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NCERT Intext Questions 10.6-10.10 (in Hindi)
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In this tutorial remaining 5 questions were covered of NCERT intext question.

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  2. NCERT INTEXT QUESTIONS 10.6 Arrange each set of compounds in order of increasing boiling points. (i)Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii)1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. Ans. (i) Chloromethane < Bromomethane< Dibromomethane < Bromoform The reason is: (a)for same alkyl group, B.Pt increases with size of halogen atom. (b)B.Pt increases as number of halogen atoms increase. (ii)Isopropyl chloride < 1- Chloropropane < 1 - Chlorobutane Reason (a)For same halogen, B.Pt. increases as size of alkyl group increases. (b)B.Pt. decreases as branching increases.

  3. 10.7. Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer. CH3 CH2 CH2 CH Bor C,CH2HCH, Br CH CH,CH2CHCH, or H,C CBr Br CH3 (i) CH,CHCH,CH2Br or CH,CH2CHCH2Br CHs CH

  4. Ans. In Sy2 mechanism, reactivity depends upon the steric hindrance around the C-atom carrying the halogen. Lesser the steric hindrance, faster the reaction. O CH,CH2 CH2CH2Br 1 alkyl halide CH, CH2 CH(Br)CH, 20 alkyl halide As steric hindrance in 2o alkyl halide is more, thus reactivity of CH,CH2CH2CH2Br> CH3CHCH(Br)CH, (i) CH, CH2 CH (Br) CH 2 alkyl halide (CH) CBr3 alkyl halide As steric hindrance in (CH3) CBris more, thus it is less reactive than CH3CH2CH (Br) CH3 (ii) Both are 2 alkyl halides but CH, group atC2 is closer to Br atom than - CH, group at C3.As a result CH,CH2 CH(CH) CH,Br suffers greater steric hindrance than CH3 CH(CH CH2 CH,Br and will thus be less reactive in SN2

  5. 10.8. In the following pairs of halogen compounds, which compound undergoes faster Sy1 reaction? Cl and CI Ans. Reactivity in S carbocations is governed by stability of Cl reacts faster due to greater stabilit,y of 3 carbocation. CI reacts faster due to' greater stability of 2 carbocation over 1 carbocation

  6. 10.9. Identify A, B, C, D, E, R and R1 in the following: Br -+Mg dry ether, A -HO, B D,0 CH3 CH, CH CH3 Na/ether g' CH3 CH Mg

  7. Mg Br . H3_CH CH 3 Br R-Br CHa CH -CH3 3-CH MgBr CH 3 CH3 CH3 Nal Mg cther C-HBr CH3 (R-Br) CH CH3 3 H2O Cra H3 H3

  8. 10.10. A hydrocarbon CjH1o does not react with chlorine in dark but gives a single monochloro compound C HgCl in bright sunlight. Identify the hydrocarbon. Ans. The hydrocarbon with molecular formula CsH10 can either a cycloalkane or an alkene.Since the compound does not react with Cl2 in the dark, therefore it cannot be an alkene but must be a cycloalkane. Since the cycloalkane reacts with Cl2 in the presence of bright sunlight to give a single monochloro compound, C5HgCI, therefore, all the ten hydrogen atoms of the cycloalkanes must be equivalent. Thus, the cycloalkane is cyclopentane Cl No reaction *dark Cyclopentane CI Cl Sunlight Monochloro-cyclopentane CHgCl)