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 = SN1
#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...???
So do these look to be as though I am on the right track with my answers to the above questions???
Except, for the FIRST question...I think that #3 should be SN2, rather than SN1 (substitution reactions).
Let's go through each question step-by-step and check if your answers are correct:
1) Match the following statements with SN1 or SN2 reactions:
#1 = SN2 - Correct. In SN2 reactions, the order of reactivity in alkyl halides is indeed methyl>primary>secondary>tertiary.
#2 = SN1 - Correct. SN1 reactions involve the formation of an intermediate carbocation.
#3 = SN1 - Correct. The rate-limiting step in SN1 reactions involves the alkyl halide and the nucleophile.
#4 = SN1 - Correct. The rate-limiting step in SN1 reactions indeed involves the formation of an intermediate carbocation.
Your answers for the first question are all correct.
2) Arrange the leaving groups in order of increasing leaving group ability:
The order from least to greatest leaving group ability is: IV (F^-), III (Cl^-), II (Br^-), I (I^-). Therefore, your order should be I, III, II, IV.
Your answer for the second question is incorrect. The correct order is I, III, II, IV.
3) Rank the anticipated order of boiling points for this set of compounds (highest to lowest):
The order from highest to lowest boiling points is: 4 (HOCH2CH2CH2OH), 2 (CH3CH2CH2CH2OH), 1 (CH3CH2OCH2CH3), 3 (CH3OCH3). Therefore, your order should be 4, 2, 1, 3.
Your answer for the third question is correct.
Overall, two out of the three answers you provided are correct.