Name/CG: 2012 Term 2 Organic Chemistry Revision (Session II) Deductive Question

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1 Name/G: 2012 Term 2 rganic hemistry Revision (Session II) Deductive Question 1(a) A yellow liquid A, 7 7 N 2, reacts with alkaline potassium manganate (VII) and on acidification gives a yellow solid B, 7 5 N 4. After treating A with tin and concentrated hydrochloric acid, the addition of sodium hydroxide releases liquid. Under suitable conditions for reduction (with hydrogen under pressure at 160 º in the presence of nickel) A is slowly converted into D, 7 15 N. D reacts with excess of iodomethane to give a quaternary salt, E, of formula 7 13 ( 3 ) x NI. (i) Deduce the possible structures of A to E giving your reasoning. Number of and atoms are comparable and no. of atoms > 6, hence A contains a benzene. A undergoes (side-chain) oxidation with alkaline KMn 4 which on acidification gives B. A contains a methyl side chain/methylbenzene B is a carboxylic acid A undergoes reduction with tin and concentrated l & Na to form liquid A is an nitrobenzene is a phenylamine A undergoes reduction with 2 at high temperature and pressure to form D, 7 15 N D is a amine D undergoes nucleophilic substitution with excess 3 I to form quaternary salt, E. 3 ence, A is B N 2 N 2

2 3 3 D N 2 N 2 (Total : 5.5 marks) (i) State the value of x in compound E and give its displayed structure. Give a reason to explain why E dissolves in water while the rest of the compounds do not. [8] Since E is a quaternary salt, x = 3. Displayed formula of E : + N I - [1] (no displayed formula no mark) E is soluble in water as it is an ionic compound which can form ion-dipole attraction with water. owever, the other compounds contain large hydrophobic benzene and cyclohexane ring which are insoluble in water. (Total : 2.5 marks)

3 (b) Draw the structural formula of the products formed when compound Y is reacted with (i) cold, alkaline potassium manganate (VII) (ii) excess of hydrogen gas in the presence of platinum catalyst. Name the functional groups present in compound Y. [2] N 2 ompound Y (i) [1] (ii) [1]

4 2 Two isomeric organic liquids A and B contain 80.6% carbon, 7.5% hydrogen and 11.9% oxygen by mass. When g of A or B was evaporated in a syringe, the volume of the vapour produced, after correction to s.t.p. was 60.0 cm 3. Both A and B gave an orange precipitate with 2,4-dinitrophenylhydrazine and silver mirror with Tollens reagent. owever, both A and B do not give a yellow precipitate on warming with aqueous alkaline iodine. With lithium aluminium hydride in dry ether, A was converted to a compound which exists in optically active forms. B under the same conditions produced a compound D which is isomeric with but cannot exist in optically active forms. Both A and B on vigorous oxidation gave compound E as an end product, having molecular formula Deduce the structural formulae of compounds A to E, giving your reasoning. Empirical formula of A and B: Mass / 100g (g) No. of mol Ratio Empirical formula of A and B: 9 10 [10] No. of mol of vapour = 22.4 = mol M r of A and B = = 134 Molecular formula of A and B: 9 10 Number of and atoms are comparable and no. of is greater than 6 A and B, both contain a benzene ring

5 A and B undergo condensation with 2,4-DNP A and B are carbonyl compounds A and B undergo oxidation with Tollens reagent A and B are aldehydes A and B do not undergo oxidation with aqueous alkaline iodine A and B do not contain or structures R 3 R 3 A and B undergo reduction with LiAl 4 to optically active and optically inactive D respectively and D are primary alcohols has a chiral carbon and D has no chiral carbon A and B undergo vigorous oxidation to form E ompound E is A and B are monosubstituted benzenes. A : 3 : B: D: 2 2

6 3 The isomers A, B, and D of the formula 5 12 can react with hot acidified K 2 r 2 7 to give products E, F, G and respectively. ut of E, F, G and, only E gives a yellow precipitate upon reaction with alkaline aqueous iodine, whilst F and does not give an orange precipitate with 2,4- dinitrophenylhydrazine. can react with D under certain conditions to give a compound that has a formula n reacting D with hot concentrated sulfuric acid, the resulting product can react with steam under certain conditions to form A. When treated with limited chlorine in the presence of uv light B forms three two monochloro-substituted products (structural isomers) with one chlorine atom while forms two three monochloro-substituted products (structural isomers). Suggest possible structures for the compounds A to, explaining your reasons. [10] A, B, and D undergo oxidation with acidified K 2 r 2 7 to form E, F, G and respectively A, B, and D are 1 o or 2 o alcohols E, F, G and are ketone or carboxylic acid E undergoes oxidation with alkaline aq I 2 E has the structure: 3 A has 3 () structure F and do not undergo condensation with 2,4-DNP F and are not carbonyl compounds F and are carboxylic acids B and D are 1 alcohols., a carboxylic acid reacts with D, an alcohol to give ester that has a formula From the structure of ester, is likely to be and D is likely to be

7 D undergoes elimination with hot conc. sulfuric acid to form alkene which undergoes electrophilic addition with steam to form A, A is () 3 Thus E is B and undergo free radical substitution with chlorine and uv light and B forms two monochloro-substituted products while forms three monochlorosubstituted products. is and B is F is and F is 3 3 3

8 4 A solid compound J ( N) is insoluble in both hydrochloric acid and dilute sodium hydroxide. n warming with aqueous sodium hydroxide, J forms a liquid K ( 7 9 N) on the surface of an alkaline mixture, X. Liquid K can be separated from the alkaline mixture X by steam distillation, and dissolves in dilute hydrochloric acid. K can also react with excess bromomethane to give a compound L which has the formula 9 14 NBr. n the other hand, upon acidification, the alkaline mixture X yields a solid M ( ) which gives rapid effervescence with aqueous sodium carbonate. xidation of M yields one of the monomers that is used to make Terylene. Terylene Deduce the structures of J, K, L and M, explaining your reasoning. Give balanced equations where appropriate. [10] Number of and are comparable and no. of is greater than 6, J contains a benzene ring. J is insoluble in both acid and base J is neutral J is an amide J thus undergoes basic hydrolysis with warm Na to give K and M. K undergoes neutralization with dilute l K is basic K is an amine

9 Bromoethane undergoes nucleophilic substitution with K to form L. Since K gains 2 3 groups upon reaction with excess bromoethane to form L, K is a secondary amine. K is L is [1] 3 N 3 Br - 3 [1] M undergoes neutralisation with sodium carbonate to give 2 gas M is a carboxylic acid Since oxidation of M gives one of the monomers that is used to make Terylene, the monomer involved is diacid monomer, i.e. M is 3 [1] J is 3 N 3 [1]

10 The relevant equations are : 3 N + Na J 3 3 Na + + N 3 K M N Br 3 N + Br 3 K Br - 3 L 3 + 3[] M + 2 Max 10m

11 5 A neutral compound T contains 46.2% carbon, 7.7% hydrogen and 46.1% oxygen by mass. When heated with aqueous sodium hydroxide, T gives compound U, 2 6 2, and sodium ethanoate. When one mole of U reacts with only one mole of hydrogen chloride, compound V, 2 5 l, is formed. ompound V gives compound W, 3 5 N, on heating with ethanolic potassium cyanide. When W is heated with dilute sulfuric acid, an optically inactive liquid X, 3 6 3, is obtained. n heating X alone, a sweet smelling liquid Y, 3 4 2, is formed. xidation of X gives a solid Z, 3 4 4, which is soluble in water. Deduce the structural formula of compound T to Z giving your reasoning. Element % Mass mole = 3.85 = 7.7 =2.89 Divide by smallest mole = 1.33 = = Ratio Empirical formula of T is [10] T undergoes basic hydrolysis with hot Na(aq) to form U and 3 Na T is an ester U is an alcohol with two carbons ne mole of U undergoes nucleophilic substitution with one mole of l to form V V is a halogenoalkane V undergoes nucleophilic substitution with ethanolic KN - to form W W is a nitrile compound (contains a N group) W undergoes acidic hydrolysis with hot dilute 2 S 4 to form optically inactive X X is a carboxylic acid X does not have a chiral carbon

12 n heating X, a sweet smelling Y is formed. Y is cyclic ester X is 2 () 2 Y is 2 2 = X undergoes oxidation to form Z Z is 2 ence, W is 2 () 2 N U is 2 () 2 () V is 2 () 2 l T is 2 () 2 3

13 6 An organic compound A, 9 11 Br, on treatment with hot aqueous potassium hydroxide gave a compound B, B responded to oxidation in three different ways. With acidified potassium dichromate it yielded, With sodium hydroxide and iodine it yielded D, Na and a yellow precipitate. With hot, acidic potassium manganate (VII) it yielded E, Identify compound A to E and explain the above reactions. [7] Number of and is comparable in A and number of atoms is greater than 6, A contains benzene A undergoes nucleophilic substitution with hot K to form B A is a halogenoalkane B is an alcohol B undergoes oxidation with acidified K 2 r 2 7 to form, 9 10 is a ketone B is a secondary alcohol B undergoes oxidation with sodium hydroxide and iodine to form D, Na and a yellow ppt yellow ppt is I 3 B contains 3 () structure B undergoes (side-chain) oxidation with acidified KMn 4 to form E, E, benzoic acid, 6 5 B is monosubstituted benzene. Br Na + A is B is is D is

14 7 A substance A, 4 9 N 2 is a optically active solid with a high melting point. It dissolves in water to give a nearly neutral solution. B which is formed from A has the formula, n warming B with ethanol and excess concentrated sulfuric acid, a liquid ester, , is formed. n exposure to light, forms slowly forms a glassy solid which is an addition polymer. B reacts with Pl 5 to give D, which rapidly reacts with water to give E, 4 7 l 2. Treatment of E with hot aqueous sodium hydroxide followed by acidification will reform B. ompound B undergoes dehydration in the presence of concentrated sulfuric acid to form compound F, Deduce the structures of the compounds of A, B,, D, E and F. Explain the chemistry of the reactions involved. [10] A, 4 9 N 2 is a optically active solid with a high melting point A could be an amino acids A contains a chiral B which is formed from A has the formula, B contains a chiral B undergoes esterification with ethanol to form liquid ester B is carboxylic acid. B also undergoes elimination with excess conc. 2 S 4, B is alcohol B is ence A is 3 2 N 2

15 undergoes addition polymerisation to form D, contains = bonds is B undergoes nucleophilic substitution with Pl 5 to give D and D undergoes nucleophilic substitution with water to form E E undergoes nucleophilic substitution with hot Na(aq) to reform B E contains l (halogenoalkane) and l (acyl chloride) groups. 3 l 2 l E is 3 l 2 D is B undergoes self-esterification with conc. sulphuric acid to form an F F is a cyclic ester. F could be 3 [Total = 10]

16 8 This question is about compound F, 9 13 N. The following information is given about F: ompound F is insoluble in water but it dissolves readily in aqueous l. When F is treated with PBr 3, F forms compound G, 9 12 NBr. When F is treated with acidified KMn 4, no decolourisation of purple KMn 4 is observed. F can be formed when compound E reacts with tin metal and concentrated hydrochloric acid. G forms, 9 11 N, when heated with alcoholic K and in turn gives compound I, 9 10 NBr 3, when reacted with aqueous Br 2 in the dark. Deduce the structures of the compounds E, F, G, and I. Explain the chemistry of the reactions described.. [10] Number of atoms and atoms are comparable and no. of is more than 6, F contains a benzene ring. F undergoes neutralisation with aqueous l F is basic F is an amine F undergoes nucleophilic substitution with PBr 3 to form G F is alcohol G is halogenoalkane F does not undergo oxidation with acidified KMn 4 F is 3 o alcohol E undergoes reduction with Sn and conc. l fo form F E is nitrobenzene F is phenylamine G undergoes elimination with alcoholic K to form. contains an alkene group. undergoes electrophilic substitution with aq Br 2 and electrophilic addition with aq Br 2 to form I, 9 10 NBr 3 contains phenylamine and alkene functional groups is either 2- or 4- substituted phenylamine

17 Structure of E: 3 2 N 3 accept 1,2-position Structure of F: 3 2 N Structure of G: N Structure of : Br N 2 Structure of I Br 3 2 N 2 Br Br [max: 10]

OCR (A) Chemistry A-level. Module 6: Organic Chemistry and Analysis

OCR (A) Chemistry A-level. Module 6: Organic Chemistry and Analysis OCR (A) Chemistry A-level Module 6: Organic Chemistry and Analysis Organic Synthesis Notes by Adam Robertson DEFINITIONS Heterolytic fission: The breaking of a covalent bond when one of the bonded atoms