Haloalkane AND Haloarenes flash cards

1. 1.Define haloalkanes and haloarenes in terms of C–X bond hybridisation.

   In haloalkanes (alkyl halides) the halogen is bonded to an $sp^3$ carbon of an alkyl group ($R-X$). In haloarenes (aryl halides) the halogen is attached directly to an $sp^2$ carbon of an aromatic ring.

   Hint: Look at the hybridisation of the C bearing X.

2. 2.Classify alkyl halides as $1^\circ$, $2^\circ$, $3^\circ$ — what is the criterion?

   Classification is by the number of carbon atoms attached to the carbon bearing the halogen: $1^\circ$ (one C), $2^\circ$ (two C), $3^\circ$ (three C).

   Hint: Count carbons on the C–X carbon.

3. 3.Distinguish allylic, benzylic and vinylic halides.

   Allylic: X on $sp^3$ C adjacent to C=C (e.g. $CH_2=CH-CH_2-X$). Benzylic: X on $sp^3$ C attached to a benzene ring. Vinylic: X directly on a C=C ($sp^2$) carbon.

   Hint: Allylic/benzylic = next to unsaturation; vinylic = on it.

4. 4.Why is the C–X bond polar, and how does polarity vary down the halogen group?

   Halogens are more electronegative than carbon, so C carries $\delta+$ and X carries $\delta-$. Electronegativity decreases F>Cl>Br>I, so bond polarity (dipole) decreases in the same order.

   Hint: Electronegativity difference drives the dipole.

5. 5.How do C–X bond length and bond enthalpy change from C–F to C–I?

   As halogen size increases (F<Cl<Br<I), the C–X bond length increases and bond enthalpy (strength) decreases: C–F strongest/shortest, C–I weakest/longest.

   Hint: Bigger halogen = longer, weaker bond.

6. 6.Order the boiling points of $CH_3Cl$, $CH_3Br$, $CH_3I$ and explain.

   $CH_3Cl < CH_3Br < CH_3I$. As halogen size/mass increases, molecular size and polarisability (van der Waals forces) increase, raising boiling point.

   Hint: Heavier, more polarisable = stronger dispersion forces.

7. 7.Compare boiling points of isomeric alkyl halides ($1^\circ$ vs $3^\circ$).

   For isomers, boiling point decreases with branching. A straight-chain $1^\circ$ halide boils higher than a branched $3^\circ$ isomer because branching lowers surface area and van der Waals contact.

   Hint: More spherical = less contact = lower b.p.

8. 8.Are haloalkanes soluble in water? Why or why not?

   They are only very slightly soluble in water. Energy released on forming new dipole–dipole attractions with water is less than that needed to break the strong H-bonds among water molecules; they dissolve readily in organic solvents.

   Hint: Cannot form H-bonds strong enough to replace water's.

9. 9.Give the IUPAC name of $(CH_3)_2CH-CH_2-Br$.

   1-Bromo-2-methylpropane (common name isobutyl bromide).

   Hint: Longest chain = propane, methyl at C2, Br at C1.

10. 10.Give IUPAC name and identify class of $CH_3-CHBr-CH_2-CH_3$.

    2-Bromobutane, a secondary ($2^\circ$) alkyl halide.

    Hint: Br on C2 of butane; two carbons attached.

11. 11.Name $CH_2=CH-CH_2Cl$ and classify the halide.

    3-Chloroprop-1-ene (allyl chloride); it is an allylic halide.

    Hint: Cl on sp3 C next to the double bond.

12. 12.Preparation of alkyl halides from alcohols using HX — what is the reactivity order of HX and of alcohols?

    $R-OH + HX \rightarrow R-X + H_2O$. Reactivity of HX: $HI > HBr > HCl$. Reactivity of alcohols: $3^\circ > 2^\circ > 1^\circ$.

    Hint: HX order follows acid strength; alcohols follow carbocation stability.

13. 13.Why is a catalyst (anhyd. $ZnCl_2$) needed to convert $1^\circ$/$2^\circ$ alcohols to chlorides with HCl, but not $3^\circ$?

    $1^\circ$ and $2^\circ$ alcohols react slowly with HCl and need Lewis acid $ZnCl_2$ to activate the C–OH; $3^\circ$ alcohols react rapidly with conc. HCl at room temperature (Lucas test basis).

    Hint: Lucas reagent = conc. HCl + $ZnCl_2$.

14. 14.Compare $SOCl_2$, $PCl_5$ and $PCl_3$ / red P + $X_2$ for making alkyl halides from alcohols. Which is best and why?

    $SOCl_2$ (thionyl chloride) is preferred: $R-OH + SOCl_2 \rightarrow R-Cl + SO_2 + HCl$, giving gaseous by-products that escape, so the product is pure. $PCl_5$/$PCl_3$ also work but give $POCl_3$/$H_3PO_3$ by-products.

    Hint: Darzens' procedure — by-products are gases.

15. 15.How are alkyl halides prepared from alkenes by addition of HX? State Markovnikov's rule.

    $H-X$ adds across C=C so that H goes to the carbon with more H's and X to the more substituted carbon (via the more stable carbocation). Order of HX addition: $HI > HBr > HCl$.

    Hint: X to the carbon giving the more stable $C^+$.

16. 16.What is the peroxide (Kharasch) effect, and for which HX does it operate?

    In the presence of peroxides, HBr adds to unsymmetrical alkenes anti-Markovnikov (Br to the less substituted carbon) via a free-radical mechanism. Only HBr shows it — not HCl or HI (bond-energy/radical-stability reasons).

    Hint: Radical addition; unique to HBr.

17. 17.Write the Finkelstein reaction and its purpose.

    $R-Cl + NaI \xrightarrow{\text{dry acetone}} R-I + NaCl\downarrow$. Used to prepare alkyl iodides; NaCl/NaBr are insoluble in acetone while NaI is soluble, driving equilibrium forward.

    Hint: Halide exchange in dry acetone.

18. 18.Write the Swarts reaction.

    Replacing Cl/Br of an alkyl halide by F using metallic fluorides: $R-Br + AgF \rightarrow R-F + AgBr$ (also $Hg_2F_2$, $CoF_2$, $SbF_3$). Used to make alkyl fluorides.

    Hint: Metallic fluoride swaps halogen for F.

19. 19.State the Wurtz reaction and its main limitation.

    $2\,R-X + 2Na \xrightarrow{\text{dry ether}} R-R + 2NaX$: couples two alkyl halides to a symmetrical alkane. Limitation: with two different halides a mixture of three alkanes forms, so it is poor for unsymmetrical alkanes.

    Hint: Na in dry ether couples R groups.

20. 20.State the Fittig reaction.

    Two aryl halides react with sodium in dry ether to give a biaryl (diaryl): $2\,C_6H_5Br + 2Na \rightarrow C_6H_5-C_6H_5 + 2NaBr$.

    Hint: Aryl analogue of Wurtz — makes biphenyl.

21. 21.State the Wurtz–Fittig reaction.

    A mixture of an alkyl halide and an aryl halide with sodium in dry ether gives an alkyl-substituted arene: $C_6H_5Br + CH_3Br + 2Na \rightarrow C_6H_5-CH_3 + 2NaBr$.

    Hint: Alkyl + aryl halide + Na → alkylarene.

22. 22.How are haloarenes prepared by electrophilic substitution (halogenation)?

    Benzene + $X_2$ with a Lewis-acid catalyst (anhyd. $FeCl_3$, $FeBr_3$): $C_6H_6 + Cl_2 \xrightarrow{FeCl_3} C_6H_5Cl + HCl$. Halogen is an o,p-director.

    Hint: Lewis acid generates $X^+$; o,p product.

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