12.1.  What are hybridisation states of each carbon atom in the following compounds? CH₂=C=O, CH₃CH=CH₂, (CH₃)₂CO, CH₂=CHCN, C₆H₆.

Answer:

12.2. Indicate the a- and n-bonds in the following molecules:
C₆H₆ , C₆H₁₂, CH₂Cl₂, CH=C=CH₂, CH₃NO₂, HCONHCH₃

Answer:

 

12.3. Write bond-line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

Answer:

12.4. Give the IUPAC names of the following compounds:

Answer: 

(a) Propylbenzene (b) 3-Methylpentanenitrite (c) 2, 5-Dimethylheptane
(d) 3-Bromo- 3-chloroheptane (e) 3-Chloropropanal (f) 2, 2-Dichloroethanol

12.5.Which of the following represents the correct TUPAC 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-methylpentane (d) But-3-yn- l-ol or But-4-ol-yne.

Answer:

(a) 2, 2-Demethylpentane (b)2, 4, 7-Trimethyloctane. For two alkyl groups on the same carbon its locant is repeated twice, 2, 4, 7-locant set is lower than 2, 5, 7.
(c) 2- Chloro-4-methylpentane. Alphabetical order of substituents, (d) But-3-yn-l-ol. Lower locant for the principal functional group, i.e., alcohol.

12.6. Draw formulas for the first five members of each homologous series beginning with the following compounds,
(a) H—COOH (b) CH₃COCH₃ (c) H—CH=CH₂

Answer: 

(a) CH₃—COOH
CH₃CH₂—COOH CH₃CH₂CH₂—COOH
CH₃CH₂CH₂CH₂—COOH
(b) CH₃COCH₃
CH₃COCH₂CH₃
CH₃COCH₂CH₂CH₃
CH₃COCH₂CH₂CH₂CH₃
CH₃CO(CH₃)₄CH₃
(c) H—CH=CH₂
CH₃CH=CH₂
CH₃CH₂CH=CH₂
CH₃CH₂CH₂CH=CH₂
CH₃CH₂CH₂CH₂CH=CH₂

12.7. 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-l, 2, 3-propanetricarboxylic acid (c) Hexanedial.

Answer:

12.8. Identify the functional groups in the following compounds:

Answer:

12.9. Which of the two: O₂NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?

Answer: 

O₂N——<——- CH₂——<——- CH₂ —<——- O^– is more stable than CH₃——<——-CH₂——<——-O- because NO₂ group has -I-effect and hence it tends to disperse the -ve charge on the O-atom. In contrast, CH₃CH₂ has +I-effect. It, therefore, tends to intensify the -ve charge and hence destabilizes it.

12.10. Explain why alkyl groups act as electron donors when attached to a π-system.

Answer: 

Due to hyperconjugation, alkyl groups act as electron donors when attached to a π- system as shown below:

12.11. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. (a) C₆H₅OH (b) C₆H₅N0₂ (c)  CH₃CH=CHCHO (d) C₆H₅—CHO (e) C₆H₅—CH₂ (f) Ch₃Ch=ChCh₂

Answer:

 

12.12. What are electrophiles and nucleophiles? Explain with examples:

Answer: 

Electrophiles: The name electrophiles means electron loving. Electrophiles are electron deficient. They may be positive ions or neutral molecules.
Ex: H⁺, Cl⁺, Br⁺, NO₂⁺, R₃C⁺, RN₂⁺, AlCl₃, BF₃
Nucleophiles: The name nucleophiles means ‘nucleus loving’ and indicates that it attacks the region of low electron density (positive centres) in a substrate molecule. They are electron rich they may be negative ions or neutral molecules.
Ex: Cl^– Br^–, CN^–, OH^–, RCR₂^–, NH₃, RNH₂, H₂O, ROH etc.

12.13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles
(a) CH₃COOH + HO^– ———–> CH₃COO^– + H₂O
(b) CH₃COCH₃ + CN ———–> (CH₃)₂ C(CN)(OH)
(c) C₆H₅ + CH₃CO ———–> C₆H₅COCH₃

Answer:

Nucleophiles: (a) and (b) and Electrophile : (c)

12.14. Classify the following reactions in one of the reaction type studied in this unit.
(a) CH₃CH₂Br + HS^– ———–> CH₃CH₂SH + Br^–
(b) (CH₃)₂C=CH₂ + HCl ———–> (CH₃)₂CCl—CH₃
(c) CH₃CH₂Br + HO^– ———–> CH₂=CH₂ + H₂O + Br^–
(d) (CH₃)₃C—CH2OH + HBr ———–> (CH₃)₂ C Br CH₂CH₂CH₃ + H₂O

Answer: 

(a) Nucleophilic substitution (b) Electrophilic addition
(c)Bimolecular elimination (d) Nucleophilic substitution with rearrangement.

12.15. What is the relationship between the members of following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Answer: 

(a) Structural isomers (actually position isomers as well as metamers)
(b) geometrical isomers
(c) resonance contributors because they differ in the position of electrons but not atoms

12.16. 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.

Answer:

12.17. Explain the terms inductive and electromeric effects. Which electron displacement effect explain the following correct orders of acidity of the carboxylic acids?
(a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂ COOH
(b) CH₃CH₂COOH > (CH₃)₂ CHCOOH > (CH₃)₃CCOOH

Answer: 

Inductive Effect: The inductive effect refers to the polarity produced in a molecule as a result of higher electronegativity of one atom compared to another.Atoms or groups which lose electron towards a carbon atom are said to have +1 Effect.
Those atoms or groups which draw electron away from a carbon atom are said to have -I Effect.
Commomexamples of -I effect are:
NO₂, F, Cl, Br, I, OH etc.
Examples of +1 effect are (Electron releasing)
(CH₃)₂C— , (CH₃)₂CH—, CH₃CH₂— CH₃— etc.
Electromeric effect: The electromeric effect refers to the polarity produced in a multiple bonded compound as it is approached by a reagent.

The atom A has lost its share in the electron pair and B has gained this share.
As a result A acquires a positive charge and B a negative charge. It is a temporary effect and takes place only in the presence of a reagent.
(a) -I-effect as shown below:
As the number of halogen atoms decreases, the overall -I- effect decreases and the acid strength decreases accordingly.

(b) +I-effect as shown below:
As the number of alkyl groups increases, the +I-effect increases and the acid strength
decreases accordingly.

12.18. Give a brief description of the principles of the following techniques taking an example in each case: (a) Crystallisation (b) Distillation (c) Chromatography

Answer:

(a) Crystallisation: In this process the impure solid is dissolved in the minimum volume of a suitable solvent. The soluble impurities pass into the solution while the insoluble ones left behind. The hot solution is then filtered and allowed to cool undisturbed till crystallisation is complete. The crystals are then separated from the mother liquor by filtraration and dried.
Example: crystallisation of sugar.
(b) Distillation: The operation of distillation is employed for the purification of liquids from non-volatile impurities. The impure liquid is boiled in a flask and the vapours so formed are collected and condensed to give back pure liquid in another vessel. Simple organic liquids such as benzene toluene, xylene etc. can be purified.
(c) Chromatography: Chromatography is based on the principle of selective distribution of the components of a mixture between two phases, a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. When the stationary phase is solid the basis is adsorption and when it is a liquid the basis is partition. Chromatography is generally used for the Reparation of coloured substances such as plant pigments or dyestuffs.

12.19. Describe the method, which can be used to separate two compounds with different solubilities in a solvent S.

Answer:

Fractional crystallisation is used for this purpose. A hot saturated solution of these two compounds is allowed to cool, the less soluble compound crystallises out while the more soluble remains in the solution. The crystals are separated from the mother liquor and the mother liquor is again concentrated and the hot solution again allowed to cool when the crystals of the second compound are obtained. These are again filtered and dried.

12.20. What is the difference between distillation, distillation under reduced pressure and steam distillation?

Answer:

Distillation is used in case of volatile liquid mixed with non-volatile impurities.
Distillation under reduced pressure: This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.
Steam distillation is used to purify steam volatile liquids associated with water immiscible impuritites.

12.21. Discuss the chemistry of Lassaigne’s test.

Answer:  

Lassaigne’s test: Nitrogen, sulphur, halogens and phosphorous present in an organic compound are detected by Lassaigne’s test.
First of all compounds are converted to ionic form by fusing the compound with sodium metal.

Cyanide, sulphide or halide of sodium are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.

12.22. Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method (ii) Kjeldahl’s method.

Answer: 

(i) Dumas method: The organic compound is heated strongly with excess of CuO ‘ (Cupric Oxide) in an atmosphere of CO₂ when free nitrogen, CO₂ and H₂O are obtained.
(ii)Kjeldahl’s method: A known mass of the organic compound is heated strongly with cone. H₂SO₄, a little potassium sulphate and a little mercury (a catalyst). As a result of reaction the nitrogen present in the organic compound is converted to ammonium sulphate.

12.23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Answer:  

Estimation of halogens: It involves oxidising the organic substance with fuming nitric acid in the presence of silver nitrate. The halogen of the substance is thus converted to silver halide which is separated and weighed:
1Weight of organic compound = W gm
weight of silver halide = x g.

Estimation of sulphur: The organic substance is heated with fuming nitric acid but no silver nitrate is added. The sulphur of the substance is oxidised to sulphuric acid which is then precipitated as barium sulphate by adding excess of barium chloride solution. From the weight of BaSO₄ so obtained the percentage of sulphur can be calculated.

Estimation of phosphorous: The organic substance is heated with fuming nitric acid whereupon phosphorous is oxidised to phosphoric acid. The phosphoric acid is precipitated as ammonium phosphomolybdate, (NH₄)₃ PO₄ .12MOO₃, by the addition of ammonia and ammonium molybdate solution which is then separated, dried and weighed.

12.24. Explain the principle of paper chromatography.

Answer: 

This is the simplest form of chromatography. Here a strip of paper acts as an adsorbent. It is based on the principle which is partly adsorption. The paper is made of cellulose fibres with molecules of water adsorbed on them. This acts as stationary phase. The mobile phase is the mixture of the components to be identified prepared in a suitable solvent.

12.25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens ?

Answer: 

Nitric acid is added to sodium extract so as to decompose
NaCN + HNO₃ ——-> NaNO₃ + HCN
Na₂S + 2HNO₃ ——> 2NaNO₃ + H₂S

12.26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Answer: 

Organic compound is fused with sodium metal so as to convert organic compounds into NaCN, Na₂S, NaX and Na₃PO₄. Since these are ionic compounds and become more reactive and thus can be easily tested by suitable reagents.

12.27. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Answer: 

Sublimation.Because camphor can sublime whereas CaSO₄ does not.

12.28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ?

Answer: 

It is because in steam distillation the sum of vapour pressure of organic compound and steam should be equal to atmospheric pressure.

12.29. Will CCl₄ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Answer: 

No. CCl₄ is a completely non-polar covalent compound whereas AgNO₃ is ionic in nature. Therefore they are not expected to react and thus a white ppt. of silver chloride will not be formed.

12.30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Answer: 

CO₂ is acidic in nature and therefore, it reacts with the strong base KOH to form K₂CO₃.
2KOH + CO₂ ——–> K₂CO₃+ H₂O.

12.31. Why is it necessary to use acetic acid and not sulphric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Answer: 

For testing sulphur sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test.

12.32. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is subjected to complete combustion.

Answer: 

12.33. 0.50 g of an organic compound was Kjeldahlished. The ammonia evolved was passed in 50 cm³ of IN H₂SO₄. The residual acid required 60 cm³ of N/2 NaOH solution. Calculate the percentage of nitrogen in the compound.

Answer: 

12.34. 0.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.

Answer: 

Mass of the compound = 0.3780 g
Mass of silver chloride = 0.5740 g

12.35. In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.

Answer: 

Mass of the compound = 0.468 g
Mass of barium sulphate= 0.668 g

12.36.

Answer:

12.37. In the Lassaigne’s test for ntrogen in an organic compound, the Prussian blue colour is obtaine d due to the formation of:
(a) Na₄[Fe(CN)₆] (b)Fe₄[Fe(CN)₆]₃
(c) Fe₂[Fe(CN)₆] (d)Fe₃[Fe(CN)₆]₄ .

Answer: 

(b) is the correct answer.

12.38. Which of the following carbocation is most stable?

Answer:

(b) is the most stable since it is a tertiary carbocation.

12.39. The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography.

Answer:

(d) is the correct answer.

12.40. The following reaction is classified as:
CH₃CH₂I + KOH (aq) ———-> CH₃CH₂OH + KI
(a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition

Answer:

(b) It is a nucleophilic substitution reaction. KOH (aq) provides OH- ion for the nucleophile attack.