Question

An organic compound A having molecular formula \(C_6\)\(H_6\)O turn blue litmus solution into red but does not react with sodium bicarbonate, but when treated with bromine water then form a white ppt of compound B. when compound A react with chloroform in presence of aqueous caustic soda solution at 340K then form two compound C and D. When compound A treated with caustic soda then form compound E compound E when treated with methyl halide then form compound F. Read the above passage carefully and answer the following questions: (i) The name of compound is: (a) 2-methyl propene-2-ol (b) 2-methyl phenol (c) 2,4,6-tribromophenol (d) Butane 1-ol (ii) Which are isomers of each other (a) A and C (b) B and C (c) C and D (d) D and E (iii) The IUPAC name of compound F is (a) Anisole (b) Methoxybenzene (c) Salicylaldehyde (d) 2-methyl propene-2-ol (iv) When compound E treated with ethyl iodide then ...... form. (a) Ehoxybenzene (b) Ethoxy hexane (c) Propoxypropane (d) Benzaldehyde (v) On oxidation with sodium dichromate and conc \(H_2\)\(SO_4\) phenol gives (a) Benzaldehyde (b) p-Benzoquinone (c) o-Benzoquinone (d) m-benzoquinone

The sequence of bases along the DNA and RNA chain establishes its primary structure which controls the specific properties of the nucleic acid. An RNA molecule is usually a single chain of ribose-containing nucleotide. On the basis of X-ray analysis of DNA, J.D., Watson and F.H.C. CYST (shared noble prize in 1962) proposed a three dimensional secondary structure for DNA. DNA molecule is a long and highly complex, spirally twisted, double helix, ladder like structure. The two polynucleotide chains or strands are linked up by hydrogen bonding between the nitrogenous base molecules of their nucleotide monomers. Adenine (purine) always links with thymine (pyrimidine) with the help of two hydrogen bonds and guanine (purine) with cytosine (pyrimidine) with the help of three hydrogen bonds. Hence, the two strands extend in opposite directions, i.e., are antiparallel and complimentary. In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. a) Assertion and reason both are correct statements and reason is correct explanation for assertion. b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. c) Assertion is correct statement but reason is wrong statement d) Assertion is wrong statement but reason is correct statement. i) Assertion - DNA molecules and RNA molecules are found in the nucleus of a cell. Reason : There are two types of nitrogenous bases, purines and pyrimidines. Adenine (A) and guanine (G)are substituted purines; cytokine (C), thymine (T) arid uracil (U) are substituted pyrimidines. ii) Assertion .- In both DNA and RNA, heterocyclic base and phosphate ester linkages are at C- 1’ and C-5’respectively of the sugar molecule. Reason : Nucleotides and nucleosides mainly differ from each other in presence of phosphate units. iii)Assertion .- The backbone of RNA molecule is a linear chain consisting of an alternating units of heterocylic base, D-ribose and a phosphate. Reason : The segment of RNA which acts as the instruction manual for the synthesis of protein is ribose. iv) Assertion.- In DNA, the complementary bases are, adenine and guanine; thymine and cytosine. Reason : The phenomenon of mutation is chemical change in DNA molecule.

Read the passage given below and answer the following questions: Alkyl halides are prepared by the free radical halogenation of alkanes, addition of halogen acids to alkenes, replacement of -OH group of alcohols with halogens using phosphorus halides, thionyl chloride or halogen acids. Aryl halides are prepared by electrophilic substitution to arene. Fluorine and iodides are best prepared by halogen exchange method. These compounds find wide applications in industry as well as in day-to-day life. These compounds are generally used as solvents and as starting material for the synthesis of a large number of organic compounds. (i) The best method for the conversion of an alcohol into analkyl chloride is by treating the alcohol with (a) \(PCl_5\) (b) dry HCl in the presence of anhydrous \(ZnCl_2\) (c) \(SOCl_2\) in presence of pyridine (d) None of these (ii) The catalyst used in the preparation of an alkyl chloride bythe action of dry HCl on an alcohol is (a) anhydrous \(AlCl_3\) (b) \(FeCl_3\) (c) anhydrous \(ZnCl_2\) (d) Cu (iii) An alkyl halide reacts with metallic sodium in dry ether. The reaction is known as : (a) Frankland’sreaction (b) Sandmeyer’sreaction (c) Wurtz reaction (d) Kolbe’s reaction (iv) Fluorobenzene (\(C_6\)\(H_5\)F) can be synthesized in the laboratory (a) by direct fluorination of benzene with \(F_2\) gas (b) by reacting bromobenzene with NaF solution (c) by heating phenol with HF and KF (d) from aniline by diazotisation followed by heating thediazonium salt with \(HBF_4\) (v) When 2-bromobutane reacts with alcoholic KOH, thereaction is called (a) halogenation (b) chlorination (c) hydrogenation (d) dehydrohalogenation

The solubility of gases increases with increase of pressure. William Henry made a systematic investigation of the solubility of a gas in a liquid. According to Henry’s law “the mass of a gas dissolved per unit volume of the solvent at constant temperature is directly proportional to the pressure of the gas in equilibrium with the solution”. Dalton during the same period also concluded independently that the solubility of a gas in a liquid solution depends upon the partial pressure of the gas. If we use the mole fraction of gas in the solution as a measure of its solubility, then Henry’s law can be modified as “the partial pressure of the gas in the vapour phase is directly proportional to the mole fraction of the gas in the solution”. Answer the following MCQs by choosing the most appropriate options: (i) Henry’s law constant for the solubility of methane in benzene at 298 K is 4.27 × \(10^5\) mm Hg. The solubility of methane in benzene at 298 K under 760 mm Hg is (a) 4.27 × \(10^{-5}\) (b) 1.78 × \(10^{-3}\) (c) 4.27 × \(10^{-3}\) (d) 1.78 × \(10^{-3}\) (ii) The partial pressure of ethane over a saturated solution containing 6.56 × \(10^{-2}\) g of ethane is 1 bar. If the solution contains 5.00 × \(10^{-2}\) g of ethane then what will be the partial pressure (in bar) of the gas? (a) 0.762 (b) 1.312 (c) 3.81 (d) 5.0 (iii) KH (K bar) values for Ar(g), \(CO_2\)(g), HCHO(g) and \(CH_4\)(g) are 40.39, 1.67, 1.83 × \(10^{-5}\) and 0.413 respectively. Arrange these gases in the order of their increasing solubility. (a) HCHO < \(CH_4\) < \(CO_2\) < Ar (b) HCHO < \(CO_2\) < \(CH_4\) < Ar (c) Ar < \(CO_2\) < \(CH_4\) < HCHO (d) Ar < \(CH_4\) < \(CO_2\) < HCHO (iv) When a gas is bubbled through water at 298 K, a very dilute solution of the gas is obtained. Henry’s law constant for the gas at 298 K is 150 K bar. If the gas exerts a partial pressure of 2 bar, the number of millimoles of the gas dissolved in 1 L of water is (a) 0.55 (b) 0.87 (c) 0.37 (d) 0.66

The properties of dilute or ideal solutions which depend only upon the concentration of the solute in the solution and no other characteristics are known as colligative properties. There are in all four such properties i.e. relative lowering in vapour pressure, osmotic pressure, elevation in boiling point temperature and depression in freezing point temperature. All of them help in calculating the observed molar mass of the solute which is inversely proportional to the colligative property involved. Out of these, osmotic pressure may be regarded as the best for the determination of molecular mass of the solute. According to Van’t Hoff theory of dilute solution, π = CRT, where ‘π’ is the osmotic pressure while ‘C’ is the molar concentration of the solution. (i) When liquids A and B are mixed, hydrogen bonding occurs. The solutions will show: a) Positive deviation from Raoult’s law b) Negative deviation from Raoult’s law c) No deviation from Raoult’s law d) Slightly increase in volume (ii) The azeotropic mixture of water and HCl boils at \(108.5^\circ\)C when the mixture is distilled. It is possible to obtain: a) Pure HCl b) Pure water c) Pure water as well as pure HCl d) Neither HCl nor water in their pure states. (iii) On freezing an aqueous solution of sugar, the solid which starts separating out is: a. Sugar b. Ice c. Solution with the same composition d. Solution with different composition (iv) The value of osmotic pressure does not depend upon: a) Concentration of the solution b) Temperature of the solution c) Number of the particles of the solute present d) Structure of the solute particles (v) Effect of adding a non-volatile solute to a solvent is : a) to lower the vapour pressure b) to increase the freezing point c) to decrease the boiling point d) to decrease the osmotic pressure

The properties of the solutions which depend only on the number of solute particles but not on the nature of the solute are called colligative properties. Relative lowering in vapour pressure is also an example of colligative properties. For an experiment, sugar solution is prepared for which lowering in vapour pressure was found to be 0.061 mm of Hg.(vapour pressure of water at \(20^\circ\)C is 17.5 mm of Hg). Answer the following MCQs by choosing the most appropriate options: (i) Relative lowering of vapour pressure for the given solution is- (a) 0.00348 (b) 0.061 (c) 0.122 (d) 1.75 (ii) The vapour pressure (mm of Hg) of Solution will be (a) 17.5 (b) 0.61 (c) 17.439 (d) 0.00348 (iii) Mole fraction of sugar in the solution is (a) 0.00348 (b) 0.9965 (c) 0.061 (d) 1.75 (iv) If weight of sugar taken is 5 g in 108 g of water then molar mass of sugar will be (a) 358 (b) 120 (c) 240 (d) 400 (v) The vapour pressure (mm of Hg) of water at 293 K when 25 g of glucose is dissolved in 450 g of water is (a) 17.2 (b) 17.4 (c) 17.120 (d) 17.02

Read the passage given below and answer the following questions: Phenol contains -OH group directly attached to carbon atoms of an aromatic system \(C_6\)\(H_5\)OH in phenol the group is attached to \(sp^2\) hybridised carbon of aromatic ring. The carbon oxygen bond length is 1:36 pm in phenol is slightly less than that in in methanol this is due to first point partial double bond character on account of the conjugation of unshared electron pair of oxygen with the aromatic ring s point \(sp^3\) hybridised state of carbon to which oxygen is attached it can be prepared by various means or methods. Some important methods are alkali fusion of sulphonates, hydrolysis of diazonium salts decarboxylation of salicylic acid and from Grignard reagent, it is prepared from Dow's process and from cumene. Aerial oxidation of human produce cumene peroxide which on hydrolysis produce phenol and acetone. In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Both assertion and reason are true and the reason is the correct explanation of assertion. (b) Both assertion and reason are true but the reason is not the correct explanation of assertion. (c) Assertion is true but reason is false. (d) Assertion is false but reason is true. (i) Assertion: C-O bond length in phenol is less than C-O bond length in ethanol. Reason: In phenol carbon atom is \(sp^2\) hybridize while in ethanol carbon atom is \(sp^3\) hybridised. (ii) Assertion: p-nitrophenol is more acidic than phenol. Reason: Nitro group helps in the stabilisation of the phenoxide ion by dispersal of negative charge due to resonance. (iii) Assertion: Phenol is prepared by the reaction of chlorobenzene with caustic soda at 623K and 300 atm pressure. Reason: The substitution of chorine atom from chlorobenzene is difficult due to resonance. (iv) Assertion: Methanol is less acidic than phenol. Reason: Due to resonance phenate ion become more stable. (v) Assertion: On nitration phenol forms 2-nitrophenol and 4- nitrophenol. Reason: The presence of –OH group in phenols activates the aromatic ring towards electrophilic substitution and directs the incoming group to ortho and para positions.