{"id":10632,"date":"2019-12-09T10:37:44","date_gmt":"2019-12-09T05:07:44","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10632"},"modified":"2021-07-02T15:43:18","modified_gmt":"2021-07-02T10:13:18","slug":"2nd-puc-chemistry-question-bank-chapter-10","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-10\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 10 Haloalkanes and Haloarenes"},"content":{"rendered":"

You can Download Chapter 10 Haloalkanes and Haloarenes Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Chemistry Question Bank Chapter 10 Haloalkanes and Haloarenes<\/h2>\n

2nd PUC Chemistry Haloalkanes and Haloarenes NCERT Textbook Questions<\/h3>\n

Question 1.
\nName, the following halides according to IUPAC system and classify as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
\n(i) (CH3<\/sub>)2<\/sub>CHCH(Cl)CH3<\/sub>
\n(ii) CH3<\/sub>CH2<\/sub>CH(CH3<\/sub>)CH(C2<\/sub>H5<\/sub>)Cl
\n(iii) CH3<\/sub>CH2<\/sub>C(CH3<\/sub>)2<\/sub>CH3<\/sub>I
\n(iv) (CH3<\/sub>)3<\/sub>CCH2<\/sub>CH(Br)C6<\/sub>H5<\/sub>
\n(v) CH3<\/sub>CH(CH3<\/sub>)CH(Br)CH3<\/sub>
\n(vi) CH3<\/sub>C(C2<\/sub>H5<\/sub>)2<\/sub>CH2<\/sub>Br
\n(vii) CH3<\/sub>C(Cl)(C2<\/sub>H5<\/sub>)CH2<\/sub>CH3<\/sub>
\n(viii) CH3<\/sub>CH=C(Cl)CH2<\/sub>CH(CH3<\/sub>)2<\/sub>
\n(ix) CH3<\/sub>CH=CHC(Br)(CH3<\/sub>)2<\/sub>
\n(x) P-ClC6<\/sub>H4<\/sub>CH2<\/sub>CH(CH3<\/sub>)2<\/sub>
\n(xi) m-CICH2<\/sub>C6<\/sub>H4<\/sub>CH2<\/sub>C(CH3<\/sub>)3<\/sub>
\n(xii) o-Br-C6<\/sub>H4<\/sub>CH(CH3<\/sub>)CH2<\/sub>CH3<\/sub>
\nAnswer:<\/p>\n

\"2nd<\/p>\n

\"2nd
\n\"2nd
\n\"2nd<\/p>\n

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Question 2.
\nGive the IUPAC names of the following compounds:
\n(i) CH3<\/sub>CH(CI)CH(Br)CHCH3<\/sub>
\n(ii) CHF2<\/sub>CBrClF
\n(iii) ClCH2<\/sub>C = CCH2<\/sub>Br
\n(iv) (CCl3<\/sub>)3<\/sub>CCl
\n(v) CHCH3<\/sub>C(p-ClCCH6<\/sub>HCH4<\/sub>)CH2<\/sub>CH(Br)CHCH3<\/sub>
\n(vi) (CH3<\/sub>)3<\/sub>CCH=ClCC6`<\/sub>H4<\/sub>I-p
\nAnswer:<\/p>\n

\"2nd
\n\"2nd<\/p>\n

Question 3.
\nWrite the structures of the following organic halogen compounds.
\n(i) 2-Chloro-3-methylpentane
\n(ii) p-Bromochlorobenzene
\n(iii) l-Chloro-4-ethylcyclohexane
\n(iv) 2-(2-Chlorophenyl)-l-iodooctane
\n(v) Perfluorobenzene
\n(vi) 4-tert-Butyl-3-iodoheptane
\n(vii) l-Bromo-4-sec-butyl-2-methyl- benzene .
\n(viii) l,4-Dibromobut-2-ene
\nAnswer:<\/p>\n

\"2nd
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nWhich one of the following has the highest dipole moment?
\n(i) CH2<\/sub>Cl2<\/sub>
\n(ii) CHCl3<\/sub>
\n(iii) CCl4<\/sub>
\nAnswer:<\/p>\n

\"2nd
\n\"2nd
\nCCl4<\/sub> is a symmetrical molecule. Therefore, the dipole moments of all four C-Cl bonds cancel each other. Hence its resultant dipole moment is zero.<\/p>\n

As shown in the above figure, in CHCl3<\/sub>, the resultant dipole moments of two C-Cl bonds is opposed by the resultant dipole moments of one C-H and one C-Cl bond. Since the resultant of one C-H and one C-Cl bond is smaller than the resultant of the two C-Cl bonds dipole moments, the opposition is to a small extent. As a result CHCl3<\/sub>\u00a0has a small net dipole moment.<\/p>\n

On the other hand, in case of CH2<\/sub>,Cl2<\/sub> the resultant of the dipole moments of two C-Cl bonds is strengthened by the resultant of the dipole moments of two C-H bonds. As a result, CH2<\/sub>Cl2<\/sub> has a higher dipole moment. Hence CH2<\/sub>Cl2<\/sub>\u00a0has the highest dipole moments among the three compounds.
\nHence, the given compounds can be arranged in the increasing order of their dipole moments as
\nCCl4<\/sub> < CHCl3<\/sub> < CH2<\/sub>Cl2<\/sub><\/p>\n

Question 5.
\nA hydrocarbon C5<\/sub>H10<\/sub> does not react with chlorine in dark but gives a single monochloro compound C5<\/sub>H9<\/sub>Cl in bright sunlight. Identify the hydrocarbon.
\nAnswer:
\nA hydrocarbon with molecular formula, C5<\/sub>H10<\/sub> belongs to the group with a general molecular form Cn<\/sub>H2n<\/sub>. therefore, it may either be an alkene or a cycloalkane since hydrocarbon does not react with chlorine in the dark, it cannot be alkene. Further, the hydrocarbon gives a single monochloro compound, C5<\/sub>H9<\/sub>Cl by reacting with chlorine in might sunshine since the formed compound is monochloro one all the C-H bonds should be equivalent. Hence the compound should be a cycloalkane. Hence the compound is C5<\/sub>H10<\/sub> (cyclopentane).<\/p>\n

\"2nd<\/p>\n

Question 6.
\nWrite the isomers of the compound having formula C4<\/sub>H9<\/sub>Br.
\nAnswer:
\nThere are four isomers of the compound having the formula C4<\/sub>H9<\/sub>Br. These isomers are given by<\/p>\n

\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nWrite the equations for the preparation of 1-iodobutane from
\n(i) 1-butanol
\n(ii) 1-chlorobutane
\n(iii) but-l-ene.
\nAnswer:<\/p>\n

\"2nd<\/p>\n

Question 8.
\nWhat are ambident nucleophiles? Explain with an example.
\nAnswer:
\nAmbident nucleophiles are nucleophiles having two nucleophilic sites. Thus ambident nucleophiles have two sites through which they can attack. For example, nitrite ion is an ambident nucleophile.<\/p>\n

\"2nd
\nNitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.<\/p>\n

\"2nd<\/p>\n

Question 9.
\nWhich compound in each of the following pairs will react faster in SN<\/sub>2 reaction with OH?
\n1. CH3<\/sub>Br or CH3<\/sub>I
\n2. (CH3<\/sub>)3<\/sub>CCl or CH3<\/sub>Cl
\nAnswer:
\n1. In the SN<\/sub>2 mechanism the reactivity of halides for the same alkyl group increase in the order. This happens because as the size increases the halide ion becomes a better leaving group.
\nR-F << R-Cl < R – Br < R-I
\nTherefore, CH3<\/sub>I will react faster than CH3<\/sub>Br in SN<\/sub>2 reaction with image 17.<\/p>\n

2. The SN<\/sub>2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. But, in case (CH3<\/sub>)3<\/sub> CCl, the attack of the nucleophile at the carbon atom is hindered by the presence of the bulky substituents on that carbon atom bearing-the leaving group in CH3<\/sub>Cl. Hence CH3<\/sub>Cl reacts faster than (CH3<\/sub>)3<\/sub> CCl in SN<\/sub>2 reaction with \"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nPredict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
\n1. 1-Bromo-l-methylcyclohexane
\n2. 2-Chloro-2-methylbutane ,
\n3. 2,2,3-Trimethyl-3-bromopentane.
\nAnswer:
\n1.<\/p>\n

\"2nd
\nIn this compound, all \u03b2 hydrogens are equivalent thus dehydrohalogenation gives only one alkene.<\/p>\n

\"2nd<\/p>\n

2.<\/p>\n

\"2nd
\nIn the given compound, there are two different sets of equivalent P – hydrogen atoms labelled as \u2018a\u2019 and \u2018b\u2019 thus dehydrohalogenation yields two alkenes.<\/p>\n

\"2nd
\nSayt zeff\u2019s rule implies that in dehydrohalogenate reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced.
\nTherefore, 2-methyl but – 2- ene is the major product in the reaction.<\/p>\n

3.<\/p>\n

\"2nd
\nIn the given compound, there are two different sets of equivalent P – hydrogen atoms labelled as \u2018a\u2019 and \u2018b Thus dehydrogenation of the compound yields two alkenes.<\/p>\n

\"2nd
\n2 -1 thyl – 3, 3 – demethyl but – tene
\nAgain according to Saytzeff\u2019s rule
\n3,4,4 – trimethylpent – 2- ene is the major product.<\/p>\n

Question 11.
\nHow will you bring about the following conversions?
\n(i) Ethanol to but-l-yne
\n(ii) Ethane to bromoethene
\n(iii) Propene to 1-nitropropane
\n(iv) Toluene to benzyl alcohol
\n(v) Propene to propyne
\n(vi) Ethanol to ethyl fluoride
\n(vii) Bromomethane to propanone
\n(viii) But-l-ene to but-2-ene
\n(ix) 1-ChIorobutane to n-octane
\n(x) Benzene to biphenyl.
\nAnswer:<\/p>\n

\"2nd<\/p>\n

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\n\"2nd<\/p>\n

Question 12.
\nExplain why
\n1. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?
\n2. Alkyl halides, though polar, are immiscible with water?
\n3. Grignard reagents should be prepared under anhydrous conditions?
\nAnswer:
\n1.<\/p>\n

\"2nd
\nIn chlorobenzene, the Cl-atom is linked to a sp2<\/sup> hybridized carbon atom. In cyclohexyl chloride, the Cl atom is linked to a sp3<\/sup> hybridized carbon. Now, sp2<\/sup> hybridized carbon has more s-character than sp3<\/sup> 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 cyclohexyl 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 C-Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.<\/p>\n

2. 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.<\/p>\n

3. Grignard reagents are very reaction. In the presence of moisture, they react to give alkanes
\nR-Mg-X + H2<\/sub>O \u2192 RH + Mg(OH)X
\nTherefore, Grignard reagents should be prepared in anhydrous conditions.<\/p>\n

Question 13.
\nGive the uses of freon 12, DDT, carbon tetrachloride and iodoform.
\nAnswer:
\nUses of freon – 12:
\nFreon -12 (dichlorodifluoromethane, CF2<\/sub>Cl2<\/sub>) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays etc. However, it damages the ozone layer. Hence its manufacture is banned.<\/p>\n

Uses of DDT:
\nDDT (p, p – dichlorodiphenylchloroethane) is one of the best known pesticides. It is very effective against mosquito and lice. However, due to its harmful effects, its manufacture is banned;
\nUses of carbon tetrachlorides (CCl4<\/sub>):<\/p>\n

    \n
  • It is used for manufacturing refrigerants and propellants for aerosol cans.<\/li>\n
  • It is used as feedstock in the synthesis of CFCs and other chemicals.<\/li>\n
  • It is used as a solvent in the manufacture of pharmaceutical products.<\/li>\n
  • CCl4<\/sub> was once widely used as a cleaning fluid, a degreasing agent in industries, a spot remover homes, and a fire extinguisher.<\/li>\n<\/ul>\n

    Uses of iodoform (CHI3<\/sub>):
    \nIodoform was used as an antiseptic, but now it has been replaced by other formulations containing iodine-due to its objectionable smell. The antiseptic property of CHI3<\/sub> is only due to the liberation of free iodine when it comes in contact with the skin.<\/p>\n

    \"KSEEB<\/p>\n

    Question 14.
    \nWrite the structure of the major organic product in each of the following reactions:<\/p>\n

    \"2nd
    \nAnswer:<\/p>\n

    \"2nd
    \n\"2nd<\/p>\n

    Question 15.
    \nWrite the mechanism of the following reaction:<\/p>\n

    \"2nd
    \nAnswer:
    \nThe given reaction is<\/p>\n

    \"2nd
    \nThe reaction is an SN<\/sub>2 reaction. In this reaction, CN–<\/sup> acts as the nucleophile and attacks the carbon atom to which Br is attacked. CN–<\/sup> is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.<\/p>\n

    \"2nd<\/p>\n

    Question 16.
    \nArrange the compounds of each set in order of reactivity towards SN2 displacement:
    \n1. 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
    \n2. l-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane
    \n3. 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, l-Bromo-2-methylbutane, l-Bromo-3-methylbutane.
    \nAnswer:
    \n1.<\/p>\n

    \"2nd
    \nAn SN<\/sub>2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards SN<\/sub>2 displacement decreases. Due to the presence of substituents, hindrance tp the approaching nucleophile increases in the following order.
    \n1- Bromopentane < 2-bromopentane < 2 – Bromo-2-methyl butane
    \nHence, the increasing order of reactivity towards SN<\/sub>2 displacement is
    \n2- Bromo – 2- methyl butane < 2-bromopentane < 1-Bromopentane<\/p>\n

    2.<\/p>\n

    \"2nd
    \nsince steric hindrance in alkyl halides increases in the order of 1\u00b0 <2\u00b0 < 3\u00b0, the increasing order of reactivity towards SN2 displacement is 3\u00b0 < 2\u00b0 <1\u00b0.
    \nHence, the given set of compounds can be arranged in the increasing order of their reactivity towards
    \nSN<\/sub>2 displacement as:
    \n2-Bromo – 2- methyl butane < 3-bromo-2-methylbutane < 1-Bromo -3- methy lbutane.<\/p>\n

    3.<\/p>\n

    \"2nd
    \nThe steric hindrance to the nucleophile in the SN<\/sub>2 mechanism increases with a decrease in the distance of the substituents form the atom containing the leaving group. Further, the steric hindrance increases in the increases with increase in the number of substituents. Therefore, the increasing order of steric hindrances in the given compounds is as below:
    \n1- Bromobutane < l-Bromo-3-methylbutane < 1 -Bromo-2-methylbutane < 1 -Bromo-2, 2- demethylbutane.
    \nHence, the increasing order of reactivity of the given compound towards SN<\/sub>2 mechanism is
    \nl-Bromo-2, 2-dimethylbutane < 1 -Bromo-2-methyl butane < l-Bromo-3-methylbutane < 1-Bromobutane.<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nOut of C6<\/sub>H5<\/sub>CH2<\/sub>Cl and C6<\/sub>H5<\/sub>CHClC6<\/sub>H5<\/sub>, which is more easily hydrolysed by aqueous KOH?
    \nAnswer:<\/p>\n

    \"2nd<\/p>\n

    Question 18.
    \np-Dichlorobenzene has higher m.p. and solubility than those of o- and m-isomers. Discuss.
    \nAnswer:<\/p>\n

    \"2nd
    \np-Dichlorobenzene is more symmetrical than o- and m- isomers. For this season, it fits more closely than o- and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has higher mp and lower solubility than o- and m- isomers.<\/p>\n

    Question 19.
    \nHow the following conversions can be carried out?
    \n(i) Propene to propan-l-ol
    \n(ii) Ethanol to but-l-yne
    \n(iii) 1-Bromopropane to 2-bromopropane
    \n(iv) Toluene to benzyl alcohol
    \n(v) Benzene to 4-bromonitrobenzene
    \n(vi) Benzyl alcohol to 2-phenylethanoic acid
    \n(vii) Ethanol to propanenitrile
    \n(viii) Aniline to chlorobenzene
    \n(ix) 2-ChIorobutane to 3, 4-dimethyIhexane
    \n(x) 2-Methyl-l-propene to 2-chloro-2-methylpropane
    \n(xi) Ethyl chloride to propanoic acid
    \n(xii) But-l-ene to n-butyliodide
    \n(xiii) 2-ChIoropropane to 1-propanol
    \n(xiv) Isopropyl alcohol to iodoform
    \n(xv) Chlorobenzene to p-nitrophenol
    \n(xvi) 2-Bromopropane to 1-bromopropane
    \n(xvii) Chloroethane to butane
    \n(xviii) Benzene to diphenyl
    \n(xix) tert-Butyl bromide to isobutyl bromide
    \n(xx) Aniline to phenylisocyanide
    \nAnswer:
    \n\"2nd
    \n\"2nd
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    \"2nd
    \n\"2nd
    \n\"2nd
    \n\"2nd
    \n\"2nd
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 20.
    \nThe 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.
    \nAnswer:
    \nIn an aqueous solution, KOH almost completely ionises to give OH–<\/sup> ions. OH–<\/sup> ions is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.<\/p>\n

    \"2nd
    \nOn the other hand, an alcoholic soln of KOH contains alkoxide (RO–<\/sup>) ion, which is a strong base. Thus, it can abstract a hydrogen from the \u03b2 – carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.<\/p>\n

    \"2nd
    \nOH–<\/sup> ion is a much weaker base than RO–<\/sup> ion. Also, OH–<\/sup> ion is highly solvated in an aqueous solution and as a result the basic character of OH’ ion decreases. Therefore, it cannot abstract a hydrogen from \u03b2 – carbon.<\/p>\n

    Question 21.
    \nPrimary alkyl halide C4<\/sub>H9<\/sub>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 gives compound (d), C8<\/sub>H18<\/sub> 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.
    \nAnswer:
    \nThere are two primary alkyl halides having the formula C4<\/sub>H9<\/sub>Br. They one n-butyl bromide and isobutyl bromide<\/p>\n

    \"2nd
    \nSince (a) reacts with Na to give a compound C8<\/sub>H18<\/sub> which is different from the compound formed
    \nwhen n – butyl bromide is reacted with Na, (a) has to be isobutyl bromide.<\/p>\n

    \"2nd<\/p>\n

    Question 22.
    \nWhat happens when
    \n(i) n-butyl chloride is treated with alcoholic KOH,
    \n(ii) bromobenzene is treated with Mg in the presence of dry ether,
    \n(iii) chlorobenzene is subjected to hydrolysis,
    \n(iv) ethyl chloride is treated with aqueous KOH,
    \n(v) methyl bromide is treated with sodium in the presence of dry ether,
    \n(vi) methyl chloride is treated with KCN?
    \nAnswer:<\/p>\n

    \"2nd<\/p>\n

    \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

    You can Download Chapter 10 Haloalkanes and Haloarenes Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Chemistry Question Bank Chapter 10 Haloalkanes and Haloarenes 2nd PUC Chemistry Haloalkanes and Haloarenes NCERT Textbook …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10632"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10632"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10632\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10632"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10632"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10632"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}