What Are the IR Spectra Indications of Products in E2 Reaction on Dihalides?

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In summary, the reaction of 3,4-dibromo-2,5-dimethylhexane with excess k t-butoxide results in the formation of 2,5-dimethyl-3-hexyne and a smaller amount of a conjugated diene product, with the alkyne being the major product formed.
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Homework Statement



3,4-dibromo-2,5-dimethylhexane treated with excess k t-butoxide will form an alkyne and diene. The IR spectra shows bands at 3010,2995,2980,1610,1450,1380 and 1050 cm^-1.

What are the predicted products? Which one actually formed? Why?

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The Attempt at a Solution



My products are 2,5-dimethyl-3-hexyne and a conjugated diene. Are my products correct?

Also, which product is formed? I searched online and it says the alkyne will form, but from the IR spectra, the diene formed?
 
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Your predicted products are correct. The alkyne product, 2,5-dimethyl-3-hexyne, is indeed formed in this reaction. The IR spectra shows peaks at 3010, 2995, and 2980 cm^-1, which are characteristic of C-H stretching in an alkyne. The peak at 1610 cm^-1 is also characteristic of a C=C stretching in a diene, which suggests the formation of a conjugated diene product. However, this peak could also be attributed to the presence of residual starting material, which could explain the presence of the diene product in the IR spectra. Therefore, based on the overall reaction conditions and the predominant peaks in the IR spectra, it can be concluded that the alkyne product is the major product formed in this reaction.
 

1. What is an E2 reaction on dihalides?

An E2 reaction on dihalides is a type of elimination reaction in which a dihalide (a molecule with two halogen atoms) loses two substituents and a double bond is formed between the two remaining carbon atoms.

2. How does an E2 reaction on dihalides occur?

An E2 reaction on dihalides occurs when a strong base, such as hydroxide or alkoxide, attacks a beta-hydrogen on the dihalide molecule, causing the halogens to leave and form a double bond between the two adjacent carbon atoms.

3. What factors influence the rate of an E2 reaction on dihalides?

The rate of an E2 reaction on dihalides is influenced by the strength of the base, the steric hindrance of the beta-hydrogen, and the stability of the resulting double bond. The reaction is faster with stronger bases, less steric hindrance, and more stable double bonds.

4. What is the difference between an E2 reaction and an SN2 reaction?

An E2 reaction and an SN2 reaction are both types of substitution reactions, but they differ in their mechanisms. In an E2 reaction, the leaving group and the nucleophile are eliminated simultaneously, while in an SN2 reaction, the nucleophile attacks the substrate and pushes off the leaving group in a single step.

5. What are some examples of dihalide compounds that undergo E2 reactions?

Some common examples of dihalides that undergo E2 reactions include 1-bromobutane, 2-chloropropane, and 1-iodobutane. These compounds have beta-hydrogens that are easily accessible for attack by a strong base.

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