Organic chemistry: dehydrohalogenation
posted by Steven on .
Hello. I know you guys don't do organic chemistry but because I got so much help from you guys in general chemistry, I was wondering if you guys can help me out again. :P
Here's my question.
In dehydrohalogenation, how do you know whether a reaction is E1 or E2?
My textbook says that E1 dehydrohalogenation occurs if there is an extremely weak base [side question: what qualifies as an extremely weak base?] and can only occur among tertiary and secondary alkyl halide. E2 dehydrohalogenation occurs among all kinds of alkyl halides (teritiary, secondary, and primary) as long as you have a strong base. Is this true?
I understand the mechanisms of E1, E2, and dehydrohalogenation. I just don't know how to find out whether an alkyl halide undergoing dehydrohalogenation will follow an E1 or an E2 reaction.
I have an OLD copy of Organic Chemistry by Morrison and Boyd which discusses E1 and E2 reactions. That book confirms what you have read and stated that E1 dehydrohalogenations occur among tertiary and secondary alkyl halides while E2 reactions occur with all three kinds of alkyl halides. As to which mechanism, E2 or E1, I quote from Morrison and Boyd on page 314 (4th edition), "For a given substrate, then, the more concentrated the base, or the stronger the base, the more E2 is favored or E1. Under the conditions typically used to bring about dehydrohalogenation--a concentrated solution of a strong base--the E2 mechanism is the path taken by elimination. In general, the E1 mechanism is encountered only with secondary or tertiary substrates, and in solutions where the base is either in low concentration or weak--typically where the base is the solvent."
A strong base would be concentrated KOH in alcohol or a similar alkoxide. A weak base would be water or alcohol (as a solvent as stated above).
My conclusions would be as follows: If dealing with a primary alkyl halide, pick E2 over E1. If dealing with secondary or tertiary alkyl halides, pick E2 when the reaction has a strong base and pick E1 when the reaction has a weak base.
You are correct that I answer very few organic chemistry questions; howwever, DrRuss checks in now and then. Perhaps he will see this and comment further.
I made a typo in which I typed the word or but I intended to type over. Here is the correction in bold.
"For a given substrate, then, the more concentrated the base, or the stronger the base, the more E2 is favored over E1.
DrBob has essentially covered it. It is not that easy to generalise, you need to consider each halide looking for steric influences, stabilisation of a carbocation and stability of the product. Some general points:
E1 is a two-step process of elimination ionization and deprotonation.
1. Ionization, Carbon-halogen breaks to give a carbocation intermediate.
2. Deprotonation of the carbocation.
Typical of tertiary and some secondary substituted alkyl halides.
The reaction rate is influenced only by the concentration of the alkyl halide because carbocation formation is the slowest, rate-determining step. Therefore first order kinetics apply.
Reaction mostly occurs in complete absence of base or presence of only weak base.
E1 reactions are in competition with SN1 reactions because both involve a carbocation.
E2 is a one-step process of elimination with a single transition state.
Typical of secondary or tertiary substituted alkyl halides. It is also observed with primary alkyl halides if a hindered base is used.
The reaction rate, influenced by both the alkyl halide and the base, is second order.
The reaction rate is influenced by halogens reactivity; iodide and bromide being favoured. There can be competition between elimination reaction and nucleophilic substitution. I.e. there is competition between E2 and SN2 and also between E1 and SN1. Generally, elimination is favoured over substitution when:
steric hindrance increases
the steric bulk of the base increases for example potassium tert-butoxide
the nucleophile is poor
Hope this helps
THANKS SO MUCH GUYS! (sorry for the delayed response; I've been pretty busy)
It's nice to know that this forum/website is filled with intelligent people!