does not undergo ozonolysis.
Assertion (A): Ethyne reacts with sodium metal to liberate hydrogen gas, but ethene and ethane do not. Reason (R): The hydrogen atoms attached to an sp hybridised carbon are more acidic than those attached to sp² or sp³ carbon.
Answer
Both A and R are true and R is the correct explanation of A. Ethyne does react with sodium metal, giving sodium acetylide and hydrogen gas, while ethene and ethane are completely unreactive towards sodium under the same conditions, so the assertion is correct. The reason explains why. An sp hybridised carbon has 50% s character, compared with about 33% for sp² and 25% for sp³. Electrons in an orbital with more s character are held closer to the nucleus and are more tightly bound, so the sp carbon behaves as though it were more electronegative. It therefore holds the electron pair of the C−H bond more firmly, and the resulting acetylide anion is a comparatively stable base. This makes the terminal alkyne hydrogen weakly but genuinely acidic, and a reactive metal such as sodium can displace it. The same effect explains why terminal alkynes give precipitates with ammoniacal silver nitrate or cuprous chloride, a standard way of telling a terminal alkyne from an internal one.
Predict the major product when hydrogen bromide is added to but-1-ene in the presence of benzoyl peroxide, and explain briefly why it differs from the product formed in the absence of peroxide.
Answer
In the presence of a peroxide the major product is 1-bromobutane, CH₃CH₂CH₂CH₂Br, which is the anti-Markovnikov product; without peroxide the product is 2-bromobutane. The peroxide changes the mechanism entirely. It breaks homolytically to give radicals, which pull a hydrogen from HBr to release a bromine radical. That bromine radical, not a proton, adds to the double bond first, and it adds to the terminal carbon because doing so leaves the unpaired electron on the more substituted carbon, giving the more stable secondary radical. That radical then abstracts a hydrogen from another HBr molecule, placing the hydrogen on the terminal carbon's neighbour and leaving the bromine at carbon 1, while regenerating a bromine radical to continue the chain. Without peroxide the reaction is ionic, the proton adds first, and the more stable secondary carbocation puts the bromide on carbon 2 instead.
Give three reasons why benzene undergoes electrophilic substitution readily but resists addition reactions, even though it is formally unsaturated.
Answer
First, benzene is stabilised by delocalisation. Its six π electrons are spread over all six carbons in a continuous cloud above and below the ring, lowering its energy by a resonance energy of roughly 150 kJ/mol relative to a hypothetical cyclohexatriene with three isolated double bonds. Any addition reaction would use up two of the ring carbons and destroy this delocalisation, so it costs a large amount of energy and is thermodynamically unattractive. Substitution restores the aromatic sextet at the end and pays no such price. Second, the delocalised π cloud is exposed on both faces of the flat ring, so it makes benzene an electron-rich target that attracts electrophiles readily. The ring donates a pair of π electrons to form a bond with the electrophile, producing a positively charged arenium ion. Third, that arenium intermediate has an easy escape route. Losing a proton from the carbon bearing the electrophile costs almost nothing and regenerates the full aromatic system, so the loss of a proton is always faster than the alternative of an anion adding to complete an addition. The whole reaction is therefore driven towards substitution.
(i) Write the complete mechanism, with all three stages named, for the chlorination of methane in the presence of ultraviolet light. (ii) Explain why the reaction is called a chain reaction. (iii) Explain why a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane is always obtained. (iv) Suggest how the yield of chloromethane can be maximised. (v) State why the reaction does not proceed in the dark at room temperature.
Answer
(i) Initiation. Ultraviolet light of the right energy is absorbed by the chlorine molecule, and the Cl−Cl bond, being the weakest bond present at about 242 kJ/mol, breaks homolytically: Cl₂ → (uv light) 2Cl• Each chlorine atom carries an unpaired electron and is extremely reactive. Propagation. Two steps repeat over and over. Step 1: Cl• + CH₄ → HCl + CH₃• — the chlorine radical abstracts a hydrogen atom, forming a strong H−Cl bond and leaving a methyl radical. Step 2: CH₃• + Cl₂ → CH₃Cl + Cl• — the methyl radical attacks a fresh chlorine molecule, giving the product and regenerating a chlorine radical. Termination. Whenever two radicals meet, the chain stops because no new radical is produced: Cl• + Cl• → Cl₂; CH₃• + Cl• → CH₃Cl; CH₃• + CH₃• → CH₃−CH₃ (the trace of ethane found in the product mixture is direct evidence for the radical mechanism). (ii) It is a chain reaction because the second propagation step regenerates the very radical consumed in the first. A single chlorine atom produced in initiation can therefore drive thousands of cycles before it is finally removed by a termination step. That is why a small amount of light produces a large amount of product, and why the reaction can become explosive in strong sunlight. (iii) Chloromethane is itself an alkane with three C−H bonds still available, and it is present in the reaction vessel alongside methane. A chlorine radical cannot distinguish between them, so it abstracts hydrogen from chloromethane too, producing CH₂Cl• and then dichloromethane, and the sequence continues to trichloromethane and tetrachloromethane. Since substitution is indiscriminate and the products stay in the mixture, all four chlorinated compounds accumulate together with unreacted methane. (iv) Use a large excess of methane relative to chlorine. Statistically a chlorine radical is then far more likely to collide with methane than with the small amount of chloromethane already formed, so multiple substitution is suppressed. Keeping the contact time short and removing the products as they form has the same effect. If the aim is instead tetrachloromethane, the opposite is done — a large excess of chlorine. (v) The reaction needs an initiation step, and that step requires enough energy to break the Cl−Cl bond homolytically. Ordinary room temperature does not supply the roughly 242 kJ/mol needed, so no chlorine radicals are generated and there is nothing to start the chain. Ultraviolet light supplies photons of exactly the right energy in a single package, which is why the mixture is stable in the dark and reacts vigorously in sunlight. Heating to a few hundred degrees Celsius does the same job thermally.
(i) Write the stepwise mechanism for the nitration of benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid, including the generation of the attacking species. (ii) Explain the role of sulphuric acid. (iii) Explain, using resonance, why the arenium ion is stabilised and why it loses a proton rather than adding a nucleophile. (iv) Predict the major product when methylbenzene is nitrated, and explain the directive influence involved. (v) Predict and justify where a second nitro group would enter nitrobenzene.
Answer
(i) Generation of the electrophile. Sulphuric acid, being the stronger acid, protonates nitric acid: HNO₃ + H₂SO₄ → H₂NO₃⁺ + HSO₄⁻ The protonated species then loses water to give the nitronium ion: H₂NO₃⁺ → NO₂⁺ + H₂O Overall: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Step 1 — attack. The π cloud of benzene donates a pair of electrons to the nitronium ion, forming a sigma bond to one ring carbon. That carbon becomes sp³ and the ring is left with a positive charge spread over the remaining five carbons; this intermediate is the arenium ion or sigma complex. This is the slow, rate-determining step, because the aromatic stabilisation is temporarily lost. Step 2 — loss of a proton. The hydrogensulphate ion removes the hydrogen from the sp³ carbon. The electrons of that C−H bond drop back into the ring, restoring the complete delocalised π system and giving nitrobenzene. This step is fast, and it also regenerates H₂SO₄. (ii) Sulphuric acid is not merely a solvent. It is a strong enough acid to protonate nitric acid, which acts as a base here, and it thereby generates the nitronium ion — the actual electrophile, since nitric acid alone produces far too little of it to nitrate benzene at a useful rate. It is also a powerful dehydrating agent, so it removes the water formed and pushes the equilibrium towards more NO₂⁺, and it is regenerated in the final step, acting in effect as a catalyst. (iii) The positive charge in the arenium ion is not localised on one atom. Three canonical structures can be drawn placing the positive charge on the two carbons ortho to and the one carbon para to the point of attack, so the charge is shared over three ring carbons. This delocalisation lowers the energy of the intermediate considerably and is why the ion forms at all. However, it is still far less stable than benzene itself, because one carbon is now sp³ and the aromatic sextet is broken. Adding a nucleophile to that ion would give a non-aromatic addition product and permanently forfeit the resonance energy of about 150 kJ/mol. Losing a proton instead costs almost nothing and returns the full aromatic stabilisation, so the substitution route always wins. (iv) Nitration of methylbenzene gives mainly 2-nitromethylbenzene and 4-nitromethylbenzene, that is the ortho and para products, and it happens faster than the nitration of benzene itself. The methyl group is electron-releasing through its +I effect and through hyperconjugation, so it pushes electron density into the ring and makes it more attractive to an electrophile. Crucially, that extra density appears specifically at the ortho and para positions: when the electrophile attacks there, one of the canonical structures of the arenium ion places the positive charge on the very carbon carrying the methyl group, where the methyl can stabilise it directly. No such structure exists for meta attack, so the ortho and para intermediates are lower in energy and form faster. (v) A second nitro group enters mainly at the meta position, giving 1,3-dinitrobenzene, and the reaction needs harsher conditions than the first nitration. The nitro group already present withdraws electron density both by a strong −I effect and by resonance, since its nitrogen bears a positive charge and its π system pulls electrons out of the ring. The ring is therefore deactivated overall. The deactivation is felt most strongly at the ortho and para positions, because attack there produces a canonical structure with the positive charge sitting on the carbon bearing the electron-withdrawing nitro group, an intolerable arrangement of two adjacent positive centres. Meta attack avoids that structure entirely, so it is the least disfavoured route and the meta product dominates.
A gaseous hydrocarbon P has the molecular formula C₄H₆. It decolourises bromine water rapidly and, when bubbled through ammoniacal silver nitrate solution, gives a white precipitate. On careful hydrogenation with one mole of hydrogen over a poisoned palladium catalyst it gives Q, C₄H₈, which on ozonolysis followed by zinc and water gives methanal and propanal. Answer: (a) Calculate the degree of unsaturation of P and say what it tells you. (b) Deduce the structures of P and Q with reasoning. (c) Explain the chemistry behind the white precipitate and why the isomer but-2-yne would not give it. (d) Write the equation for the reaction of P with an excess of hydrogen bromide and name the final product.
Answer
(a) Degree of unsaturation = [(2 × 4) + 2 − 6] ÷ 2 = (8 + 2 − 6) ÷ 2 = 4 ÷ 2 = 2. A value of 2 means the molecule contains the equivalent of two double bonds, so it could be a diene, an alkyne, a compound with one double bond and one ring, or a compound with two rings. The rapid decolourisation of bromine water rules out a purely cyclic saturated structure and confirms genuine carbon–carbon multiple bonding. (b) The white precipitate with ammoniacal silver nitrate is the diagnostic test for a terminal alkyne, so P contains a −C≡CH group. With four carbons and one terminal triple bond, P must be but-1-yne, CH≡C−CH₂−CH₃. Adding one mole of hydrogen over a poisoned palladium catalyst stops at the alkene stage, giving Q = but-1-ene, CH₂=CH−CH₂−CH₃. Ozonolysis of but-1-ene cuts the double bond between carbon 1 and carbon 2: the terminal CH₂ becomes methanal, HCHO, and the remaining CH−CH₂−CH₃ fragment becomes propanal, CH₃CH₂CHO. This matches the products stated, confirming both structures. (c) The hydrogen atom on an sp hybridised carbon is weakly acidic, because the sp orbital has 50% s character and holds the bonding electrons close to the nucleus, so the resulting acetylide anion is comparatively stable. The silver ion in the ammoniacal reagent replaces that acidic hydrogen, precipitating silver but-1-ynide as a white solid: CH≡C−CH₂−CH₃ + [Ag(NH₃)₂]⁺ → AgC≡C−CH₂−CH₃ (white) + NH₄⁺ + NH₃ But-2-yne, CH₃−C≡C−CH₃, has its triple bond in the middle of the chain, so both alkyne carbons carry methyl groups and there is no hydrogen attached to an sp carbon at all. With nothing acidic to replace, no precipitate forms, and the test therefore distinguishes terminal from internal alkynes. (d) An alkyne adds two molecules of hydrogen bromide, and each addition follows Markovnikov's rule, so the bromine goes to the carbon that already carries the fewer hydrogens. CH≡C−CH₂−CH₃ + HBr → CH₂=CBr−CH₂−CH₃ CH₂=CBr−CH₂−CH₃ + HBr → CH₃−CBr₂−CH₂−CH₃ Overall: CH≡C−CH₂−CH₃ + 2HBr → CH₃−CBr₂−CH₂−CH₃. Both bromine atoms end up on the same carbon, giving a geminal dihalide, and the product is named 2,2-dibromobutane.
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