1. How do you account for the formation of ethane during chlorination of methane?

Transcription

1 1. How do you account for the formation of ethane during chlorination of methane? The formation of ethane is due to the side reaction in termination step by the combination of two CH 3 free radicals. 2. Write IUPAC names of the following compounds: (a) 2-Methyl - but-2-ene (b) Pent-1-ene-3-yne (c) Buta 1, 3-diene. (d) 4-Phenyl-but-1-ene (e) 2-Methyl phenol (f) 5-(2-Methyl propyl)-decane (g) 4-Ethyl-deca-1, 5, 8-triene. 3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated : (a) C 4 H 8 (one double bond) (b) C 5 H 8 (one triple bond) 1

2 (a) (i) CH 2 = CH - CH 2 - CH 3 But - 1-ene (ii) CH 3 - CH 2 = CH - CH 3 But - 2-ene (iii) 2-Methyl propene (b) (i) HC C - CH 2 - CH 2 - CH 2 Pent - 1-yne (ii) CH 3 - C C - CH 2 - CH 2 Pent - 2-yne (iii) 3-Methyl-but- 1-yne. 4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (i) Pent-2-ene (ii) 3,4-Dimethyl-hept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene (i) Ethanal and propanal. (ii) Butan-2-one and pentan-2-one. (iii) Methanal and pentan-3-one. (iv) Propanal and benzaldehyde. 5. An alkene A on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the and IUPAC name of A. The IUPAC name of A is 3-Ethyl-pent-2-ene. 6. An alkene A contains three C C, eight C H bonds and one C C p bond. A on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of A. The IUPAC name of A is But 2 ene 7. Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the 2

3 structural formula of the alkene? 4-Ethyl-hex-3-ene 8. Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene (a) C 4 H 10 (g) + 4CO 2 (g) +5H 2 O(g) (b) C 5 H 10 (g) + 5CO 2 (g) +5H 2 O(g) (c) C 6 H 10 (g) + 6CO 2 (g) +5H 2 O(g) (d) C 7 H 8 (g) +5O 2 (g) 7CO 2 (g) +4H 2 O(g). 9. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why? The cis and trans structures of hex-2-ene The cis form will have higher boiling point. This is due to more polar nature leading to strong intermolecular dipole-dipole interaction, thus requiring more heat energy to separate them. 10. Why is benzene extra ordinarily stable though it contains three double bonds? 3

4 Benzene is extra ordinarily stable due to resonance. 11. What are the necessary conditions for any system to be aromatic? The essential conditions for any system to be aromatic are Planar, conjugated ring system with delocalisation of (4π + 2) p electrons, where n is an integer. 12. Explain why the following systems are not aromatic? The compounds are not aromatic due to lack of cyclic cloud p-electrons having (4n + 2) p electrons. 13. How will you convert benzene into (i) p-nitrobromobenzene (ii) m- nitrochlorobenzene (iii) p - nitrotoluene (iv) acetophenone? 4

5 14. In the alkane H 3 C CH 2 C(CH 3 ) 2 CH 2 CH(CH 3 ) 2, identify 1,2,3 carbon atoms and give the number of H atoms bonded to each one of these. 6H attached to 1 0 carbons 4H attached to 2 0 carbons 1H attached to 3 0 carbons 15. What effect does branching of an alkane chain has on its boiling point? The boiling point will be lower if there are more branching in alkane. 16. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene? The result tells us that all the three products cannot be obtained by any one of the Kekule s structures showing that benzene is a resonance hybrid of the two resonating structure. 17. Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. 5

6 H C C H > C 6 H 6 > C 6 H 14 The acidic property is maximum in ethyne due to maximum s orbital character in ethyne (50 percent) as compared to 33 percent in benzene and 25 percent in n-hexane. 18. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? Due to the presence of p -electrons in benzene it can undergo electrophilic substitution reactions easily. In Nucleophilic substitutions the nucleophile will be repelled by p -electrons. So benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty. 19. How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane 20. Write structures of all the alkenes which on hydrogenation give 2-methylbutane. 21. Out of benzene, m dinitrobenzene and toluene which will undergo nitration most easily and why? 6

7 Due to electron releasing nature of the methyl group toluene undergoes nitration most easily. 22. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene. Another Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene is FeCl Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer bytaking one example. Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms because of the formation of side products. For example, by starting with 1-bromo propane and 1- bromo butane, hexane and octane are the side products besides heptane. 7