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???
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)?????
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
Let's go through each question and evaluate your answers:
1) Match the following statements with SN1 or SN2 reactions:
1) The order of reactivity in alkyl halides is methyl>primary>secondary>tertiary
Your answer: SN2
2) There is an intermediate carbocation
Your answer: SN1
3) The rate-limiting step involves the alkyl halide and the nucleophile
Your answer: SN2
4) The rate-limiting step involves the formation of an intermediate carbocation
Your answer: SN1
Your answers are correct for this question. Good job!
2) Arrange the leaving groups in order of increasing leaving group ability.
I) Cl (with negative charge)
II) I (with negative charge)
III) Br (with negative charge)
IV) F (with negative charge)
Your answer: IV, I, III, II
Your answer is correct. The leaving group ability generally follows the trend F > Cl > Br > I.
3) Rank the anticipated order of boiling points for this set of compounds (highest to lowest):
1) CH3CH2OCH2CH3 (i.e., CH3CH2OCH2CH3)
2) CH3CH2CH2CH2OH
3) CH3OCH3
4) HOCH2CH2CH2OH
Your answer: 4, 2, 1, 3
Your answer is correct. The boiling point generally increases with increasing molecular weight and hydrogen bonding capabilities.
Overall, your answers are correct for all the questions. Great job!