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Haloalkane & Haloarene flash cards

Master Haloalkane & Haloarene through 124 NEET-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Haloalkane & Haloarene, question and answer

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  1. 1.What are haloalkanes and haloarenes?

    Haloalkanes (alkyl halides): halogen atom bonded to an sp3sp^3 carbon of an alkyl group. Haloarenes (aryl halides): halogen atom bonded to an sp2sp^2 carbon of an aromatic ring. General formula for monohaloalkanes is CnH2n+1XC_nH_{2n+1}X.

    Hint: sp3 vs sp2 carbon bearing X

  2. 2.How are halogen compounds classified by number of halogen atoms?

    By the number of X atoms: monohalogen (one X, e.g. CH3ClCH_3Cl), dihalogen (two X), trihalogen (three X, e.g. CHCl3CHCl_3) and polyhalogen compounds. Dihalides may be gem (same C) or vic (adjacent C).

    Hint: count the X atoms

  3. 3.Distinguish gem-dihalides and vic-dihalides.

    gem-dihalides (geminal): two halogen atoms on the same carbon, e.g. CH3CHCl2CH_3CHCl_2 (ethylidene chloride). vic-dihalides (vicinal): two halogens on adjacent carbons, e.g. ClCH2CH2ClClCH_2CH_2Cl (ethylene dichloride).

    Hint: gem = same C, vic = neighbouring C

  4. 4.Classify alkyl halides as primary, secondary and tertiary.

    Based on the carbon bearing X: — X-carbon attached to one other carbon; — attached to two; — attached to three. This classification governs SN1S_N1 vs SN2S_N2 reactivity.

    Hint: how many carbons on the C-X carbon

  5. 5.How are allylic and benzylic halides defined?

    Allylic halide: X on an sp3sp^3 carbon adjacent to a C=C double bond, e.g. CH2=CHCH2ClCH_2=CH-CH_2Cl. Benzylic halide: X on an sp3sp^3 carbon attached directly to a benzene ring, e.g. C6H5CH2ClC_6H_5CH_2Cl.

    Hint: sp3 carbon next to a pi system

  6. 6.Distinguish vinylic and aryl halides.

    Vinylic halide: X bonded to an sp2sp^2 carbon of a C=C (e.g. CH2=CHClCH_2=CHCl). Aryl halide: X bonded to an sp2sp^2 carbon of an aromatic ring (e.g. C6H5ClC_6H_5Cl). Both have partial double-bond character in the C-X bond.

    Hint: sp2 carbon of alkene vs ring

  7. 7.Give the IUPAC name of CH3CH2CH2ClCH_3CH_2CH_2Cl and its common name.

    IUPAC: 1-chloropropane. Common name: n-propyl chloride. Halogen is treated as a substituent (chloro-) in IUPAC nomenclature.

    Hint: chloro prefix, lowest locant

  8. 8.IUPAC name of (CH3)3CBr(CH_3)_3CBr?

    2-bromo-2-methylpropane (common name tert-butyl bromide). It is a tertiary alkyl halide.

    Hint: branch at C2, tert-butyl

  9. 9.IUPAC name of CH3CHClCH3CH_3CHClCH_3?

    2-chloropropane (isopropyl chloride). Secondary alkyl halide.

    Hint: X on middle carbon

  10. 10.State the rule for numbering the chain when naming haloalkanes.

    Number the parent chain to give the lowest set of locants to substituents (halogens and alkyl groups treated with equal priority). If there is a choice, the substituent cited first alphabetically gets the lower number.

    Hint: lowest locants, then alphabetical

  11. 11.How are substituents cited when both halogens and alkyl groups are present?

    All substituents are listed alphabetically as prefixes (bromo, chloro, methyl...), each with a locant. Di/tri multiplying prefixes are ignored for alphabetising.

    Hint: alphabetical order of substituent names

  12. 12.IUPAC name of ClCH2CH2CH2ClClCH_2CH_2CH_2Cl?

    1,3-dichloropropane. Two chlorines, locants as low as possible.

    Hint: two Cl, three-carbon chain

  13. 13.Why is the common name of CHCl3CHCl_3 chloroform and its IUPAC name trichloromethane?

    Common name chloroform is historic. IUPAC name derives from methane with three chloro substituents: trichloromethane. Similarly CHI3CHI_3 is triiodomethane (iodoform).

    Hint: methane + 3 Cl

  14. 14.How are dihaloalkanes named (gem vs vic) in IUPAC?

    They are named as dihalo derivatives with proper locants: CH3CHCl2CH_3CHCl_2 = 1,1-dichloroethane (gem); ClCH2CH2ClClCH_2CH_2Cl = 1,2-dichloroethane (vic). Locants show whether X atoms are on same or adjacent carbons.

    Hint: 1,1 vs 1,2 locants

  15. 15.Describe the nature and geometry of the C-X bond in haloalkanes.

    The carbon is sp3sp^3 hybridised; the C-X bond is a polar covalent sigma bond. Halogen is more electronegative than carbon, so carbon bears δ+\delta+ and halogen δ\delta-, making the molecule polar with a dipole moment.

    Hint: polar sigma, C is delta+

  16. 16.How do C-X bond length and bond enthalpy vary from C-F to C-I?

    As the halogen size increases down the group, the bond length increases (CF<CCl<CBr<CIC-F < C-Cl < C-Br < C-I) and the bond enthalpy decreases. Thus C-I is longest and weakest, C-F shortest and strongest.

    Hint: bigger halogen = longer, weaker bond

  17. 17.Why is C-I bond the most easily broken among C-X bonds?

    C-I is the longest and has the lowest bond dissociation enthalpy because the large iodine atom overlaps poorly with carbon 2p2p. Hence iodides are the most reactive in substitution/elimination.

    Hint: longest, weakest C-X

  18. 18.Compare dipole moments of CH3F,CH3Cl,CH3Br,CH3ICH_3F, CH_3Cl, CH_3Br, CH_3I.

    Order: CH3Cl>CH3F>CH3Br>CH3ICH_3Cl > CH_3F > CH_3Br > CH_3I. Though F is most electronegative, its small C-F bond length gives a smaller dipole; CH3ClCH_3Cl has the maximum dipole moment as electronegativity and bond length balance best.

    Hint: CH3Cl is the maximum, not CH3F

  19. 19.Why are alkyl halides only slightly soluble in water despite being polar?

    To dissolve, energy is needed to break the strong hydrogen bonds between water molecules, but alkyl halides cannot form strong H-bonds with water. The energy released on solvation is insufficient, so solubility is low. They dissolve well in organic solvents.

    Hint: can't replace water-water H-bonds

  20. 20.How do boiling points of haloalkanes vary with halogen and molecular size?

    For the same alkyl group, b.p. rises RF<RCl<RBr<RIR-F < R-Cl < R-Br < R-I (increasing mass and van der Waals forces). For the same halogen, b.p. increases with chain length. Branching lowers b.p. (smaller surface area).

    Hint: heavier + bigger = higher b.p.; branching lowers

  21. 21.How does density of haloalkanes vary?

    Density increases with number and mass of halogen atoms and decreases with increasing length of the alkyl chain. Bromo, iodo and polychloro compounds are denser than water.

    Hint: heavier halogen = denser

  22. 22.Name three methods to prepare haloalkanes from alcohols.

    (1) With halogen acids HXHX (often with ZnCl2ZnCl_2 catalyst for HCl); (2) with phosphorus halides (PCl3PCl_3, PCl5PCl_5, red P + X2X_2); (3) with thionyl chloride SOCl2SOCl_2.

    Hint: HX, PX3/PX5, SOCl2

  23. 23.Why is SOCl2SOCl_2 (thionyl chloride) the preferred reagent to convert alcohols to chloroalkanes?

    Because the by-products SO2SO_2 and HClHCl are both gases that escape, leaving pure alkyl chloride. Reaction: ROH+SOCl2RCl+SO2+HClR-OH + SOCl_2 \to R-Cl + SO_2 + HCl. Best (Darzen's) method giving pure product.

    Hint: gaseous by-products, pure product

  24. 24.Give the Lucas test order of reactivity of alcohols with HCl/ZnCl2HCl/ZnCl_2.

    Reactivity of alcohols with HX: 3° > 2° > 1°. Tertiary alcohols react immediately (turbidity at once), secondary in ~5 min, primary only on heating. Basis of the Lucas test.

    Hint: 3° fastest with HX

  25. 25.How is an alkyl halide prepared from an alkene by addition of HX (Markovnikov)?

    HX adds across C=C following Markovnikov's rule: H goes to the carbon with more H, X to the more substituted carbon. e.g. CH3CH=CH2+HBrCH3CHBrCH3CH_3CH=CH_2 + HBr \to CH_3CHBrCH_3 (2-bromopropane).

    Hint: X to more substituted C

  26. 26.State the peroxide (anti-Markovnikov) effect for HBr addition to alkenes.

    In presence of peroxides, HBr adds to unsymmetrical alkenes anti-Markovnikov (Kharasch effect) via a free-radical mechanism, giving the 1-bromo product. Only HBr shows this; HCl and HI do not.

    Hint: peroxide + HBr only

  27. 27.How does halogenation of an alkene give a vic-dihalide?

    X2X_2 (Cl2_2 or Br2_2) adds across the double bond to give a vicinal dihalide: CH2=CH2+Br2BrCH2CH2BrCH_2=CH_2 + Br_2 \to BrCH_2CH_2Br. Decolourisation of bromine water is a test for unsaturation.

    Hint: X2 adds across C=C

  28. 28.What is the Finkelstein reaction?

    Conversion of an alkyl chloride/bromide to an alkyl iodide using NaINaI in dry acetone: RCl+NaIRI+NaClR-Cl + NaI \to R-I + NaCl\downarrow. Works because NaClNaCl/NaBrNaBr are insoluble in acetone, driving the equilibrium forward.

    Hint: NaI / dry acetone → R-I

  29. 29.What is the Swarts reaction?

    Preparation of alkyl fluorides by heating an alkyl chloride/bromide with a metallic fluoride such as AgFAgF, Hg2F2Hg_2F_2, CoF2CoF_2 or SbF3SbF_3: CH3Br+AgFCH3F+AgBrCH_3Br + AgF \to CH_3F + AgBr.

    Hint: metal fluoride → R-F

  30. 30.Why is free-radical halogenation of alkanes not a good preparative method for pure haloalkanes?

    It gives a mixture of mono- and polyhalogenated products and isomers that are hard to separate. So direct halogenation is synthetically poor for a single pure haloalkane.

    Hint: mixtures, poor selectivity

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