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?
P.S.
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
basicity increases
temperature 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!
Steven
Hello! I'm here to help you with your question on dehydrohalogenation reactions.
Determining whether a dehydrohalogenation reaction is E1 or E2 depends on several factors, including the strength of the base, the nature of the alkyl halide, and the reaction conditions. Let's break it down further:
1. E1 Dehydrohalogenation:
- E1 reactions occur when the rate-determining step of the reaction involves the formation of a carbocation intermediate.
- The reaction usually occurs with tertiary and secondary alkyl halides, as these carbocations are more stable due to electron-donating alkyl groups.
- The strength of the base used in an E1 reaction is typically extremely weak, such as neutral solvents like water or alcohols.
- An extremely weak base does not favor elimination but instead promotes the formation of a carbocation intermediate that can undergo subsequent elimination.
2. E2 Dehydrohalogenation:
- E2 reactions occur when the base directly removes a proton from the alkyl halide, resulting in the simultaneous loss of a halide ion and the formation of a C=C double bond.
- E2 reactions can occur with tertiary, secondary, and primary alkyl halides as long as a strong base is present.
- A strong base, such as hydroxide ion (OH⁻) or alkoxide ions (RO⁻), is required for E2 reactions as they are capable of removing a proton from the alkyl halide.
To answer your side question, an extremely weak base refers to a base that has poor affinity for protons. For example, water (H₂O) or alcohols, which are weak bases, are typically used in E1 reactions because they have a minimal tendency to deprotonate the alkyl halide directly.
Overall, your textbook's statement is generally accurate. E1 dehydrohalogenation typically occurs with tertiary and secondary alkyl halides using an extremely weak base, while E2 dehydrohalogenation can occur with various types of alkyl halides, including tertiary, secondary, and primary, provided a strong base is present.
I hope this explanation helps! If you have any more questions, feel free to ask.