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Textbook Exercise Questions 10.20-10.22 (in Hindi)
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In this tutorial we will cover the conversions and determine the product of reactions by solving NCERT questions.

Pursuing M.Pharm,Completed B.PHARM From Jamia Hamdard, New Delhi. Completed XII from KGSBV SCHOOL. My hobbies are teaching, #YOUTUBER

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Thank you so much as these topics are explained in English. There are some people who don't understand Hindi still Hindi is mother tongue of india.iam the one. I think it is better to explain in English as everyone know very well.

  2. NCERT Exercise 10.20 The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain. Ans. If aqueous solution, KOH is almost completely ionized to give OH- ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. Further in the aqueous solution, OH- ions are highly solvated (hydrated). This solvation reduces the basic character of OH ions which, therefore, fails to abstract a hydrogen from the P-carbon of the alkyl chloride to form alkenes. In contrast, an alcoholic solution of KOH contains alkoxide (RO-) ion which being a much stronger base than OH-ions perferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

  3. 10.21 Primary alkyl halide CaH Br (a) reacted with alcoholic KOH to give compound (b) Compound (b) is reacted, with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it give compound (d), CgH18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. Ans. (i) There are two primary alkyl halides having the molecular formula, C,H9Br. CH, CHC22CH2Br and CH,CH-CH2Br -Butylbromide lsobutyl bromide (ii) Since compound (a) when reacted with Na metal gave a compound (d) with molecular formula CgH18which was different from die compound obtained when n-butyl bromide was reacted with Na metal, therefore, (a) must be isobutyl bromide and compound (d) must be 2,3- dimethylhexane

  4. 2CH, CH2 CH, CH2 Br +2Na Wt rtin CH,CH2 CH,CH,CH2CH,CH,CH, n-Octane CH CH CH, 2CH CH-CH2Br +2Na Wrtz reaction CH CH CH2 CH2-CH CH, 2,5-dimethylhexane (d (ii) If compound (a) is isobutyl bromide, than the compound (b) which it gives on treatment with alcoholic KOH must be 2-methyl-1-propane. CH, CH CH,-CH-CH,Br Dehydrohalogenation 2-Methyl-1-propne(t) CH,-C Isobutyl bromide

  5. (iv) The compound (b) on treatment with HBr gives compound (c) in accordance with Markownikoff rule. Therefore, compound (c) is tert-butyl bromide which is an isomer of compound (a) ,i.e., isobutyl ' bromide. CH CH 3 3 CH3CH3 Br tert-Butylbromide (c) (anisomerof compound (a) Thus (a)is isobutyl bromide, (b)is 2-methyl-1 -propane, (c)is tert-butylbromide, and (d)is 2,5-dimethylhexane.

  6. 10.22 What happens when. (i)n-butyi chloride is treated with alcoholic KOH. (ii)bromobenzene is treated with Mg in the presence of dry ether. (ii)chlorobenzene is subjected to hydrolysis. (iv)ethyl chloride is treated with aqueous. KOH. (v)methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN.

  7. Ans. Dehydrohalogenation n- Butylchloride CH,CH2CH CH2 + KCl + H2O But-1-ene MgBr Bromobenzene Phenylmagnesium bromide (i )6-8% , 623K, 300 atm CI + NaOH (aq) (ii) Dil. HCT Chlorobenzene (v)CH,CH2CKOH (aq) HydrosCH,CH2-oH+KCI+ H2o (v) CH,-Br2Na +BrCHCH2NaBr (v)CH,-CI+KCNot0,CH,C N + KCI Ethykchloride Ethyl alcohol ether 3 (wurtzreaction) Nucleophilicsubstitution Methyl cyanide Methyl chloride