RowQ
The Vault
RowQ
The Vault
CBSE Class 12 Chemistry · 11 questions · 26 marks
Replacing one hydrogen of an alkane with a halogen turns an unreactive molecule into a versatile starting point for almost every functional group you know. This chapter follows the C–X bond through SN1 and SN2 substitutions, elimination governed by Saytzeff's rule, and the striking unreactivity of aryl halides, and it explains why chirality means the mechanism you choose decides the product you get.
Which compound will undergo hydrolysis by an SN1 mechanism most readily?
Answer
(CH₃)₃CCl. SN1 is controlled by how easily the carbocation forms, and the tertiary carbocation (CH₃)₃C⁺ is the most stable of the four because three alkyl groups release electron density by the inductive effect and provide nine α-hydrogens for hyperconjugation. A more stable intermediate means a lower activation energy for the rate-determining ionisation step.
Chlorobenzene is far less reactive than chlorocyclohexane towards nucleophilic substitution mainly because:
Answer
The C–Cl bond in chlorobenzene has partial double-bond character due to resonance and the carbon is sp² hybridised. The chlorine lone pair delocalises into the ring, shortening and strengthening the bond so that chloride is a poor leaving group. In addition, an sp² carbon is more electronegative than sp³ and holds the bonding pair more tightly, and the π electron cloud repels the approaching nucleophile.
2-Bromobutane is heated with alcoholic KOH. The major organic product is:
Answer
But-2-ene. Alcoholic KOH promotes β-elimination rather than substitution, so a molecule of HBr is removed. Loss of hydrogen from C-3 gives the disubstituted but-2-ene, while loss from C-1 gives the monosubstituted but-1-ene. Saytzeff's rule states that the more highly substituted alkene, being more stable through hyperconjugation, is the major product.
The IUPAC name of (CH₃)₂CHCH₂CH₂Br is:
Answer
1-Bromo-3-methylbutane. The longest chain containing the carbon bearing bromine has four carbons, so the parent is butane. Numbering starts from the end nearer the substituent that gets the lowest locant set, giving bromine at C-1 and the methyl branch at C-3. Substituents are then cited alphabetically, bromo before methyl.
Assertion (A): The hydrolysis of optically active 2-bromooctane with aqueous KOH under SN2 conditions gives a product whose configuration is inverted. Reason (R): In an SN2 reaction the nucleophile attacks the carbon from the face opposite the leaving group, so the other three bonds turn inside out like an umbrella in the wind.
Answer
Both A and R are true and R is the correct explanation of A. In the single-step SN2 process the incoming hydroxide approaches along the line of the C–Br bond, 180° from bromine, passing through a transition state in which carbon is bonded to five groups. As bromide departs the remaining three bonds invert their spatial arrangement, so the product is the enantiomer of what retention would have given. This stereochemical outcome is called Walden inversion and is the classic experimental proof of the SN2 mechanism.
Distinguish between SN1 and SN2 mechanisms on the basis of molecularity, kinetics, stereochemistry and the effect of substrate structure.
Answer
Molecularity and steps: SN2 is bimolecular and takes place in one concerted step through a transition state; SN1 is unimolecular in its rate-determining step and takes place in two steps through a carbocation intermediate. Kinetics: SN2 follows Rate = k[RX][Nu⁻], second order overall; SN1 follows Rate = k[RX], first order, and the nucleophile concentration does not affect the rate. Stereochemistry: SN2 gives complete inversion of configuration at a stereocentre; SN1 gives a largely racemic product because the planar carbocation can be attacked from either face. Substrate structure: SN2 reactivity falls 1° > 2° > 3° as steric crowding blocks backside attack; SN1 reactivity rises 1° < 2° < 3° as the carbocation becomes more stable.
Why is it preferable to prepare chloroethane from ethanol using thionyl chloride rather than concentrated hydrochloric acid? Write the equation.
Answer
CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂↑ + HCl↑. Both by-products, sulphur dioxide and hydrogen chloride, are gases that escape from the reaction mixture, so the chloroethane is left essentially pure and needs no elaborate separation. With concentrated HCl the reaction is reversible and slow for a primary alcohol, requires a ZnCl₂ catalyst, and leaves the product mixed with water and unreacted acid from which it must be separated.
Explain why haloarenes such as chlorobenzene are ortho- and para-directing towards electrophilic substitution even though the halogen deactivates the ring.
Answer
The halogen exerts two opposing effects. Its high electronegativity withdraws electron density from the ring by the inductive effect, which lowers the electron density everywhere and makes the ring less reactive than benzene — hence deactivation and the need for harsher conditions. At the same time a lone pair on the halogen is donated into the ring by resonance, and the resonance structures place the negative charge specifically at the ortho and para positions. Those positions therefore retain more electron density than the meta position, and the intermediate carbocation formed by ortho or para attack is resonance stabilised by the halogen. The result is a deactivated ring that still directs the incoming electrophile to the ortho and para sites.
(i) Describe the SN2 mechanism for the reaction of bromoethane with aqueous potassium hydroxide, showing the transition state and the rate law. (ii) Explain why the rate of SN2 hydrolysis follows the order CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr. (iii) Predict, with reasons, the major product when 1-bromopropane is treated with (a) aqueous KOH and (b) alcoholic KOH.
Answer
(i) Hydroxide ion approaches the carbon bearing bromine from the side directly opposite the C–Br bond. As the O–C bond begins to form, the C–Br bond begins to break, and the molecule passes through a single transition state in which carbon is partially bonded to five groups: the three hydrogens (or H and CH₃) lie in a plane, with the incoming OH and departing Br on opposite sides. HO⁻ + CH₃CH₂Br → [HO···CH₂(CH₃)···Br]‡ → CH₃CH₂OH + Br⁻. Because bond formation and bond breaking happen together in the single rate-determining step, both species appear in the rate law: Rate = k[CH₃CH₂Br][OH⁻], second order overall and bimolecular. (ii) The transition state is crowded, so anything that blocks the backside approach slows the reaction. Methyl bromide has three small hydrogens around the reacting carbon and the nucleophile has a clear path. Each replacement of a hydrogen by a methyl group adds bulk on the face being attacked, raising the energy of the already crowded transition state. By the time three methyl groups are present in (CH₃)₃CBr the backside is completely shielded and the SN2 route is effectively shut down, which is why tertiary halides react by SN1 instead. Electronic effects reinforce this, since alkyl groups also push electron density towards the carbon and make it less attractive to a nucleophile. (iii) (a) Aqueous KOH supplies a high concentration of the strongly nucleophilic hydroxide ion in a polar protic solvent that favours substitution, so 1-bromopropane gives propan-1-ol, CH₃CH₂CH₂OH. (b) Alcoholic KOH provides ethoxide/hydroxide acting as a base rather than a nucleophile, and the alcoholic medium favours elimination, so a β-hydrogen and bromine are removed to give propene, CH₃CH=CH₂, together with KBr and water.
(i) Write the equations and conditions for converting propan-1-ol into 1-iodopropane in two steps. (ii) Explain what happens when propene reacts with HBr in the presence and in the absence of benzoyl peroxide, naming the rule and the mechanism in each case. (iii) State two harmful environmental effects of chlorofluorocarbons and one useful medicinal halogen compound.
Answer
(i) Step 1: treat the alcohol with red phosphorus and bromine (or PBr₃) to make the bromide. 3CH₃CH₂CH₂OH + PBr₃ → 3CH₃CH₂CH₂Br + H₃PO₃. Step 2: carry out a Finkelstein reaction by refluxing the bromide with sodium iodide in dry acetone. CH₃CH₂CH₂Br + NaI → CH₃CH₂CH₂I + NaBr↓. The reaction is driven forward because sodium bromide is insoluble in dry acetone and precipitates, removing a product from the equilibrium. (ii) In the absence of peroxide the reaction is electrophilic addition. The proton adds first to give the more stable secondary carbocation CH₃–CH⁺–CH₃ rather than the primary one, and bromide then attacks it, so the product is 2-bromopropane. This is Markovnikov's rule: the hydrogen goes to the carbon already carrying more hydrogens. In the presence of benzoyl peroxide the mechanism changes to free-radical addition. The peroxide breaks to give radicals that abstract hydrogen from HBr, producing a bromine radical. This adds to the terminal carbon because doing so generates the more stable secondary carbon radical, and the chain is completed by abstraction of hydrogen from HBr. The product is therefore 1-bromopropane, the anti-Markovnikov product, and the effect is called the peroxide or Kharasch effect. Note that it is observed only with HBr, not with HCl or HI. (iii) Chlorofluorocarbons are chemically inert and long-lived, so they drift into the stratosphere where ultraviolet light cleaves the C–Cl bond and releases chlorine radicals; a single radical destroys thousands of ozone molecules in a chain reaction, thinning the ozone layer and letting more harmful ultraviolet radiation reach the ground. They are also potent greenhouse gases that contribute to global warming. A useful medicinal halogen compound is chloroquine, used to treat malaria; iodoform has also long been used as a mild antiseptic.
A student has a sample of optically active 2-bromobutane. Portion A is warmed with a dilute solution of sodium hydroxide in an aqueous polar solvent and gives a product with essentially no optical rotation. Portion B is treated with a high concentration of sodium ethoxide, and the rate of the reaction is found to double when the ethoxide concentration is doubled. Portion C is warmed with alcoholic KOH. (a) Explain why the product from portion A has no optical rotation. (b) State the rate law and the mechanism operating in portion B. (c) Name the major organic product formed from portion C and state the rule that predicts it. (d) Why does 2-bromobutane show optical activity in the first place?
Answer
(a) Portion A reacts by an SN1 mechanism, because the dilute nucleophile and the ionising polar solvent favour prior ionisation of the C–Br bond. The intermediate is a planar sp² carbocation whose two faces are equally accessible to water or hydroxide. Attack from either face is equally likely, so equal amounts of the two enantiomeric alcohols are formed. Their rotations are equal and opposite, so the racemic mixture shows no net optical rotation. (b) The rate depends on the concentration of the nucleophile as well as the substrate, so Rate = k[2-bromobutane][C₂H₅O⁻]. This second-order kinetics indicates a bimolecular, single-step SN2 process, with the nucleophile attacking from the side opposite bromine and inverting the configuration. (c) Portion C gives but-2-ene as the major product, formed by β-elimination of HBr. Saytzeff's rule predicts it: the alkene with the greater number of alkyl groups on the doubly bonded carbons is the more stable through hyperconjugation and therefore predominates over but-1-ene. (d) In 2-bromobutane the second carbon is attached to four different groups — a hydrogen, a bromine, a methyl group and an ethyl group. Such a carbon is a stereocentre, so the molecule and its mirror image cannot be superimposed. The two enantiomers rotate the plane of plane-polarised light through equal angles in opposite directions, which is what optical activity means.
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