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CH3Br, CH3CH2CH2I, (CH3)3Cl, CH3CHBrCH3, or CH3CHICH3

2006-11-27 08:00:09 · 3 answers · asked by lovely_antionette 1 in Science & Mathematics Chemistry

3 answers

The rate detemining step in SN1 reactions is the first step. Step one in SN1 reaction involves carbocation formation by ionisation. Any positive feature of the alkyl halide or leaving group will increase the reaction rate:
In general the rate of reaction is:
Methyl > 1° > 2° >>>3°
Similarly leaving group (LG) ability is a function of stability of the LG. The weaker the conjugate base, the better the LG
For the acids: HF, HCl, HBr, HI the acidity increases as you go down the period, so then the anion stability also increases as you go down the period. Then the general stability is:
I- > Br- > Cl- >> F-

Putting it all together you get (slowest to fastest):
CH3Br < CH3CH2CH2I < CH3CHBrCH3 < CH3CHICH3 < (CH3)3Cl

2006-11-28 04:24:00 · answer #1 · answered by Peter B 3 · 0 0

CH3Br is least reactive in SN1 (though it could be quite reactive in SN2) because the CH3+ carbonium ion (carbocation) is not very stable, and Br- is not as good a leaving group as some others. CH3CH2CH2Cl is slightly more reactive, because a C-C+ carbonium ion is slightly more stable than CH3+. CH3CHBrCH3 is next, because the isopropyl carbonium ion is fairly stable, and Br- is a good leaving group. CH3CHICH3 is next, because I- isan even better leaving group. (CH3)3C-Cl is most reactive, because the tertiary (CH3)3C+ carbonium ion is the most stable on the list.

2006-11-27 08:11:11 · answer #2 · answered by steve_geo1 7 · 0 0

In SN2 reactions, chloromethane > a million-chloropentane > chlorocyclopentane because of increasing crowding of communities around the carbon atom. a million-Chloro-a million-pentene must be unreactive in route of SN1 or SN2 reaction, by way of the indisputable fact that's a vinyl-form by-product. 3-Chloro-a million-pentene ought to move SN1 in preference to SN2, because it is going with the help of an very solid allylic carbonium ion. Chloromethylbenzene likewise, because the carbonium ion is benzylic. 2-Chloro-2-methylbutane likewise, because the carbonium ion is tertiary. Chlorocyclopentane ought to move SN1 or SN2 relying on solvent polarity, because the carbonium ion can be a a lot less solid secondary.

2016-11-27 02:11:33 · answer #3 · answered by ? 4 · 0 0

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