Chemical Properties

Haloalkanes are among the most reactive classes of organic compound, precisely because of the polar C-X bond -- the electron-poor carbon readily attracts nucleophiles, and the halide ion is a reasonably good leaving group. Their reactions fall into four broad groups, each covered as its own sub-topic below: (i) nucleophilic substitution (the halogen is replaced by a nucleophile, via either the SN1 or SN2 mechanism), (ii) elimination (a beta-hydrogen and the halogen are removed together to form a new C=C double bond, via either the E1 or E2 mechanism), (iii) reaction with metals (forming organometallic compounds such as t …

Because the carbon of C(delta+)-X(delta-) is electron-poor, nucleophiles are drawn to it and displace X-, giving substitution. (i) Hydrolysis: haloalkane + aqueous KOH (boil) gives the alcohol + KBr; haloalkane + moist Ag2O/H2O likewise gives the alcohol + AgBr. (ii) Ammonolysis: haloalkane + alcoholic ammonia gives a primary alkylamine + HBr; with excess haloalkane, the amine keeps alkylating further to secondary and tertiary amines and finally a quaternary ammonium salt (ethylamine -> diethylamine -> triethylamine -> tetraethylammonium bromide). (iii) Reaction with alcoholic KCN gives an alkyl cyanide (nitrile), attack through carbon; (iv) reaction with alcoholic AgCN instead gives the alkyl isocyanide, attack through nitrogen -- cyanide and nitrite are 'ambident nucleophiles', able to attack from either of two sites depending on conditions and reagent. (v)/(vi) Reaction with sodium or potassium nitrite gives an alkyl nitrite (O-attack), while reaction with silver nitrite instead gives a nitroalkane (N-attack) -- the same ambident-nucleophile idea applied to NO2-. (vii) Reaction with sodium/potassium hydrogen sulphide gives a thioalcohol (thiol). (viii) Williamson ether …

SN2 rate depends on the concentration of both the alkyl halide and the nucleophile: Rate = k2[alkyl halide][nucleophile]; it is second order overall and proceeds in a single step. The nucleophile attacks from the side directly opposite the leaving halogen (backside attack), passing through a transition state in which both the incoming nucleophile and the departing halogen are partially bonded to carbon. The carbon being attacked turns inside-out, exactly as an umbrella inverts in a strong wind -- this inversion of configuration is called Walden inversion, after Paul Walden, who first observed it. An optically active haloalkane undergoing SN2 is therefore always accompanied by inversion at the reacting (asymmetric) ca …

SN1 ('S' substitution, 'N' nucleophilic, '1' one species in the rate-determining step) has a rate that depends only on the alkyl halide's concentration and is independent of the nucleophile's concentration: Rate = k[alkyl halide]; it follows first-order kinetics and proceeds in two steps. Worked through for tert-butyl bromide + aqueous KOH: Step 1 (slow, rate-determining) -- the polar C-Br bond breaks heterolytically to give a planar carbocation (the empty 2p orbital has two equivalent lobes, so it can be attacked equally fast from either face) plus Br-. Step 2 (fast) -- the nucleophile OH- attacks the carbocation from either face to give tert-butyl alcohol. Because the starting tert-butyl bromide here is not itself optically active, the product is optically inactive too; but if the starting haloalkane IS optically active (chiral, with the halogen on the reacting carbon), the planar carbocation intermediate is attacked from both faces in equal amounts, giving an optically inactive racemic mixture -- e.g. hydrolysis of optically active 2-bromobutane gives a racemic (+/-) mixture of butan-2-ol. Overall reactivity order: SN1 favours t …

When a haloalkane carrying a hydrogen on its beta-carbon is treated with ethanolic KOH, an alkene forms: the halogen leaves the alpha-carbon and a hydrogen leaves the adjacent beta-carbon, and a new double bond forms between them -- called beta-elimination or dehydrohalogenation, e.g. bromoethane + ethanolic KOH gives ethylene + KBr + H2O. Where more than one beta-hydrogen is available, a mixture of alkenes can form; Saytzeff's rule predicts the major product is the alkene with more alkyl groups on the doubly-bonded carbons (the more substituted, more stable double bond), e.g. 2-bromobutane + alcoholic KOH gives mainly 2 …

E2 (bimolecular elimination) rate depends on the concentration of both the alkyl halide and the base: Rate = k[alkyl halide][base]; second order overall. Primary alkyl halides generally undergo E2 with alcoholic KOH. It is a single-step process: the base abstracts a proton from the beta-carbon and the halide leaves the alpha-carbon simultaneously, so the new double bond and the departure of both groups happen together with no discrete int …

E1 (unimolecular elimination) generally occurs for tertiary alkyl halides with alcoholic KOH and follows first-order kinetics. Worked through for tert-butyl chloride: Step 1 (slow) -- heterolytic fission of the C-Cl bond gives the tert-butyl carbocation. Step 2 (fast) -- a base removes a proton from a beta-carbon of the carbocation, and the electron pair from that C-H bond becomes the new pi bond, giving isobutylene (2-methylpropene) + H2O + KCl. Because the slow step generates a carbocation exactly as in SN1, a given tertiary substrate under basic/protic conditi …

(a) Grignard reaction: a solution of haloalkane in dry ether treated with magnesium gives an alkyl magnesium halide, the Grignard reagent, e.g. ethyl bromide + Mg (dry ether) gives ethylmagnesium bromide. (b) Reaction with other active metals: haloalkane + 2 Li (dry ether) gives an alkyllithium + LiBr, e.g. ethyl lithium; and 4 haloalkane + a sodium-lead alloy gives a tetraalkyl lead compound + NaBr + …

Haloalkanes are reduced to the parent alkane by either of two routes. Catalytic hydrogenation: haloalkane + H2 over a nickel or palladium catalyst (523 K) gives the alkane + HX, e.g. bromoethane + H2 (Ni/Pd) gives ethane + HBr. Chemical (ionic) reduction: haloalkane + hydroiodic acid in the presence of red phosphorus gives the alkane + I2, e.g. iodoethane + HI (red P) gives ethane + I2 -- here the red phosphorus's role is to keep regenerating HI in situ (from the I2 produced, by reacting it with more red phosphorus and trace water) so the reduction can run through to completion using only a limited, catalytic supply of iodine overall, rather than needing a full stoichiometric excess of expensive HI. Both routes remove the halogen entirely and replace it with hydr …