2015 JC2 H2 Carboxylic Acids and Derivatives Part 1 Tutorial (Teachers) Updated

SERANGOON JUNIOR COLLEGE 9647 H2 CHEMISTRY 2015 JC2 Carboxylic Acids and Derivatives (Part 1) Tutorial 1. J97/I/8 One of the products from the vigorous hydrolysis of oil of bitter almonds is mandelic acid, H. It can be synthesised from benzaldehyde in two steps: step I step II C6H5CHO J C6H5CH(OH)CO2H H (a) Suggest reagents and conditions for steps I and II, and draw the structural formula of the intermediate J. step I: HCN in trace amt of NaOH(aq) / HCN in trace amt of NaCN (aq), 10 – 20 C step II: HCl (aq) / H2SO4(aq), heat CN C

OH

H

J

(b)

Suggest structural formulae for the products of the reaction of mandelic acid

H

O

C C OH (i) (ii) (iii) (iv) (v)

OH

with the following reagents. HBr PCl5 CH3COCl C2H5OH/H+ NaOH(aq)

(i)

Br

H COCl C

(ii)

H

2. Regents & conditions given. 3. Identify type of reaction undergone.

No reaction with –COOH unless PBr3 added.

COOH C

Thinking process… 1. Identify all functional groups in structure.

Cl

COOH C

O

O C CH3

H

(iii)

O

O

CH 2CH 3

C C

(iv)

H -

COO Na C

(v)

Reversible reaction. Lower yield of esters compared with (iii)

OH

+

OH

H

MONOBASIC acid! How do you tell?? (c) A sample of mandelic acid isolated from bitter almonds was contaminated with a neutral impurity. A 0.100 g sample of the impure acid required 6.00 cm3 of 0.100 mol dm-3 NaOH to neutralise it. Calculate the percentage purity of the mandelic acid. nNaOH  nmandelic acid = (6/1000) x 0.100 = 6 x 10-4 mol mmandelic acid = 6 x 10-4 [8(12) + 8(1) + 3(16)] = 0.0912 g % purity = (0.0912/0.100) x 100 = 91.2 %

2. (a)

J92/I/10(a) Explain the following observations: (i) ethanoic acid is more acidic than ethanol CH3COOH

CH3COO¯ + H+

C2H5OH C2H5O¯ + H+  The carboxylate ion is stabilised by charge delocalization or (resonance stabilized)

O-

O R C        (ii)





(iii)

O-

R C

O

This disperses the negative charge over the C atom and the 2 oxygen atoms on the carboxylate ion. Greater stabilisation of the carboxylate ion, promoting greatest loss of H+. CH3CO2H is more acidic than CH3CH2OH Alkoxide ion is not stabilised by charge delocalisation. Presence of electron-donating alkyl group intensifies the negative charge on the alkoxide ion. This destabilises the alkoxide ion, least H+ released Hence CH3CH2OH is a weaker acid. chloroethanoic acid is stronger than ethanoic acid. CH2ClCOOH CH2ClCOO¯ + H+ In chloroethanoic acid, the electron withdrawing group (Cl) disperse the negative charge on the carboxylate ion hence stabilise the conjugate base and increasing the stability of the carboxylate ions. Acidity:

CH2ClCOOH > CH3COOH

Ethanoic acid (b.p 118ºC) has a higher boiling point than ethyl ethanoate (77ºC) Ethanoic acid exists as polar molecules held together by stronger hydrogen bonds while ethyl ethanoate exists as polar molecules held together by weaker permanent dipole-permanent dipole interactions

(iv) Ethanoic acid and sodium ethanoate are soluble in water but ethyl ethanoate is insoluble in water Ethanoic acid is a simple molecular compound which can form hydrogen bonds with water. Favourable hydrogen bonds between (Ethanoic acid) solute and water molecules are formed via hydration. Sodium ethanoate, an ionic compound, formation of favourable ion-dipole interactions which results in the release of energy that overcome giant ionic structure for hydration. Ethyl ethanoate is a simple molecular compound that has weaker van der Waal’s forces between solute molecules which are not strong enough to displace the stronger hydrogen bonding between water molecules for hydration to occur. (b)

How would you expect the acidity of 3-chloropropanoic acid 2-fluoropropanoic acid to compare with that of 2-chloropropanoic acid?

and

of

Comparing 2-chloropropanoic acid and 3-chloropropanoic acid CH3CHClCOOH CH2ClCH2COOH

CH3CHClCOO¯ + H+ CH2ClCH2COO¯ + H+

General effect : The electron-withdrawing chloro group disperses the negative charge on the carboxylate anion hence stabilises the carboxylate anion relative to the acid Specfic effect: 

Distance of electron withdrawing group from COOH group: CH3CHClCOOH < CH2ClCH2COOH



Extent of electron-withdrawing effect: CH3CHClCOOH > CH2ClCH2COOH



Extent of stability of carboxylate ion relative to acid:: CH3CHClCOOH > CH2ClCH2COOH



Acidity: CH3CHClCOOH > CH2ClCH2COOH

Comparing 2-chloropropanoic acid and 2-fluoropropanoic acid CH3CHClCOOH CH3CHFCOOH

CH3CHClCOO¯ + H+ CH3CHFCOO¯ + H+

General effect: The electron-withdrawing chloro and fluoro groups disperses the negative charge on the carboxylate anion hence stabilises the carboxylate anion relative to the acid

  

Specfic effect: F in CH3CHFCOOH is more electronegative than compared to Cl in CH3CHClCOOH Extent of electron-withdrawing effect from COOH group: CH3CHFCOOH > CH3CHClCOOH



Extent of stability of carboxylate ion relative to acid:: CH3CHFCOOH > CH3CHClCOOH



Acidity: CH3CHFCOOH > CH3CHClCOOH

3.

N08/III/3(f) 2-hydroxyphenylethanoic acid, N, can be prepared from compound M by the following route. O H H OH OH H+(aq) C C KOH(aq) C CHO CO2H CO2K M N Suggest two tests (stating reagent and observations) that would enable the compounds M and N to be distinguished from each other. Test 1: Add Na2CO3(aq) to test tubes containing M & N respectively. Pass any gas produced via a delivery tube to a test tube containing Ca(OH)2 (aq) Observation for test 1 M: No gas produced, no white ppt observed in test tube containing Ca(OH)2 (aq) N: Gas produced formed white ppt in Ca(OH)2 (aq) Test 2: Add 2, 4-DNPH, warm to the test tubes containing M & N respectively (Comment: accept Tollen’s or Fehling’s as well as PCl5) Observation for test 2 M: orange ppt is seen N: no orange ppt

Presence of 2 carboxylic acids group in the same structure

4.

(a)

J92/I/11 Compound B, a diacid that occurs in apples and fruits has the following composition by mass: C, 35.8% H, 4.5% O, 59.7% B reacts with ethanol, in the presence of concentrated sulfuric acid under reflux to give C, C8H14O5. Compound C evolves hydrogen gas when treated with sodium metal and reacts with acidified potassium dichromate (VI) to give compound D. Compound D produces an orange precipitate with 2,4dinitrophenylhydrazine but has no reaction with Fehling’s or Tollen’s reagent. Calculate the empirical formula of B. C % mass 35.8 moles 2.98 1 x 4 to get whole no. 4 Empirical formula of B: C4H6O5

(b)

H 4.5 4.5 1.5 6

O 59.7 3.73 1.25 5

Deduce structures for compounds B, C and D, explain the reactions described.  B undergoes nucleophilic substitution/esterification with CH3CH2OH in the presence of conc.H2SO4 under reflux to form compound C (C8H14O5) Since B gained 4 carbon atoms (4C  8C)  C may be a diester C undergoes redox reaction with Na to form H2  C can be an alcohol or carboxylic acid C undergoes oxidation with hot acidified K2Cr2O7 to form D  C is a primary or secondary alcohol  D is a carboxylic acid or ketone D undergoes condensation with 2,4-DNPH to give orange ppt, but is not oxidised by Fehling’s or Tollens’ reagent.  D is a ketone  C is a secondary alcohol

Structures

O H H O

B:

HO C C C C OH H OH

O H H O C:

CH3CH2O C C C C O CH2CH3 H OH

D:

O

H

O O

HO C C C C OH

CH3CHOOH

H

5.

J2000/I/8 The limescale that collects in kettles in hard water areas is mostly calcium carbonate. It can be removed fairly harmlessly by a warm solution of vinegar, which contains ethanoic acid. The limescale dissolves with fizzing and a solution of calcium ethanoate remains.

(a)

Write a balanced equation for the reaction between ethanoic acid and calcium carbonate. 2CH3COOH + CaCO3  Ca(CH3COO)2 + CO2 + H2O

When the resulting solution produced in part (a) is evaporated, and the solid residue calcium ethanoate heated strongly in a test tube, an organic compound G is formed. Compound G condenses to a colourless liquid and the residue in the test tube consists of calcium carbonate. (b) When 0.10 g of G was injected into a gas syringe at a temperature of 383 K and a pressure of 1.0 x 105 Pa (1 atm), 55 cm3 of vapour were produced. Calculate the relative molecular mass of G. PV = mRT/M M = mRT/PV = 0.1(8.31)(383)/(105 x 55 x 10-6) = 57.9 g mol-1 Mr of G = 57.9

(c)

Compound G is neutral and water-soluble. G does not react with sodium metal nor with Fehling’s solution but it does react with alkaline aqueous iodine. Suggest a structural formula for G. Justify your answer by reference to these properties of G. G is neutral and so is not carboxylic acid. G is water-soluble, showing that it forms hydrogen bond with water molecules. G does not undergo redox reaction with sodium metal and so is neither alcohol nor carboxylic acid. G is not aldehyde since it does not undergo oxidation reaction with Fehling’s solution. G reacts with alkaline aqueous iodine via oxidation giving yellow precipitate of CHI3

O showing that it contains the

C CH3

group.

O C => Hence G is propanone (d)

CH3

Construct a balanced equation for the formation of G by the action of heat on calcium ethanoate. heat

Ca(CH3COO)2 (e)

CH3

CH3COCH3

+ CaCO3

Suggest a simple one-step chemical test that you could carry out to confirm the identity of the functional group present in G. State the reagents and the observations you would make. Add 2,4-DNPH to G and warm. An orange ppt is obtained showing that G is a carbonyl compound.

(f)

Suggest the structural formula of the organic product you might expect when calcium propanoate, (CH3CH2CO2)2Ca, is heated strongly. Inferring from above: Ca(CH3COO)2  (CH3CH2CO2)2Ca 

O C CH3CH2

CH2CH3

Recognising patterns in Organic Reactions CH3COCH3 + CaCO3 CH3CH2COCH2CH3 + CaCO3

6.

Oxalic acid is an organic compound with the formula H2C2O4. This colourless solid is a dicarboxylic acid. In terms of acid strength, it is about 3,000 times stronger than acetic acid. Its conjugate base, known as oxalate (C2O42−), is a reducing agent as well as a chelating agent for metal cations. Oxalic acid dissociates in water according to the following equations HOOC-COOH + H2O HOOC-COO- + H3O+ I Ka1= 5.6 x 10-2 mol dm-3 HOOC-COO- + H2O OOC-COO- + H3O+ II Ka2 = 5.4 x 10-5 mol dm-3

(a)

(i) Explain why oxalic acid is more acidic than acetic acid. The conjugate base of oxalic acid is more stable than that of acetic acid. Stabilisation of the HOOC-COO- via intramolecular hydrogen bonding with the unionised –COOH group will result in favouring the dissociation of the first acid proton in oxalic acid. (ii) Explain why the value of Ka1 is larger than Ka2. 

After the first dissociation, the ‘second’ acid proton in the remaining COOH group is held more tightly by the monoanion via intramolecular hydrogen bonding, resulting in a very stable monoanion. i.e the second acid proton does not dissociate easily.



Also, the removal of the second H+ from a species that already contain a negative charge is electrostatically unfavourable.

(iii) Write expressions for acid dissociation constants for equation I and II above.

Ka1=

[H 3 O  ][HOOC  COO  ] [HOOC  COOH]

Ka2 =

[H 3 O  ][  OOC  COO  ] [HOOC  COO  ]

(b)

(i)

The neutralisation between oxalic acid and sodium hydroxide corresponds to the two equations given HOOC-COOH + NaOH → HOOC-COONa + H2O step 1 + HOOC-COONa + NaOH→ NaOOC-COONa + H2O step 2 A 25 cm3 sample of oxalic acid of concentration 0.100 mol dm-3 was titrated with sodium hydroxide of concentration 0.100 mol dm-3. The titration curve for the above titration was given below. pH

Y

X

25

50

Vol of NaOH/cm3

At the first equivalent point, X, the species formed is HOOC-COO-(aq), which is both an acid and a base where the relevant equilibriums are: HOOC-COO-(aq) + H2O(l) H3O+(aq) + (COO)22-(aq) HOOC-COO-(aq) + H2O(l) OH-(aq) + (COOH)2 (aq) It can be shown that in such an instance that the [H+] at the first equivalent point, X, can be given by expression, [H3O+] =

K a1 K a 2

Using the expression, determine the pH value at point X. [H3O+] = 5.6x102 x5.4x105 = 1.74 x 10-3 pH =2.76

Calculate the pH value at the second equivalent point, Y, given that the [OH-] can be assumed to be entirely due to the hydrolysis:

(ii)

-

OOC-COO- + H2O

Kb2 = 

HOOC-COO- + OH-

[OH  ][HOOC  COO  ] [  OOC  COO  ]

K w [OH  ][HOOC  COO  ] = K a2 [  OOC  COO  ]

Since [OH-] = [HOOC-COO-], Kw [OH  ] 2 =  K a 2 [ OOC  COO  ]

[-OOC-COO-] = 25/1000 x 0.1 (25+50)/1000 = 0.0333 mol dm-3 [OH  ] 2 0.0333 [OH ] = 2.48 x 10-6 mol dm-3

Hence, 1.85 x 10-10 =

pOH =5.61  pH = 8.39 (c)

The concentration of oxalic acid in a solution can be determined by an acid-base titration. State another way by which it’s concentration can be determined volumetrically. Another method will be to titrate an acidified solution of the oxalic acid with potassium manganate (VII) through redox titration.

(d)

Oxalic acid was one of the products formed when an aromatic organic compound, A, with molecular formula C10H10O2 undergoes oxidation with acidified manganate(VII) to form another organic product, B, with the molecular formula C8H8O2. No other organic compound was formed in the oxidation. Compound B reacts readily with 2 mole of Br2(aq) to form compound E, C8H6O2Br2. Compounds A and B are both soluble in NaOH and both A and B reacts with 2,4 - DNPH. Compound A reacts with acidified dichromate to give an acid, C, C10H10O3. Compound C reacts with SOCl2 to form a sweet-smelling compound D, C10H8O2. Deduce the structures of A,B,C,D and E. Explain your deductions -

A and B have comparable no. of C and H atoms and no. of C ≥ 6, => A and B contains a benzene ring

-

B undergoes easy electrophilic substitution with 1 mole of Br2(aq)  B contains phenol.

-

A and B are soluble in NaOH  A and B consists of carboxylic acids or phenol

-

A and B undergoes condensation with 2,4-DNPH  A and B contains the carbonyl functional group.

-

A undergoes oxidation with acidified dichromate to form an acid, C  A contains aldehyde.

-

C undergoes nucleophilic substitution with SOCl2 to form acyl chloride which will then undergo nucleophilic substitution with the phenol to form a cyclic ester.

HO

HO

H

H

C C C O

O

CH3

CH3

Compound A

Compound B

HO

O H C O

C

H

C

O C

C

C

CH3

CH3

Compound C

Br

O

C

HO HO

C

Compound D

CH3 Br Compound E

7.

J90/I/28 The same carboxylic acid is obtained either by the hydrolysis of a nitrile P or by the oxidation of an alcohol Q. Which of the following pairs could be P and Q? Explain your choice. P Q A CH3CH2CN CH3CH2OH B (CH3)2CHCN (CH3)3COH C C6H5CH(CH3)CN C6H5CH2CH(OH)CH3 Note: Use mild D C6H5CH2CN C6H5CH2CH2OH E C6H5CN C6H5OH oxidation agent like Answer: D Acidic hydrolysis of P: C6H5CH2CN + HCl + 2H2O  C6H5CH2COOH + NH4Cl

K2Cr2O7 . Cannot use KMnO4 otherwise benzoic acid will be formed.

Oxidation of Q: C6H5CH2CH2OH + 2[O]  C6H5CH2COOH + H2O

8.

Lower pKa stronger acid

J96/III/28 In which sequence is it correctly stated that the value of pKa decreases continuously? Explain your choice. CH3CO2H > CCl3CO2H > C2H5OH > C6H5OH A CCl3CO2H > CH3CO2H > C2H5OH > C6H5OH B Make sure you can identify C2H5OH > C6H5OH > CH3CO2H > CCl3CO2H C relative acidity of C6H5OH > C2H5OH > CH3CO2H > CCl3CO2H D respective organic

compounds given to you

Answer: C CCl3COOH

CCl3COO¯ + H+

CH3COOH

CH3COO¯ + H+

C6H5OH

C6H5O¯ + H+

C2H5OH C2H5O¯ + H+  Both CCl3CO2H and CH3CO2H are more acidic than C6H5OH as carboxylate ion is stabilised by charge delocalisation.  Lone pair of electrons on oxygen interacts with the carbonyl group.  This disperses the negative charge on the two oxygen on the carboxylate ion and stabilises it to a greater extent. The loss of H+ is greatly promoted. 



CCl3CO2H is the most acidic due to the presence of electron-withdrawing Cl groups which disperses the negative charge on the carboxylate ion to an even greater extent and stabilises it more. C2H5OH is the least acidic due to the presence of electron-donating alkyl group which intensifies the negative charge on the ethoxide ion and destabilizes it. Therefore, least H+ is released.

Assignment 1. N93/I/10 2-hydroxybenzoic acid, A, is a useful intermediate for making aspirin (an analgesic) and “oil of wintergreen” (used in ointments). COOH

COOH

I

II

OH

OCOCH 3

CO 2CH 3

OH

(a)

aspirin A oil of wintergreen Suggest suitable reagents and conditions for reactions I and II.

[2]

(b) (i) (ii) (iii)

Draw the structural formulae of the organic molecules produced when A reacts with Na2CO3(aq), NaOH(aq), dilute HNO3

[3]

(c)

(d) (i) (ii)

When an aspirin tablet was crushed up in water and titrated with 0.1 mol dm-3 sodium hydroxide, 13.9 cm3 of alkali were required to neutralize its acidity. What was the mass of aspirin in the tablet? [2] A “soluble aspirin” contains the calcium salt of aspirin. What reagent could you use to convert aspirin into its calcium salt? Suggest why aspirin is insoluble in water whereas its calcium salt is soluble. (a) reaction I : reagents and conditions: CH3COCl, r.t.p [1] reaction II: reagents and conditions: CH3OH, conc. H2SO4, heat [1] -

COO Na

(b) (i)

+

OH

[1] -

COO Na

(ii)

-

[3]

O Na

+

+

[1]

COOH O 2N

COOH

OH NO 2 OH (iii) or [1] -3 (c) naspirin = (13.9/1000) x 0.1 = 1.39 x 10 mol [1] Maspirin = 1.39 x 10-3 x [9(12) + 8(1) + 4(16)] = 0.250 g [1] (d) (i) Ca(OH)2 [1] (ii) Aspirin has more extensive intramolecular hydrogen bond.This reduces the extent of hydrogen bond with water molecules. [1] Its calcium salt is ionic and forms favourable ion-dipole interactions with the water molecules, thus easily solvated. [1]

2.

Compound B has the structure as shown below:

OH CHC

OH

O (a)

Show how compound B can be synthesized starting from benzoic acid. Your answer should include the intermediate compound(s) formed and the reagents and conditions required for each of the step(s). [4]

(b)

Compound B reacts with concentrated sulphuric acid under reflux to produce cyclic compound C with molecular formula C16H12O4. C is neutral to litmus and has no reaction with 2,4-DNPH. Draw the structure of compound C. [1] Solutions: (a) Step 1 Reagents and conditions: LiAlH4 in dry ether, r.t.p

COOH + 4[H]

LiAlH4 in dry ether

CH2OH

rtp

+ H 2O [1]

Step 2 Reagents and conditions: K2Cr2O7 in H2SO4 (aq), distillation O K2Cr2O7/H+ + [O] C + CH2OH distillation

H2O

H

[1]

Step 3 Reagents and conditions: HCN in NaOH (aq), 10 – 20 oC [1]

CN CHO + HCN

NaOH (aq) o

10-20 C

CH OH

Step 4 Reagents and Conditions: HCl (aq), heat

CN +

CH + 2H2O + H OH

reflux

CHCOOH

+ NH4+

OH Compound B

[1]

(b) O C O CH

Self Esterification CH O

C

O Compound C

3.

[1]

Values of the acid dissociation constants, Ka, for some weak acids are shown below: acid Formula Ka / mol dm-3 Benzoic acid C6H5COOH 6.5 x 10-5 Carbonic acid H2CO3 4.5 x 10-7 Ethanoic acid CH3COOH 1.8 x 10-5 Methanol CH3OH 3.0 x 10-16 Phenol C6H5OH 1.3 x 10-10 (a) Which of these weak acids are stronger than carbonic acids?

[1]

(b) Calculate the pH of 0.10 mol dm-3 of phenol

[1]

(c) Explain in terms of their molecular structures why methanol and phenol have significantly different Ka values. [2] (d) Methyl benzoate, C6H5CO2CH3 and phenyl ethanoate, CH3CO2C6H5 are isomers. (i) Outline how methyl benzoate may be formed from benzoic acid. (ii) Outline how phenyl ethanoate may be produced from ethanoic acid.

[2]

[Total: 6]

(a) Ethanoic acid and benzoic acid (The larger the Ka, the stronger the acid and both of these acids have a Ka > 4.5 x 10-7) [1] (b) pH = 5.44 [1]

(c) For Phenol: [1] for phenol explanation  Phenoxide ion is resonance stabilised, when the p-orbitals of oxygen overlaps with the  electron cloud of the benzene ring. 

The negative charge is dispersed over six C atoms which have lower electronegativity than the O atom.



Stabilises the phenoxide ion, promoting the loss of H+ resulting in a higher Ka value as compared to methanol.

For methanol: [1] for methanol explanation 

Alkoxide ion is not stabilised by charge delocalisation. Presence of electron-donating alkyl group.



This intensifies the negative charge on the alkoxide ion.



Destabilises the alkoxide ion, least H+ released.

(d) (i) Reagents and Conditions: CH3OH, concentrated H2SO4, heat [1] (ii) Step 1: PCl3 (s) or PCl5 (s), rtp [1] Step 2: Phenol, r.t.p.

END