1. What are the hybridisation states of each carbon atom in the following compounds? CH 2 =C=O, CH 3 CH=CH 2, (CH 3 ) 2 CO, CH 2 =CHCN, C 6 H 6

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1 1. What are the hybridisation states of each carbon atom in the following compounds? CH 2 =C=O, CH 3 CH=CH 2, (CH 3 ) 2 CO, CH 2 =CHCN, C 6 H 6 The hybridization state of each carbon in the given compounds, starting from left is given as: (i) CH 2 =C=O sp 2,sp 2 (ii) CH 3 - CH =CH 2 sp 3,sp 2, sp 2 (iii) (CH 3 ) 2 CO sp 3, sp 2 (iv) CH 2 =CHCN sp 2 sp 2, sp (v) C 6 H 6 sp Indicate the σ and π bonds in the following molecules : C 6 H 6, C 6 H 12, CH 2 Cl 2, CH 2 =C=CH 2, CH 3 NO 2, HCONHCH 3. C 6 H 6 C 6 H 12 CH 2 Cl 2 CH 2 =C=CH 2 CH 3 NO 2 HCONHCH 3 12 σ (6 C-C,6C-H) and 3 π bonds 18 σ bonds (6 C-C,12C-H) 4 σ bonds 6 σ (2C-C,4 C-H) bonds and 2 π bonds 6 σ (3C H, 1C Ν, 2N-O) bonds and 1 π bond 8 σ (4C-H, 1C-O,1N-H,2N-C) bonds and 1 π bond. 3. Give the IUPAC names of the following compounds (a) propyl benzene (b) 3-methyl pentane nitrile (c) 2,5- dimethyl heptane (d) 2- bromo 3- chloroheptane 1

2 (e) 3- chloro propanal (f) 2,2- dichloroethanol 4. Which of the following represents the correct IUPAC name for the compounds concerned? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7- Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methyl pentane (d) But-3-yn-1-ol or But-4-ol-1-yne. The correct IUPAC names from each pair is given below: (a) 2,2, dimethyl pentane, (b) 2,4,7-trimethyl octane, (c) 2-chloro-4 methyl pentane. (d) But-3-yn-1-ol 5. Write the formulas for the first five members of each homologous series beginning with the following compounds. (a) H COOH (b) CH 3 COCH 3 Carboxylic acid: H-COOH, CH 3 -COOH, CH 3 -CH 2 -COOH, CH 3 -CH 2 -CH 2 -COOH, CH 3 -CH 2 -CH 2 -CH 2 -COOH Ketones : CH 3 -CO-CH 3, CH 3 -CH 2 -CO-CH 3, CH 3 -CH 2 -CH 2 - CO-CH 3, CH 3 -CH 2 -CH 2 -CH 2 -CO-CH 3, CH 3 - CH 2 -CH 2 -CH 2 -CH 2 -CO-CH 3 6. Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for : (a) 2,2,4-Trimethylpentane (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid Hexanedial Hexanedial BOND LINE FORMULAE CONDENSED FORMULAE FUNCTIONAL GROUP (a) It is Alkane 2

3 Carboxylic acid group and Hydroxyl group (b) Aldehyde group (c) 7. Identify the functional groups in the following compounds: a) 1) Aldehyde group 2) Ether group 3) Hydroxyl group b) 1) Amino group 2) Ester group 3) Substituted Amino group c) 1) Alkenes group 2) Nitro group 8. Which of the two: O 2 NCH 2 CH 2 O or CH 3 CH 2 O is expected to be more stable and why? 3

4 O 2 NCH 2 CH 2 O - is more stable than CH 3 CH 2 O -. The stability of a molecule depends on the resonance stabilization energy which is in turn depends on the more number of contributing structures which a molecule can provide. The molecule O 2 N CH 2 CH 2 O - has more contributing structures. Further O 2 NCH 2 CH 2 O - has more number of unpaired electrons and more number of covalent bonds. Hence it is more stable. 9. Explain why alkyl groups act as electron donors when attached to a π system. Carbon is slightly more electronegative than hydrogen. Thus, carbon atom in an alkyl (say methyl) group has higher electron density around it as compared with an H atom. As a result, alkyl group are able to donate electrons inductively when attached to a π system. 10. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. 4

5 11. What are electrophiles and nucleophiles? Explain with examples. Nucleophiles A nucleophile is a reagent containing an atom having unshared or lone pair of electrons. A nucleophile is electron rich and supplies electrons to electron deficient sites, i.e., (nucleus loving). According to Lewis concepts of acid and base, nucleophiles behave as Lewis bases. Nucleophiles are electron rich species. They carry a negative charge and they are nucleus loving. They attack electron deficient centers. Cl -, I -, OH -, CN -, NO - 2 CH 3 O - etc. are examples of nucleophiles. 5

6 Neutral molecules like H 2 O; R 3 N etc. are also nucleophiles due to the presence of lone pair of electrons. NH 3, H 2 O, ROH, ROR (Neutral species). Electrophiles A positively charged on neutral species which are deficient of electrons and can accept a pair of electrons are called electrophiles. These are also called electron loving (philic) species for examples: H +, H 3 O +, Cl +, CH 3, NO 2 + (positively charged) AlCl 3, BF 3, SO 3 (Neutral species) Both aluminium and boron have total of six electrons i.e., two less than octet. Therefore, they try to complete their octet and act as electrophile. According to Lewis concepts of acids and bases they are called Lewis acids. 12. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) CH 3 COOH + HO - Negative nucleophile CH 3 -COO - + H 2 O (b) CH 3 - CO - CH 3 + NC - CH 3 - C (CN) - OH CH 3 CN- is a negative nucleophile CH 3 CO is a positive electrophile 13. Classify the following reactions in one of the reaction type studied. (a) CH 3 CH 2 Br + HS - CH 3 CH 2 SH + Br - (b) (CH 3 ) 2 C = CH 2 + HCl (CH 3 ) 2 Cl C CH 3 (c) (CH 3 ) 3 C CH 2 OH + HBr (CH 3 ) 2 CBr. CH 2 CH 3 (d) CH 3 CH 2 Br + HO - CH 2 = CH 2 6

7 (a) Nucleophilic substitution (b) Addition reaction (c) Nucleophilic substitution followed by rearrangement reaction. (d) Elimination reaction. 14. What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors? a) They are positional isomers. b) They are geometrical isomers. c) Resonance contributors. d) Geometrical isomers 15. For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion. 7

8 16. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? Inductive effect Inductive effect involves electrons in bonds. It occurs in saturated compounds. The electron pair is only slightly displaced and there are only partial charges. It is transmitted only a quite short distance. Mesomeric effect Mesomeric effect occurs only in unsaturated compounds. It involves electrons in bonds. The electrons pair is transferred completely with the result full positive and negative charges are created. It is transmitted from one end to the other of quite large molecules. In the first case the inductive effect in di and tri halogen substituted acids is more marked with the result these acids are progressively more stronger than the corresponding mono halogeno substituted acid. In the second case, the decreasing acidic strength is due to increase in +I effect due to alkyl groups. 17. Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography Crystallisation This is the most common method for the purification of solid organic compounds. It is based on the fact that certain organic compounds are partly soluble in a solvent at room temperature and solubility increases with increase in temperature. Example: Separation of sugar from a mixture of sugar and common salt by using C 2 H 5 OH. 8

9 Distillation This method is based on the principle that at constant pressure every pure liquid boils at a definite temperature called its boiling point. The method is used for the purification of those liquids which boil without decomposition provided the impurities are non- volatile. The method is applied for the purification when the two liquids differ in their boiling points by K. Example: a mixture of hexane (b.p.342k) and (b.p.384k) is separated by distillation. Chromatography The technique depends on the distribution of the mixture between two phases, one fixed and the other mobile (moving). The fixed phase may bean absorbent column (a solid chemical compound) or a paper strip. The moving phase may be a liquid or gas. The mixture to be separated is dissolved in the moving phase and passed over the fixed phase. There are three types of chromatography. Adsorption Chromatography The fixed phase is a solid, e.g., alumina, magnesium oxide, silica gel, etc. and mobile phase is a liquid. Different constituents are adsorbed in different parts of the adsorbent, the component which has maximum adsorption affinity for the absorbent is adsorbed at the nearest starting point and the component having minimum affinity is adsorbed at the farthest point of the adsorbent. Thus different bonds, zones or chromatograms are formed at different parts of the column. The adsorbed components are then extracted (eluted) with the help of a suitable solvent. Partition Chromatography Here the fixed phase may be solid or liquid strongly adsorbed on a solid support while the mobile phase is liquid. Paper chromatography is a special case of adsorption chromatography in which adsorbent column is a paper strip (solid). Gas Chromatography Here the fixed phase may be solid or liquid while the mobile phase is a mixture of gases. 18. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S. The two compounds with different solubilities in a solvent S can be separated by the method of crystallisation. Crystallization is used to separate two compounds with different solubilities in a solvent S. Less soluble component will crystallize first and more soluble component will crystallize on heating again and then cooling it. 19. What is the difference between distillation, distillation under reduced pressure and steam distillation? Distillation This is used to separate either (a) Volatile liquid from nonvolatile liquid or solid separately or Distillation under reduced pressure This is used to purify liquids which decompose at or below their boiling points. Steam distillation This is used for purifying substances which are steam volatile and immiscible with water. 9

10 (b) Two volatile liquids having their boiling points differing appreciably (greater than 20 0 C). a a 20. Discuss the chemistry of Lassaigne s test. Preparation of Lassaigne's filtrate or sodium fusion extract: This step is common in all the Lassigne's tests,i.e. for nitrogen, sulphur and halogens. A small piece (pea-size) of dry metallic sodium is heated in an ignition (fusion) tube till it melts. A small quantity of the compound is now added to molten metal and the tube is again heated very strongly till it becomes red hot. The red hot tube is broken into pieces by immersing it in cold distilled water in a china dish (if necessary, break the tube with the help of tongs). The contents of the china dish are boiled and then filtered. The filtrate obtained is known as Lassaigne's extract and can be used for testing the presence of N, S and halogens in the given compound. Chemical reactions involved in the formation of Lassaigne's extract. If N is present: Na + C+N NaCN Metallic fusion from organic compound Sodium cyanide If S is present : fusion 2Na + S Na 2 S (sod. Sulphide) If both N and S are present: fusion Na + C + N+ S NaCNS From organic compound Sod. Sulphocyanide If X is present: Na + X NaX (Sod. Halide) where X = Cl, Br or I (a) (i) Test of nitrogen from Lassaigne's filtrate. The filtrate is tested for alkaline nature, if it is not alkaline it is made so by adding a few drops of sodium hydroxide solution. Now 1-2 ml of freshly prepared ferrous sulphate solution is added, the contents are boiled and then cooled. Dilute sulphuric or hydrochloric acid is added to disslove ferrous hydroxide. Appearance of green or prussian blue colour confirms the presence of nitrogen in the organic compound. 10

11 Chemistry of the test: Sodium cyanide formed during the fusion of the compound with sodium reacts with ferrous sulphate to form sodium ferrocyanide, Na 4 [Fe(CN) 6 ]. The latter reacts with ferric sulphate (produced by the oxidation of ferrous sulphate by air in presence of OH - ) and give a precipitate of prussian blue coloured compound, ferric ferrocyanide. 2NaCN + FeSO 4 Fe (CN) 2 + Na 2 SO 4 Ferrous cyanide Fe(CN) 2 + 4NaCN Na 4 [Fe(CN) 6 ] Sod. Ferrocyanide 3Na 4 [Fe(CN) 6 ] + 2Fe 2 (SO 4 ) 3 Fe 4 [Fe(CN) 6 ] 3 + 6Na 2 SO 4 Ferric ferrocyanide (Prussian blue) When compound contains both N and S then sodium sulphocyanide is formed which gives a blood red colour on the addition of ferric chloride solution. 3NaCNS + FeCl 3 Fe(CNS) NaCl Sod. Sulphocyanide Ferric sulphocyanide(blood red) But the absence of blood red colouration does not necessarily mean that sulphur is absent. This is due to the fact that in the presence of excess of sodium, sodium thiocyanate decomposes to form sodium cyanide and sodium sulphide. 2Na + NaCNS Na 2 S + NaCN. 21. Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl s method. (i) Principal of Duma's method A known mass of the given organic compound is heated strongly with excess of cupric oxide in an atmosphere strongly with excess of cupric oxide in an atmosphere of CO 2. Carbon and hydrogen are oxidized to CO 2 and H 2 O respectively while nitrogen present in the compound is set free. A small amount of the oxides of nitrogen if formed are reduced copper gauze. Heat C + 2CuO CO 2 + Cu (Compound) Heat 2H + CuO H 2 O + Cu (Compound) Heat 2N + CuO N 2 + small amount of oxides of -nitrogen. Heat Oxides of nitrogen + Cu N 2 + CuO 11

12 (Reduced) The gaseous mixture is passed through the rough concentrated KOH solution which absorb both CO 2 and H 2 O vapours. Nitrogen (N 2 ) is not absorbed by KOH and gets collected over it. the volume of nitrogen evolved is noted and from this the amount of nitrogen or percentage can be calculated. (ii) Principal of Kjeldahl's method A known mass of the organic compound is digested (heated strongly) with conc. H 2 SO 4 and a little potassium sulphate (to increase the b.p. of H 2 SO 4 ) and a little mmercury or copper sulphate (a catalyst) in a long necked flask called Kjeldahl's flask. As a result of digestion, the nitrogen present in the organic compound is converted into ammonium sulphate. The compound so obtained in the form of a solution ammonia goes thus, liberated is absorbed in a known volume of a standard solution (molarity known) of an acid such as H 2 SO 3 Digertiony N + H 2 SO 4 (NH 4 ) 2 SO 4 (from organic compound) (NH 4 ) 2 SO 4 + 2NaOH Na 2 SO 4 + 2H 2 O +2NH 3 2NH 3 + H 2 SO 4 (NH 4 ) 2 SO 4 The volume of the acid left unused is determined by titrating it against a standard alkali solution such as NaOH. H 2 SO 4 + 2NaOH Na 2 SO 4 + H 2 O. 22. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound. Principle of estimation of halogens Carius method involves conversion of the halogen present in any organic compound into the corresponding silver halide. From the mass of the organic compound taken and the mass of silver halide formed, the percentage of halogen in the compound can be calculated. Principle of estimation of sulphur Sulphur present in any organic compound is oxidized to sulphuric acid by disgesting with fuming nitric acid. Sulphuric acid so formed is then quantitatively precipitated as barium sulphate by the addition of excess barium chloride. The precipitate of barium sulphate is filtered, washed, dried and weighed. Principle of estimation of phosphorus The phosphorous present in an organic compound is oxidized to phosphoric acid by heating with fuming nitric acid. The phosphoric acid so obtained is precipiitated as MgNH 4 PO 4, which on ignition is convered into Mg 2 P 2 O 7. Then, from the mass of Mg 2 P 2 O 7, the mass of phosphorous present in the sample of the organic compound can be calculated. 23. Explain the principle of paper chromatography. 12

13 Paper chromatography is a special type of partition chromatography. It is based on the differential (continuous) partitioning of components of a mixture between stationary phase and mobile phase. In paper chromatography, a strip of special paper spotted at the base with the solution of the mixture is suspended in a suitable solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in the two phases. The paper strip so developed is known as chromatogram. The spots of the separate coloured compounds are visible at different heights from the position of initial spot on the chromatogram. The spot of the separate colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent. 24. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens? When we test halogens from sodium extract, a small amount of dil. HNO 3 is added. When we test from Lassaigne s extract, it contains halogens. But if nitrogen and sulphur are also present along with halogens in the organic compound, the Lassaigne s extract contains sodium sulphide and sodium cyanide (Na 2 S and NaCN) along with sodium halide. Nitric acid decomposes sodium cyanide and sodium sulphide which otherwise form precipitate with silver nitrate and thus interfere with the test. NaCN + HNO 3 NaNO 3 + HCI Na 2 S + 2HNO 3 2NaNO 3 + H 2 S Therefore, dil nitric acid is added before testing halogens to expel all the gases if evolved. 25. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens. The elements, nitrogen, sulphur and halogens are covalently bonded to carbon atoms in the organic compounds. To make their detection easy, these are to be first converted into the ionic forms. This is done by fusing the organic compound with sodium metal. The ionic compounds formed during the fusion are extracted in aqueous solution, and can be detected by simple chemical tests. Fusion extracted C+N+ Na NaCN +Na + (aq) + CN - (aq) From organic compound sodium in water cyanide Fusion X (Cl, Br, I) + Na NaX (X= Cl, BrI) Na + (aq) + X - (aq) Extracted in water Fusion extracted 13

14 S + 2Na Na 2 S 2Na + (aq) + S 2- (aq) From organic in Water compound sodium sulphide If nitrogen and sulphur both are present in any organic compound, sodium thiocyanate (NaSCN) is formed during fusion which is the presence of excess sodium, forms sodium cyanide and sodium sulphide. Fusion extracted Na + C + N + S NaCNS Na + (aq) + CNS - (aq) In the organic Compound Sodium in water thiocyanate NaCNS + 2Na NaCN + Na 2 S 3Na + (aq) + CN - (aq) + S 2- (aq) extracted in water Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor. Calcium sulphate and camphor present in a mixture can be separated by sublimitation. Camphor sublimes in heating and can be recovered in crystalline form as sublimate. 26. Explain, why an organic liquid vaporises at a temperature below its boiling point in steam distillation? In steam distillation, the liquid starts boiling when sum of the vapour pressures of the organic liquid (p 2 ) and that due to water vapour (p 1 ) becomes equal to the atmosphere (p) i.e., P= p 1 + p 2 Since a liquid boils as soon as its vapour pressure becomes equal to the atmospheric pressure, a mixture of two immiscible liquids will boil at a temperature lower than the normal boiling points of both the liquids. The mixture will continue to boil at the same temperature until one of the liquids is completely distilled out. 27. Will CCl 4 give white precipitate of AgCl on heating it with silver nitrate? Give a reason for your answer. CCl 4 will not give white precipitate of AgCl due to the absence of electron releasing group on the carbon atom bonded to the Cl atoms (CCl 4 is a non-polar solvent.) 28. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound? During the estimation of percentage of carbon in the organic compound, carbon is oxidised by heating the substance in excess of oxygen. 14

15 C + O 2 CO 2 CO 2, being acidic oxide is absorbed by KOH. So CO 2 formed in the estimation of carbon, is absorbed in a known weight of KOH. From the increase in the weight of KOH, percentage of carbon in the organic substance is estimated. 29. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test? For testing sulphur from sodium extract, it is acidified with acetic acid but not sulphuric acid. Because SO 4 2- ions from H 2 SO 4 reacts with Pb ++ from Lead Acetate, giving a white ppt of PbSO 4, though sulphur is not present in the organic compound. H 2 SO 4 2H + + SO 4 2- Pb SO 4 (from Pb acetate) PbSO An organic compound contains 69% carbon and 4.8% hydrogen and the remaining percentage of oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion. Percentage of carbon = 69% Percentage of hydrogen = 4.8% Mass of substance = 0.2g Mass of CO 2, formed C + O 2 CO 2 12g 44g Amount of carbon present in 0.2g of the substance = = 0.138g 12g of carbon on complete combustion gives = 44 g of CO g of carbon gives = x = g of CO 2 Mass of water formed Amount of hydrogen, present in 0.2 g of the substance = = g 2H 2 + O 2 4g 2H 2 O 36g 4g of H 2 on complete combustion gives = 36g of water g of H 2.. = g of water. 31. A sample of 0.50 g of an organic compound was treated according to Kjeldahl s 15

16 method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 ml of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. Mass of organic compound = 0.50 g Initial volume of acid = 50 ml Molarity of acid = 0.5 M Volume of NaOH of 0.5 M molarity required for neutralization of residual acid = 60 ml = 30 ml of 0.5 M H 2 SO 4 Volume of H 2 SO 4 (0.5 M) consumed = (50-30) = ml of 0.5 M H 2 SO 4 = 40 ml of 0.5 M NH 3 or 20 ml of 1 M NH ml of 1 M NH 3 contains 14 g Nitrogen (1 mol of N 2 ) 20 ml of 1 M NH 3 contains g Nitrogen 0.5g of organic compound contains g N 100 g of organic compound contains = g= 56% of N g of an organic chloro compound gave g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. Mass of silver chloride = g Molecular mass of silver chloride = = g of silver chloride contains = 35.5 g.5740 g of AgCl will contain = =.142 g Percentage of chlorine = x 100 = % 33. In the estimation of sulphur by Carius method, g of an organic sulphur compound afforded g of barium sulphate. Find out the percentage of sulphur in the given compound. Mass of barium sulphate = g Molecular mass of barium sulphate = 233 g 233 g of barium sulphate contains sulphur = 32g g of barium sulphate contains = 16

17 = = 19.66% 34. In the organic compound CH 2 = CH - CH 2 - CH 2 - C = CH, the pair of hybridised orbitals involved in the formation of: C2 - C3 bond is: (a) sp - sp 2 (b) sp - sp 3 (c) sp 2 - sp 3 (d) sp 3 - sp 3 sp 3 sp In the Lassaigne s test for nitrogen in an organic compound the Prussian blue colour is obtained due to the formation of: Na 4 [Fe(CN) 6 ] Fe 4 [Fe(CN) 6 ] 3 Fe 2 [Fe(CN) 6 ] Fe 3 [Fe(CN) 6 ] 4 Fe 4 [Fe(CN) 6 ] Which of the following carbocation is most stable? a) (CH 3 ) 3 C.+CH 2 b) (CH 3 ) 3 C + c) CH 3 CH 2 +CH 2 d) CH 3 +CHCH 2 CH 3 b) (CH 3 ) 3 C The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation 17

18 (d) Chromatography (d) Chromatography. 38. The reaction is classified as (a) electrophilic substitution reaction (b) nucleophilic substitution reaction (c) elimination reaction (d) addition reaction (b) nucleophilic substitution reaction. 18