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JC2 H2 CHEMISTRY (9647) CHEMISTRY REVISION PROGRAMME WORKSHEET (Ans) Name: Class: TOPICS: Carboxylic Acids and Derivatives, Nitrogen Compounds 1. Summary of reactions of Carboxylic Acids and Derivatives No. Reactants Carboxylic Acid 1 (e.g. CH3COOH) Carboxylic Acid 2 Type of reaction / Remarks Reagents & Conditions Products Reactive metals Carboxylate salt - (e.g. Na), (e.g. CH3COO ) r.t.p. + ½ H2(g) Carboxylate salt Aqueous NaOH, r.t.p. (e.g. CH3COO ) 3 3 Carbonate or hydrogen carbonate (e.g. Na2CO3, NaHCO3), r.t.p. Carboxylate salt - (e.g. CH COO ) + CO2 + H2O Carboxylic Acid 4 5 (e.g. CH3COOH) Carboxylic Acid (e.g. CH3COOH) Neutralisation - (e.g. CH3COOH) Carboxylic Acid (e.g. CH3COOH) Acid-metal displacement ½ mol of H2 formed for each –OH group present Primary Alcohol LiAlH4 in dry ether, r.t.p. 1. Anhydrous PCl3, heat or (e.g. CH3CH 2OH) Acyl Chloride 2. Anhydrous PCl5, r.t.p. or (e.g. CH3COCl) Neutralisation Only carboxylic acids are acidic enough to react with carbonates and hydrogen carbonates. (used as a chemical test – effervescence of CO2 that forms a white ppt with Ca(OH)2. Heating is required to produce more CO2 to make white ppt more visible ) Reduction Not reduced by the other reagents that can reduce carbonyl compounds Nucleophilic Substitution (PCl5 used as a chemical test – white fumes of HCl formed. Alcohols also give similar observation) 3. Anhydrous SOCl2, heat Acyl Chloride 6 (e.g. CH3COCl) Carboxylic Acid H2O, r.t.p. (e.g. CH3COOH) + HCl 1 Nucleophilic Substitution (used as a chemical test – white fumes of HCl formed) Acyl chlorides produce very acidic solutions in water due to the HCl formed. No. 7 Reactants Reagents & Conditions Products Carboxylic Acid Alcohol (e.g. CH3OH) conc. H2SO4 catalyst, heat Ester (e.g. CH3COOH) Type of reaction / Remarks e.g. CH3COOCH3 Alcohol or phenol 8 Acyl Chloride Ester (e.g. ROH), r.t.p. e.g. CH3COOR (e.g. CH3COCl) Nucleophilic Substitution Phenols do NOT react with carboxylic acids Nucleophilic Substitution When a phenol is used, aqueous NaOH is often used to produce the phenoxide ion, which is a stronger nucleophile than the phenol (since the electron density on the O atom is increased). Choice of alcohol or phenol depends on the type of ester to be produced. Ammonia or amines 9 Acyl Chloride (e.g. CH3COCl) Amide (e.g. RNH2), (e.g. CH3CONHR) r.t.p. Nucleophilic Substitution When synthesizing amides in the lab, acyl chlorides are used, NOT carboxylic acids because the yield will be poor due to the neutralization reaction between the acid and the basic reagent. Choice of ammonia or amine depends on the type of amide to be produced. Carboxylic Acid 10 Ester (e.g. CH3COOH) e.g. CH3COOR + Alcohol or phenol (depends on the ester given) Aqueous HCl (or H2SO4), heat Carboxylic Acid Amide 11 (e.g. CH3COOH) e.g. CH3CONH2 + NH4 + Carboxylate Salt - 12 (e.g. CH3COO ) Ester + Alcohol or phenoxide salt (depends on the ester given) e.g. CH3COOR Aqueous NaOH, heat Carboxylate Salt 13 Acidic Hydrolysis For acidic hydrolysis of 2° and 3° amides, an amine salt is + formed instead of NH4 . Amide - (e.g. CH3COO ) e.g. CH3CONH2 + Ammonia 2 Basic Hydrolysis The NH3 produced from basic hydrolysis of a 1° amide can be used as a chemical test – gas evolved turns moist red litmus paper blue. For basic hydrolysis of 2° and 3° amides, an amine is formed instead of NH3. 2. Summary of reactions of Nitrogen Compounds No. Reactants Reagents & Conditions Products Amine e.g. CH3NH2 Aqueous HCl (or H2SO4), r.t.p. Amine salt 1 e.g. CH3NH3 Type of reaction / Remarks Neutralisation Actually other acids e.g. carboxylic acids can also neutralize amines but are not often used (e.g. you don’t want to have the carboxylate salt being formed also otherwise it’s difficult to obtain the amine salt) + Nucleophilic substitution 2 Amide Amine Acyl chloride e.g. CH3NH2 (e.g. CH3CH2COCl), r.t.p. e.g. CH3NHCOCH2CH3 Carboxylic acids NOT used for laboratory synthesis of amides due to neutralization reaction that can take place instead Electrophilic substitution NH2 NH2 X X Aqueous X2, r.t.p. 3 (because of the strong electron-donating –NH2 group which activates the ring) (aqueous (X = Cl or Br) Br2 used as a chemical test – decolourisation X of reddish-brown Br2 solution and white ppt formed) Reduction NO 2 (i) Sn in conc. HCl, heat 4 (ii) aqueous NaOH Amino Acids 5 NH2 e.g. H2NCH2COOH Same reagents and conditions needed and same type of products formed as for the individual amines and carboxylic acids 3 After step (i), the salt is formed instead of the amine. Hence in step (ii), excess NaOH is added to liberate the phenylamine Neutralisation Structured Questions Time allocation: 15 mins 1 Trichloroethanal (chloral), CCl3CHO, was much used up to the middle of this century as a drug and a stimulant. (a) (i) When it is warmed with aqueous sodium hydroxide, one organic product is trichloromethane (chloroform). Construct an equation for this reaction and name the other organic product. CCl3CHO + NaOH → CHCl3 + HCO2-Na+ [1] Sodium methanoate [1] This reaction is linked to the iodoform test for carbonyl compounds e.g. for CH3CHO, the intermediate is CI3CHO, which gives the final products CHI3 and HCO2-Na+ (ii) Hence, or otherwise, suggest how chloral can be converted into ethyl methanoate, stating any other required reagents and conditions used in the conversion. [4] Cl Cl O C C Cl O NaOH (aq) H + C H heat O-Na+ H2SO4 (aq) r.t.p. [½] [½] O O CH3CH2OH H C H O CH2 CH 3 CHCl 3 C conc H2SO4, heat [½] OH [½] (b) Chloral is easily oxidized to trichloroethanoic acid. (i) Draw a displayed formula of trichloroethanoic acid Cl Cl C O C O Cl H 4 Remember to show ALL bonds! [1] (ii) Explain why this acid is a much stronger acid than ethanoic acid CH3CO2CCl3CO2- CH3CO2H CCl3CO2H H+ H+ + + In trichloroethanoic acid, there are 3 electron-withdrawing chloro groups [1] that disperses the negative charge on the carboxylate anion hence stabilising the carboxylate anion relative to the acid [1] [3] (c) Arrange the following in order of increasing acid strength and explain your reasoning. O OH C OH OH CH3 [3] Acid strength: O OH OH OH C CH 3 < < [1] Benzoic acid O OH O C O C + H+ The carboxylate anion is resonance stabilized [½] by the delocalization of the negative charge over the C atom and both oxygen atoms [½] in the carboxylate anion, hence stabilizing the carboxylate anion relative to acid. As such, benzoic acid has a greater tendency to ionize compared to phenol, thus is the strongest acid. 5 CH3 2-methyl phenol OH OCH 3 CH3 + H+ The electron-donating alkyl group decreases the intensity of the negative charge on the phenoxide anion, hence destabilizing the phenoxide anion relative to acid. As such, benzoic acid has a greater tendency to ionize compared to phenol, thus 2-methyl phenol is the weakest acid. (Modified N96/III/3a, b(i) and (ii)) [Total: 10] Time allocation: 15 mins 2 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. [1] 2CH3COOH + CaCO3 → (CH3COO)2Ca + CO2 + H2O [1] (b) When the solution produced in (a) is evaporated, and the resulting solid calcium ethanoate heated strongly in a test-tube, an organic compound G is formed which condenses into a colourless liquid. The residue in the tube consists of calcium carbonate. 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. G has a Mr of about 57.9. Suggest a structural formula for G. Justify your answer by reference to these properties of G. [5] G does NOT undergo acid-metal displacement with Na ⇒ G is NOT an alcohol. [1] G does NOT undergo oxidation with Fehling’s solution ⇒ G is NOT an aliphatic aldehyde [1] G undergoes oxidation with alkaline aqueous iodine ⇒ G has the H3C C group in the compound [1] O Based on the above properties and the given Mr, G is likely to be CH3COCH3, [1] propanone, which would be neutral and is water-soluble since it can form hydrogen bonds with water. [1] 6 (c) Construct a balanced equation for the formation of G by the action of heat on calcium ethanoate. [1] (CH3CO2)2Ca → CH3COCH3 + CaCO3 [1] (d) Suggest a simple one-step test you could carry out to confirm the identity of the functional group present in G. You should give the reagent and the observation you would make. [2] Add 2,4-DNPH to G and heat. [1] An orange ppt [1] should be observed. (e) Suggest the structural formula of the organic product you might expect when calcium propanoate, (CH3CH2CO2)2Ca, is heated strongly. [1] CH3CH2COCH2CH3 [1] (Similar to example in (c), since (CH3CO2)2Ca gives CH3COCH3) (Modified from J2000/I/8) Time allocation: 15 mins 3 Compound H, C8H11NO, is weakly basic. It dissolves in dilute aqueous hydrochloric acid to give a solution from which a crystalline solid J can be isolated. H decolourises aqueous bromine with the formation of a white precipitate K. No orange crystals are observed when 2,4-dinitrophenyihydrazine is added to H. However when H is heated with alkaline aqueous iodine and then followed by careful acidification, some yellow crystals are produced together with L, C7H7NO2. L can be produced from 4-nitromethylbenzene in a 2-step reaction via an intermediate M. 4-nitromethylbenzene is first heated with tin in the presence of concentrated hydrochloric acid to form M, and then followed by adding hot acidified potassium manganate(Vll) to produce L. Deduce the structures of the compounds H to M. Explain your reasoning and write balanced equations for the reactions involved where appropriate. [10] [Total: 10] H undergoes neutralization with dil. aq. HCl to give a crystal J [½] ⇒ H may contain an amine group [½] ⇒ J is a salt H undergoes electrophilic substitution with aq. bromine to give white ppt, K [½] ⇒ H may be a phenylamine [½] ⇒ K may be a bromosubstituted phenylamine [½] 7 H does not undergo condensation with 2,4-DNPH ⇒ H does not contain a carbonyl group [½] H undergoes oxidation with alkaline aq. iodine and acidification to form a yellow crystals of CHI3 and L. [½] ⇒ H may contains the CH3CH(OH)- group [½] ⇒ L is a carboxylate salt [½] 4-nitromethylbenzene undergoes reduction with tin and conc. HCl to form M [½] ⇒ M could be CH3 NH2 [½] M undergoes oxidation with hot potassium manganate to form L [½] ⇒ L could be COOH NH2 [½] H could be CH3 HO C NH2 H [½] 8 J could be CH3 HO C H NH3+Cl- [½] K could be CH3 HO C H Br Br NH2 [½] Equations CH3 HO C CH3 H HO + C HCl NH2 NH3+Cl- CH3 HO C H [1] CH3 H HO + C H 2Br2 + 2HBr Br NH2 Br NH2 9 [1] Remember to show ALL bonds! CH3 CH3 + 4[H] + NO2 NH2 2H2O [1] CH3 HO C COO-Na+ H + 3I2 + 4NaOH + CHI3 + 3NaI + 3H2O NH2 NH2 [1] [Total: 12, max = 10] Time allocation: 4(a) 9 mins Draw a displayed formula of aminoethanoic acid (glycine) H H N H (b) C H [1] O C O H [1] Penicillin is widely used to kill bacteria. The general structure of a penicillin molecule is given below. R = an aryl group (i) Mark with an asterisk * on the above structure any carbon atoms which are chiral centres. [2] 10 * * For a complete answer, you need to show understanding that you recognize the amide groups in penicillin are neutral (as compared to amines which are basic) * [3 = 2 marks; 2 = 1 mark; 1 or 0 = 0 mark] (ii) Bacteria mistake penicillin for protein molecules and bond on to the amino acid sequence present in penicillin. This sequence is also present in the skeleton of the aminoethanoic acid molecule you have drawn in (a). Draw a ring around the part of the penicillin molecule which bacteria mistake for an amino acid. [1] | Circle at the bottom right portion of the molecule – N – CH – CO2H (c) [1] Would you expect penicillin to be acidic, neutral or basic? Explain your answer as fully as you can. [2] (N95/III/5) Acidic [1] because of the acidic –COOH group present and the amide groups are neutral. [1] Time allocation: 5(a) 14 mins How would you expect the melting point and the solubility in water, of an unionised covalent form of glycine (aminoethanoic acid) to compare with the actual properties of the zwitterionic form? [2] Melting point for the unionised form < zwitterionic form because of the weaker hydrogen bonding between the amino acid molecules compared to the stronger electrostatic forces of attraction between the zwitterions. [1] Solubility for the unionised form < zwitterionic form because the zwitterions can form ion-dipole interactions with water which releases energy for the detachment of the ions from the crystal lattice for hydration. [1] 11 (b) The following scheme of reactions illustrates the synthesis of a dipeptide, E. (i) What type of reaction is step I? Nucleophilic substitution [1] (ii) Suggest a reagent for step II. PCl5, r.t.p. [1] (iii) During which step is the peptide bond formed? Step III [1] (iv) What are the products of hydrolysis of E, and how may the reaction be carried out? Dilute H2SO4, heat [1] + + → Products: H3N–CH(CH3)COOH [1] + H3NCH2CO2H [1] OR Dilute NaOH, heat [1] → Products: H2N–CH(CH3)COO- [1] + H2NCH2CO2- [1] (v) Suggest the type of reaction in step IV. Reduction [1] This is an unusual reaction which involves the addition of H2 (reduction) with the cleavage of C-N bond occurring simultaneously. [7] (Modified N96/I/10) 12