MODEL ANSWER
PRACTICE TEST
VOLUME II
1. (i)
SN1 reaction proceeds via the formation of carbocation. The alkyl halide (I) is 3° while (II) is 2°. Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. Greater the stability of the carbocation, faster is the rate of SN1 reaction. Since 3° carbocation is more stable than 2° carbocation. (I), i.e. 2−chloro-2-methylpropane, undergoes faster SN1 reaction than (II) i.e., 3-chloropentane.
(ii)
The alkyl halide (I) is 2° while (II) is 1°. 2° carbocation is more stable than 1° carbocation. Therefore, (I), 2−chloroheptane, undergoes faster SN1 reaction than (II), 1-chlorohexane.
2.
Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is
Therefore, the compound R − Br is
When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz reaction. Therefore, the halide, R1−X, is
Therefore, compound D is
And, compound E is
3. (i)
2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an SN2 mechanism.
(ii)
2-Bromobutane is 2° alkylhalide whereas 2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers of substituents are present in 3° alkyl halide than in 2° alkyl halide to hinder the approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism.
(iii)
Both the alkyl halides are primary. However, the substituent −CH3 is at a greater distance to the carbon atom linked to Br in 1-bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by SN2 mechanism.
3. (i)
In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom. In cyclohexyl chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. Now, sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Therefore, the former is more electronegative than the latter. Therefore, the density of electrons of C−Cl bond near the Cl-atom is less in chlorobenzene than in cydohexyl chloride.
Moreover, the −R effect of the benzene ring of chlorobenzene decreases the electron density of the C−Cl bond near the Cl-atom. As a result, the polarity of the C−Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) To be miscible with water, the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules and so held together by dipole-dipole interactions. Similarly, strong H-bonds exist between the water molecules. The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, alkyl halides (though polar) are immiscible with water.
(iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.
Therefore, Grignard reagents should be prepared under anhydrous conditions.
4.
(ii)
(viii)
5. (i)
(ii)
6. (i)
(ii)
(iii)
7. (i)
In aryl alkyl ethers, due to the +R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.
Thus, benzene is activated towards electrophilic substitution by the alkoxy group.
(ii) It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.
8. (i)
(ii)
(iii)
(iv)
9. The −OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.
As a result, the benzene ring is activated towards electrophilic substitution.
10. (i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.
(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.
(iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol on hydrolysis.
(iv) When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.
11
(iv)
12. (i)
The +I effect of the alkyl group increases in the order:
Ethanal < Propanal < Propanone < Butanone
The electron density at the carbonyl carbon increases with the increase in the +I effect. As a result, the chances of attack by a nucleophile decrease. Hence, the increasing order of the reactivities of the given carbonyl compounds in nucleophilic addition reactions is:
Butanone < Propanone < Propanal < Ethanal
(ii)
The +I effect is more in ketone than in aldehyde. Hence, acetophenone is the least reactive in nucleophilic addition reactions. Among aldehydes, the +I effect is the highest in p-tolualdehyde because of the presence of the electron-donating −CH3 group and the lowest in p-nitrobezaldehyde because of the presence of the electron-withdrawing −NO2 group. Hence, the increasing order of the reactivities of the given compounds is:
Acetophenone < p-tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde
13. (i)
The +I effect of −CH3 group increases the electron density on the O−H bond. Therefore, release of proton becomes difficult. On the other hand, the −I effect of F decreases the electron density on the O−H bond. Therefore, proton can be released easily. Hence, CH2FCO2H is a stronger acid than CH3CO2H.
(ii)
F has stronger −I effect than Cl. Therefore, CH2FCO2H can release proton more easily than CH2ClCO2H. Hence, CH2FCO2H is stronger acid than CH2ClCO2H.
(iii)
Inductive effect decreases with increase in distance. Hence, the +I effect of F in CH3CHFCH2CO2H is more than it is in CH2FCH2CH2CO2H. Hence, CH3CHFCH2CO2H is stronger acid than CH2FCH2CH2CO2H.
(iv)
Due to the −I effect of F, it is easier to release proton in the case of compound (A). However, in the case of compound (B), release of proton is difficult due to the +I effect of −CH3 group. Hence, (A) is a stronger acid than (B).
14 (i)
(ii)
(iii)
(iv)
(v)
15 (i) Cyclohexanones form cyanohydrins according to the following equation.
In this case, the nucleophile CN− can easily attack without any steric hindrance. However, in the case of 2, 2, 6 trimethylcydohexanone, methyl groups at α-positions offer steric hindrances and as a result, CN− cannot attack effectively.
For this reason, it does not form a cyanohydrin.
(ii) Semicarbazide undergoes resonance involving only one of the two −NH2 groups, which is attached directly to the carbonyl-carbon atom.
Therefore, the electron density on −NH2 group involved in the resonance also decreases. As a result, it cannot act as a nucleophile. Since the other −NH2 group is not involved in resonance; it can act as a nucleophile and can attack carbonyl-carbon atoms of aldehydes and ketones to produce semicarbazones.
(iii) Ester along with water is formed reversibly from a carboxylic acid and an alcohol in presence of an acid.
If either water or ester is not removed as soon as it is formed, then it reacts to give back the reactants as the reaction is reversible. Therefore, to shift the equilibrium in the forward direction i.e., to produce more ester, either of the two should be removed
16 % of carbon = 69.77 %
% of hydrogen = 11.63 %
% of oxygen = {100 − (69.77 + 11.63)}%
= 18.6 %
Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic compound can be given as:
Therefore, the empirical formula of the compound is C5H10O. Now, the empirical formula mass of the compound can be given as:
5 × 12 + 10 ×1 + 1 × 16
= 86
Molecular mass of the compound = 86
Therefore, the molecular formula of the compound is given by C5H10O.
Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again, the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.
The given compound also gives a mixture of ethanoic acid and propanoic acid.
Hence, the given compound is pentan−2−ol.
The given reactions can be explained by the following equations:
17. (i)
(ii)
(iii)
18 (i) Methylamine and dimethylamine can be distinguished by the carbylamine test.
Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test, but dimethylamine does not.
(ii) Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl).
Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali. For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N, N−diethylbenzenesulphonamide, which is insoluble in an alkali. Tertiary amines, however, do not react with Hinsberg’s reagent.
(iii) Ethylamine and aniline can be distinguished using the azo-dye test. A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dil.HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol. The dye is usually yellow, red, or orange in colour. Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions.
(iv) Aniline and benzylamine can be distinguished by their reactions with the help of nitrous acid, which is prepared in situ from a mineral acid and sodium nitrite. Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas.
On the other hand, aniline reacts with HNO2 at a low temperature to form stable diazonium salt. Thus, nitrogen gas is not evolved.
(v) Aniline and N-methylaniline can be distinguished using the Carbylamine test. Primary amines, on heating with chloroform and ethanolic potassium hydroxide, form foul-smelling isocyanides or carbylamines. Aniline, being an aromatic primary amine, gives positive carbylamine test. However, N-methylaniline, being a secondary amine does not.
19 (i) In C2H5NH2, only one −C2H5 group is present while in (C2H5)2NH, two −C2H5 groups are present. Thus, the +I effect is more in (C2H5)2NH than in C2H5NH2. Therefore, the electron density over the N-atom is more in (C2H5)2NH than in C2H5NH2. Hence, (C2H5)2NH is more basic than C2H5NH2.
Also, both C6H5NHCH3 and C6H5NH2 are less basic than (C2H5)2NH and C2H5NH2 due to the delocalization of the lone pair in the former two. Further, among C6H5NHCH3 and C6H5NH2, the former will be more basic due to the +T effect of −CH3 group. Hence, the order of increasing basicity of the given compounds is as follows:
C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH
We know that the higher the basic strength, the lower is the pKb values.
C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH
(ii) C6H5N(CH3)2 is more basic than C6H5NH2 due to the presence of the +I effect of two −CH3 groups in C6H5N(CH3)2. Further, CH3NH2 contains one −CH3 group while (C2H5)2NH contains two −C2H5 groups. Thus, (C2H5)2 NH is more basic than C2H5NH2.
Now, C6H5N(CH3)2 is less basic than CH3NH2 because of the−R effect of −C6H5 group.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6H5NH2 < C6H5N(CH3)2 < CH3NH2 < (C2H5)2NH
(iii) (a)
In p-toluidine, the presence of electron-donating −CH3 group increases the electron density on the N-atom.
Thus, p-toluidine is more basic than aniline.
On the other hand, the presence of electron-withdrawing
−NO2 group decreases the electron density over the N−atom in p-nitroaniline. Thus, p-nitroaniline is less basic than aniline.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
p-Nitroaniline < Aniline < p-Toluidine
(b) C6H5NHCH3 is more basic than C6H5NH2 due to the presence of electron-donating −CH3 group in C6H5NHCH3.
Again, in C6H5NHCH3, −C6H5 group is directly attached to the N-atom. However, it is not so in C6H5CH2NH2. Thus, in C6H5NHCH3, the −R effect of −C6H5 group decreases the electron density over the N-atom. Therefore, C6H5CH2NH2 is more basic than C6H5NHCH3.
Hence, the increasing order of the basic strengths of the given compounds is as follows:
C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2.
(iv) In the gas phase, there is no solvation effect. As a result, the basic strength mainly depends upon the +I effect. The higher the +I effect, the stronger is the base. Also, the greater the number of alkyl groups, the higher is the +I effect. Therefore, the given compounds can be arranged in the decreasing order of their basic strengths in the gas phase as follows:
(C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3
(v) The boiling points of compounds depend on the extent of H-bonding present in that compound. The more extensive the H-bonding in the compound, the higher is the boiling point. (CH3)2NH contains only one H−atom whereas C2H5NH2 contains two H-atoms. Then, C2H5NH2 undergoes more extensive H-bonding than (CH3)2NH. Hence, the boiling point of C2H5NH2 is higher than that of (CH3)2NH.
Further, O is more electronegative than N. Thus, C2H5OH forms stronger H−bonds than C2H5NH2. As a result, the boiling point of C2H5OH is higher than that of C2H5NH2 and (CH3)2NH.
Now, the given compounds can be arranged in the increasing order of their boiling points as follows:
(CH3)2NH < C2H5NH2 < C2H5OH
(vi) The more extensive the H−bonding, the higher is the solubility. C2H5NH2 contains two H-atoms whereas (C2H5)2NH contains only one H-atom. Thus, C2H5NH2 undergoes more extensive H−bonding than (C2H5)2NH. Hence, the solubility in water of C2H5NH2 is more than that of (C2H5)2NH.
Further, the solubility of amines decreases with increase in the molecular mass. This is because the molecular mass of amines increases with an increase in the size of the hydrophobic part. The molecular mass of C6H5NH2 is greater than that of C2H5NH2 and (C2H5)2NH.
Hence, the increasing order of their solubility in water is as follows:
C6H5NH2 < (C2H5)2NH < C2H5NH2
20 (i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
21 (i) On hydrolysis, sucrose gives one molecule of ∝-D glucose and one molecule of β- D-fructose.
(ii) The hydrolysis of lactose gives β-D-galactose and β-D-glucose.
22 (i) When D-glucose is heated with HI for a long time, n-hexane is formed.
(ii) When D-glucose is treated with Br2 water, D- gluconic acid is produced.
(iii) On being treated with HNO3, D-glucose get oxidised to give saccharic acid.
23
Fibrous protein Globular protein 1. It is a fibre-like structure formed by the polypeptide chain. These proteins are held together by strong hydrogen and disulphide bonds. 1. The polypeptide chain in this protein is folded around itself, giving rise to a spherical structure. 2. It is usually insoluble in water. 2. It is usually soluble in water. 3. Fibrous proteins are usually used for structural purposes. For example, keratin is present in nails and hair; collagen in tendons; and myosin in muscles. 3. All enzymes are globular proteins. Some hormones such as insulin are also globular proteins.
24 (i) Hexamethylenediamine and adipic acid
(ii)
(iii) Tetrafluoroethene
25 (i) Vinyl chloride (CH2=CHCl)
(ii) Tetrafluoroethylene (CF2 = CF2)
(iii) Formaldehyde (HCHO) and phenol (C6H5OH)
26 The monomeric repeating unit of nylon 6 is, which is derived from Caprolactam.
The monomeric repeating unit of nylon 6, 6 is, which is derived from hexamethylene diamine and adipic acid.
27 The condensation polymerisation of ethylene glycol and terephthalic acid leads to the formation of dacron.
28 The main constituents of dettol are chloroxylenol and α-terpineol.
29 (i) Cationic detergent
Cationic detergents are quaternary ammonium salts of acetates, chlorides, or bromides. These are called cationic detergents because the cationic part of these detergents contains a long hydrocarbon chain and a positive charge on the N atom.
For example: cetyltrimethylammonium bromide
(ii) Anionic detergents
Anionic detergents are of two types:
1.Sodium alkyl sulphates: These detergents are sodium salts of long chain alcohols. They are prepared by first treating these alcohols with concentrated sulphuric acid and then with sodium hydroxide. Examples of these detergents include sodium lauryl sulphate (C11H23CH2OSO3−Na+) and sodium stearyl sulphate (C17H35CH2OSO3−Na+).
2.Sodium alkylbenzenesulphonates: These detergents are sodium salts of long chain alkylbenzenesulphonic acids. They are prepared by Friedel-Crafts alkylation of benzene with long chain alkyl halides or alkenes. The obtained product is first treated with concentrated sulphuric acid and then with sodium hydroxide. Sodium 4-(1-dodecy) benzenesulphonate (SDS) is an example of anionic detergents.
(iii) Non-ionic detergents
Molecules of these detergents do not contain any ions. These detergents are esters of alcohols having high molecular mass. They are obtained by reacting polyethylene glycol and stearic acid.
30
Antiseptics and disinfectants are effective against micro-organisms. However, antiseptics are applied to the living tissues such as wounds, cuts, ulcers, and diseased skin surfaces, while disinfectants are applied to inanimate objects such as floors, drainage system, instruments, etc. Disinfectants are harmful to the living tissues.
Iodine is an example of a strong antiseptic. Tincture of iodine (2 − 3 percent of solution of iodine in alcohol − water mixture) is applied to wounds. 1 percent solution of phenol is used as a disinfectant.