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Alkanes, halogenoalkanes, alkenes and alcohols

This chapter develops the reactions and mechanisms of the core homologous series of aliphatic organic chemistry. It covers the alkanes (distillation, cracking, combustion and free-radical substitution), the halogenoalkanes (nucleophilic substitution and elimination), the alkenes (electrophilic addition, Markownikoff's rule and addition polymerisation) and the alcohols (production, oxidation and dehydration).

5 sections·~15 min reading time·3 competencies·Level Standard 3 · Advanced 2

T·121212 / 18
Exam profile
AO1 · Recall the reactions, reagents and conditions for alkanes, halogenoalkanes, alkenes and alcoholsAO2 · Write the free-radical substitution, nucleophilic substitution, elimination and electrophilic addition mechanismsAO3 · Explain reactivity trends and evaluate environmental issues such as pollutants and ozone depletion
Operators:statedescribeexplainoutline the mechanismpredictjustify

basic level

AS-Level requires the reactions and the four mechanisms (free-radical substitution, nucleophilic substitution, elimination, electrophilic addition) for these four series.

higher level

The full A-Level uses these transformations in multi-step synthesis and combines them with the analytical and further organic topics.

Depth

Reading depth: In depth

Text

Text size: Standard

Contents · 5 sections▾
  1. Alkanes, halogenoalkanes, alkenes and alcohols
    • 01Alkanes: distillation, cracking, combustion and free-radical substitution◐
    • 02Halogenoalkanes: nucleophilic substitution and elimination●
    • 03Alkenes: electrophilic addition and Markownikoff's rule●
    • 04Addition polymers◐
    • 05Alcohols: production, oxidation and dehydration◐
§ 01

Alkanes: distillation, cracking, combustion and free-radical substitution#

●●○StandardLPAQA 7405 3.3.2.1LPAQA 7405 3.3.2.2LPDfE GCE Chemistry - alkanes

Fractional distillation of crude oil

Fractionating columnSchematic diagram with 7 elements, column (hot at base), gases / petrol (low bp), kerosene, diesel, bitumen (high bp), crude oil in, smaller molecules rise highercolumn (hot atbase)gases / petrol(low bp)kerosenedieselbitumen (highbp)crude oil insmallermolecules rise …
Fig. 1Crude oil is vaporised and separated by boiling point: heavy fractions condense low down, light fractions high up.

Key points

Alkanes are saturated hydrocarbons (only C-C and C-H single bonds), the main constituents of crude oil. Crude oil is separated into useful fractions by fractional distillation: the oil is vaporised and passed up a fractionating column that is hotter at the bottom than the top, so that fractions condense at different heights according to their boiling points. Larger molecules (higher boiling point) condense low down; smaller, more volatile molecules rise higher before condensing.
Cracking breaks larger, less useful alkanes into smaller, more valuable molecules, meeting the demand for petrol and for alkenes. Thermal cracking (high temperature and pressure) tends to produce a high proportion of alkenes for the polymer industry, while catalytic cracking (a zeolite catalyst, slight pressure, lower temperature) favours branched alkanes, cycloalkanes and aromatics for high-grade fuels. Cracking always produces at least one alkene because there are not enough hydrogen atoms to saturate all the fragments.
Alkanes are used as fuels because their combustion is strongly exothermic. Complete combustion in plenty of oxygen gives carbon dioxide and water; incomplete combustion in a limited supply gives carbon monoxide (a toxic gas) and carbon (soot). Combustion of impure fuels also releases pollutants: sulfur dioxide (from sulfur impurities, causing acid rain) and oxides of nitrogen (formed at the high temperatures in engines), which are removed from vehicle exhausts by catalytic converters.
Alkanes react with halogens (chlorine or bromine) in the presence of ultraviolet light by free-radical substitution, in three stages. Initiation is the homolytic splitting of the halogen molecule by UV light into two radicals (Cl2→2Cl∙\text{Cl}_2 \rightarrow 2\text{Cl}\bulletCl2​→2Cl∙). Propagation is a chain of two steps in which a radical reacts to make a product and a new radical, so the reaction sustains itself (Cl∙+CH4→HCl+CH3∙\text{Cl}\bullet + \text{CH}_4 \rightarrow \text{HCl} + \text{CH}_3\bulletCl∙+CH4​→HCl+CH3​∙, then CH3∙+Cl2→CH3Cl+Cl∙\text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bulletCH3​∙+Cl2​→CH3​Cl+Cl∙). Termination is when two radicals combine, removing radicals from the system. A limitation is that further substitution gives a mixture of products (di-, tri- and tetra-substituted).
Cl2→  UV  2 Cl∙\text{Cl}_2 \xrightarrow{\;UV\;} 2\,\text{Cl}\bulletCl2​UV​2Cl∙

Initiation

Ultraviolet light homolytically splits the chlorine molecule into two radicals.

Cl∙+CH4→HCl+CH3∙,CH3∙+Cl2→CH3Cl+Cl∙\text{Cl}\bullet + \text{CH}_4 \rightarrow \text{HCl} + \text{CH}_3\bullet, \quad \text{CH}_3\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\bulletCl∙+CH4​→HCl+CH3​∙,CH3​∙+Cl2​→CH3​Cl+Cl∙

Propagation

Each step consumes one radical and makes another, sustaining the chain.

Worked example

Mechanism of free-radical substitution

Give the mechanism for the reaction of methane with chlorine in ultraviolet light to form chloromethane.

  1. 01Initiation

    UV light splits chlorine into radicals: Cl2 -> 2Cl(.).

  2. 02Propagation

    Cl(.) + CH4 -> HCl + CH3(.); then CH3(.) + Cl2 -> CH3Cl + Cl(.).

  3. 03Termination

    Two radicals combine, e.g. CH3(.) + Cl(.) -> CH3Cl, removing radicals from the chain.

Result: Initiation (UV homolysis), a two-step propagation chain forming chloromethane, and termination by radical combination.

Exam focus

  • Write the initiation, propagation and termination steps of free-radical substitution, using fishhook/radical notation.
  • Explain fractional distillation and cracking, and identify the pollutants from combustion and how they are controlled.

Typical mistakes

  • Omitting the UV light for initiation, or writing ionic (heterolytic) rather than radical steps.
  • Forgetting that free-radical substitution gives a mixture of further-substituted products.

Active revision

Write the three types of step (initiation, propagation, termination) for the reaction of methane with chlorine in UV light, giving one termination step.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

§ 02

Halogenoalkanes: nucleophilic substitution and elimination#

●●●AdvancedLPAQA 7405 3.3.3.1LPAQA 7405 3.3.3.2LPDfE GCE Chemistry - halogenoalkanes

Nucleophilic substitution by hydroxide

OH- attacks the delta+ carbon; Br- leavesSkeletal structure with 3 atoms and 1 bonds, 3 lone pairs, 2 mechanism arrows, Data: HO, C, Br, C–BrHO-CBrnucleophile attacksC-Br breaks
Fig. 2The hydroxide lone pair attacks the delta-positive carbon; the C-Br bond breaks heterolytically and bromide leaves.

Key points

Halogenoalkanes contain a carbon-halogen bond that is polar, because the halogen is more electronegative than carbon, leaving the carbon slightly positive (δ+\delta+δ+). This δ+\delta+δ+ carbon is attacked by nucleophiles, so halogenoalkanes undergo nucleophilic substitution: the nucleophile donates a lone pair to the carbon, forming a new bond, while the carbon-halogen bond breaks heterolytically and the halide leaves as an ion. With hydroxide ions this gives an alcohol, with cyanide ions a nitrile (adding a carbon to the chain), and with ammonia an amine.
The rate of nucleophilic substitution depends on the strength of the carbon-halogen bond, not on its polarity. Although the C-F bond is the most polar, it is also the strongest, so fluoroalkanes react slowest; the C-I bond is the weakest, so iodoalkanes hydrolyse fastest. This is a key experimental point: the rate of hydrolysis increases in the order R-Cl < R-Br < R-I, following the decreasing bond enthalpy down the group.
Halogenoalkanes can also undergo elimination, in which a hydrogen halide is removed to form an alkene. Here the hydroxide ion acts as a base rather than a nucleophile: it removes a hydrogen from the carbon next to the C-X carbon, the electrons form a C=C double bond, and the halide leaves. Whether substitution or elimination dominates depends on the conditions - warm aqueous hydroxide (a solvent that favours the ion) gives mainly substitution, whereas hot ethanolic (alcoholic) hydroxide favours elimination.
Halogenoalkanes have important environmental impacts. Chlorofluorocarbons (CFCs), once used as refrigerants and propellants, are so unreactive that they reach the stratosphere, where UV light generates chlorine radicals that catalyse the breakdown of ozone: Cl∙+O3→ClO∙+O2\text{Cl}\bullet + \text{O}_3 \rightarrow \text{ClO}\bullet + \text{O}_2Cl∙+O3​→ClO∙+O2​, then ClO∙+O→Cl∙+O2\text{ClO}\bullet + \text{O} \rightarrow \text{Cl}\bullet + \text{O}_2ClO∙+O→Cl∙+O2​, regenerating the chlorine radical so one radical destroys many ozone molecules. Understanding this catalytic cycle, and why CFCs have been replaced, is examinable.
CH3CH2Br+OH−→CH3CH2OH+Br−\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^-CH3​CH2​Br+OH−→CH3​CH2​OH+Br−

Nucleophilic substitution

Warm aqueous hydroxide converts a halogenoalkane to an alcohol by substitution.

Worked example

Substitution versus elimination

State the conditions and product when 2-bromopropane reacts with (a) warm aqueous sodium hydroxide and (b) hot ethanolic sodium hydroxide.

  1. 01Aqueous hydroxide

    OH- acts as a nucleophile in nucleophilic substitution, giving propan-2-ol.

  2. 02Ethanolic hydroxide

    OH- acts as a base in elimination, removing HBr to form propene.

  3. 03Explain the difference

    The aqueous, ionising solvent favours the nucleophile (substitution); the hot ethanolic conditions favour the base (elimination).

Result: Warm aqueous NaOH gives propan-2-ol (substitution); hot ethanolic NaOH gives propene (elimination).

Exam focus

  • Draw the nucleophilic substitution mechanism with correct curly arrows and identify the nucleophile and leaving group.
  • Explain the trend in hydrolysis rate (R-Cl < R-Br < R-I) in terms of carbon-halogen bond enthalpy, and describe how conditions select substitution or elimination.

Typical mistakes

  • Explaining the reactivity order by bond polarity rather than bond enthalpy (the weakest bond, C-I, reacts fastest).
  • Confusing the conditions: aqueous hydroxide favours substitution, ethanolic hydroxide favours elimination.

Active revision

Explain why 1-iodobutane is hydrolysed faster than 1-chlorobutane, and give the mechanism for the reaction of 1-bromopropane with aqueous sodium hydroxide.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 03

Alkenes: electrophilic addition and Markownikoff's rule#

●●●AdvancedLPAQA 7405 3.3.4.1LPAQA 7405 3.3.4.2LPDfE GCE Chemistry - alkenes

Electrophilic addition of HBr to propene (first step)

propene + HBr: first (rate-determining) stepSkeletal structure with 5 atoms and 3 bonds, 2 mechanism arrows, Data: CH2, CH, CH3, H, Br, CH2–CH (double), CH–CH3, H–BrCH2CHCH3HBrπ attacks HH-Br breaks
Fig. 3The pi bond attacks the delta-positive H of HBr; the H-Br bond breaks to give bromide and a carbocation (on the more substituted carbon).

Key points

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond, which is made of a strong σ\sigmaσ bond and a weaker π\piπ bond. The π\piπ bond is an exposed region of high electron density above and below the plane of the molecule, so it is attacked by electrophiles (electron-pair acceptors). Alkenes therefore undergo electrophilic addition: the double bond opens and two groups add across it, converting the unsaturated alkene to a saturated product.
The mechanism has two steps. First, the π\piπ electrons attack the electrophile (for example the δ+\delta+δ+ hydrogen of H-Br\text{H-Br}H-Br, or an induced dipole in Br2\text{Br}_2Br2​), forming a new bond and generating a positively charged carbocation, while the rest of the electrophile leaves as a negative ion. Second, the carbocation is attacked by the negative ion (the nucleophile), completing the addition. The addition of bromine is used as the test for a C=C double bond: bromine water is decolourised from orange to colourless.
When an unsymmetrical reagent such as HBr\text{HBr}HBr adds to an unsymmetrical alkene such as propene, two products are possible, and the major one is predicted by Markownikoff's rule: the hydrogen adds to the carbon that already has more hydrogens. The reason is carbocation stability - the reaction goes through the more stable carbocation. Tertiary carbocations are more stable than secondary, which are more stable than primary, because the surrounding alkyl groups are electron-releasing and spread out (delocalise) the positive charge.
So HBr adds to propene mainly to give 2-bromopropane (via the more stable secondary carbocation) rather than 1-bromopropane (which would need a less stable primary carbocation). Being able to draw the mechanism, identify the carbocation intermediate, and use its relative stability to predict the major product is a central A-level skill, and the same reasoning explains the addition of steam (hydration) and of sulfuric acid to alkenes.
CH3CH=CH2+HBr→CH3CHBrCH3\text{CH}_3\text{CH}{=}\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CHBrCH}_3CH3​CH=CH2​+HBr→CH3​CHBrCH3​

Markownikoff addition

The H adds to the carbon with more hydrogens, giving 2-bromopropane via the more stable secondary carbocation.

Worked example

Predicting the major addition product

When HBr adds to propene, explain why 2-bromopropane is the major product rather than 1-bromopropane.

  1. 01Two possible carbocations

    The H+ can add to either double-bond carbon, giving a secondary carbocation (positive on the middle carbon) or a primary one (positive on the end carbon).

  2. 02Compare stability

    The secondary carbocation is more stable, because two electron-releasing alkyl groups spread out the positive charge.

  3. 03Predict the product

    The reaction goes through the more stable secondary carbocation, so Br- adds to the middle carbon, giving 2-bromopropane.

Result: 2-bromopropane predominates because it forms via the more stable secondary carbocation (Markownikoff's rule).

Exam focus

  • Draw the two-step electrophilic addition mechanism, showing the carbocation intermediate and correct curly arrows.
  • Use carbocation stability (tertiary > secondary > primary) and Markownikoff's rule to predict the major product.

Typical mistakes

  • Drawing the curly arrow from the atoms rather than from the pi bond, or forgetting the carbocation intermediate.
  • Predicting the minor product by ignoring carbocation stability.

Active revision

Predict the major product of the addition of HBr to but-1-ene and justify it using carbocation stability.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 04

Addition polymers#

●●○StandardLPAQA 7405 3.3.4.3LPDfE GCE Chemistry - addition polymers

Addition polymerisation of ethene

ethene monomer -> poly(ethene) repeat unit (n units)Skeletal structure with 4 atoms and 2 bonds, reaction scheme, Data: CH2, CH2, CH2, CH2, CH2–CH2 (double), CH2–CH2polymeriseCH2CH2CH2CH2
Fig. 4The C=C double bond of ethene opens to form single bonds, giving the poly(ethene) repeat unit -CH2-CH2-.

Key points

Alkenes undergo addition polymerisation, in which many small alkene molecules (monomers) join together to form one very large molecule (the polymer), with no other product. The C=C double bond of each monomer opens and forms single bonds to its neighbours, so ethene becomes poly(ethene) and propene becomes poly(propene). Because nothing is lost, the empirical formula of the polymer is the same as that of the monomer.
The structure of an addition polymer is described by its repeat unit - the smallest section that repeats along the chain - drawn with a bond extending through each bracket to show the chain continues, and a subscript nnn for a large number of units. To find the monomer from a polymer, or vice versa, you convert the single bond in the repeat unit back to the double bond of the monomer (and remove the extending bonds), or the reverse. The repeat unit of poly(ethene) is −CH2−CH2−-\text{CH}_2-\text{CH}_2-−CH2​−CH2​−.
Addition polymers are chemically inert and non-biodegradable, because their strong carbon-carbon backbone is not attacked by the enzymes of microorganisms or by water. This makes them durable and useful, but creates a serious waste problem: discarded polymers persist in the environment for a very long time. Disposal options include recycling (sorting and reprocessing), incineration for energy (but this can release toxic gases such as HCl from PVC), and the development of biodegradable or photodegradable alternatives.
The properties of a polymer depend on its monomer and on how the chains pack. Poly(ethene) made under different conditions gives low-density (branched, flexible) and high-density (linear, rigid) forms, and adding a plasticiser to PVC makes it flexible for uses such as cable insulation. When you draw a repeat unit from a monomer, keep the carbon backbone and the side groups exactly as in the monomer, changing only the C=C double bond to a C-C single bond with continuing bonds.
n CH2=CH2→−(CH2−CH2)n−n\,\text{CH}_2{=}\text{CH}_2 \rightarrow -(\text{CH}_2-\text{CH}_2)_n-nCH2​=CH2​→−(CH2​−CH2​)n​−

Addition polymerisation

Many ethene monomers add together; the repeat unit has the same atoms as the monomer with the double bond opened.

Worked example

Repeat unit from a monomer

Chloroethene is CH2=CHCl. Draw the repeat unit of the polymer it forms (PVC) and name the polymer.

  1. 01Open the double bond

    The C=C becomes a C-C single bond, with bonds continuing at each end through the brackets.

  2. 02Keep the side group

    One carbon keeps its chlorine and hydrogen, the other keeps its two hydrogens, giving the repeat unit -CH2-CHCl-.

  3. 03Name it

    The polymer is poly(chloroethene), commonly PVC.

Result: The repeat unit is -(CH2-CHCl)- and the polymer is poly(chloroethene), PVC.

Exam focus

  • Draw the repeat unit of an addition polymer from its monomer (and deduce the monomer from a repeat unit).
  • Discuss the disposal problems of addition polymers arising from their unreactive, non-biodegradable backbone.

Typical mistakes

  • Leaving the C=C double bond in the repeat unit instead of opening it to a single bond with continuing bonds.
  • Forgetting the bonds that extend through the brackets to show the chain continues.

Active revision

Draw the repeat unit of the polymer formed from propene (CH3CH=CH2\text{CH}_3\text{CH}{=}\text{CH}_2CH3​CH=CH2​) and name it.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

§ 05

Alcohols: production, oxidation and dehydration#

●●○StandardLPAQA 7405 3.3.5.1LPAQA 7405 3.3.5.2LPDfE GCE Chemistry - alcohols

Oxidation routes of the alcohols

Oxidation of alcoholsGraph, primary alcohol → aldehyde (distil), aldehyde (distil) → carboxylic acid (reflux), secondary alcohol → ketone, tertiary alcohol → not oxidisedprimary alcoholaldehyde(distil)carboxylic acid(reflux)secondaryalcoholketonetertiary alcoholnot oxidised[O], distil[O], reflux[O]no H on C-OH
Fig. 5Primary alcohols oxidise to aldehydes (distil) then carboxylic acids (reflux); secondary to ketones; tertiary are not oxidised.

Key points

Alcohols contain the hydroxyl (-OH) functional group. They are made industrially in two ways. Hydration of an alkene (steam and a phosphoric acid catalyst at high temperature and pressure) is a fast, continuous, high-atom-economy process using ethene from cracking. Fermentation of sugars by yeast enzymes (C6H12O6→2C2H5OH+2CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2C6​H12​O6​→2C2​H5​OH+2CO2​) is a slower, batch process using renewable plant material at low temperature - a comparison of the two routes (rate, conditions, feedstock, sustainability) is a common exam theme.
Alcohols are classified as primary, secondary or tertiary according to how many carbon atoms are attached to the carbon bearing the -OH group (one, two or three respectively). This classification controls their oxidation. Oxidation uses an oxidising agent such as acidified potassium dichromate(VI), which turns from orange to green when it oxidises an alcohol; a tertiary alcohol is not oxidised, so the dichromate stays orange - the basis of a test to distinguish the classes.
The oxidation products differ by class and by conditions. A primary alcohol is oxidised first to an aldehyde and then to a carboxylic acid; the aldehyde is obtained by gently heating and distilling it off as it forms (so it does not oxidise further), while the carboxylic acid is obtained by heating under reflux with excess oxidising agent. A secondary alcohol is oxidised to a ketone, which resists further oxidation. A tertiary alcohol cannot be oxidised in this way because the carbon bearing the -OH has no hydrogen to remove.
Alcohols also undergo elimination (dehydration): heating with a concentrated acid catalyst (such as concentrated sulfuric or phosphoric acid) removes a water molecule to form an alkene - the reverse of hydration. This is a useful way of making alkenes from renewable alcohols. When you describe an alcohol's oxidation, always state the reagent (acidified potassium dichromate(VI)), the colour change (orange to green), and the conditions (distil for the aldehyde, reflux for the acid).
C6H12O6→2 C2H5OH+2 CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\,\text{C}_2\text{H}_5\text{OH} + 2\,\text{CO}_2C6​H12​O6​→2C2​H5​OH+2CO2​

Fermentation

Yeast enzymes convert glucose to ethanol and carbon dioxide at low temperature.

CH3CH2OH→  conc. H2SO4, Δ  CH2=CH2+H2O\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\;conc.\ H_2SO_4,\ \Delta\;} \text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O}CH3​CH2​OHconc. H2​SO4​, Δ​CH2​=CH2​+H2​O

Dehydration

Concentrated acid removes water from an alcohol to form an alkene.

Worked example

Distinguishing alcohol classes by oxidation

You have samples of propan-1-ol (primary) and 2-methylpropan-2-ol (tertiary). Describe a test using acidified potassium dichromate(VI) to distinguish them.

  1. 01Add the reagent and warm

    Add acidified potassium dichromate(VI) to each sample and warm gently.

  2. 02Primary alcohol

    Propan-1-ol is oxidised, so the dichromate turns from orange to green.

  3. 03Tertiary alcohol

    2-methylpropan-2-ol is not oxidised, so the dichromate stays orange.

Result: The primary alcohol turns the dichromate orange to green; the tertiary alcohol leaves it orange, distinguishing the two.

Exam focus

  • State the oxidation product of a primary, secondary and tertiary alcohol, with the reagent, colour change and conditions (distil vs reflux).
  • Compare hydration and fermentation as routes to ethanol in terms of rate, conditions, feedstock and sustainability.

Typical mistakes

  • Saying a tertiary alcohol oxidises - it does not, because the C-OH carbon has no hydrogen to lose.
  • Forgetting that distilling gives the aldehyde while refluxing gives the carboxylic acid from a primary alcohol.

Active revision

Describe how you would obtain (a) ethanal and (b) ethanoic acid from ethanol, giving the reagent, conditions and colour change.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Contents

Section -- / 05

    • 01Alkanes: distillation, cracking, combustion and free-radical substitution◐
    • 02Halogenoalkanes: nucleophilic substitution and elimination●
    • 03Alkenes: electrophilic addition and Markownikoff's rule●
    • 04Addition polymers◐
    • 05Alcohols: production, oxidation and dehydration◐

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Alkanes, halogenoalkanes, alkenes and alcohols

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References & sources

Sources

Department for Education

  • GCE AS and A level subject content for the sciences

AQA

  • AQA A-level Chemistry 7405 specification

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