Grade XII Organic Reasoning Questions with Solutions PDF

Summary

This document is a collection of organic chemistry questions and solutions. Questions cover various topics, including reactions of alcohols with KI, solubility of alkyl halides in water, reactions of haloalkanes with KCN and AgCN, SN2 reactions, and the preparation of Grignard reagents.

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GRADE XII ORGANIC REASONING QUESTIONS WITH SOLUTIONS 1. Why is sulphuric acid not used during the reaction of alcohols with KI? Ans. Sulphuric acid oxides KI to HI then to iodine. Hence it cannot be used during the reaction of alcohols with KI. 2. Alkyl halides though polar are immiscible...

GRADE XII ORGANIC REASONING QUESTIONS WITH SOLUTIONS 1. Why is sulphuric acid not used during the reaction of alcohols with KI? Ans. Sulphuric acid oxides KI to HI then to iodine. Hence it cannot be used during the reaction of alcohols with KI. 2. Alkyl halides though polar are immiscible with water. Give reason. Ans. In order for halo alkanes to dissolve in water, energy is required to overcome the attractions between the haloalkane molecules and break the hydrogen bonds between the water molecules. Less energy is released when new attractions are set up between the haloalkane and the water molecules as these are not as strong as the original hydrogen bonds in water. Hence alkyl halides are less soluble in water. OR Alkyl halides cannot form hydrogen bonds with water molecules, hence they are immiscible in water. 3. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain. Ans. KCN is ionic in nature and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through carbon atom and not through nitrogen atom since C-C bond is more stable than C-N bond. Whereas, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product. 4. In the following pairs of halogen compounds, which would undergo SN2 reaction faster and why? (i) CH2Cl OR Cl (a) (b) Ans. (a) will undergo SN2 reaction faster since it is a primary halide and has less steric hindrance. The order of reactivity is 1o > 2o > 3o (ii) OR Cl (a) I (b) Ans. (a) will undergo SN2 reaction faster since iodine is a better leaving group due to its large size. 5. Grignard reagents should be prepared only under anhydrous condition. Why? Ans. Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. 1 RMgX + H2O RH + Mg(OH)X 6. Explain why the dipole moment of chlorbenzene is lower than that of cyclo hexyl chloride. Ans. Cl Cl Dipole moment = Charge X distance (i) Cl is bonded to more electronegative sp2 hybridised carbon in chloro benzene, due to which the partial negative charge on chlorine decreases. (ii) The lone pair of electons on chlorine in chloro benzene is in conjugation with pi electrons of the ring. Due tothis C-Cl bond acquires partial double bond character and thereby decreasing the bond length. 7. Out of C6H5CH2Cl and C6H5CHClC6H5 which is more easily hydrolysed by aqueous KOH? + Ans. C6H5CH2Cl C6H5CH2 ----------------- 1 + C6H5CHClC6H5 C6H5CHC6H5 --------------- 2 The second carbo cation is stabilized by two phenyl groups compared to first carbo cation, which is stabilised by one phenyl group. Therefore C6H5CHClC6H5 is more easily hydrolysed. 8. P-Chloro benzene has higher melting point than those of ortho and meta isomers. Discuss. Ans. This is due to symmetry which fit into the crystal lattice better. 9. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are the major products. Explain. Ans. Aqueous solution of KOH is almost completely ionised to give OH- ions. OH- ions are strong nucleophiles as well as bases. As OH- ions in aqueous solution are highly hydrated, the basic character of OH- ions are reducded, hence they bring about nucleophilic substitution reaction and form alcohols. Whereas alcoholic solution of KOH contains, alkoxide ions (RO-) which being much stronger base than OH- ions eliminates a molecule of HX from alkyl halide to form alkene. 10. Allylic and benzylic halides show high reactivity towards SN1 reaction. Explain. Ans. SN1 reaction proceeds through the formation of carbo cation. Any factor which results in stabilising the carbo cation, increases the reactivity through SN1 path way. The carbo cation formed from allylic and benzylic halide viz allylic carbo cation and benzyl carbo cation are stabilised by resonance. Therefore they show greater reactivity. 11. Aryl halides are less reactive towards nucleophilic substitution reaction. Why? 2 Ans. This is due to the following reasons: (i) The lone pair of electrons on halogen is in conjugation with pi electrons of the ring. Due to this C-Cl bond acquires partial double bond character. (ii) In halo arene, the halogen atom is bonded to more electro negative sp2 hybridised carbon atom, which holds the electron pair of C-X bond more tightly, thereby decreasing the bond length. (iii) In halo arene, the phenyl carbo cation formed will not be stabilised by resonance. Therefore SN1 mechanism is ruled out. (iv) It is difficult for an electron rich nucleophile to approach an electron rich arene due to repulsion. 12. Although chlorine is electron with drawing group, yet it is ortho, para directing in electrophilic aromatic substitution reactions. Why? Ans. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution. Through resonance, halogen tends to stabilise the carbocation and the effect is ore pronounced at ortho and para- positions. The inductive effect is stronger than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and para- positions and hence makes the deactivation less for ortho- and para- attack. Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect. 13. NO2 group at meta position does not increase reactivity in halo arene. Explain. Ans. When nucleophile OH- attacks chloro benzene – carbanion is formed which is stabilised by resonance. The negative charge appearing at ortho and para positions is stabilised by – NO2 group, while in the case where - NO2 group in meta position, none of the resonating structures bear the negative charge on carbon atom bearing - NO2 group. Therefore the presence of - NO2 group at meta position does not stabilise the negative charge and hence no effect on reactivity is observed. 14. Why is chloroform stored in brown bottles? Ans. It is stored in brown bottles because it gets oxidised in the presence of sunlight to form poisonous gas phosgene. 2 CHCl3 + O2 2COCl2 + 2HCl 15. Explain why propanol has higher boiling point than that of the hydrocarbon butane. Ans. Propanol molecules are held by intermolecular hydrogen bonding whereas hydrocarbons are held by Vander waals forces. Hence propanol has higher boiling point. 16. Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain. 3 Ans. Alcohols are soluble in water as they form hydrogen bond with water molecules. 17. Phenols are more acidic than alcohols. Give reason. Ans. Phenol is more acidic than alcohol due to the following reasons: (i) The ionisation of an alcohol and phenol takes place as follows: ROH R-O- + H+ Alkoxide ion OH O- + H+ Phenoxide ion -OH group in phenol is connected to more electronegative sp2 hybridised carbon atom, due to which the electron density decreases on oxygen. This increases the polarity of O-H bond and results in an increase in ionisation of phenol, thereby increasing the acidity in phenol. (ii) In Alkoxide ion, the negative charge is localised on oxygen while in Phenoxide ion, the charge is delocalised. As the Phenoxide ion is stabilised by resonance, Phenoxide ion is more stable than Alkoxide ion and hence phenol is more acidic. 18. Compare the acidity of ortho nitro phenol and ortho methoxy phenol. Ans. Presence of electron withdrawing group like nitro group increases the polarity of O-H bond and thereby increases acidity in nitro phenol. Whereas electron donating methoxy group decreases the polarity of O-H bond, thereby decreasing acidity in methoxy phenol. 19. Ortho nitro phenol is steam volatile while para nitro phenol is less volatile. Give reason. Ans. Ortho nitro phenol is steam volatile due to intra molecular hydrogen bonding while para nitro phenol is less volatile due to inter molecular hydrogen bonding which causes the association of molecules. 20. Give two reactions that show the acidic nature of phenol. Ans. (i) Phenol on reaction with active metals such as sodium, potassium yields corresponding pheoxide liberating hydrogen. C6H5OH + Na C6H5ONa + H2 (ii)Phenol on reaction with aqueous sodium hydroxide forms, sodium phenoxide. C6H5OH + NaOH C6H5ONa + H2O The above reactions show that phenol is acidic in nature. 4 21. Explain how does the –OH group attached to a carbon of benzene ring activate it towards electrophilic substitution? Ans. The lone pair of electrons on –OH group is involved in resonance with the pi electrons of the ring (+R effect). Due to this electron density increases at ortho and para positions thereby facilitating electrophilic substitution at ortho and para positions. 22. Give reason for the higher boiling point of ethanol in comparison to methoxymethane. Ans. Ethanol undergoes intermolecular hydrogen bonding whereas methoxy methane does not undergo hydrogen bonding. Hence the boiling point of ethanol is higher than methoxy methane. 23. Illustrate with example the limitations of Williamson synthesis for the preparation of certain types of ethers. Ans. (i) Williamson’ synthesis cannot be used for tertiary alkyl halides as they yield alkenes instead of ethers. CH3 CH3-C-Br + Na-O-CH3 CH3-C=CH2 + NaBr CH3 CH3 (ii) Aryl halides and vinyl halides cannot be used as substrates because of their low reactivity in nucleophilic substitution. 24. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain. Ans. The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance and hence it is less reactive than propanal. 25. Aldehydes are generally more reactive than ketones in nucleophilic addition reactions. Why? Ans. Aldehydes are more reactive than ketones due to steric and electronic reasons. 26. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why? Ans. (i) Phenoxide ion is stabilised by unequal resonating structures while carboxylate ion is stabilised by two equivalent resonating structures. (ii) In the resonating structures of phenoxide ion, the negative charge is on less electronegative carbon atom, while in the resonating structures of carboxylate ion, the negative charge is on more electronegative oxygen atom. Therefore carboxylate ion is more stabilised than phenoxide ion. Hence carboxylic acid is stronger acid than phenol. 27. Give plausible explanation for each of the following: (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-Trimethylcyclohexanone does not. 5 Ans. Due to the presence of three methyl groups at α-positions, the nucleophilic attack by CN- ion does not occur due to steric hindrance. While in cyclohexanone, there is no steric hindrance and hence it forms cyanohydrin in good yield. (ii) There are two –NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazones. Ans.Semicarbazide has two –NH2 groups but one of them is involved in resonance. Thus, electron density on –NH2 group decreases hence it does not act as a nucleophile. While the lone pair of electron on other –NH2 group is not involved in resonance. Hence it is available for nucleophilic attack on C=O group of aldehyde and ketones. (iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester formed should be removed as soon as it is formed. Ans. The formation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst is a reversible reaction. In order to shift the equilibrium in the forward direction, the water or the ester formed should be removed as fast as it is formed. H2SO4 RCOOH + R’OH RCOOR’ + H2O 28. Mention the disadvantage of ammonolysis process. Ans. This method of preparing amines has a disadvantage of yielding a mixture of primary, secondary, tertiary amines and quaternary ammonium salts. 29. Aniline or aromatic amine cannot be prepared by Gabriel phthalimide synthesis. Give reason. Ans. Aniline cannot be prepared because of the fact that the aryl halides do not undergo nucleophilic substitution reaction under ordinary conditions. Hence, aryl halides do not react with potassium phthalimide to give N-phenylphthalimide. 30. Compare the basicity of alkanamine and ammonia. Ans. Alkyl amines are stronger bases than ammonia. The reaction of alkyl amine and ammonia with proton can be given as: H H + R N: + H R-N+-H H H H H H N: + H+ R-N+-H H H The electron releasing alkyl group due to +I effect stabilises the cation formed and also increases the availability of electron on nitrogen, thereby making it available for protonation. 31. Aniline is less basic than ammonia. Give reason. Ans. The less basic nature of aniline can be explained on the basis of resonance. 6 (i) The lone pair of electron on nitrogen is in conjugation with the aromatic ring and thus making it less available for protonation. (ii) Anilinium ion formed by protonation of aniline can be stabilised by only two resonating structures while aniline is stabilised by more no. Of resonating structures, aniline is more stable than anilinium ion. 32. Compare the basic character of primary, secondary and tertiary amines in gas phase. Ans. The electron releasing +I effect is more in tertiary amine due to the presence of three alkyl groups. This increases the protonation of tertiary amines. Hence the order of basicity in gas phase is, 3o>2o>1o 33. Give reason for the following basic strength order in gas phase: (i) (CH3)2NH > CH3NH2 >(CH3)3N Ans. The basic strength of amine is decided by overall effect of +I effect, hydration effect, and steric effect. The combination of all these effects favour, secondary amine to be more basic. In methyl amine series, the steric hinderance to hydrogen bonding is least. In this case, hydrogen bonding predominates over the stability due to +I effect and therefore the order of basic strength is, 2o > 1o >3o (ii) (C2H5)2NH > (C2H5)3N > C2H3NH2 Ans. The basic strength of amine is decided by overall effect of +I effect, hydration effect, and steric effect. The combination of all these effects favour, secondary amine to be more basic. In ethyl amine, there is considerable steric hinderance to hydrogen bonding. Thus, in this case +I effect predominates over the stability due to hydrogen bonding. Hence in this case the basic strength is, 2o > 3o >1o 34. Aniline does not undergo Friedel-Crafts reaction. Give reason. Ans. This is because aniline reacts with aluminium chloride and forms salt. Due to this, nitrogen atom in aniline acquires positive charge and hence acts as a strong deactivating group for further reaction. 35. Diazonium salts play a important role. Give reason. Ans. The replacement of diazo group by other groups is helpful in preparing those substituted aromatic compounds which cannot be prepared by direct substitution in benzene or substituted benzene. 7

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