O.Chemistry *****PLEASE CHECK*****!!!!

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Please check these out for me, if I am correct:

Match the following statements with SN1 or SN2 reactions:

1) The order of reactivity in alkyl halides is methyl>primary>secondary>tertiary

2) There is an intermediate carbocation

3) The rate-limiting step involves the alkyl halide and the nucleophile

4) The rate-limiting step involves the formation of an intermediate carbocation


#1 = SN2

#2 = SN1

#3 = SN2

#4 = SN1


Arrange the leaving groups in order of increasing leaving group ability.(least first):

I) Cl (with neg. charge)

II) I (with neg. charge)

III) Br (with neg. charge)

IV) F (with neg. charge)


I think that the order should be:

I^-, Br^-, Cl^-, F^- (or: IV, I, III, II)?????



Rank the anticipated order of boiling points for this set of compounds.( highest to lowest)):


1) CH_3CH_2OCH_2CH_3 (ie.CH3CH2OCH2CH3)

2) CH_3CH_2CH_2CH_2OH

3) CH_3OCH_3

4) HOCH_2CH_2CH_2OH

I think that the order should be: 4,2,1,3...???



Does it look to be that I am on the right track with my answers to the above questions???

  • O.Chemistry *****PLEASE CHECK*****!!!! -

    1) The order of reactivity in alkyl halides is methyl>primary>secondary>tertiary

    2) There is an intermediate carbocation

    3) The rate-limiting step involves the alkyl halide and the nucleophile

    4) The rate-limiting step involves the formation of an intermediate carbocation


    #1 = SN2

    #2 = SN1

    #3 = SN2

    #4 = SN1

    all fine

    Arrange the leaving groups in order of increasing leaving group ability.(least first):

    I) Cl (with neg. charge)

    II) I (with neg. charge)

    III) Br (with neg. charge)

    IV) F (with neg. charge)


    I think that the order should be:

    I^-, Br^-, Cl^-, F^- (or: IV, I, III, II)?????

  • O.Chemistry *****PLEASE CHECK*****!!!! -

    1) The order of reactivity in alkyl halides is methyl>primary>secondary>tertiary

    2) There is an intermediate carbocation

    3) The rate-limiting step involves the alkyl halide and the nucleophile

    4) The rate-limiting step involves the formation of an intermediate carbocation


    #1 = SN2

    #2 = SN1

    #3 = SN2

    #4 = SN1

    all fine

    Arrange the leaving groups in order of increasing leaving group ability.(least first):

    I) Cl (with neg. charge)

    II) I (with neg. charge)

    III) Br (with neg. charge)

    IV) F (with neg. charge)


    I think that the order should be:

    I^-, Br^-, Cl^-, F^- (or: IV, I, III, II)?????
    F is less likely to leave because it is so electronegative (look at periodic table) (The leaving group ability to leave follows the group 7 order for the elements noted here [with the least to the most willing to leave going down the table]) Iodine is very large so it is the most willing to leave out of all of the listed elements here. So your second guess was correct

    Rank the anticipated order of boiling points for this set of compounds.( highest to lowest)):


    not sure about the last one

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