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Haloalkanes and Haloarenes Test - 6

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Haloalkanes and Haloarenes Test - 6
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  • Question 1
    1 / -0

    CH= CHCH3 + H – I → CHCHCH2I + CHCHICH3 (major) This reaction is

    Solution

    This is electrophilic addition reaction following Markovnikoffs rule.and negative part that is I will go to the carbon having lesser number of hydrogen.

  • Question 2
    1 / -0

    Finkelstein reaction is

    Solution

    Halide exchange reaction is Finkelstein reaction in which alkyl chloride is converted to alkyl iodide.
    The classic Finkelstein reaction entails the conversion of an alkyl chloride or an alkyl bromide to an alkyl iodide by treatment with a solution of sodium iodide in dry acetone. Sodium iodide is soluble in acetone and sodium chloride and sodium bromide are not. The reaction is driven toward products according to Le chatelier's principle due to the precipitation of the salt insoluble in acetone. For example, in this case chloropropane can be converted to iodopropane:
    CH3CH2CH2Cl(acetone) + NaI(acetone) → CH3CH2CH2I(acetone) + NaCl (s)

  • Question 3
    1 / -0

    In the following reaction, the compound used in the reaction for synthesizing ethyl fluoride is  – +AgF → H3C – F + AgBr

    Solution

    Alkyl fluorides is only formed by Swartz reaction.in swartz reaction alkyl chlorides or bromide reacts with metallic flouride AgF Hg2Fto form alkyl flourides.

  • Question 4
    1 / -0

    Which compound in the following pair reacts faster in SN2 reaction with OH ?

    1. CH3Br or CH3
    2. CH3Cl, (CH3)3CCl
    Solution

    For SN2 reaction leaving group should be good and alkyl halide should be primary.so CH3Br and CH3Cl are 10organic compounds hence rate of  sn2 reaction will be more.

  • Question 5
    1 / -0

    Arrange the following organic compounds in descending order of their reactivity towards SN1 reaction
    C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)(C6H5)Br

    Solution

    The nucleophilic substitution in SN1 occurs in two steps. In step I, the polarised C—Br bond undergoes slow cleavage to produce a carbocation and a bromide ion. This is the slow rate determining step. Thus, the stability of the carbocation formed determines the rate of reaction. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction.
    In C6H5C(CH3)(C6H5)Br carbocation will be tertiary and will be resonance stablised. So the rate of SN1 reaction is greatest in this case. In case of C6H5CH(C6H5)Br, the carbocation formed will be secondary and hence less stable than the previous one, so the reaction will be slower in this case. In C6H5CH(CH3)Br, the carbocation formed will be less stable than that formed in case of C6H5CH(C6H5)Br because in the latter the carbocation is stabilised by two phenyl groups due to resonance. In C6H5CH2Br, the carbocation formed will be primary and hence least stable causing the rate of reaction to be least in this case.

  • Question 6
    1 / -0

    Upon oxidation, one of the following halogen containing compounds forms an extremely poisonous gas  – + O → 2 COCl2 + 2HCl

    Solution

    CHCl3 on reaction with oxygen forms phosgene. Chloroform decomposes when it exposed to sunlight.it decomposes to a harmful gas phosgene(COCl2) Chloroform decomposes when it exposed to sunlight.it decomposes to a harmful gas phosgene(COCl2)

  • Question 7
    1 / -0

    Carbon – halogen bond of alkyl halides is responsible for their nucleophilic substitution, elimination and their reaction with metal atoms to form organometallic compounds because of its

    Solution

    Since halogen atoms are more electronegative than carbon, the carbon-halogen bond of alkyl halide is polarised; the carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge. This is responsible for nucleophilic substitution reactions, elimination reactions and reaction of alkyl halides with a metal to form organometallic compounds.

  • Question 8
    1 / -0

    A mixture containing two enantiomers in equal proportions

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