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Advanced A Level Organic Chemistry: The preparation of carboxylic acids

Part 6. The Chemistry of  Carboxylic Acids and their Derivatives

Doc Brown's Chemistry Advanced Level Pre-University Chemistry Revision Study Notes for UK KS5 A/AS GCE IB advanced level organic chemistry students US K12 grade 11 grade 12 organic chemistry preparation of carboxylic acids

Part 6.3 Synthetic routes and methods of preparing aliphatic and aromatic carboxylic acids

Sub-index for this page (some overlap with section 6.6)

6.3.1 Oxidation of primary alcohols (aliphatic/aromatic acids)

6.3.2 Oxidation of aldehydes (to aliphatic/aromatic acids)

6.3.3 Oxidation of aromatic hydrocarbons to make aromatic acids

6.3.4 Hydrolysis of nitriles (aliphatic/aromatic acids)

6.3.5 Hydrolysis of esters

6.3.6 The iodoform reaction

6.3.7 The hydrolysis of amides

6.3.8 Examples of synthetic routes to substituted carboxylic acids

INDEX of all My carboxylic acids and derivatives notes

All My Advanced A Level Organic Chemistry Notes

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6.3.1 Oxidation of primary alcohols (aliphatic or aromatic acids)

When oxidised they form aldehydes and further oxidation gives the relatively stable carboxylic acid e.g.

  1. aldehydes and ketones nomenclature (c) doc b ===>aldehydes and ketones nomenclature (c) doc b

    • ethanol  ==> ethanoic acid

  2. aldehydes and ketones nomenclature (c) doc b ===>(c) doc b

    • 2-methylpropan-1-ol ===> 2-methylpropanoic acid

  3. aldehydes and ketones nomenclature (c) doc b ===>aldehydes and ketones nomenclature (c) doc b

    • butan-1-ol ===> butanoic acid

  4. aldehydes and ketones nomenclature (c) doc b===>aldehydes and ketones nomenclature (c) doc b

    • pentan-1-ol ==> pentanal ==> pentanoic acid

 

reflux diagram oxidation of primary alcohols ethanol to carboxylic acid ethanoic acid potassium dichromate(VI) sulfuric acid reaction conditions equations advanced organic chemistry revision notes doc brownOxidation of primary alcohols to carboxylic acids

The technique illustrated on the right (diagram PD5) is called heating under reflux, a method which enables a reaction to be carried out at a higher temperature than room temperature to speed up the reaction AND retain the solvent (reaction medium e.g. water) and any volatile reactant or product (e.g. an alcohol/aldehyde/ketone).

As the mixture boils, the vapours of the solvent or volatile reactant/product are condensed back into the flask in the vertical condenser, so any volatile reactant is used up and no volatile product lost (at least at this stage in a preparation!).

Refluxing the alcohol with excess of the potassium dichromate(VI)/sulfuric acid mixture.

 

The oxidation is ...

primary alcohol ==> carboxylic acid

 

2Cr2O72-(aq) + 3RCH2OH(aq) + 16H+(aq) ===> 3RCOOH(aq) + 4Cr3+(aq) + 11H2O(l)

oxidation half-reaction: RCH2OH(aq) + H2O(l) ==> RCOOH(aq) + 4H+(aq) + 4e-(aq)  (R = alkyl or aryl)

reduction half reaction: Cr2O72–(aq) + 14H+(aq) + 6e ===> 2Cr3+(aq) + 7H2O(l)

The orange dichromate(VI) ion is reduced to the green chromium(III) ion.

 

Examples using simplified symbol equations:

ethanol ==> ethanoic acid

CH3CH2OH + 2[O] ===> CH3COOH + H2O

 

propan-1-ol (1-propanol, n-propyl alcohol, n-propanol) ==> propanoic acid

CH3CH2CH2OH + 2[O] ===> CH3CH2COOH + H2O

 

Aromatic primary alcohols (NOT phenols) can be similarly oxidised e.g. phenylmethanol to benzoic acid.

(c) doc b +   2[O]  ===> (c) doc b  + H2O

 

reflux diagram oxidation of primary alcohols to carboxylic acid potassium manganate(VII)sodium hydroxide reaction conditions equations advanced organic chemistry revision notes doc brownOxidation of primary alcohols with alkaline potassium manganate(VII)

Heating a primary alcohol with a aqueous sodium hydroxide and potassium manganate(VII) mixture under reflux (diagram PD2) will give the sodium salt of the carboxylic acid and it is not possible to isolate the intermediate aldehyde.

However, the acid/dichromate(VI) method under reflux is better, because the carboxylic acid is less liable to further degradative oxidation. The complex reaction can be summarised as:

RCH2OH(aq) + NaOH(aq) + 2[O] ==> RCOO-Na+(aq) + 2H2O(l)

(R = alkyl or aryl)

After removing the excess KMnO4/MnO2 the weak acid is freed from its sodium salt  by adding strong dilute hydrochloric acid.

RCOO-(aq) + H+(aq) ==> RCOOH(aq/l/s)


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6.3.2 Oxidation of aldehydes (aliphatic or aromatic acids)

You can use exactly the same method described in 6.3.1 using a mixture potassium dichromate(VI) and moderately concentrated sulfuric acid.

aldehyde ==> carboxylic acid

Cr2O72-(aq) + 3RCHO(aq) + 8H+(aq) ===> 3RCOOH(aq) + 2Cr3+(aq) + 4H2O(l) (R = alkyl or aryl)

oxidation half-reaction: RCHO(aq) + H2O(l) ===> RCOOH(aq) + 2H+(aq) + 2e-(aq)

reduction half reaction: Cr2O72–(aq) + 14H+(aq) + 6e ===> 2Cr3+(aq) + 7H2O(l)

The orange dichromate(VI) ion is reduced to the green chromium(III) ion.

Examples using simplified symbol equations:

ethanal ==> ethanoic acid:  CH3CHO + [O] ===> CH3COOH

propanal (propionaldehyde) ==> propanoic acid (propionic acid)

CH3CH2CHO + [O] ===> CH3CH2COOH

butanal (butyraldehyde) ==> butanoic acid (butyric acid)

CH3CH2CH2CHO + [O] ===> CH3CH2CH2COOH


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6.3.3 Oxidation of aromatic hydrocarbons to make aromatic acids

Acidified potassium dichromate(VI) might oxidise alkyl benzene compounds to benzoic acid, but if it does, it will be much slower than with the alkaline manganate(VII) method described below - NaOH/KMnO4 is a much stronger oxidising agent.

reflux diagram oxidation of aromatic hydrocarbons to aromatic carboxylic acid potassium manganate(VII)sodium hydroxide reaction conditions equations methylbenzene to benzoic acid dimethybenzene to benzenedicarboxylic acidsOverall change is represented by the equations for the conversion:

(c) doc b  ===>  (c) doc b

C6H5CH3 + 3[O] ===> C6H5COOH + H2O

Aromatic are not easily oxidised and longish reflux times are necessary (illustrated, fig. PD2 right).

Hydrocarbons are difficult to oxidise with typical organic oxidising agents compared to compounds like alcohols.

However, aromatic hydrocarbons with an alkyl side chain can be oxidised with strong reagents such as aqueous potassium manganate(VII)/sodium hydroxide.

Whatever the length of the alkyl group on a benzene ring it gets whittled down to carbon of the carboxylic acid group e.g. propylbenzene ends up as benzoic acid, the same product obtained by oxidising the shorter methylbenzene.

The more stable aromatic benzene ring is left intact.

The overall process for producing benzoic acid from methylbenzene can be summarised ..

C6H5CH3 (l) + NaOH(aq) + 3[O] ===> C6H5COO-Na+ (aq) + 2H2O(l)

After removing the excess KMnO4/MnO2 with a reducing agent, the weak acid, benzoic acid is freed from its sodium salt  by adding dilute hydrochloric acid.

C6H5COO-(aq) + H+(aq) ===> C6H5COOH

and you can oxidise the three isomeric dimethyl benzene to make the three isomeric aromatic dicarboxylic acids:

(i) benzene-1,2-dicarboxylic acid from 1,2-dimethylbenzene

ea02 + 6[O] ===> 1847 +  2H2O

(i) benzene-1,3-dicarboxylic acid from 1,3-dimethylbenzene

(c) doc b +  6[O] ===> (c) doc b +  2H2O

(i) benzene-1,4-dicarboxylic acid from 1,2-dimethylbenzene

(c) doc b +  6[O] ===> 1847 +  2H2O

After recrystallisation of the products e.g. using aqueous ethanol, you can determine the melting point and confirm the identity from data tables of melting points of carboxylic acids.

A sharp melting point indicates a reasonable pure sample of the prepared acid.

The product should also give effervescence of carbon dioxide (limewater test) when added to sodium hydrogen carbonate solution - a simple test for a carboxylic acid.

RCOOH  +  NaHCO3  ===>  RCOO-Na+  +  H2O  +  CO2


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preparing by refux a carboxylic acid by hydrolysing a nitrile by refluxing with dilute acid or alkali advanced organic chemistry6.3.4 Hydrolysis of nitriles to give aliphatic or aromatic carboxylic acids

The triple bonded nitrile group CN is hydrolysed to the carboxylic acid group COOH.

The nitrogen ends up as ammonia or the ammonium ion.

 If the nitrile is refluxed with dilute hydrochloric/sulfuric acid (strong acids) or sodium hydroxide (strong base - alkali) the corresponding carboxylic acid or its sodium salt is formed.

The hydrolysis with pure water is to slow, but the reaction is speeded up by a strong acid or strong alkali and reflux conditions.

Strictly speaking all the reactants and products should be suffixed by (aq)

 

(a) Nitriles synthesised from halogenoalkane: 

RX  +  KCN  ===>  RCN  +  KX

Nucleophilic substitution reaction where X = Cl, Br or I  and  R = H, alkyl or aryl.

The nucleophilic substitution reaction between halogenoalkanes (haloalkanes) and potassium cyanide

(i) Equations for the dilute mineral acid hydrolysis of a nitrile to give the free (weaker) acid

In this case converting propanenitrile to propanoic acid or its salt, sodium propanoate

(c) doc b  +  2H2O  +  H+  (c) doc b  +  NH4+ 

Here the free acid and an ammonium ion are formed.

(more detailed structured formula hydrolysis equation)

(c) doc b  +  2H2O  +  H+   (c) doc b +   NH4+

(less detailed structured formula hydrolysis equation)

(c) doc b  +  2H2O  +  H+  (c) doc b +   NH4+ 

(skeletal formula hydrolysis equation)

 

(ii) Equations for the alkaline hydrolysis of a nitrile to give the sodium salt (if aqueous sodium hydroxide is used), in the equations you write out the product as the carboxylate anion.

In this case converting propanenitrile to its salt, e.g. sodium propanoate

(c) doc b  +  H2O  +  OH-    +  NH3 

Here the carboxylate anion (propanoate ion) and free ammonia are formed.

(structured formula hydrolysis equation)

(c) doc b  +  H2O  +  OH-   +  NH3   

(abbreviated structured formula hydrolysis equation)

(c) doc b  +  H2O  +  OH-  +   NH3   

(skeletal formula hydrolysis equation)

 

The acid, here propanoic acid, is set free by the addition of a strong mineral acid e.g. dilute sulfuric acid.

CH3CH2COO-(aq) + H+(aq) ===> CH3CH2COOH

 

 

(c) doc b(c) doc b(iii) The hydrolysis of 2-methylpropanenitrile

2-methylpropanenitrile === hydrolysis ===> free 2-methylpropanoic acid or its salt

(CH3)2CHCN  +   2H2O  +  H+    (CH3)2CHCOOH  +  NH4+    (acid hydrolysis, free acid)

(CH3)2CHCN  +   H2O  +  OH-    (CH3)2CHCOO-  +  NH3      (alkaline hydrolysis, salt of acid)

 

(b) Hydroxynitriles prepared from aldehydes and ketones by addition of hydrogen cyanide

R2C=O  + HCN  ===> R2C(OH)CN  (R = H, alkyl or aryl)

Aldehydes and ketones and nucleophilic addition of hydrogen cyanide

The general equation for the hydrolysis of a hydroxy nitrile to a hydroxy-carboxylic acid.

R2C(OH)CN  +  2H2O  ===> R2C(OH)COOH  +  NH3

The nitrile is converted to a carboxylic acid by hydrolysis.

The reaction is slow, so it is speeded up by heating the nitrile with the hydrolysis reagent (e.g. dilute sulfuric acid or aqueous sodium hydroxide) under reflux.

 

(a) Hydrolysis equation with acid - good yield under reflux conditions

(i) +  H+  + 2H2O  ===>    +  NH4+

2-hydroxypropanenitrile  +  hydrogen ion ===>  2-hydroxypropanoic acid  +  ammonium ion

Note that this produces the free acid.

 

(ii) +  H+  + 2H2O  ===>    +  NH4+

2-hydroxy-2-methylpropanenitrile  +  hydrogen ion ===>  2-hydroxy-2-methylpropanoic acid  +  ammonium ion

Note that this produces the free acid.

 

(iii) + H+ + H2O  ===>+ NH4+

2-hydroxy-2-methylbutanenitrile  +  water  ===>  2-hydroxy-2-methylbutanoic acid  +  ammonia

Note that this produces the free acid.

 

(b) Hydrolysis with alkali - good yield under reflux conditions

(i) +  OH-  +  H2O  ===>    +  NH3

2-hydroxypropanenitrile  +  hydroxide in  ===>  2-hydroxypropanoate ion  +  ammonia

Note that this produces the salt of the acid and the ammonia would be boiled off under reflux conditions.

e.g. from sodium hydroxide you would get sodium 2-hydroxypropanoate.

You add dilute mineral acid (sulfuric or hydrochloric) to free the acid - strong acid displacing a weak one.

 

(ii) +  OH-  +  H2O  ===>    +  NH3

2-hydroxy-2-methylpropanenitrile  +  hydroxide in  ===>  2-hydroxy-2-methylpropanoate ion  +  ammonia

Note that this produces the salt of the acid and the ammonia would be boiled off under reflux conditions.

e.g. from sodium hydroxide you would get sodium 2-hydroxy-2-methylpropanoate.

You add dilute mineral acid (sulfuric or hydrochloric) to free the acid - strong acid displacing a weak one.

 

(iii) + OH-  + H2O  ===>   +  NH3

2-hydroxy-2-methylbutanenitrile  +  water  ===>  2-hydroxy-2-methylbutanoate ion  +  ammonia

Note that this produces the salt of the acid and the ammonia would be boiled off under reflux conditions.

e.g. from potassium hydroxide you would get potassium 2-hydroxy-2-methylbutanoate

You add dilute mineral acid (sulfuric or hydrochloric) to free the acid - strong acid displacing a weak one.

The acid, here 2-hydroxy-2-methylbutanoic acid, is set free by the addition of a strong mineral acid e.g. dilute sulfuric acid.

CH3CH2C(CH3)(OH)COO-(aq) + H+(aq) ===> CH3CH2C(CH3)(OH)COOH

 

Note on R/S isomers (optical stereoisomers)

If the nitrile is derived from aldehydes from ethanal (CH3CHO) onwards and all unsymmetrical ketones (e.g. butanone CH3COCH2CH3), R/S isomers are formed on equal probability basis - a racemic mixture (racemate).

In other words, if the resulting nitriles from these carbonyl compounds are hydrolysed, you will also get a racemic (50 : 50)mixture of the mirror image forms of the hydroxynitrile and hence a racemic mixture of the acid obtained by the hydrolysis of the nitrile.

 


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preparing by refux a carboxylic acid by hydrolysing an ester by refluxing with dilute acid or alkali advanced organic chemistry6.3.5 Hydrolysis of esters

If an ester is warmed with a dilute aqueous strong acid or alkali (much faster than pure water), it changes back into the original carboxylic acid and alcohol from which the ester was made.

This reaction is called hydrolysis and this particular hydrolysis is also called saponification if alkali is used

(i) Hydrolysis with acid (dilute sulfuric or hydrochloric acids)

ethyl ethanoate + water ===> ethanoic acid + ethanol

+ H2O ===> 

(ii) Hydrolysis with alkali (dilute sodium hydroxide)

CH3COOCH2CH3  +  OH-  ===> CH3COO-Na+  +  CH3CH2OH

With alkali, you have to add a strong dilute mineral acid (hydrochloric or sulfuric) to free the weak organic carboxylic acid.

CH3COO-Na+  +  H+  ===>  CH3COOH

(Note that esterification is the opposite of the hydrolysis: ethanoic acid + ethanol ==> ethyl ethanoate + water)

 

Hydrolysis of esters and soap production

One type of soap is made by hydrolysing triglyceride esters from natural plant oils or fats using sodium/potassium hydroxide.

For example hydrolysing a saturated fat molecule - a triglyceride ester e.g. using sodium hydroxide

diagram equation for hydrolysis of triglyceride triester with sodium hydroxide to give sodium palmitate soap advanced organic chemistryforming sodium palmitate

Another hydrolysis of a fat molecule might yield the salt sodium stearate,

diagram of molecular ionic structure of sodium stearate advanced organic chemistry sodium stearate

On adding a strong dilute mineral acid like sulfuric or hydrochloric acid, the free long chain fatty acid can be extracted,

diagram of molecular structure of stearic acid doc brown's advanced organic chemistry revision notes stearic acid

The soap is the sodium or potassium salt of a long chain saturated or unsaturated fatty acid.

Soaps are made by hydrolysing esters from naturally occurring plant oils and animal fat


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6.3.6 The iodoform reaction

This reaction is given by all secondary alcohols with the 2-ol structure -R-CHOH-CH3 and all ketones with the 2-one structure R-CO-CH3  ('methyl ketones') to yield carboxylic acids. R = alkyl or aryl

To make the reagent, sodium hydroxide solution is added to a solution of iodine in potassium iodide solution until most of the colour has gone.

You can also use a mixture of potassium iodide dissolved in sodium chlorate(I) solution.

The organic compound is warmed with this solution.

It is a complicated reaction, in which the hydrogen atoms of the methyl group of the CH3CO grouping are replaced by hydrogen atoms. The CO group becomes part of the carboxylic acid group.

The sodium hydroxide/iodine reagent causes the C-C bond of the CH3CO grouping to break in the process of forming the pale yellow precipitate of CHI3 (triiodomethane, 'iodoform'), that smells like an antiseptic.

The following equilibrium is found in the NaOH/I2 reagent:

I2  +  2OH-    IO-  +  I-  H2O

The iodate(I) ion, IO-, substitutes I for H forming the I3CCO grouping.

The electron withdrawing effect of the three iodine atoms weakens the C-C sigma bond allowing for the formation of CHI3.

For R = alkyl or aryl e.g. CH3, CH2CH3, C6H5 etc., the overall equations are:

The simple 'molecular' general equation:

CH3COR  +  3I2  +  4NaOH  ===>  CHI3  +  RCOONa  +  3NaI  +  3H2O

The general ionic equation with state symbols:

CH3COR(aq)  +  3I2(aq)  +  4OH-(aq)  ===>  CHI3(s)  +  RCOO-(aq)  +  3I-(aq)  +  3H2O(l)

The weak carboxylic acid must be freed by adding a strong dilute mineral acid e.g. hydrochloric or sulfuric/

RCOO-(aq)  +  H+(aq)  ===> RCOOH(aq/s)

 

Examples of conversions

(i) butan-2-ol  OR  butanone will yield propanoic acid

CH3CH2CH(OH)CH3  OR  CH3CH2COCH3  ===>  CH3CH2COOH

For the full equations R would be CH3CH2

(ii) 1-phenylethanol would give benzoic acid

(c) doc b  ===> (c) doc b

For the full equations R would be C6H5

 

This would not normally be considered a useful synthetic route to prepare carboxylic acids.


6.3.7 The hydrolysis of amides

Amides are another derivative of carboxylic acids that can be hydrolysed back to the parent carboxylic acid.

If amides are refluxed with strong acid (hydrochloric or sulfuric) or alkali potassium/sodium hydroxide, the product is the original carboxylic acid or its salt.

(a) Acid Hydrolysis of amides

e.g.  the formation of the free weaker carboxylic acid

RCONH2(aq)  +  HCl(aq)  +  H2O(l)  ===>  RCOOH(aq)  + NH4Cl(aq)

and the more correct ionic equation

RCONH2(aq)  +  H+(aq)  +  H2O(l)  ===>  RCOOH(aq)  + NH4+(aq)

e.g. propanamide hydrolysed to propanoic acid

(c) doc b (aq)  +  H+(aq)  +  H2O(l)  ===>  (c) doc b (aq)  + NH4+(aq)

(b) Alkaline hydrolysis of amides

e.g. the formation of the salt of the parent carboxylic acid.

RCONH2(aq)  +  NaOH(aq)   ===>  RCOONa(aq)  +  NH3(aq/g)

and the more correct ionic equation

RCONH2(aq)  +  OH-(aq)  ===>  RCOO-(aq)  +  NH3(aq/l)

e.g. hydrolysis of benzamide to benzoic acid

(c) doc b (aq)  +  OH-(aq)   ===>  (aq)  +  NH3(aq/l)

The carboxylic acid is freed by adding a stronger mineral acid

 RCOO-(aq)  + H+(aq)  ===>  RCOO-(aq)

e.g.   (aq)  + H+(aq)  ===>  (c) doc b (aq)

 


6.3.8 Examples of synthetic routes to substituted carboxylic acids

Just a few brief notes to give some ideas of how synthetic sequences can be put together from reactions you are likely to have, or will, study.

 

(a) Chlorination - chlorinated carboxylic acids

You can chlorinate carboxylic acids in the same way as for alkanes e.g. chlorine plus uv light.

e.g.  chlorine  +  ethanoic acid  ===>  chloroethanoic acid  +  hydrogen chloride

CH3COOH  +  Cl2  ===>  ClCH2COOH  +  HCl

The chlorination may continue to give dichloroethanoic acid and trichloroethanoic acid.

Cl2CHCOOH  and  Cl3COOH

It is a free radical chain substitution reaction mechanism.

For propanoic acid, there are two possible isomers, 2-chloropropanoic acid or 3-chloropropanoic acid.

CH3CH2COOH  +  Cl2  ===>  {CH3CHClCOOH  or  ClCH2CH2COOH}  +  HCl

 

(b) Hydroxy carboxylic acid formation

You can hydrolyse the chlorinated acids by refluxing with sodium hydroxide.

e.g. chloroethanoic acid  +  sodium hydroxide  ===> hydroxyethanoic acid  +  sodium chloride

ClCH2COOH  +  NaOH  ===>  HOCH2COOH  +  NaCl

CH3COOH  +  OH-  ===>  HOCH2COOH  +  Cl-

This will a nucleophilic substitution reaction mechanism.

With excess alkali, the carboxylic acid is formed as the sodium salt (not shown) and freed by adding a stronger mineral acid (HCl or H2SO4)

Similarly for the chloropropanoic acids you 2-hydroxypropanoic acid and 3-hydroxypropanoic acid, though you just get one product from each isomer (no or involved!).

CH3CHClCOOH  +  OH-  ===>  CH3CH(OH)COOH  +  Cl-

ClCH2CH2COOH  +  OH-  ===> HOCH2CH2COOH  +  Cl-

 

(b) Amine formation - amino acid synthesis

You can react the chlorinated carboxylic acid with ammonia under high pressure and temperature to insert the amine group substituent forming a synthetic route to amino acids.

e.g. chloroethanoic acid  +  ammonia  ===>   aminoethanoic acid 

ClCH2COOH  +  2NH3  ===>  H2NCH2COOH  +  NH4Cl

This will also be nucleophilic substitution reaction mechanism.

Similarly for the chloropropanoic acids you 2-aminopropanoic acid and 3-aminopropanoic acid you just get one product from each isomer.

CH3CHClCOOH  +  2NH3  ===>  CH3CH(NH2)COOH  +  NH4Cl

ClCH2CH2COOH  +  2NH3  ===> H2NCH2CH2COOH  +  NH4Cl


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Doc Brown's Advanced Level Chemistry Revision Notes

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INDEX of all My carboxylic acids and derivatives notes

All My Advanced A Level Organic Chemistry Notes

 

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