Halogenoalkanes

Class 11 Chemistry, unit 17, book pages 339–348

Practise 45 MCQs Take the unit quiz Review 16 cards

Halogenoalkanes (alkyl halides) are alkanes in which a hydrogen atom has been replaced by a halogen atom. The unit classifies them as primary, secondary and tertiary, links their polar C-X bond to their physical properties, and shows how they are made from alcohols, alkanes and alkenes. It then explains why they are reactive, and studies their two typical reactions: nucleophilic substitution (with NaOH, KCN, ammonia and silver nitrate) and elimination of a hydrogen halide to give an alkene.

Learning outcomes (4)

Classify halogenoalkanes based on the type of halogen atom and its position in the carbon chain, and explain how the molecular structure affects their reactivity.
Explain the organic functional groups involved in a simple halogenoalkane synthesis, and predict the major product(s) based on the reaction conditions. This includes: (a) the free radical substitution of alkanes by Cl2 or Br2 in the presence of ultraviolet light, as exemplified by the reaction of ethane. (b) electrophilic addition of an alkene with a halogen, X2, or hydrogen halide, HX(g), at room temperature. (c) substitution of an alcohol, e.g. by reaction with HX or HBr with H2SO4 or H3PO4, or with PCl3 and heat or PCl5 or SOCl2.
Describe nucleophilic substitution reactions specifically (a) the reaction with NaOH(aq) and heat to produce an alcohol (b) the reaction with KCN in ethanol and heat to produce a nitrile (c) the reaction with ammonia in ethanol heated under pressure to produce an amine (d) the reaction with aqueous silver nitrate in ethanol as a method of identifying the halogen present as exemplified by bromoethane.
Describe the elimination reaction with NaOH in ethanol and heat to produce an alkene as exemplified by bromoethane.

What are halogenoalkanes?

Halogenoalkanes are very common organic compounds. They are used as non-polar solvents and in the synthesis of many other organic compounds.

Halogenoalkane (alkyl halide): A compound in which one hydrogen atom of an alkane has been replaced by a halogen atom. Halogenoalkanes are also called alkyl halides, or halogen derivatives of alkanes.

The general formula of an alkyl halide is $R - X$ , where $R$ is an alkyl group and $X$ is a halogen atom ( $F$ , $Cl$ , $Br$ or $I$ ). For a monohalogen derivative the general formula is $C X_{n} H X_{2 n + 1} X$ .

R - X or C X_{n} H X_{2 n + 1} X

General formula of an alkyl halide

How an alkyl halide is related to its alkane

Start from an alkane, for example ethane.
$CH X_{3} - CH X_{3}$
Replace one hydrogen atom by a halogen atom, here chlorine.
$CH X_{3} - CH X_{2} - Cl$
The result is an alkyl halide: chloroethane (ethyl chloride). Check the formula: $n = 2$ gives $C X_{2} H X_{5} Cl$ , which is $C X_{n} H X_{2 n + 1} X$ .

17.1Types of halogenoalkanes

Halogenoalkanes may be mono, di, tri or poly haloalkanes, depending on the number of halogen atoms present in the molecule.

Figure 17.1: Mono, di and tri haloalkanes.

X

is a halogen atom, shown in the accent colour; the remaining bonds of the central carbon go to hydrogen.

Mono, di and tri haloalkanes of methane
Type	Halogen atoms	Example formula
Monohaloalkane	one	$CH X_{3} X$
Dihaloalkane	two	$CH X_{2} X X_{2}$
Trihaloalkane	three	$CHX X_{3}$

Classification of halogenoalkanes

Halogenoalkanes are classified into primary, secondary and tertiary alkyl halides, based on the type of halogen atom and its position in the carbon chain. What decides the class is the carbon atom that carries the halogen.

1. Primary alkyl halides

An alkyl halide in which the halogen atom is attached to a primary carbon is called a primary alkyl halide. A primary carbon is a carbon to which one or no alkyl group is attached.

Methyl chloride (chloromethane)

Ethyl chloride (chloroethane)

n-Propyl chloride (chloropropane)

In each of these the carbon bonded to chlorine is the primary carbon.

2. Secondary alkyl halides

An alkyl halide in which the halogen atom is attached to a secondary carbon is called a secondary alkyl halide. A secondary carbon is a carbon to which two alkyl groups are attached.

2-Chloropropane (iso-propyl chloride)

2-Chlorobutane (sec-butyl chloride)

In each of these the carbon bonded to chlorine is the secondary carbon.

3. Tertiary alkyl halides

An alkyl halide in which the halogen atom is attached to a tertiary carbon is called a tertiary alkyl halide. A tertiary carbon is a carbon to which three alkyl groups are attached.

2-Methyl-2-chloropropane (t-butyl chloride)

The carbon bonded to chlorine carries three methyl groups, so it is the tertiary carbon.

Classification of alkyl halides
Class	Carbon carrying the halogen has	Examples
Primary	one or no alkyl group	$CH X_{3} Cl$ , $CH X_{3} CH X_{2} Cl$ , $CH X_{3} CH X_{2} CH X_{2} Cl$
Secondary	two alkyl groups	$CH X_{3} CHClCH X_{3}$ , $CH X_{3} CHClCH X_{2} CH X_{3}$
Tertiary	three alkyl groups	$(CH X_{3}) X_{3} CCl$

Worked example: classifying the isomers of C4H9Br

There are four structural isomers of

C X_{4} H X_{9} Br

that have a bromine atom on a saturated carbon chain. Draw them and classify each as a primary, secondary or tertiary alkyl halide.

1-bromobutane (primary)

2-bromobutane (secondary)

1-bromo-2-methylpropane (primary)

2-bromo-2-methylpropane (tertiary)

Start with the two carbon skeletons of $C X_{4}$ : the straight chain (butane) and the branched chain (2-methylpropane). Place the bromine on each different kind of carbon.
Classify by counting the carbon atoms joined to the carbon that carries $Br$ (ignore the halogen itself). One carbon: primary. Two carbons: secondary. Three carbons: tertiary.
1-Bromobutane: the $CBr$ carbon is joined to one carbon, so it is primary. 2-Bromobutane: joined to two carbons, so secondary.
1-Bromo-2-methylpropane: the $CBr$ carbon is joined to one carbon (the $CH$ ), so it is primary. 2-Bromo-2-methylpropane: joined to three methyl groups, so tertiary.

two primary, one secondary and one tertiary isomer

17.2Physical properties

Alkyl halides have higher melting and boiling points than alkanes.

Reason: the $C - X$ bond is polar, because a halogen is more electronegative than carbon. The shared pair of electrons is pulled towards the halogen, so the carbon carries a partial positive charge ( $δ +$ ) and the halogen a partial negative charge ( $δ -$ ). The bond pair electron density is more concentrated on the halogen, $X$ .

Figure 17.2: Polarity of the

C - X

bond. The arrow points from the

δ +

carbon towards the

δ -

halogen, the direction in which the bond electrons are drawn.

Structure

Halogenoalkanes ( $R - X$ ) consist of an $s p^{3}$ hybridised carbon atom which is bonded to a halogen ( $X$ ) by a $σ$ bond.
The carbon-halogen bond is quite polar because of the difference in electronegativity between carbon and the halogen.

17.3Preparations of alkyl halidesOften examined

The book gives three ways of making alkyl halides: from alcohols, from alkanes and from alkenes.

1. From alcohol

Alcohols are converted to the corresponding alkyl halides when treated with $HCl$ in the presence of anhydrous zinc chloride, or with thionyl chloride in the presence of pyridine as a solvent, or with phosphorus trihalides or phosphorus pentahalides. The $- OH$ group of the alcohol is replaced by a halogen. (The book refers to section 18.5 for more.)

Conversion of ethanol into ethyl halides
Reagent (conditions)	Reaction of ethanol	Alkyl halide
$HCl$ , anhydrous $ZnCl X_{2}$	$CH X_{3} CH X_{2} OH + HCl CH X_{3} CH X_{2} Cl + H X_{2} O$	chloroethane
$SOCl X_{2}$ , pyridine as a solvent	$CH X_{3} CH X_{2} OH + SOCl X_{2} CH X_{3} CH X_{2} Cl + SO X_{2} + HCl$	chloroethane
$PBr X_{3}$	$3 CH X_{3} CH X_{2} OH + PBr X_{3} 3 CH X_{3} CH X_{2} Br + H X_{3} PO X_{3}$	bromoethane
$PCl X_{3}$ or $PCl X_{5}$	$CH X_{3} CH X_{2} OH + PCl X_{5} CH X_{3} CH X_{2} Cl + POCl X_{3} + HCl$ and $3 CH X_{3} CH X_{2} OH + PCl X_{3} 3 CH X_{3} CH X_{2} Cl + H X_{3} PO X_{3}$	chloroethane
$PI X_{3}$	$3 CH X_{3} CH X_{2} OH + PI X_{3} 3 CH X_{3} CH X_{2} I + H X_{3} PO X_{3}$	iodoethane

2. From alkanes

When an alkane is treated with chlorine or bromine in the presence of diffused sunlight or ultraviolet light, a hydrogen atom of the alkane is replaced by a halogen and a halogenoalkane is produced. The reaction proceeds by a free radical mechanism.

CH X_{3} - CH X_{3} + Cl X_{2} h ν CH X_{3} CH X_{2} Cl + HCl

Ethane and chlorine in ultraviolet light

3. From alkenes

(a) Addition of halogen acid (hydrohalogenation)

Addition of a halogen acid to an alkene is known as hydrohalogenation. Alkenes react with a solution of a halogen acid ( $HX$ ) at room temperature to form alkyl halides. The reaction involves breaking of the carbon-carbon double bond, followed by electrophilic addition of an H atom and a halogen. The addition follows Markonikov's rule (the hydrogen adds to the carbon of the double bond that already has more hydrogen atoms, so the halogen goes to the other carbon).

The order of reactivity of the halogen acids is $HI > HBr > HCl$ .

R - CH = CH X_{2} + HX R - CH (X) - CH X_{3}

Alkene and halogen acid give an alkyl halide

H X_{2} C = CH X_{2} + HCl H X_{3} C - CH X_{2} - Cl

Ethene gives ethyl chloride

(b) Addition of halogen

Alkenes react with a halogen in an inert solvent such as carbon tetrachloride at room temperature to give vicinal dihalides, that is 1,2-dihaloalkanes (the two halogen atoms are on adjacent carbons).

H X_{2} C = CH X_{2} + X X_{2} CCl X_{4} XCH X_{2} - CH X_{2} X

Ethene gives a vicinal dihalide

1,2-Dibromoethane (a vicinal dihalide)

$Br X_{2}$ and $Cl X_{2}$ are effective electrophilic reagents. Fluorine is too reactive to control the reaction. Iodine does not react.

Preparation of halogen compounds from alkenes
Alkene reacts with	Conditions	Product
$HX$ (halogen acid)	room temperature	alkyl halide (Markovnikov addition)
$X X_{2}$ ( $Cl X_{2}$ or $Br X_{2}$ )	inert solvent such as $CCl X_{4}$ , room temperature	vicinal dihalide (1,2-dihaloalkane)

Worked example: three routes to chloroethane

Show how chloroethane can be prepared from (a) ethene, (b) ethane and (c) ethanol. Which route gives the purest product, and why?

Three preparations of chloroethane. The route from the alcohol (accent) is the best because it gives the alkyl chloride without a mixture of products (with SOCl₂ the by-products are gases that leave).

(a) Ethene: hydrogen chloride adds across the double bond at room temperature.
$CH X_{2} = CH X_{2} + HCl CH X_{3} CH X_{2} Cl$
(b) Ethane: chlorine and ultraviolet light replace one hydrogen by a free radical reaction.
$CH X_{3} CH X_{3} + Cl X_{2} h ν CH X_{3} CH X_{2} Cl + HCl$
(c) Ethanol: thionyl chloride with pyridine replaces the $- OH$ group by $- Cl$ . The by-products $SO X_{2}$ and $HCl$ are gases and leave.
$CH X_{3} CH X_{2} OH + SOCl X_{2} CH X_{3} CH X_{2} Cl + SO X_{2} + HCl$
Compare. Route (b) cannot be stopped at one chlorine, so dichloro and more highly chlorinated compounds form as well. Route (c) leaves only gaseous by-products, so the alkyl chloride is easy to obtain pure. This is why the book calls alcohols with inorganic halides the best method.

route (c): ethanol with SOCl X_{2} (or PCl X_{3}, PCl X_{5}) gives the purest chloroethane

17.4ReactivityOften examined

Two main factors control the reactivity of halogenoalkanes:

bond polarity of the \ce{C-X} bond
bond energy of the \ce{C-X} bond

1. Bond polarity

The molecule of an alkyl halide is polarised because halogens are more electronegative than carbon. Carbon therefore acquires a partial positive charge, whereas the halogen acquires a partial negative charge.

The carbon is electron deficient, so it can be attacked by a species rich in electrons. The halogen, with its extra electron density, can be replaced by another nucleophile. On the basis of bond polarity, the reactivity of alkyl halides decreases in the order:

\ce{R-F} > \ce{R-Cl} > \ce{R-Br} > \ce{R-I}

Electronegativity values given in the book
Atom	Electronegativity	Atom	Electronegativity
F	3.98	I	2.66
Cl	3.16	H	2.2
Br	2.96	C	2.55

The bigger the electronegativity difference, the more polar the bond
Bond	Electronegativity of X	Difference from C (2.55)	Polarity
\ce{C-F}	3.98	1.43	most polar
\ce{C-Cl}	3.16	0.61	less polar
\ce{C-Br}	2.96	0.41	less polar
\ce{C-I}	2.66	0.11	least polar

2. Bond energy

The energy required to break one mole of bonds of the same type is called the bond energy. The greater the bond energy, the more energy is required to break the bond.

The reactivity of alkyl halides depends on the strength of the \ce{C-X} bond. The \ce{C-F} bond is the strongest, while the \ce{C-I} bond is the weakest. So the overall order of reactivity of alkyl halides is:

\ce{R-I} > \ce{R-Br} > \ce{R-Cl} > \ce{R-F}

In fact the \ce{C-F} bond is so strong that alkyl fluorides do not react under ordinary conditions.

Polarity and bond strength act in opposite directions; bond strength wins
Alkyl halide	Polarity of \ce{C-X}	Strength of \ce{C-X}	Overall reactivity
\ce{R-F}	highest	strongest	lowest (no reaction under ordinary conditions)
\ce{R-Cl}	high	strong	low
\ce{R-Br}	lower	weaker	high
\ce{R-I}	lowest	weakest	highest

Worked example: polarity and reactivity of chloro, bromo and iodo ethane

Using the electronegativities C = 2.55, Cl = 3.16, Br = 2.96 and I = 2.66, find the electronegativity difference in \ce{C-Cl}, \ce{C-Br} and \ce{C-I}. Arrange the three ethyl halides (i) by polarity of the bond and (ii) by reactivity, and say which gives a precipitate first with \ce{AgNO3}.

The electronegativity difference measures the polarity of the bond. Subtract the value of carbon from that of the halogen.
\begin{aligned}\Delta EN(\ce{C-Cl})&=3.16-2.55=0.61\\ \Delta EN(\ce{C-Br})&=2.96-2.55=0.41\\ \Delta EN(\ce{C-I})&=2.66-2.55=0.11\end{aligned}
(i) The larger the difference, the more polar the bond.
\ce{C2H5Cl} > \ce{C2H5Br} > \ce{C2H5I}\quad\text{(polarity)}
(ii) Reactivity depends mainly on the strength of the \ce{C-X} bond, which is weakest for iodine. The order is the reverse of the polarity order.
\ce{C2H5I} > \ce{C2H5Br} > \ce{C2H5Cl}\quad\text{(reactivity)}
With \ce{AgNO3} the halogen leaves as the halide ion and forms \ce{AgX}. The most reactive halide releases its halide ion fastest, so iodoethane gives the yellow precipitate of \ce{AgI} first.

\text{polarity: Cl > Br > I; reactivity: I > Br > Cl; precipitate first from iodoethane}

17.5Reactions of alkyl halidesOften examined

Alkyl halides undergo two types of reactions:

nucleophilic substitution reactions (S_N reactions)
elimination reactions

Nucleophilic substitution (S_N) reactions of alkyl halides

Reactions in which the halogen of an alkyl halide is substituted, or replaced, by an attacking nucleophile are known as nucleophilic substitution (S_N) reactions. A substitution means that one group replaces another. The general S_N reaction is shown below.

\ce{Nu:^- + R-X -> R-Nu + :X^-}

General nucleophilic substitution

Figure 17.3: General nucleophilic substitution. The lone pair of the attacking nucleophile forms a new bond to carbon, and at the same time the \ce{C-X} bond pair goes to \ce{X}, which leaves as \ce{X^-}. The curved arrows (accent) show the movement of the electron pairs.

What happens in an S_N reaction

The carbon of \ce{R-X} carries a partial positive charge (\delta+) because \ce{X} is more electronegative.
The nucleophile donates a lone pair of electrons to this electrophilic carbon and forms a new bond.
The \ce{C-X} bond breaks and its electron pair goes to the halogen, which departs as the halide ion, the leaving group.
The product \ce{R-Nu} is formed. The halogen has been replaced by the nucleophile.
\ce{Nu:^- + R-X -> R-Nu + :X^-}

Substrate

The alkyl halide molecule on which the nucleophile attacks is called the substrate molecule.

Nucleophile

A nucleophile is a species having a lone pair of electrons, which it donates to the electrophilic carbon of the alkyl halide. It may be neutral or negatively charged.

Examples of nucleophiles
Nucleophile (formula)	Name
\ce{HO^-}	hydroxide ion
\ce{C2H5O^-}	ethoxide ion
\ce{HS^-}	hydrogen sulphide ion
\ce{SCN^-}	thiocyanate ion
\ce{H2O} (with lone pairs on oxygen)	water
\ce{NH2^-}	amino group (amide ion)
\ce{Cl^-}	chloride ion
\ce{Br^-}	bromide ion
\ce{:NH3}	ammonia

Leaving group (LG)

The halogen of an alkyl halide is called the leaving group. A leaving group is also a nucleophile: it departs with an unshared pair of electrons. The incoming nucleophile must be stronger than the departing one.

Good leaving groups: \ce{Cl^-}, \ce{Br^-}, \ce{I^-} and \ce{HSO4^-}.
Poor leaving groups: \ce{OH^-}, \ce{OR^-} and \ce{NH2^-}.
Iodide ion is a good nucleophile as well as a good leaving group.

17.5Nucleophilic substitution reactions of alkyl halidesOften examined

Four substitution reactions of alkyl halides are studied. In each, a different nucleophile replaces the halogen.

1. Reaction with aqueous sodium hydroxide

When an alkyl halide is heated with an aqueous solution of sodium hydroxide, an alcohol is produced. The nucleophile is the hydroxide ion, which replaces the halogen.

\ce{R-X + NaOH ->[\Delta] R-OH + NaX}

\ce{CH3-Cl + NaOH ->[\Delta] CH3-OH + NaCl}

Chloromethane gives methanol

2. Reaction with potassium cyanide

When an alkyl halide is heated with \ce{KCN} in ethanol, the halogen atom is replaced by \ce{CN} and a nitrile is produced. The nucleophile is the cyanide ion, \ce{CN^-}.

\ce{R-X + KCN ->[\text{ethanol},\ \Delta] R-CN + KX}

\ce{CH3-Cl + KCN ->[\text{ethanol},\ \Delta] CH3-CN + KCl}

Chloromethane gives ethanenitrile

3. Reaction with ammonia

Primary amines are formed when an alkyl halide is heated with excess ammonia in ethanol (under pressure). The nucleophile is ammonia, \ce{:NH3}.

\ce{R-X + NH3 ->[\text{ethanol},\ \Delta] R-NH2 + HX}

\ce{CH3-Cl + NH3 ->[\text{ethanol},\ \Delta] CH3-NH2 + HCl}

Chloromethane gives methylamine

4. Reaction with silver nitrate

When an alkyl halide is heated with \ce{AgNO3} in ethanol, the halogen atom is replaced by the \ce{NO3} group. This reaction can be used to find which halogen is present in an alkyl halide.

\ce{R-X + AgNO3 -> R-NO3 + AgX}

\ce{CH3-CH2-Br + AgNO3 -> CH3-CH2-NO3 + AgBr v}

Bromoethane gives ethyl nitrate and silver bromide

In this reaction the halogen is displaced as the halide ion, which combines with \ce{Ag+} ions to form a precipitate of the silver halide.

\ce{Ag+ + X- -> AgX v}

The colour of the precipitate, and its solubility in aqueous ammonia, identify the halogen present in the alkyl halide.

Identifying the halogen of an alkyl halide
Halide ion in the alkyl halide	Precipitate with \ce{AgNO3}	Solubility in aqueous ammonia
\ce{Cl^-}	white	soluble
\ce{Br^-}	cream	partially soluble
\ce{I^-}	yellow	insoluble

Worked example: identifying a halogen with silver nitrate, and the mass of precipitate

(a) An alkyl halide is heated with \ce{AgNO3} in ethanol. A yellow precipitate forms, which does not dissolve in aqueous ammonia. Which halogen does the compound contain? (b) A sample of 1.09\,\mathrm{g} of bromoethane reacts completely with excess \ce{AgNO3}. Find the mass of the precipitate.

Given

m(\ce{C2H5Br}) = 1.09\,\mathrm{g}
A_r = C = 12,\ H = 1,\ Br = 80,\ Ag = 108

To find

\text{halogen in (a)}
m(\ce{AgBr})

(a) Compare with the book's table: \ce{AgCl} is white and soluble, \ce{AgBr} is cream and partially soluble, \ce{AgI} is yellow and insoluble in ammonia. A yellow, insoluble precipitate is \ce{AgI}, so the compound is an alkyl iodide.
(b) Write the equation. One mole of bromoethane gives one mole of \ce{AgBr}.
\ce{CH3CH2Br + AgNO3 -> CH3CH2NO3 + AgBr v}
Molar mass of bromoethane, \ce{C2H5Br}.
M = 2(12)+5(1)+80 = 109\,\mathrm{g\,mol^{-1}}
Moles of bromoethane.
n = \dfrac{1.09}{109} = 0.0100\,\mathrm{mol}
Moles of \ce{AgBr} equal moles of bromoethane. Molar mass of \ce{AgBr} is 108+80=188\,\mathrm{g\,mol^{-1}}.
m = 0.0100\times 188 = 1.88\,\mathrm{g}

\text{(a) iodine; (b) } 1.88\,\mathrm{g\ of\ AgBr}

Summary of the nucleophilic substitution reactions of alkyl halides
Reagent (nucleophile)	Conditions	Product	Type of product
\ce{NaOH} (\ce{OH^-})	aqueous, heat	\ce{R-OH}	alcohol
\ce{KCN} (\ce{CN^-})	ethanol, heat	\ce{R-CN}	nitrile
\ce{NH3} (excess)	ethanol, heat under pressure	\ce{R-NH2}	primary amine
\ce{AgNO3}	ethanol, heat	\ce{R-NO3} and \ce{AgX}	silver halide precipitate

17.5Elimination reactionsOften examined

Elimination reaction: The chemical reaction in which two groups (the halogen and a \beta-hydrogen from an alkyl halide) are eliminated from two adjacent carbon atoms to form a carbon-carbon double bond. Since a \beta-hydrogen is necessary for elimination, it is also called \beta-elimination.

In bromoethane the carbon that carries bromine is the \alpha carbon, and the carbon next to it is the \beta carbon. The hydrogen on the \beta carbon is the \beta-hydrogen.

Figure 17.4: Elimination from bromoethane by a base. The base takes the \beta-hydrogen, the electrons of the \ce{C-H} bond form the new carbon-carbon bond, and \ce{Br^-} leaves with the \ce{C-Br} bond pair.

\ce{CH2(H)-CH2Br + OH^- ->[heat] CH2=CH2 + H2O + Br^-}

Bromoethane gives ethene

Steps of the elimination

The strong base, hydroxide ion, uses a lone pair to attract the \beta-hydrogen.
\ce{OH^- + H-CH2-CH2-Br}
The \ce{C-H} bond pair shifts to form a double bond between the two carbons.
Bromine leaves with the bond pair of \ce{C-Br} as the bromide ion. The hydrogen and the hydroxide ion form water.
\ce{H-CH2-CH2-Br + OH^- -> CH2=CH2 + H2O + Br^-}

Explanation

The \beta-hydrogen atom in alkyl halides is slightly acidic, because of the electron withdrawing effect (inductive effect) of the halogen. A base can therefore remove it.

Strong bases such as \ce{NaOH} in ethanol cause elimination reactions. The alkyl halide is heated with the base.

\ce{CH3-CH2-Br + NaOH ->[\text{ethanol},\ \Delta] CH2=CH2 + NaBr + H2O}

Bromoethane and ethanolic sodium hydroxide

Substitution and elimination compared
Reaction	Reagent and solvent	Product from bromoethane
Substitution	\ce{NaOH} in water, heat	ethanol, \ce{CH3CH2OH}
Elimination	\ce{NaOH} in ethanol, heat	ethene, \ce{CH2=CH2}

Worked example: same reagent, different solvent: 1-bromopropane

Write the products when 1-bromopropane is heated with (a) \ce{NaOH} in water and (b) \ce{NaOH} in ethanol. Why can chloromethane not give an alkene?

1-Bromopropane. The α carbon carries the halogen; the hydrogens on the β carbon (accent) are the ones removed in elimination.

(a) In water the hydroxide ion acts as a nucleophile. It attacks the \delta+ carbon that carries bromine and replaces \ce{Br}: substitution.
\ce{CH3CH2CH2Br + NaOH(aq) ->[\Delta] CH3CH2CH2OH + NaBr}
(b) In ethanol the strong base takes a \beta-hydrogen from the carbon next to the \ce{C-Br} carbon. Bromide leaves and a double bond forms: elimination.
\ce{CH3CH2CH2Br + NaOH(ethanolic) ->[\Delta] CH3CH=CH2 + NaBr + H2O}
Elimination needs a \beta-hydrogen. In \ce{CH3Cl} the only carbon is the one bonded to chlorine, so there is no \beta carbon and no \beta-hydrogen. Only substitution is possible.

\text{(a) propan-1-ol (substitution); (b) propene (elimination)}

Concept assessment exercise 17.1

Complete the following reactions.

Concept assessment exercise 17.1 (a): propene and hydrogen bromide

Complete: \ce{CH3-CH=CH2 + HBr ->}

This is the addition of a halogen acid to an alkene (hydrohalogenation) at room temperature. The double bond breaks and \ce{H} and \ce{Br} add to the two carbons.
Apply Markovnikov's rule: the hydrogen adds to the carbon of the double bond that already has more hydrogen atoms (\ce{CH2}), so bromine goes to the middle carbon.
\ce{CH3-CH=CH2 + HBr -> CH3-CH(Br)-CH3}
The product is 2-bromopropane, a secondary alkyl halide.

\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}

Concept assessment exercise 17.1 (b): bromoethane and aqueous sodium hydroxide

Complete: \ce{CH3-CH2-Br + NaOH_{(aq)} ->}

Aqueous \ce{NaOH} with heat gives nucleophilic substitution. The hydroxide ion replaces the bromine.
The sodium ion and the bromide ion form sodium bromide.
\ce{CH3-CH2-Br + NaOH_{(aq)} ->[\Delta] CH3-CH2-OH + NaBr}
The product is ethanol, an alcohol.

\ce{CH3CH2Br + NaOH_{(aq)} -> CH3CH2OH + NaBr}

Concept assessment exercise 17.1 (c): chloroethane and potassium cyanide

Complete: \ce{C2H5-Cl + KCN ->}

\ce{KCN} in ethanol with heat gives nucleophilic substitution. The cyanide ion replaces the chlorine.
Potassium and chloride ions form potassium chloride.
\ce{C2H5-Cl + KCN ->[\text{ethanol},\ \Delta] C2H5-CN + KCl}
The product is a nitrile, propanenitrile (\ce{CH3CH2CN}).

\ce{C2H5Cl + KCN -> C2H5CN + KCl}

Concept assessment exercise 17.1 (d): chloroethane and ammonia

Complete: \ce{C2H5-Cl + NH3 ->}

Ammonia is the nucleophile. Heated with excess ammonia in ethanol it replaces the chlorine and gives a primary amine.
The hydrogen chloride that is lost with the amine is written as the second product.
\ce{C2H5-Cl + NH3 ->[\text{ethanol},\ \Delta] C2H5-NH2 + HCl}
The product is ethylamine, a primary amine.

\ce{C2H5Cl + NH3 -> C2H5NH2 + HCl}

Formula sheet

General formula of an alkyl halide

\ce{R-X}\quad\text{or}\quad \ce{C_{n}H_{2n+1}X}

\ce{R} is an alkyl group and \ce{X} is a halogen atom; n is the number of carbon atoms (monohalo derivative).

Unit: none

General nucleophilic substitution

\ce{Nu:^- + R-X -> R-Nu + :X^-}

\ce{Nu:^-} is the attacking nucleophile, \ce{R-X} the substrate and \ce{X^-} the leaving group.

Unit: none

Key terms

Halogenoalkane (alkyl halide): An alkane in which one hydrogen atom has been replaced by a halogen atom, with general formula \ce{R-X} (or \ce{C_nH_{2n+1}X} for one halogen atom).
Primary alkyl halide: An alkyl halide in which the halogen is attached to a primary carbon, a carbon with one or no alkyl group attached.
Secondary alkyl halide: An alkyl halide in which the halogen is attached to a secondary carbon, a carbon with two alkyl groups attached.
Tertiary alkyl halide: An alkyl halide in which the halogen is attached to a tertiary carbon, a carbon with three alkyl groups attached.
Bond polarity: The unequal sharing of the bonding pair of electrons in the \ce{C-X} bond because the halogen is more electronegative than carbon, giving \delta+ on carbon and \delta- on the halogen.
Bond energy: The energy required to break one mole of bonds of the same type. The greater the bond energy, the more energy is needed to break the bond.
Hydrohalogenation: The addition of a halogen acid, \ce{HX}, to an alkene to form an alkyl halide, following Markovnikov's rule.
Vicinal dihalide: A 1,2-dihaloalkane, with a halogen on each of two adjacent carbon atoms, formed when an alkene reacts with a halogen in an inert solvent such as \ce{CCl4}.
Nucleophile: A species with a lone pair of electrons, neutral or negatively charged, that donates the pair to an electrophilic (electron deficient) carbon.
Substrate: The alkyl halide molecule on which the nucleophile attacks.
Leaving group: The halogen of an alkyl halide, which departs with an unshared pair of electrons. \ce{Cl^-}, \ce{Br^-}, \ce{I^-} and \ce{HSO4^-} are good leaving groups; \ce{OH^-}, \ce{OR^-} and \ce{NH2^-} are poor.
Nucleophilic substitution (S_N) reaction: A reaction in which the halogen of an alkyl halide is replaced by an attacking nucleophile.
Elimination reaction: A reaction in which a halogen and a \beta-hydrogen are removed from two adjacent carbon atoms of an alkyl halide to form a carbon-carbon double bond (also called \beta-elimination).
\beta-Hydrogen: A hydrogen atom on the carbon next to the carbon that carries the halogen. It is slightly acidic because of the inductive effect of the halogen.

Multiple choice questionsp. 346

In primary alkyl halides, the halogen atom is attached to a carbon which is further attached to how many carbon atoms?

ATwo
BThree
COneAnswer
DFour

Why

A primary carbon has one or no alkyl group attached (book, section 17.1). In \ce{CH3CH2Cl} the carbon that carries the chlorine is bonded to only one other carbon, the \ce{CH3} group, so the answer is one. Two describes a secondary alkyl halide and three describes a tertiary one, so both are wrong. Four is not used to classify alkyl halides.

Alkyl halides are considered to be very reactive compounds towards nucleophiles, because:

Athey have an electrophilic carbon
Bthey have an electrophilic carbon and a good leaving groupAnswer
Cthey have an electrophilic carbon and a bad leaving group
Dthey have a nucleophilic carbon and a good leaving group

Why

The carbon bonded to the halogen is partly positive (\delta^+), so it is electrophilic and attracts nucleophiles. The halogen is also a good leaving group (\ce{Cl-}, \ce{Br-}, \ce{I-}, book section 17.5), which departs with the bonding pair of electrons. Option A leaves out the leaving group, which is the second half of the reason. Option C calls the leaving group bad, which is wrong. Option D gives the carbon the wrong character, since it is electrophilic, not nucleophilic.

iii

Which one of the following is not a nucleophile?

A\ce{H2O}
B\ce{H2S}
C\ce{BF3}Answer
D\ce{NH3}

Why

A nucleophile must have a lone pair of electrons to donate to an electrophilic carbon. \ce{H2O}, \ce{H2S} and \ce{NH3} all have lone pairs, so they are nucleophiles. In \ce{BF3} the boron has no lone pair and has an empty orbital, so it accepts electron pairs. It is an electrophile (a Lewis acid), not a nucleophile.

Double bond is formed as a result of:

Asubstitution reactions
Belimination reactionsAnswer
Caddition reactions
Drearrangement reactions

Why

In an elimination reaction two atoms or groups are removed from adjacent carbon atoms, and the electrons left behind form a new \pi bond, so a C=C double bond is formed (book, Key Points). Substitution replaces one group by another without creating a double bond. Addition does the opposite: it uses up a double bond. Rearrangement only moves groups within the molecule.

Which of the following alkyl halides cannot be formed by direct reaction of alkanes with halogen?

A\ce{RBr}
B\ce{RCl}
C\ce{RF}
D\ce{RI}Answer

Why

Chlorine and bromine react with alkanes in diffused sunlight or ultraviolet light to give \ce{RCl} and \ce{RBr} (book, section 17.3). The book states that iodine does not react, so \ce{RI} cannot be formed by this route at all. Fluorine reacts too violently to control, which gives an uncontrolled mixture rather than no reaction, so it is a different kind of problem.

Note. No key printed in the book. Worked out from section 17.3 and the statement on p. 342 that iodine does not react. C (RF) is the alternative reading.

Short questionsp. 347

Each short question is answered as a complete 3-mark answer. Questions (i) to (viii) of the book's question 2.

i3 marks

What are primary, secondary and tertiary alkyl halides?

Solution

Alkyl halides are classified by the carbon that carries the halogen, and by how many alkyl groups are attached to that carbon. In a primary alkyl halide the carbon has one or no alkyl group attached, for example \ce{CH3CH2Cl} (chloroethane). In a secondary alkyl halide it has two alkyl groups, for example \ce{CH3CHClCH3} (2-chloropropane). In a tertiary alkyl halide it has three alkyl groups, for example \ce{(CH3)3CCl} (2-chloro-2-methylpropane, t-butyl chloride).

Chloroethane (primary)

2-Chloropropane (secondary)

2-Chloro-2-methylpropane (tertiary)

ii3 marks

What are nucleophilic substitution reactions or SN reaction?

Solution

A nucleophilic substitution (SN) reaction is one in which a nucleophile attacks the electrophilic carbon of an alkyl halide and replaces the halogen. The nucleophile is an electron-rich species with a lone pair, which may be neutral or negatively charged. It donates a pair of electrons to the carbon, which is the substrate. The halogen leaves as a halide ion, so it is the leaving group.

\ce{R-X + Nu- -> R-Nu + X-}

General SN reaction (R = alkyl group, X = halogen, Nu = nucleophile)

For example, \ce{CH3Cl} heated with aqueous \ce{NaOH} gives methanol, and \ce{CH3Cl} heated with \ce{KCN} gives \ce{CH3CN}. In each case the nucleophile takes the place of the halogen.

iii3 marks

Tertiary alkyl halides show SN1 reactions mostly, why?

Solution

SN1 takes place in two steps. The slow first step breaks the C-X bond and gives a carbocation and a halide ion. The fast second step is the attack of the nucleophile on the carbocation.
The slow step involves only the alkyl halide, so the rate depends only on the alkyl halide.
\text{Rate} = k\,[\text{alkyl halide}]
Step 1 (slow): ionisation.
\ce{(CH3)3C-Cl ->[slow] (CH3)3C+ + Cl-}
Step 2 (fast): attack by the nucleophile.
\ce{(CH3)3C+ + OH- ->[fast] (CH3)3C-OH}

A tertiary carbon is bonded to three alkyl groups, and these groups release electron density towards the positive carbon (the inductive effect; hyperconjugation also helps). This spreads and reduces the positive charge, so a tertiary carbocation is the most stable of the three classes and forms most easily. Primary and secondary carbocations are less stable, so those halides do not follow the SN1 route as readily.

Tertiary alkyl halides therefore react mostly by SN1, through the stable carbocation.

Note. SN1 is not named or taught in this unit of the book (it describes only the general SN reaction). The answer gives the mechanism at the level the question needs.

iv3 marks

What are elimination reactions?

Solution

An elimination reaction is one in which two atoms or groups are removed from adjacent carbon atoms, and a carbon-carbon double bond is formed. For an alkyl halide the halogen leaves one carbon and a hydrogen is removed from the next carbon (the \beta-carbon), so it is also called \beta-elimination. A nucleophile that acts as a base, such as \ce{OH-}, removes the \beta-hydrogen.

\ce{CH3CH2Br + NaOH ->[ethanol, heat] CH2=CH2 + NaBr + H2O}

Bromoethane gives ethene with NaOH in ethanol and heat

A \beta-hydrogen must be present. The halogen withdraws electron density by the inductive effect, which makes the \beta-hydrogen slightly acidic, so a strong base removes it. The products are the alkene, the halide ion and water.

v3 marks

Which factor decides the reactivity of alkyl halides?

Solution

The reactivity of an alkyl halide is controlled by two factors: the polarity of the C-X bond and the bond energy (strength) of that bond. The overall order is set by bond energy.

Bond polarity. The carbon bonded to the halogen is partly positive, so nucleophiles attack it. Polarity decreases from R-F to R-I, because the electronegativity falls from fluorine (3.98) to iodine (2.66). On polarity alone, fluorides would be the most reactive.

Bond energy. This is the energy needed to break one mole of the bond. C-F is the strongest and C-I the weakest, and the weaker the bond, the more easily it breaks. The C-F bond is so strong that alkyl fluorides do not react under ordinary conditions.

\ce{R-I > R-Br > R-Cl > R-F}

Overall order of reactivity of alkyl halides

vi3 marks

What are the diazonium salt?

Solution

A diazonium salt is a compound containing the group \ce{-N2+} attached to a benzene ring, with an anion (usually chloride) to balance the charge. An example is benzenediazonium chloride, \ce{C6H5N2+ Cl-}. It is made from a primary aromatic amine, such as aniline (\ce{C6H5NH2}), with nitrous acid (\ce{HNO2}), which is formed in the mixture from sodium nitrite and hydrochloric acid, at 0 to 5 °C. Diazonium salts are unstable and are used as intermediates to make other aromatic compounds.

\ce{C6H5NH2 + HNO2 + HCl -> C6H5N2+ Cl- + 2H2O}

Aniline to benzenediazonium chloride

Benzenediazonium ion

Note. Question as printed. Diazonium salts are not taught in this unit, so the answer gives the definition and preparation only.

vii3 marks

How can nucleophilic addition of a primary amine giving an imine?

Solution

A primary amine, \ce{RNH2}, is a nucleophile because its nitrogen has a lone pair. It attacks the electrophilic carbon of the \ce{C=O} group of an aldehyde or ketone. This nucleophilic addition gives an unstable compound with \ce{-OH} and \ce{-NHR} groups on the same carbon.

\ce{R'CHO + RNH2 -> R'CH(OH)NHR -> R'CH=NR + H2O}

Aldehyde and primary amine give an imine

The loss of a water molecule forms the \ce{C=N} double bond. The product, \ce{R'CH=NR}, is an imine (also called a Schiff base).

Note. Question as printed, with its wording incomplete. Carbonyl compounds and imines are not taught in this unit, so the answer gives only the mechanism needed.

viii3 marks

Amines are more basic than analogous alcohols why?

Solution

Basicity is the ability of a species to accept a proton (\ce{H+}) using a lone pair. The more available the lone pair, the stronger the base.

In an amine the lone pair is on nitrogen, which is less electronegative than oxygen. It is held less tightly, so it is more available for sharing with \ce{H+}. The oxygen of an alcohol holds its lone pairs more strongly. Alkyl groups on nitrogen also release electron density towards it, which makes its lone pair more available still.

\ce{R-NH2 + H+ <=> R-NH3+}\qquad\ce{R-OH + H+ <=> R-OH2+}

Protonation of an amine and of an alcohol

The protonated alcohol, \ce{ROH2+}, is a much stronger acid than \ce{RNH3+}, so it gives up its proton more easily. The alcohol therefore accepts a proton less readily, and the amine is the stronger base.

Note. Not taught in this unit; answer kept to the reasoning needed.

Long questionsp. 347–348

Questions 3 to 12 of the book's exercise. The book does not sub-head them, so they are grouped here as long questions.

Discuss the reactivity of alkyl halides.

Solution

Introduction

The reactivity of an alkyl halide is controlled by two factors (book, section 17.4): the polarity of the C-X bond, which makes the carbon electrophilic, and the strength of the C-X bond, which decides how easily it breaks. The two factors pull in different directions, and the overall order is set by bond energy.

1. Bond polarity

The halogen is more electronegative than carbon, so the bonding pair of the C-X bond is drawn towards the halogen. The carbon becomes partly positive (\delta^+) and the halogen partly negative (\delta^-). The electrophilic carbon is attacked by nucleophiles, and the halogen then departs as a good leaving group.

Electronegativity of the halogens (book)
Atom	Electronegativity
F	3.98
Cl	3.16
Br	2.96
I	2.66

Because fluorine is the most electronegative, the polarity of the C-X bond decreases in the order \ce{R-F > R-Cl > R-Br > R-I}. On polarity alone, fluorides would react most readily.

2. Bond energy

Bond energy is the energy needed to break one mole of a given bond. The more energy needed, the less reactive the halide. The C-F bond is the strongest and the C-I bond the weakest, so the energy needed to break the C-X bond rises from C-I to C-F.

3. Overall order of reactivity

Because the C-X bond must break for a substitution to occur, bond energy is the decisive factor. The overall order is:

\ce{R-I > R-Br > R-Cl > R-F}

Polarity and bond strength of the C-X bond, and the reactivity they give
Halide	Polarity of C-X	Strength of C-X	Reactivity
\ce{R-F}	highest	strongest	lowest; does not react under ordinary conditions
\ce{R-Cl}	high	strong	moderate
\ce{R-Br}	lower	weaker	high
\ce{R-I}	lowest	weakest	highest

The C-F bond is so strong that alkyl fluorides do not react under ordinary conditions. Iodide is both a good leaving group and a good nucleophile, so \ce{R-I} reacts most readily.

4. The leaving group

The halogen is the leaving group and departs with a pair of electrons. The incoming nucleophile must be stronger than the leaving group. \ce{Cl-}, \ce{Br-}, \ce{I-} and \ce{HSO4-} are good leaving groups, while \ce{OH-}, \ce{OR} and \ce{NH2-} are poor ones (book).

Note. Misprint on p. 342: 'Halogen becomes nucleophilic in character'. The carbon is the electrophilic (partly positive) site, so the answer uses that.

Give two methods for the preparation of alkyl halides.

Solution

Method 1: from alcohols (the best method)

Alcohols are converted into alkyl halides by treatment with inorganic halides. The reagents are \ce{SOCl2} (with pyridine as solvent), \ce{PX3} or \ce{PX5}, or \ce{HCl} with anhydrous \ce{ZnCl2}. This is the best method, because the by-products are gases or easily separated (book, Key Points).

\ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 + HCl}

Ethanol with thionyl chloride (pyridine as solvent) gives chloroethane

\ce{3CH3CH2OH + PBr3 -> 3CH3CH2Br + H3PO3}

Ethanol with phosphorus tribromide gives bromoethane

Three molecules of alcohol are needed for each \ce{PBr3}, because \ce{PBr3} has three bromine atoms to give.

Method 2: from alkanes

When an alkane is treated with chlorine or bromine in diffused sunlight or ultraviolet light, a hydrogen atom is replaced by a halogen. The reaction proceeds by a free radical mechanism.

\ce{CH3CH3 + Cl2 ->[h\nu] CH3CH2Cl + HCl}

Ethane with chlorine in ultraviolet light

52025 Annual I, Set B, Q.2 (xii)

What are \beta-elimination reactions? Explain them with detail.

Solution

Definition

A \beta-elimination reaction is an elimination in which a halogen atom, on one carbon (the \alpha-carbon), and a hydrogen atom, on the adjacent \beta-carbon, are removed together. A C=C double bond is formed between the two carbons. It is called \beta-elimination because the hydrogen comes from the \beta-carbon, the carbon next to the one that carries the halogen. A base or nucleophile is needed to remove the hydrogen.

Why the \beta-hydrogen is needed

The halogen withdraws electron density through the inductive effect. This makes the \beta-hydrogen slightly acidic, so a strong base such as \ce{OH-} can remove it. Without a \beta-hydrogen the reaction cannot occur.

Figure: the base takes the beta-hydrogen, the C-H pair forms the C=C bond and the bromide leaves, giving ethene.

Mechanism

The strong base \ce{OH-} approaches the \beta-hydrogen and removes it as a proton, forming water. The electrons of the C-H bond move to form a new \pi bond between the \alpha- and \beta-carbons.
At the same time the C-Br bond breaks and the bromide ion leaves with the pair of electrons. Both changes happen in one step, so the double bond is formed as the halogen leaves.
Overall equation for bromoethane.
\ce{CH3CH2Br + OH- -> CH2=CH2 + H2O + Br-}

Conditions

A strong base, such as NaOH or KOH, in ethanol as solvent, with heating, favours elimination. Aqueous NaOH, by contrast, gives the alcohol by substitution (book, sections 17.5 and 17.5.1).

\ce{CH3CH2Br + OH- -> CH2=CH2 + H2O + Br-}

How will you convert ethyl chloride to the (i) ethyl cyanide (ii) ethanol?

Solution

(i) Ethyl chloride to ethyl cyanide

Nucleophilic substitution by the cyanide ion. The chlorine is replaced by \ce{CN-}. Heat the ethyl chloride with KCN in ethanol.
Ethyl cyanide is propanenitrile. The CN group adds one carbon atom to the chain.
\ce{CH3CH2Cl + KCN ->[ethanol, heat] CH3CH2CN + KCl}

Ethyl cyanide (propanenitrile)

\ce{CH3CH2Cl + KCN -> CH3CH2CN + KCl}

(ii) Ethyl chloride to ethanol

Nucleophilic substitution by the hydroxide ion. The chlorine is replaced by \ce{OH-}. Use aqueous NaOH and heat; ethanolic NaOH would cause elimination instead.
Overall reaction.
\ce{CH3CH2Cl + NaOH(aq) -> CH3CH2OH + NaCl}

Ethanol

\ce{CH3CH2Cl + NaOH(aq) -> CH3CH2OH + NaCl}

Evaluate the role of haloalkanes in the synthesis of organic compounds.

Solution

Introduction

Halogenoalkanes (haloalkanes) are important intermediates in organic synthesis (book, Do You Know?). The C-X bond is polar and the halogen is a good leaving group, so the halogen can be replaced by many other groups, and the molecule can also be made to eliminate HX to give an alkene.

1. Conversion by substitution

A single halogenoalkane can be converted by nucleophilic substitution into several classes of compounds, as the table shows.

Substitution products of a halogenoalkane (book, section 17.5)
Reagent and conditions	Product	Class of product
\ce{NaOH} (aqueous), heat	\ce{CH3CH2OH}	alcohol
\ce{KCN} in ethanol, heat	\ce{CH3CH2CN}	nitrile
Excess \ce{NH3} in ethanol, heated under pressure	\ce{CH3CH2NH2}	primary amine
\ce{AgNO3} in ethanol	\ce{CH3CH2NO3}	alkyl nitrate

2. Conversion by elimination

With \ce{NaOH} in ethanol and heat, the same halogenoalkane loses HX and gives an alkene, \ce{CH2=CH2} from bromoethane. This is a useful route to alkenes.

3. Control by reaction conditions

The same reagent can lead to different products, so the choice of solvent and temperature is part of the synthesis. Aqueous NaOH gives an alcohol, while NaOH in ethanol gives an alkene. Without the right conditions the wrong class of product is formed.

4. Ways of making halogenoalkanes

Alkyl halides are made from alcohols, alkanes and alkenes, so they link these classes of compounds. The alcohol route gives the purest product; the alkane route gives a mixture that must be separated.

5. Limitations

Halogenation of alkanes gives a mixture of compounds with more than one halogen atom.
Alkyl fluorides do not react under ordinary conditions, so they are poor intermediates.
Iodine does not react directly with alkanes.
Substitution and elimination compete, so the reagent and conditions must be chosen with care.

Conclusion

Halogenoalkanes are valuable in organic synthesis because the C-X bond is easily broken and the halogen can be replaced by many groups. Their use depends on choosing the right reagent and conditions, and on the reactivity order \ce{R-I > R-Br > R-Cl > R-F}.

Note. The book (p. 344) says the silver nitrate reaction gives a 'nitroalkane', but its equation gives R-NO3, an alkyl nitrate. The answer follows the equation.

82026 Annual I, Set D, Q.6 (a)2026 Annual I, Set D, Q.6 (b)2026 Annual I, Set D, Q.6 (c)

Design a synthetic route for the preparation of bromoethane. What starting materials would you use?

Solution

Starting materials

The preferred starting material is ethanol, with phosphorus tribromide, \ce{PBr3}, as the reagent. An alternative is ethene with hydrogen bromide. Ethanol and ethene are the two starting materials.

Ethanol (starting material)

Bromoethane (target)

Route 1: alcohol to bromoethane (preferred)

Convert ethanol into bromoethane with \ce{PBr3}. The alcohol is the best starting material for an alkyl halide, and the by-product \ce{H3PO3} is easily separated.
\ce{3CH3CH2OH + PBr3 -> 3CH3CH2Br + H3PO3}

Route 2: alkene to bromoethane

Ethene adds hydrogen bromide at room temperature (hydrohalogenation). Ethene is symmetrical, so only one product, bromoethane, can form.
\ce{H2C=CH2 + HBr -> CH3CH2Br}

Route 3: alkane (not suitable)

Ethane treated with bromine in sunlight gives bromoethane by a free radical reaction. The product is a mixture with dibromoethanes, which must be separated, so this route is not chosen.

\ce{CH3CH3 + Br2 ->[h\nu] CH3CH2Br + HBr}

Ethane with bromine in ultraviolet light (not preferred)

\ce{3CH3CH2OH + PBr3 -> 3CH3CH2Br + H3PO3}

Write the structural formulae of a primary, a secondary and a tertiary halogenoalkane that have molecular formula \ce{C4H9Cl}.

Solution

All three compounds have the molecular formula \ce{C4H9Cl}. They differ in how many alkyl groups are attached to the carbon that carries the chlorine.

1-Chlorobutane (primary)

2-Chlorobutane (secondary)

2-Chloro-2-methylpropane (tertiary)

In 1-chlorobutane the chlorine-bearing carbon is bonded to one carbon, so it is primary. In 2-chlorobutane it is bonded to two carbons, so it is secondary. In 2-chloro-2-methylpropane it is bonded to three carbons, so it is tertiary.

How can you convert?

Solution

(i) \ce{CH3CH2OH} to chloroethane, and then to a nitrile

Ethanol to chloroethane: use \ce{SOCl2} with pyridine as solvent (or \ce{PCl5}, or HCl with anhydrous \ce{ZnCl2}).
\ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 + HCl}
Chloroethane to the nitrile: heat with KCN in ethanol. The CN group adds one carbon to the chain, giving ethyl cyanide (propanenitrile).
\ce{CH3CH2Cl + KCN -> CH3CH2CN + KCl}

Ethyl cyanide (a nitrile)

\ce{CH3CH2OH ->[SOCl2] CH3CH2Cl ->[KCN] CH3CH2CN}

(ii) \ce{CH3CH2OH} to \ce{CH3CH2Cl} and back to ethanol

Ethanol to chloroethane with \ce{SOCl2} and pyridine, as above.
\ce{CH3CH2OH + SOCl2 -> CH3CH2Cl + SO2 + HCl}
Chloroethane back to ethanol: heat with aqueous NaOH, which replaces Cl by OH.
\ce{CH3CH2Cl + NaOH(aq) -> CH3CH2OH + NaCl}

\ce{CH3CH2OH ->[SOCl2] CH3CH2Cl ->[NaOH(aq)] CH3CH2OH}

(iii) \ce{CH3CH2OH} to bromoethane, and then to ethyl nitrate

Ethanol to bromoethane with \ce{PBr3} (three ethanol molecules to one \ce{PBr3}).
\ce{3CH3CH2OH + PBr3 -> 3CH3CH2Br + H3PO3}
Bromoethane with \ce{AgNO3} in ethanol. The bromide is replaced by the nitrate group, and silver bromide precipitates.
\ce{CH3CH2Br + AgNO3 -> CH3CH2NO3 + AgBr}

Ethyl nitrate

\ce{CH3CH2Br + AgNO3 -> CH3CH2NO3 + AgBr}

Note. Parts (i) and (ii) use the same first step (ethanol to chloroethane) as in the book's own working.

Which is an isomer of 2-chloropropane? Propane or i-chloropropane or both, give reason.

Solution

Isomers have the same molecular formula but different arrangements of atoms. 2-Chloropropane is \ce{C3H7Cl}, with the structure shown below.

2-Chloropropane (i-chloropropane)

Propane is not an isomer. It is \ce{C3H8}, with no chlorine, so its molecular formula is different.

i-Chloropropane is another name for 2-chloropropane (isopropyl chloride). It is the same compound, so it is not an isomer either.

The true isomer is 1-chloropropane (n-propyl chloride), which is also \ce{C3H7Cl} but has the chlorine on a terminal carbon.

1-Chloropropane (an isomer of 2-chloropropane)

Neither propane nor i-chloropropane is an isomer. The isomer is 1-chloropropane.

Note. No key printed in the book. The answer follows the definition of an isomer: propane has a different formula, and i-chloropropane is the same compound.

Give chemical reactions to produce the following starting with a halogenoalkane:

Solution

(i) A nitrile

Heat the halogenoalkane with KCN in ethanol. The halogen is replaced by \ce{CN-}.
\ce{CH3CH2Cl + KCN ->[ethanol, heat] CH3CH2CN + KCl}

Propanenitrile (a nitrile)

\ce{CH3CH2Cl + KCN -> CH3CH2CN + KCl}

(ii) An amide

Make the nitrile first, as in part (i).
\ce{CH3CH2Cl + KCN -> CH3CH2CN + KCl}
Partial hydrolysis of the nitrile (for example with alkaline hydrogen peroxide, or warm concentrated sulphuric acid under controlled conditions) stops at the amide.
\ce{CH3CH2CN + H2O ->[H2O2, OH-] CH3CH2CONH2}

Propanamide (an amide)

\ce{CH3CH2Cl -> CH3CH2CN -> CH3CH2CONH2}

(iii) A vicinal dihalide

Eliminate HCl: heat with NaOH in ethanol to give ethene.
\ce{CH3CH2Cl + NaOH ->[ethanol, heat] CH2=CH2 + NaCl + H2O}
Add bromine to the alkene in an inert solvent such as \ce{CCl4} at room temperature to give the vicinal (1,2-) dibromide.
\ce{CH2=CH2 + Br2 ->[CCl4] BrCH2CH2Br}

1,2-Dibromoethane (a vicinal dihalide)

\ce{CH3CH2Cl -> CH2=CH2 -> BrCH2CH2Br}

(iv) An amine

Heat the halogenoalkane with excess ammonia in ethanol under pressure. A primary amine is formed, with HX as a by-product.
\ce{CH3CH2Cl + NH3 -> CH3CH2NH2 + HCl}

Ethylamine (a primary amine)

\ce{CH3CH2Cl + NH3 -> CH3CH2NH2 + HCl}

(v) A nitrate

Heat the halogenoalkane with AgNO3 in ethanol. The nitrate group replaces the bromine, and AgBr precipitates.
\ce{CH3CH2Br + AgNO3 -> CH3CH2NO3 + AgBr}

Ethyl nitrate

\ce{CH3CH2Br + AgNO3 -> CH3CH2NO3 + AgBr}

(vi) An alkene

Heat the halogenoalkane with NaOH in ethanol. Elimination removes HBr and forms the C=C double bond.
\ce{CH3CH2Br + NaOH ->[ethanol, heat] CH2=CH2 + NaBr + H2O}

Ethene (an alkene)

\ce{CH3CH2Br + NaOH -> CH2=CH2 + NaBr + H2O}

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2025 Annual I, Set B, MCQ 121 mark

Identify the product formed when ethyl bromide is warmed with \ce{NaOH} in ethanol:

A\ce{CH2=CH2}Answer
B\ce{CH3-CH2-ONa}
C\ce{CH3-CHO}
D\ce{CH2=CH-Br}

Why

Ethyl bromide warmed with \ce{NaOH} in ethanol (a strong base in an alcoholic solvent) undergoes \beta-elimination. \ce{HBr} is removed from adjacent carbons and ethene forms: \ce{CH3CH2Br + NaOH ->[ethanol] CH2=CH2 + NaBr + H2O}.

The substitution product, ethanol, forms with aqueous \ce{NaOH}, not with ethanolic \ce{NaOH}. Sodium ethoxide (B) and ethanal (C) are not formed in this reaction, and vinyl bromide (D) would need \ce{Br2} to be lost from ethene, not added.

2026 Annual I, Set B, Q.2 (iii)1.5+1.5 = 3 marks

Complete the following reactions: (a) \ce{CH3-CH=CH2 + HBr ->} (b) \ce{C2H5-Cl + KCN ->}

Solution

(a) \ce{CH3-CH=CH2 + HBr ->}

This is the addition of a halogen acid to an alkene (hydrohalogenation) at room temperature.
Apply Markovnikov's rule. The hydrogen adds to the doubly bonded carbon that already has more hydrogen atoms (\ce{CH2}), so the bromine goes to the middle carbon. This follows the more stable secondary carbocation.
\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}
The major product is 2-bromopropane. A little 1-bromopropane forms by the primary carbocation.

2-Bromopropane (major product)

\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}

(b) \ce{C2H5-Cl + KCN ->}

Potassium cyanide in ethanol, heated, gives nucleophilic substitution. The cyanide ion \ce{CN-} replaces the chlorine.
The potassium ion and the chloride ion form potassium chloride.
\ce{C2H5-Cl + KCN ->[\text{ethanol},\ \Delta] C2H5-CN + KCl}
The product is the nitrile propanenitrile. It has one more carbon atom than the alkyl halide.

Propanenitrile

\ce{C2H5-Cl + KCN -> C2H5-CN + KCl}

Note. Part (a) is hydrohalogenation of an alkene (unit 16) and part (b) is nucleophilic substitution of an alkyl halide (unit 17).

2025 Annual I, Set B, Q.2 (iv) OR3 marks

Write down the reaction of ethyl bromide with \ce{KOH(aq)}, \ce{NH3} and \ce{AgNO3}.

Solution

(a) Ethyl bromide with \ce{KOH(aq)}

Aqueous \ce{KOH} with heat gives nucleophilic substitution: \ce{OH-} replaces \ce{Br} and ethanol is formed.
\ce{CH3CH2Br + KOH(aq) ->[\Delta] CH3CH2OH + KBr}

\ce{CH3CH2Br + KOH(aq) -> CH3CH2OH + KBr}

(b) Ethyl bromide with \ce{NH3}

Heated with excess ammonia in ethanol under pressure, \ce{NH3} is the nucleophile and replaces \ce{Br}. The product is ethylamine, a primary amine.
\ce{CH3CH2Br + NH3 ->[\text{ethanol},\ \Delta] CH3CH2NH2 + HBr}

\ce{CH3CH2Br + NH3 -> CH3CH2NH2 + HBr}

Heated with \ce{AgNO3} in ethanol, the bromine is replaced by \ce{NO3} and the bromide ion forms a cream precipitate of \ce{AgBr}, which is partially soluble in aqueous ammonia.
\ce{CH3CH2Br + AgNO3 -> CH3CH2NO3 + AgBr v}

\ce{CH3CH2Br + AgNO3 -> CH3CH2NO3 + AgBr v}

2025 Annual I, Set B, Q.2 (xii)3 marks

Briefly describe \beta-elimination reaction by giving a suitable example.

Solution

Definition

A **\beta-elimination** is an elimination in which a halogen on one carbon (the \alpha-carbon) and a hydrogen on the neighbouring \beta-carbon are removed together, and a \ce{C=C} double bond forms between the two carbons. A base is needed, and the \beta-hydrogen must be present.

Example and mechanism: bromoethane

The strong base \ce{OH-} removes the slightly acidic \beta-hydrogen as a proton. The halogen's inductive effect makes this hydrogen acidic.
The \ce{C-H} bond pair forms the new \pi bond between the carbons, and at the same time the \ce{C-Br} bond breaks and \ce{Br-} leaves.
Overall reaction.
\ce{CH3CH2Br + OH- -> CH2=CH2 + H2O + Br-}

\ce{CH3CH2Br + NaOH ->[\text{ethanol},\ \Delta] CH2=CH2 + NaBr + H2O}

2026 Annual I, Set B, Q.5 OR6 marks

Classify Halogenoalkanes based on the type of halogen atom and its position in carbon chain.

Solution

Halogenoalkanes (\ce{R-X}, general formula \ce{C_nH_{2n+1}X} for a mono derivative) are classified in the following ways.

1. By the number of halogen atoms

Type	Halogen atoms	Example formula
Monohaloalkane	one	\ce{CH3X}
Dihaloalkane	two	\ce{CH2X2}
Trihaloalkane	three	\ce{CHX3}

2. By the type of halogen atom

They are fluoro-, chloro-, bromo- or iodoalkanes according to whether \ce{X} is \ce{F}, \ce{Cl}, \ce{Br} or \ce{I}, for example \ce{CH3CH2Cl} (chloroethane) and \ce{CH3CH2Br} (bromoethane).

3. By the position of the halogen in the carbon chain

What decides the class is the carbon atom that carries the halogen, and how many alkyl groups are attached to that carbon.

Primary alkyl halide: the halogen is on a carbon with one or no alkyl group.
\ce{CH3Cl},\ \ce{CH3CH2Cl},\ \ce{CH3CH2CH2Cl}
Secondary alkyl halide: the halogen is on a carbon with two alkyl groups.
\ce{CH3CHClCH3},\ \ce{CH3CHClCH2CH3}
Tertiary alkyl halide: the halogen is on a carbon with three alkyl groups.
\ce{(CH3)3CCl}

Chloroethane (primary)

2-Chloropropane (secondary)

2-Chloro-2-methylpropane (tertiary)

Class	Carbon carrying the halogen has	Examples
Primary	one or no alkyl group	\ce{CH3Cl}, \ce{CH3CH2Cl}
Secondary	two alkyl groups	\ce{CH3CHClCH3}
Tertiary	three alkyl groups	\ce{(CH3)3CCl}

2026 Annual I, Set B, Q.6 (a)2 marks

How can ethyl chloride be prepared from the following: ethene. Write complete balanced equations with conditions involved in each reaction.

Solution

Ethene adds hydrogen chloride at room temperature (hydrohalogenation, electrophilic addition). Ethene is symmetrical, so only one product forms.
\ce{CH2=CH2 + HCl -> CH3-CH2-Cl}

\ce{CH2=CH2 + HCl -> CH3CH2Cl}

2026 Annual I, Set B, Q.6 (b)2 marks

How can ethyl chloride be prepared from the following: ethane. Write complete balanced equations with conditions involved in each reaction.

Solution

Ethane is treated with chlorine in ultraviolet light or diffused sunlight. One hydrogen is replaced by chlorine by a free radical substitution (initiation, propagation, termination).
\ce{CH3-CH3 + Cl2 ->[h\nu] CH3-CH2-Cl + HCl}
The mechanism: \ce{Cl2 -> 2Cl.}; then \ce{CH3CH3 + Cl. -> CH3CH2. + HCl} and \ce{CH3CH2. + Cl2 -> CH3CH2Cl + Cl.}.

\ce{CH3CH3 + Cl2 ->[h\nu] CH3CH2Cl + HCl}

2026 Annual I, Set B, Q.6 (c)2 marks

How can ethyl chloride be prepared from the following: ethanol. Write complete balanced equations with conditions involved in each reaction.

Solution

The best method is to treat ethanol with thionyl chloride, \ce{SOCl2}, with pyridine as a solvent. The \ce{-OH} group is replaced by \ce{-Cl} and the by-products \ce{SO2} and \ce{HCl} are gases, so the ethyl chloride is easy to collect.
\ce{CH3CH2OH + SOCl2 ->[pyridine] CH3CH2Cl + SO2 + HCl}
Other reagents also work: \ce{PCl5}, \ce{PCl3}, or \ce{HCl} with anhydrous \ce{ZnCl2} on heating.
\begin{gathered}\ce{CH3CH2OH + PCl5 -> CH3CH2Cl + POCl3 + HCl}\\ \ce{3CH3CH2OH + PCl3 -> 3CH3CH2Cl + H3PO3}\end{gathered}

\ce{CH3CH2OH + SOCl2 ->[pyridine] CH3CH2Cl + SO2 + HCl}

2025 Annual I, Set D, MCQ 131 mark

An alkyl halide is reacted with \ce{AgNO3} in ethanol, light yellow precipitate insoluble in \ce{NH4OH} are formed. Identify the alkyl halide:

A\ce{CH3-CH2-F}
B\ce{CH3-CH2-Cl}
C\ce{CH3-CH2-Br}
D\ce{CH3-CH2-I}Answer

Why

With \ce{AgNO3} in ethanol the halide ion forms a silver halide. The colour and solubility in ammonia identify it: chloride is white and soluble, bromide is cream and partially soluble, iodide is yellow and insoluble. A yellow precipitate insoluble in \ce{NH4OH} is \ce{AgI}, so the alkyl halide is the iodide, option D.

Bromoethane (C) is the tempting option, since the \ce{AgBr} is pale yellow or cream, but it is partially soluble in ammonia. Chloroethane (B) gives a white precipitate that dissolves, and fluoroethane (A) gives no precipitate because the \ce{C-F} bond is too strong.

2026 Annual I, Set D, Q.2 (iii)3 marks

Complete the following equations: (a) \ce{C2H5Cl + NH3 ->} (b) \ce{CH3CH2Br + NaOH ->}

Solution

(a) \ce{C2H5Cl + NH3 ->}

Ammonia has a lone pair on nitrogen, so it is a nucleophile. Heated with excess ammonia in ethanol under pressure, it replaces the chlorine of the alkyl halide (nucleophilic substitution).
\ce{C2H5-Cl + NH3 ->[\text{ethanol},\ \Delta] C2H5-NH2 + HCl}
The product is ethylamine, a primary amine. Excess ammonia keeps the product a primary amine.

Ethylamine

\ce{C2H5Cl + NH3 -> C2H5NH2 + HCl}

(b) \ce{CH3CH2Br + NaOH ->}

Heated with aqueous \ce{NaOH}, the hydroxide ion replaces bromine (nucleophilic substitution) and an alcohol is formed.
\ce{CH3-CH2-Br + NaOH(aq) ->[\Delta] CH3-CH2-OH + NaBr}
The product is ethanol. If the \ce{NaOH} were in ethanol instead, elimination would give ethene, \ce{CH2=CH2}.

\ce{CH3CH2Br + NaOH(aq) -> CH3CH2OH + NaBr}

Note. Part (a) also gives the first step of the preparation of amines (unit 20). Part (b) is written with aqueous NaOH, since the question gives no ethanol as solvent.

2026 Annual I, Set D, Q.5 OR2x3 = 6 marks

Describe nucleophilic substitution reactions. How silver nitrate in ethanol identifies the halogen in alkyl halides? Explain with reaction.

Solution

(a) Describe nucleophilic substitution reactions. (3 marks)

A reaction in which the halogen of an alkyl halide is replaced by an attacking nucleophile is a nucleophilic substitution (S_N) reaction. The carbon of the polar \ce{C-X} bond is \delta+, and the halogen is a good leaving group.

\ce{Nu:^- + R-X -> R-Nu + :X^-}

The nucleophile (a species with a lone pair, neutral or negative) donates its lone pair to the \delta+ carbon of the substrate (the alkyl halide) and forms a new bond.
The \ce{C-X} bond breaks and its pair goes to \ce{X}, which leaves as the halide ion (the leaving group).
The product \ce{R-Nu} is formed. The incoming nucleophile must be stronger than the leaving group.

Nucleophile	Conditions	Product
\ce{OH-} (\ce{NaOH})	aqueous, heat	alcohol, \ce{R-OH}
\ce{CN-} (\ce{KCN})	ethanol, heat	nitrile, \ce{R-CN}
\ce{NH3} (excess)	ethanol, heat under pressure	primary amine, \ce{R-NH2}
\ce{AgNO3}	ethanol, heat	\ce{R-NO3} and \ce{AgX} precipitate

\ce{CH3CH2Br + NaOH(aq) ->[\Delta] CH3CH2OH + NaBr}

(b) How silver nitrate in ethanol identifies the halogen in alkyl halides? Explain with reaction. (3 marks)

When an alkyl halide is heated with \ce{AgNO3} in ethanol, the halide ion leaves as \ce{X-} and combines with \ce{Ag+} to form a precipitate of the silver halide. The colour of the precipitate, and whether it dissolves in aqueous ammonia, shows which halogen is present.

\ce{R-X + AgNO3 -> R-NO3 + AgX v}

\ce{CH3-CH2-Br + AgNO3 -> CH3-CH2-NO3 + AgBr v}

Halide ion	Precipitate with \ce{AgNO3}	Solubility in aqueous ammonia
\ce{Cl-}	white	soluble
\ce{Br-}	cream	partially soluble
\ce{I-}	yellow	insoluble

Note. The marks are printed as '2x3' and are partly cut off at the edge of the page. The question is answered as two parts of 3 marks each.

2026 Annual I, Set D, Q.6 (a)2 marks

How can ethyl Bromide be prepared from the following: ethene. Write complete balanced equations with conditions involved in each reaction.

Solution

Ethene adds hydrogen bromide at room temperature (hydrohalogenation, electrophilic addition). Ethene is symmetrical, so bromoethane is the only product.
\ce{CH2=CH2 + HBr -> CH3-CH2-Br}

\ce{CH2=CH2 + HBr -> CH3CH2Br}

2026 Annual I, Set D, Q.6 (b)2 marks

How can ethyl Bromide be prepared from the following: ethane. Write complete balanced equations with conditions involved in each reaction.

Solution

Ethane is treated with bromine in ultraviolet light or diffused sunlight (free radical substitution).
\ce{CH3-CH3 + Br2 ->[h\nu] CH3-CH2-Br + HBr}
Initiation: \ce{Br2 -> 2Br.}. Propagation: \ce{CH3CH3 + Br. -> CH3CH2. + HBr} and \ce{CH3CH2. + Br2 -> CH3CH2Br + Br.}.

\ce{CH3CH3 + Br2 ->[h\nu] CH3CH2Br + HBr}

2026 Annual I, Set D, Q.6 (c)2 marks

How can ethyl Bromide be prepared from the following: ethanol. Write complete balanced equations with conditions involved in each reaction.

Solution

Heat ethanol with potassium bromide and concentrated sulphuric acid. The acid makes \ce{HBr} from \ce{KBr}, and this replaces the \ce{-OH} group.
\ce{CH3CH2OH + KBr + H2SO4 ->[heat] CH3CH2Br + KHSO4 + H2O}
Another reagent is phosphorus tribromide. Three molecules of alcohol react with one \ce{PBr3}, because each \ce{PBr3} has three bromine atoms to give.
\ce{3CH3CH2OH + PBr3 -> 3CH3CH2Br + H3PO3}

\ce{3CH3CH2OH + PBr3 -> 3CH3CH2Br + H3PO3}

Exam focus

Prepare ethyl chloride or ethyl bromide from ethene, ethane and ethanol, with complete equations and conditions. (short)
Asked in both 2026 sets as three 2-mark parts: Set B Q.6 (a) to (c) (ethyl chloride) and Set D Q.6 (a) to (c) (ethyl bromide). Same as book long question 6.
Go to answer
Describe nucleophilic substitution reactions of alkyl halides with \ce{NaOH(aq)}, \ce{KCN}, \ce{NH3} and \ce{AgNO3}. (long)
2026 Set D Q.5 OR (6 marks), 2026 Set B Q.2 (iii) (b), 2026 Set D Q.2 (iii) (a) and (b), 2025 Set B Q.2 (iv) OR. Book long question 4.
Go to answer
Identify the halogen of an alkyl halide with \ce{AgNO3} in ethanol: colour of the precipitate and solubility in ammonia. (mcq)
2025 Set D MCQ 13 and part (b) of 2026 Set D Q.5 OR, and the cream \ce{AgBr} in 2025 Set B Q.2 (iv) OR.
Go to answer
Explain \beta-elimination of alkyl halides with ethanolic \ce{NaOH}, and how it differs from substitution with aqueous \ce{NaOH}. (short)
2025 Set B Q.2 (xii) and MCQ 12, and part (b) of both 2026 Q.2 (v) questions (solved in unit 16). Book long question 3.
Go to answer
Classify halogenoalkanes into primary, secondary and tertiary (and mono, di, tri), with examples. (long)
2026 Set B Q.5 OR (6 marks). Book short question 1.
Go to answer
Carbocation stability and the SN1 reaction of tertiary alkyl halides. (short)
2025 Set D Q.2 (x) (factors affecting the stability of carbocations) is solved in unit 16 with Markovnikov's rule. Book short question 3.
Go to answer
Reactivity of alkyl halides: polarity and bond energy of the \ce{C-X} bond. (short)
Not in the 2025 or 2026 paper, but it is an SLO; book short question 5 and long question 1 ask it.
Go to answer
Convert an alkyl halide into other compounds and design a route to a given compound (the role of halogenoalkanes in synthesis). (long)
Not in the 2025 or 2026 paper as a whole question; book long questions 4, 5, 8 and 10 practise it.
Go to answer