Haloalkanes

When an alkyl halide is heated with sodium metal in dry ether, two alkyl groups join to form a higher alkane (Wurtz reaction). For example, bromomethane gives ethane:

$$\mathrm{2\,CH_3Br + 2\,Na \xrightarrow{\text{dry ether},\ \Delta} CH_3{-}CH_3 + 2\,NaBr}$$

When bromoethane is heated with alcoholic KOH, a molecule of HBr is removed from two neighbouring carbon atoms (β-elimination, or dehydrohalogenation), and ethene is formed:

$$\mathrm{CH_3CH_2Br + KOH(alc) \xrightarrow{\Delta} CH_2{=}CH_2 + KBr + H_2O}$$

i) Markovnikov's rule

When an unsymmetrical reagent (such as HX) adds to an unsymmetrical alkene, the negative part of the reagent goes to the double-bonded carbon that has fewer hydrogen atoms. The positive part ($H^+$) goes to the carbon that has more hydrogen atoms.

ii) Identification of A to E

The hydrocarbon $C_3H_6$ is propene, $CH_3CH{=}CH_2$.

(A) Propene adds HBr following Markovnikov's rule:

$$\mathrm{CH_3CH{=}CH_2 + HBr \longrightarrow \underset{\text{2-bromopropane (A)}}{CH_3CHBrCH_3}}$$

(B) Hydrolysis with aqueous alkali:

$$\mathrm{CH_3CHBrCH_3 + KOH(aq) \xrightarrow{\Delta} \underset{\text{propan-2-ol (B)}}{CH_3CH(OH)CH_3} + KBr}$$

(C) Reaction with $PBr_3$:

$$\mathrm{3\,CH_3CH(OH)CH_3 + PBr_3 \longrightarrow \underset{\text{2-bromopropane (C)}}{3\,CH_3CHBrCH_3} + H_3PO_3}$$

(D) Reaction with alcoholic AgCN. AgCN is mainly covalent, so it attacks through nitrogen and gives an isocyanide:

$$\mathrm{CH_3CHBrCH_3 + AgCN \xrightarrow{alc.,\ \Delta} \underset{\text{isopropyl isocyanide (D)}}{(CH_3)_2CH{-}N{\equiv}C} + AgBr}$$

(E) Reduction with $LiAlH_4$ gives a secondary amine:

$$\mathrm{(CH_3)_2CH{-}NC + 4[H] \xrightarrow{LiAlH_4} \underset{\text{N-methylpropan-2-amine (E)}}{(CH_3)_2CH{-}NH{-}CH_3}}$$

iii) Reaction of E with nitrous acid

E is a secondary amine. With nitrous acid ($NaNO_2$ + HCl) it gives a yellow oily N-nitrosoamine (Liebermann's nitroso reaction):

$$\mathrm{(CH_3)_2CH{-}NH{-}CH_3 + HO{-}N{=}O \longrightarrow (CH_3)_2CH{-}N(CH_3){-}N{=}O + H_2O}$$

iv) Conversion of B into $C_3H_8$ (propane)

Propan-2-ol is reduced by heating with HI and red phosphorus at about 423 K:

$$\mathrm{CH_3CH(OH)CH_3 + 2\,HI \xrightarrow{\text{red P},\ 423\,K} CH_3CH_2CH_3 + H_2O + I_2}$$

(Red P turns the $I_2$ back into HI.)

Chloroform reacts with acetone to give the sleep-inducing drug chloretone, so A = chloroform ($CHCl_3$).

a) Preparation from ethanol (an isomer of $C_2H_6O$)

Ethanol (the other isomer is methoxymethane) is distilled with a paste of bleaching powder and water. The reaction goes in steps:

1. Bleaching powder gives chlorine and slaked lime:

$$\mathrm{CaOCl_2 + H_2O \longrightarrow Ca(OH)_2 + Cl_2}$$

2. Chlorine oxidises ethanol to ethanal:

$$\mathrm{CH_3CH_2OH + Cl_2 \longrightarrow CH_3CHO + 2\,HCl}$$

3. Ethanal is chlorinated to chloral:

$$\mathrm{CH_3CHO + 3\,Cl_2 \longrightarrow CCl_3CHO + 3\,HCl}$$

4. Chloral is hydrolysed by $Ca(OH)_2$ to chloroform:

$$\mathrm{2\,CCl_3CHO + Ca(OH)_2 \longrightarrow 2\,CHCl_3 + (HCOO)_2Ca}$$

b) i) Exposed to air

In the presence of air and sunlight, chloroform is slowly oxidised to phosgene (carbonyl chloride), which is very poisonous:

$$\mathrm{2\,CHCl_3 + O_2 \xrightarrow{\text{light}} 2\,COCl_2 + 2\,HCl}$$

ii) Heated with phenol and aqueous NaOH (Reimer–Tiemann reaction)

A –CHO group enters the ortho position of phenol. The product is the sodium salt of salicylaldehyde, and acidification gives salicylaldehyde (o-hydroxybenzaldehyde):

$$\mathrm{C_6H_5OH + CHCl_3 + 4\,NaOH \xrightarrow{\Delta,\ 340\,K} \textit{o-}NaOC_6H_4CHO + 3\,NaCl + 3\,H_2O}$$ $$\mathrm{\textit{o-}NaOC_6H_4CHO + HCl \longrightarrow \underset{\text{salicylaldehyde}}{\textit{o-}HOC_6H_4CHO} + NaCl}$$

(a) Compound B

Aqueous KOH replaces Br with OH, so B = propan-1-ol (n-propyl alcohol):

$$\mathrm{CH_3CH_2CH_2Br + KOH(aq) \longrightarrow \underset{(B)}{CH_3CH_2CH_2OH} + KBr}$$

(b) Mechanism: $S_N2$

1-Bromopropane is a primary alkyl halide, so its hydrolysis goes by the $S_N2$ (bimolecular nucleophilic substitution) mechanism, for two reasons:

• The carbon carrying Br has only one alkyl group, so there is little steric hindrance. The $OH^-$ ion can easily attack from the side opposite to Br (backside attack).
• A primary carbocation is very unstable, so the $S_N1$ route (which needs a carbocation) is not favoured.

The reaction takes place in one step through a transition state in which the C–OH bond forms as the C–Br bond breaks:

            H                     H   H                       H
             \                     \ /                       /
  HO(-)  +  H-C-Br   --->   [ HO.....C.....Br ](-)  --->  HO-C-H   +  Br(-)
             /                      |                       \
          C2H5                     C2H5                      C2H5
  (OH- attacks from        transition state:            configuration
   the back, opposite      C-OH forming, C-Br           inverted (umbrella
   to Br)                  breaking (one step)          turned inside out)

Rate $= k[CH_3CH_2CH_2Br][OH^-]$, which is second order. The configuration is inverted (Walden inversion).

(c) With alcoholic KOH

Dehydrohalogenation (elimination) takes place and propene is formed:

$$\mathrm{CH_3CH_2CH_2Br + KOH(alc) \xrightarrow{\Delta} CH_3CH{=}CH_2 + KBr + H_2O}$$

(d) Nitropropane from A

A is heated with alcoholic silver nitrite. $AgNO_2$ is covalent, so it attacks through nitrogen:

$$\mathrm{CH_3CH_2CH_2Br + AgNO_2 \xrightarrow{alc.,\ \Delta} \underset{\text{1-nitropropane}}{CH_3CH_2CH_2NO_2} + AgBr}$$

Ethanol is warmed with a paste of bleaching powder and water and the product is distilled (haloform reaction). Bleaching powder supplies $Cl_2$ (oxidising and chlorinating agent) and $Ca(OH)_2$ (base):

$$\mathrm{CaOCl_2 + H_2O \longrightarrow Ca(OH)_2 + Cl_2}$$ $$\mathrm{CH_3CH_2OH + Cl_2 \longrightarrow CH_3CHO + 2\,HCl}$$ $$\mathrm{CH_3CHO + 3\,Cl_2 \longrightarrow \underset{\text{chloral}}{CCl_3CHO} + 3\,HCl}$$ $$\mathrm{2\,CCl_3CHO + Ca(OH)_2 \longrightarrow 2\,CHCl_3 + (HCOO)_2Ca}$$

Chloroform distils over. It is washed, dried over anhydrous $CaCl_2$ and redistilled (b.p. 334 K).

(d) Propene

Propene adds HBr by Markovnikov's rule to give 2-bromopropane (B). Wurtz reaction joins two isopropyl groups to give 2,3-dimethylbutane (C):

$$\mathrm{2\,CH_3CH{=}CH_2 \xrightarrow{2\,HBr} 2\,(CH_3)_2CHBr \xrightarrow[\Delta]{2\,Na/\text{ether}} (CH_3)_2CH{-}CH(CH_3)_2}$$

When chloroethane is boiled with alcoholic KOH, HCl is removed from neighbouring carbon atoms and ethene is formed:

$$\mathrm{CH_3CH_2Cl + KOH(alc) \xrightarrow{\Delta} CH_2{=}CH_2 + KCl + H_2O}$$

1. Saytzeff's rule

In dehydrohalogenation, when more than one alkene can form, the more substituted alkene (the one with fewer H atoms on the double-bonded carbons) is the major product. For example, 2-bromobutane with alcoholic KOH:

$$\mathrm{CH_3CH_2CHBrCH_3 + KOH(alc) \xrightarrow{\Delta} \underset{\text{but-2-ene (major, about 80\%)}}{CH_3CH{=}CHCH_3} + \underset{\text{but-1-ene (minor, about 20\%)}}{CH_3CH_2CH{=}CH_2} + KBr + H_2O}$$

2. Wurtz reaction

Bromoethane heated with sodium in dry ether gives butane:

$$\mathrm{2\,C_2H_5Br + 2\,Na \xrightarrow{\text{dry ether},\ \Delta} C_2H_5{-}C_2H_5 + 2\,NaBr}$$

3. Chloroform exposed to air

In air and sunlight, chloroform is slowly oxidised to phosgene (carbonyl chloride), which is a very poisonous gas:

$$\mathrm{2\,CHCl_3 + O_2 \xrightarrow{\text{sunlight}} 2\,COCl_2 + 2\,HCl}$$

Prevention

• Chloroform is stored in dark brown (amber) bottles that are filled up to the brim and tightly closed. Light and air are kept out.
• About 1% ethanol is added. Ethanol turns any phosgene formed into harmless diethyl carbonate:

$$\mathrm{COCl_2 + 2\,C_2H_5OH \longrightarrow (C_2H_5O)_2CO + 2\,HCl}$$

4. Uses of chloroform (any two)

• As a solvent for fats, oils, waxes, rubber, resins, iodine and alkaloids.
• In the preparation of chloretone (a hypnotic drug) and chloropicrin (an insecticide and tear gas).
• As a laboratory reagent, for example in the carbylamine test for primary amines.
• As a preservative for anatomical specimens. (It was once used as an anaesthetic.)

(c) $S_N1$ and $S_N2$ Mechanism

Isopropyl chloride, $(CH_3)_2CHCl$, is a secondary alkyl halide. It is not too crowded for backside attack ($S_N2$), and its secondary carbocation is fairly stable ($S_N1$). So it can react by both mechanisms. Which one dominates depends on the nucleophile and the solvent.

Each compound can be made from the one before it, so the five compounds form the chain:

$$\text{Ethoxyethane} \rightarrow \text{Ethanol} \rightarrow \text{Iodoethane} \rightarrow \text{Ethyl cyanide} \rightarrow \text{Propanamine}$$

Step 1: Ethoxyethane → Ethanol (hydrolysis of ether)

Ethoxyethane is heated with water and dilute $H_2SO_4$ under pressure:

$$\mathrm{C_2H_5{-}O{-}C_2H_5 + H_2O \xrightarrow{\text{dil. }H_2SO_4,\ \Delta,\ \text{pressure}} 2\,C_2H_5OH}$$

Step 2: Ethanol → Iodoethane

Ethanol is heated with red phosphorus and iodine (which form $PI_3$ in situ):

$$\mathrm{2P + 3I_2 \longrightarrow 2PI_3}$$ $$\mathrm{3\,C_2H_5OH + PI_3 \xrightarrow{\Delta} 3\,C_2H_5I + H_3PO_3}$$

Step 3: Iodoethane → Ethyl cyanide (propanenitrile)

Iodoethane is boiled with alcoholic KCN. KCN is ionic, so $CN^-$ attacks through carbon:

$$\mathrm{C_2H_5I + KCN \xrightarrow{alc.,\ \Delta} C_2H_5CN + KI}$$

Step 4: Ethyl cyanide → Propanamine (Mendius reduction)

Ethyl cyanide is reduced with sodium and ethanol (or with $LiAlH_4$):

$$\mathrm{C_2H_5C{\equiv}N + 4[H] \xrightarrow{Na/C_2H_5OH} \underset{\text{propan-1-amine}}{CH_3CH_2CH_2NH_2}}$$

Reaction chain with reagents

$$\mathrm{C_2H_5OC_2H_5 \xrightarrow[\Delta,\ \text{pressure}]{H_2O/H^+} C_2H_5OH \xrightarrow{\text{red P}/I_2} C_2H_5I \xrightarrow{\text{alc. KCN}} C_2H_5CN \xrightarrow{Na/C_2H_5OH} C_3H_7NH_2}$$

(The chain can also start with ethoxyethane and cold HI, which gives iodoethane and ethanol directly: $\mathrm{C_2H_5OC_2H_5 + HI \rightarrow C_2H_5I + C_2H_5OH}$.)

Chloroform adds to the C=O group of acetone in the presence of aqueous KOH. The product is chloretone (1,1,1-trichloro-2-methylpropan-2-ol), a hypnotic (sleep-inducing) drug:

$$\mathrm{(CH_3)_2C{=}O + CHCl_3 \xrightarrow{KOH} \underset{\text{chloretone}}{(CH_3)_2C(OH)CCl_3}}$$

(d) $CH_3CH_2Cl$

This is the Wurtz reaction, in which two alkyl groups join together:

$$2CH_3CH_2Cl + 2Na \xrightarrow{dry\ ether} CH_3CH_2CH_2CH_3 + 2NaCl$$

Butane, $C_4H_{10}$, has twice the carbon count of the alkyl group, so A must have two carbons. $CH_3Cl$ would give ethane, and $CH_3CH_2CH_2Cl$ or $CH_3CH_2CH_2CH_2Cl$ would give hexane or octane.

Preparation from ethanol (haloform reaction). Ethanol is heated with bleaching powder and water. Chlorine is produced, oxidises ethanol to ethanal, and then chlorinates it:

$$CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$$ $$CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl$$ $$CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO + 3HCl$$ $$2CCl_3CHO + Ca(OH)_2 \rightarrow 2CHCl_3 + (HCOO)_2Ca$$

Actions of trichloromethane

(a) Silver powder: On heating, chlorine is removed to give ethyne:

$$2CHCl_3 + 6Ag \xrightarrow{\Delta} HC\equiv CH + 6AgCl$$

(b) Conc. $HNO_3$: It gives chloropicrin (trichloronitromethane), a tear gas and insecticide:

$$CHCl_3 + HNO_3 \rightarrow CCl_3NO_2 + H_2O$$

(c) Propanone: In the presence of a base (KOH) it adds to give chloretone, a hypnotic:

$$CH_3COCH_3 + CHCl_3 \xrightarrow{KOH} (CH_3)_2C(OH)CCl_3$$

(d) Aqueous KOH: It is hydrolysed to potassium formate:

$$CHCl_3 + 4KOH \rightarrow HCOOK + 3KCl + 2H_2O$$

Compound (A) is chloroform, $CHCl_3$. Its vapour is an anaesthetic, and it gives chloropicrin ($CCl_3NO_2$, the tear gas) with conc. $HNO_3$.

i) Preparation from an isomer of $C_2H_6O$ [2]

The isomer used is ethanol. It is heated with bleaching powder (which supplies chlorine and acts as the base):

$$CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$$ $$CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl \quad(\text{oxidation})$$ $$CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO + 3HCl \quad(\text{chlorination})$$ $$2CCl_3CHO + Ca(OH)_2 \rightarrow 2CHCl_3 + (HCOO)_2Ca \quad(\text{hydrolysis})$$

ii) Reactions of chloroform [1+1+1]

a) Exposed in air (in the presence of light): it is oxidised to poisonous phosgene gas. This is why it is stored in dark bottles filled to the brim.

$$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$$

b) Heated with Ag powder: chlorine is removed and ethyne is formed.

$$2CHCl_3 + 6Ag \xrightarrow{\Delta} HC\equiv CH + 6AgCl$$

c) Heated with aqueous alkali: hydrolysis gives potassium formate (the end product is a formate).

$$CHCl_3 + 4KOH(aq) \rightarrow HCOOK + 3KCl + 2H_2O$$

(a) Nitric acid

Chloroform is nitrated by conc. $HNO_3$ to give chloropicrin (trichloronitromethane):

$$CHCl_3 + HNO_3 \rightarrow CCl_3NO_2 + H_2O$$

Compound (A) is chloroform ($CHCl_3$). It condenses with acetone to give chloretone, a hypnotic (sleep-inducing) drug.

i) Preparation from ethanol (an isomer of $C_2H_6O$; the other is dimethyl ether). Ethanol is heated with bleaching powder and water. Bleaching powder gives chlorine, which oxidises ethanol to ethanal, then chlorinates it to chloral, and chloral is hydrolysed by $Ca(OH)_2$:

$$CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$$ $$CH_3CH_2OH \xrightarrow{Cl_2} CH_3CHO \xrightarrow{3Cl_2} CCl_3CHO\ (\text{chloral})$$ $$2CCl_3CHO + Ca(OH)_2 \rightarrow 2CHCl_3 + (HCOO)_2Ca$$

ii) Storage. In air and sunlight chloroform is slowly oxidised to phosgene ($COCl_2$), a very poisonous gas:

$$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$$

It is kept in a tightly closed, dark (brown) bottle filled to the top. About 1% ethanol is added because ethanol converts any phosgene formed into harmless diethyl carbonate:

$$COCl_2 + 2C_2H_5OH \rightarrow (C_2H_5O)_2CO + 2HCl$$

iii) Boiling with aqueous alkali. Chloroform is hydrolysed to potassium formate:

$$CHCl_3 + 4KOH \rightarrow HCOOK + 3KCl + 2H_2O$$

Identifying the compounds

• T ($C_3H_8O$) is a secondary alcohol, so T = propan-2-ol, $CH_3CH(OH)CH_3$.
• P ($C_3H_7Br$) is made from T, so P = 2-bromopropane (isopropyl bromide), $CH_3CHBrCH_3$.
• Q ($C_6H_{14}$) has two methyl groups on adjacent carbons, so Q = 2,3-dimethylbutane, $(CH_3)_2CH{-}CH(CH_3)_2$.
• R ($CH_2O$) = methanal (formaldehyde), HCHO; S ($C_2H_4O$) = ethanal (acetaldehyde), $CH_3CHO$.

1. P from T. Propan-2-ol is heated with $PBr_3$ (or with HBr):

$$\mathrm{3\,CH_3CH(OH)CH_3 + PBr_3 \longrightarrow 3\,CH_3CHBrCH_3 + H_3PO_3}$$

2. P to Q (Wurtz reaction). 2-Bromopropane is heated with sodium metal in dry ether. Two isopropyl groups join together:

$$\mathrm{2\,(CH_3)_2CHBr + 2\,Na \xrightarrow{\text{dry ether},\ \Delta} (CH_3)_2CH{-}CH(CH_3)_2 + 2\,NaBr}$$

3. P to R and S.

(a) Dehydrohalogenation: P is boiled with alcoholic KOH to give propene.

$$\mathrm{CH_3CHBrCH_3 + KOH(alc) \xrightarrow{\Delta} CH_3CH{=}CH_2 + KBr + H_2O}$$

(b) Ozonolysis: ozone adds to propene to form propene ozonide.

$$\mathrm{CH_3CH{=}CH_2 + O_3 \longrightarrow \underset{\text{propene ozonide}}{CH_3CH(O_3)CH_2}}$$

Structure of propene ozonide:

              O ---- O
             /        \
   CH3 --- CH          CH2
             \        /
                 O

(c) The ozonide is hydrolysed with Zn and water. The C=C bond is broken, giving ethanal (S) and methanal (R):

$$\mathrm{CH_3CH(O_3)CH_2 + H_2O \xrightarrow{Zn} \underset{(S)}{CH_3CHO} + \underset{(R)}{HCHO} + H_2O_2}$$

(Zn removes the $H_2O_2$ as ZnO + $H_2O$, so the aldehydes are not oxidised further.)

Summary: T (propan-2-ol) → P (2-bromopropane) → Q (2,3-dimethylbutane) by Wurtz reaction; P → propene → R (HCHO) + S ($CH_3CHO$) by ozonolysis.

(b) Methane

Compound A is chloroform (phosgene in air; chloretone with acetone). When it is reduced with zinc dust and water (neutral medium), all three Cl atoms are replaced by H:

$$\mathrm{CHCl_3 + 6[H] \xrightarrow{Zn/H_2O} CH_4 + 3\,HCl}$$

In acidic medium (Zn/HCl), it gives methylene chloride ($CH_2Cl_2$) instead.

a) Compound X

X gives a hypnotic drug (chloretone) with acetone and forms carbonyl chloride (phosgene) in air, so X = chloroform (trichloromethane), $CHCl_3$.

b) Formation of X from ethanol

Ethanol is distilled with bleaching powder and water:

$$\mathrm{CaOCl_2 + H_2O \longrightarrow Ca(OH)_2 + Cl_2}$$ $$\mathrm{CH_3CH_2OH + Cl_2 \longrightarrow CH_3CHO + 2\,HCl}$$ $$\mathrm{CH_3CHO + 3\,Cl_2 \longrightarrow CCl_3CHO + 3\,HCl}$$ $$\mathrm{2\,CCl_3CHO + Ca(OH)_2 \longrightarrow 2\,CHCl_3 + (HCOO)_2Ca}$$

c) X with conc. nitric acid

Chloroform is nitrated to chloropicrin (trichloronitromethane), which is used as an insecticide and a tear gas:

$$\mathrm{CHCl_3 + HO{-}NO_2 \xrightarrow{\Delta} \underset{\text{chloropicrin}}{CCl_3NO_2} + H_2O}$$

d) X into acetylene

Chloroform is heated with silver powder:

$$\mathrm{2\,CHCl_3 + 6\,Ag \xrightarrow{\Delta} HC{\equiv}CH + 6\,AgCl}$$

(b) Saytzeff's rule

In elimination (dehydrohalogenation or dehydration), the hydrogen is removed from the carbon bearing the fewer hydrogens, so the more substituted (more stable) alkene is the major product. For example, 2-bromobutane with alcoholic KOH gives mainly but-2-ene rather than but-1-ene.

(c) elimination

Heating an alkyl halide with alcoholic KOH removes H and X from adjacent carbons (dehydrohalogenation), forming a $C=C$ bond:

$$CH_3CH_2Br + KOH(alc) \xrightarrow{\Delta} CH_2=CH_2 + KBr + H_2O$$

The preparation is based on the haloform reaction. A compound containing the $CH_3CO-$ group is exhaustively halogenated at the methyl group. The resulting trihalo ketone is then cleaved by alkali into a haloform and a carboxylate salt.

With bleaching powder, which acts as both chlorinating and oxidising agent:

1. Bleaching powder in water supplies chlorine: $CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$
2. Chlorination of acetone: $CH_3COCH_3 + 3Cl_2 \rightarrow CCl_3COCH_3 + 3HCl$
3. Hydrolysis of trichloroacetone by $Ca(OH)_2$:

$$2CCl_3COCH_3 + Ca(OH)_2 \rightarrow 2CHCl_3 + (CH_3COO)_2Ca$$

The $CHCl_3$ is distilled off from the mixture.