JEE (Main) Chemistry Offline Solved Paper 2017

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1 JEE (Main) Chemistry ffline Solved Paper 07. Which of the following compounds will form significant amount of meta product during mono-nitration reaction? C 3 N NCC 3 (d). DU is equal to isochoric work isobaric work adiabatic work (d) isothermal work 3. The increasing order of the reactivity of the following halides for the S N reaction is C -- C-- C C Ω Cl 3 3 C 3 C C Cl p 3 C C 6 4 C Cl (I) (II) (III) (III) < (II) < (I) (II) < (I) < (III) (I) < (III) < (II) (d) (II) < (III) < (I) 4. The radius of the second Bohr orbit for hydrogen atom is: (Planck constant, h = J s; mass of electron, m e = kg; charge of electron, e = C; and permittivity of vacuum, e = kg m A.).65 Å 4.76 Å 0.59 Å (d). Å 5. pk a of a weak acid (A) and pk b of a weak base (B) are 3. and 3.4, respectively. The p of their salt (AB) solution is (d).0 6. The formation of which of the following polymers involves hydrolysis reaction? Nylon 6 Bakelite Nylon 6, 6 (d) Terylene 7. The most abundant elements by mass in the body of a healthy human adults are: xygen (6.4%); Carbon (.9%); ydrogen (0.0%) and Nitrogen (.6%). The mass which a 75 kg person would gain if all atoms are replaced by atoms is 5 kg 37.5 kg 7.5 kg (d) 0 kg. 8. Which of the following, upon treatment with tert-buna followed by addition of bromine water, fails to decolourize the colour of bromine?

2 JEEC. Complete Chemistry JEE Main 9. In the following reactions, Zn is respectively acting as a/an I. Zn + Na Æ Na Zn II. Zn + C Æ ZnC 3 base and acid base and base acid and acid (d) acid and base 0. Both lithium and magnesium display several similar properties due to the diagonal relationship. owever, the one which is incorrect is: Both form basic carbonates Both form soluble bicarbonates Both form nitrides (d) Nitrates of both Li and Mg yield N and an heating.. 3-Methylpent--ene on reaction with in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is: Six Zero Two (d) Four. A metal crystallises in a face-centred cubic structure. If the edge length of its unit cell is a, the closest approach between two atoms in the metallic crystal will be a a a (d) a / 3. Two reactions R and R have identical pre-exponentital factors. Activation energy of R exceeds that of R by 0 kj mol. If k and k are rate constants for the reactions R and R, respectively at 300 K, then ln (k /k ) is equal to: (R = 8.34 J K mol ) 8 6 (d) 4 4. The correct sequence of reagents for the following conversion will be C 3 (d) C 6 5 C C 3 [Ag(N 3 ) ] +, + /C 3, C 3 Mg C 3 Mg, + /C 3, [Ag(N 3 ) ] + C 3 Mg, [Ag(N 3 ) ] +, + /C 3 (d) [Ag(N 3 ) ] +, C 3 Mg, + /C 3 5. The Tyndall effect is observed only when following conditions are observed: (I) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (II) The diameter of the dispersed particles is not much smaller than the wavelength of the light used. (III) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude (IV) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude (I) and (IV) (II) and (IV) (I) and (III) (d) (II) and (III) C 3

3 Chemistry ffline Solved Paper 07 JEEC.3 6. Which of the following compounds will behave as a reducing sugar in aqueous K solution? 7. Given: C C C CC 3 C C 3 (d) C C C C C 3 C(graphite) + (g) Æ C (g); D r = kj mol (g) + (g) Æ (l); D r = 85.8 kj mol C (g) + (l) Æ C 4 (g) + (g); D r = kj mol Based on the above thermochemical equations, the value of D r at 98 K for the reaction C(graphite) + (g) Æ C 4 (g) will be kj mol kj mol 74.8 kj mol (d) 44.0 kj mol 8. Which of the following reactions is an example of a redox reaction? XeF 4 + F Æ XeF 6 + XeF + PF 5 Æ [XeF] + PF 6 XeF 6 + Æ XeF 4 + F (d) XeF 6 + Æ Xe F + 4F 9. The product obtained when chlorine gas reacts with cold and dilute aqueous Na are: Cl and Cl 3 Cl and Cl 3 Cl and Cl (d) Cl and Cl 3 0. The major product obtained in the following reaction is: BuK C 6 5 ææææ D C 6 5 (±) C 6 5 C(Bu)C C 6 5 C 6 5 C == CC 6 5 (+) C 6 5 C(Bu)C C 6 5 (d) ( ) C 6 5 C(Bu)C C 6 5. Sodium salt of an organic acid X produces effervescence with conc. S 4. X reacts with acidified aqueous CaCl solution to give a white precipitate which decolourises acidic solution of KMn 4. The compound X is C 6 5 CNa CNa C 3 CNa (d) Na C 4. Which of the following species is not paramagnetic? N C (d) B 3. The freezing point of benzene decreases by 0.45 C when 0. g of acetic acid is added to 0 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (Given: K f (benzene) = 5. K kg mol ): 64.6% 80.4% 74.6% (d) 94.6% 4. Which of the following molecule is least resonance stabilized. N (d)

4 JEEC.4 Complete Chemistry JEE Main 5. n treatment of 00 ml of 0. M solution of CoCl 3 6 with excess AgN 3,. 0 ions are precipitated. The complex is [Co( ) 4 Cl ]Cl [Co( ) 3 Cl 3 ] 3 [Co( ) 6 ]Cl 3 (d) {Co( ) 5 Cl]Cl 6. The major product obtained in the following reaction is DIBAL- æææææ C C C C C C (d) C C 7. A water sample has ppm level concentrations of following anions F = 0; S 4 = 00; N 3 = 50 The anion/anions that make/makes the water sample unstable for drinking is/are only N 3 both S 4 and N 3 only F (d) only S 4 8. ne gram of a carbonate (M C 3 ) on treatment with excess Cl produces mol of C, the molar mass of M C 3 in g mol is (d) Given: E Cl Cl =.36 V E 3+ Cr Cr = 0.74 V E Cr 7 Cr 3+ =.33 V E Mn Mn+ =.5 V 4 Among the following, the strongest reducing agent is Cr Mn + Cr 3+ (d) Cl 30. The group having isoelectronic species is, F, Na +, Mg +, F, Na, Mg +, F, Na, Mg + (d), F, Na +, Mg + C ANSWERS (d) (d) 0.. (d). (d) 3. (d) (d). 3. (d) 4. (d) 5. (d)

5 Chemistry ffline Solved Paper 07 JEEC.5 INTS AND SLUTINS. Nitration is carried out in the presence of conc. S 4 and conc. N 3. Aniline being a base forms anilinum ion with acid. The -- + N 3 ion being a deactivating group results in the formation of meta mono-nitration product. N + N 3 N 3 + æææ conc. N 3 ææææææ conc.s4 47% N. Adiabatic condition implies q = 0. Thus, from the first law of thermodynamics, we have DU = q + w = 0 + w = w 3. The S N reaction proceeds through ionic mechanism. The larger the stability of the resultant carbocation, larger the reactivity of the halide. The stability of the carbocations formed from the given compounds is C C C < C --C-- C C < p -C--C -- C (II) (I) (III) The carbocation from the compound (III) is most stable due to conjugation with phenyl group. The primary carbocation (II) is lesser stable than the secondary carbocation (I). 4. Recall that the Bohr radius (i.e. radius of n = orbit of hydrogen atom) is 5.9 pm ( = m = 0.59 Å). Also, ence r n = n r r = () (0.59 Å) =.6 Å. Å 5. The p of a solution containing a salt of weak acid and weak base is given by the expression p = (pk w + pk a pk b ) ence, at 5 C, we have p = ( ) = Nylon 6 is obtained by hydrolysis of caprolactam: N (i)hydrolysis (ii) polymerization Ω æææææææ C-- N -- (C ) 5 n Caprolactam Nylon-6 7. Mass of in 75 kg of the person = kg = 7.5 kg. Since the replacement of by causes doubling the mass, the gain in mass of the person would be 7.5 kg. 8. Treatment with tert-buna will cause elimination reaction resulting into the formation of alkene. The compound of choice does not undergo elimination reaction and thus will not decolourize bromine water.

6 JEEC.6 Complete Chemistry JEE Main 9. xygen acceptor is an acid and oxygen gainer is a base. In the reaction I, Zn gains while in reaction II, Zn loses. 0. Magnesium forms basic carbonate (4MgC 3 Mg() 5 ) whereas Li does not. All the other choices (namely, b, c and d) are correct.. in the presence of peroxide follows anti-markownikov addition which states that the hydrogen of acid attaches to the carbon containing lesser number of hydrogen. The addition follows free-radical mechanism. Tertiary radical is formed by the addition of followed by the addition of. C 3 C 3 C C C C C ææææ 3 C 3 C C CC + 3 ææææ 3 3-methyl pent--ene C 3 C 3 C C C C * * 3 There are two asymmetric carbon leading to (i.e. four) optical isomers.. In a face-centred cubic unit cell, atoms touch each other along the face diagonal of the cube. There are atoms at the corners and one at the face of the cube. Thus, the face-diagonal a will be equal to 4r. The closest approach between two atoms is r and thus will be equal to a/. 3. k = A e E /RT and k = A e (E + 0 kj mol )/RT ence k k = e e -E / RT - -( E + 0 kj mol )/ RT Ê ln k ˆ Ë Á k = E RT - ( E + 0 kj mol ) + = RT Jmol = ( JK mol )( 300 K) 4. [Ag(N 3 ) ] + is Tollens reagent. This causes oxidation of C to C. + /C 3 causes esterification of C group. C 3 Mg adds to C and ester groups. The reactions proceed as follows. [Ag(N 3) ] æææææææ + /C3 ææææææ C C C 3 C Mg 3 C C 3 C 3 C 3 5. The choice is correct. 6. The compound of choice contains an ester group which is hydrolysed by K resulting into group. Due to hemiacetal group, the resulting compound acts as a reducing sugar.

7 Chemistry ffline Solved Paper 07 JEEC.7 C C C C CC 3 aq. K æææææ C C C C C C Reducing sugar 7. The following manipulations gives the desired reaction. C(graphite) + (g) Æ C (g); D r = kj mol [ (g) + (g) Æ (l)]; D r = 85.8 kj mol Add C (g) + (l) Æ C 4 (g) + (g); D r = kj mol C(graphite) + (g) Æ C 4 (g); D r = ( ) kj mol = 74.8 kj mol. 8. nly in the choice, there occurs change in oxidation number of Xe and XeF + F Æ XeF ere Xe is oxidised whereas is reduced. 9. With cold and dilute aqueous Na, Cl reacts as Cl + Æ Cl + Cl + 0. Elimination of occurs resulting into the formation of double bond. C 6 5 C 6 5 BuK ææææ C D C -- C -- C 6 5. The compound X is Na C 4. The given reactions are: Na C 4 + S 4 Æ Na S 4 + C + C + conc. Na C 4 + CaCl Æ CaC 4 Ø + NaCl white ppt. Mn C 4 Æ Mn C (pink) (colourless). The electronic configurations of given species are: N valence electrons (s s) (s*s) (pp x ) (pp y ) (s p z ) (p*p x ) C 0 valence electrons (ss) (s*s) (pp x ) (pp y ) (sp z ) valence electrons (ss) (s*s) (sp z ) (pp x ) (pp y ) (p*p x ) (p*p y ) B 6 valence electrons (ss) (s*s) (pp x ) (pp y ) The species C does not have unpaired electron(s) and thus will not be paramagnetic. 3. Molality of acetic acid in benzene is m = n m = ( m/ M) 0 60 m = (. g/ g mol ) 00. kg = - 6 molkg -

8 JEEC.8 Complete Chemistry JEE Main In the solution, we have C 3 C (C 3 C) m( a) m(a/) Total molality of solution, m total = m( a) + m(a/) = m( a/) Since DT f = K f m, we have (0.45 K) = (5. K kg mol ) {( mol kg )/6} ( a/) This gives a = Ê ˆ Ë Á 5 = The percentage association of acetic acid is 00 = 94.53% 4. The molecule is least-resonance stabilized as it does not involve 4n + electrons (aromatic requirement). 5. Amount of complex, n = MV = (0. mol L )(0. L) = 0.0 mol N. 0 Amount of chloride precipitated, n = = = 0.0 mol 3 - N A ( mol ) The complex of choice (d) will produce twice the precipitable chloride ions. 6. DIBAL- is diisobutylaluminium hydride. It is an electrophilic reducing agent as it reacts more quickly with electron-rich compound. It reduces carboxylic acids, their derivatives and nitriles to aldehyde. By contrast, the use of LiAl 4 reduces esters and acyl chloride to primary alcohol and nitriles to primary amines. In the given reaction, the product is given by the choice. 7. The maximum concentrations of the given ions in the drinkable water are N 3 = 50 ppm; S 4 = 500 ppm; F = ppm 8. The reaction is M C 3 + Cl Æ MCl + C + The stoichiometric numbers of M C 3 and C are identical, the amount of M C 3 will also be equal to mol. If M is the molar mass of M C 3, then g M = mol i.e M = g = 84.3 g mol mol 9. Lesser the E of the couple, lesser the reduction tendency of the species being reduced and larger the oxidation tendency of the species being oxidized. Larger oxidation tendency implies the strongest reducing agent. E Cr 3+ Cr has the least value and thus Cr is the strongest reducing agent. 30. The species in the choice have the same number of valence electrons, namely, 0 electrons.

JEE Main C CHEMISTRY CO 3. ) on treatment with excess HCl produces mole of CO gram of a carbonate (M 2

JEE Main C CHEMISTRY CO 3. ) on treatment with excess HCl produces mole of CO gram of a carbonate (M 2 61. 1 gram of a carbonate (M C ) on treatment with excess H produces 0.01186 mole of C. The molar mass of M C in g mol 1 is : 84. () 118.6 () 11.86 1186 84. 6. given C ( g) C ( g); ( graphite) 9.5 kj mol