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When tetra hydra furan is treated with excess HI,the product formed is a)2,4-diiodobutane b)1,4-butanediol c)2-iodotetrahydrofuran d)4-iodo-1- butanol?
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When tetra hydra furan is treated with excess HI,the product formed is...
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When tetra hydra furan is treated with excess HI,the product formed is...
When tetrahydrofuran (THF) is treated with excess hydrogen iodide (HI), the product formed is 2-iodotetrahydrofuran. Let's understand the reaction and the mechanism involved in detail.

Reaction:
Tetrahydrofuran + Excess HI → 2-Iodotetrahydrofuran

Mechanism:
1. Protonation of Tetrahydrofuran:
In the presence of excess HI, the acidic hydrogen of HI protonates the oxygen atom of the tetrahydrofuran ring. This leads to the formation of a more reactive oxonium ion intermediate.

2. Nucleophilic Attack:
The iodide ion (I-) acts as a nucleophile and attacks the electrophilic carbon in the oxonium ion. This results in the formation of a new bond between the carbon and iodine, while the oxygen receives a positive charge.

3. Proton Transfer:
To regain stability, a proton transfer occurs from the positively charged oxygen atom to the iodide ion. This forms a new C-I bond and regenerates the HI catalyst.

4. Deprotonation:
The final step involves the deprotonation of the positively charged oxygen atom. This generates the product, 2-iodotetrahydrofuran.

Explanation:
The product formed, 2-iodotetrahydrofuran, is obtained due to the addition of iodide ion at the electrophilic carbon of the tetrahydrofuran ring. This occurs because the iodide ion is a stronger nucleophile compared to the oxygen atom of the tetrahydrofuran ring. The reaction takes place under acidic conditions, which promote the protonation of the tetrahydrofuran ring and facilitate the nucleophilic attack.

The other options mentioned in the question are not the major products of this reaction. Option (a), 2,4-diiodobutane, would require an additional step involving the addition of a butyl group to the tetrahydrofuran ring, which is not mentioned. Option (b), 1,4-butanediol, would require the reduction of the C-I bond, which is not the expected outcome under excess HI conditions. Option (d), 4-iodo-1-butanol, would require the addition of a hydroxyl group to the tetrahydrofuran ring, which is not mentioned.

In conclusion, the correct answer is option (c), 2-iodotetrahydrofuran, which is formed when tetrahydrofuran is treated with excess hydrogen iodide.
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When tetra hydra furan is treated with excess HI,the product formed is a)2,4-diiodobutane b)1,4-butanediol c)2-iodotetrahydrofuran d)4-iodo-1- butanol?
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When tetra hydra furan is treated with excess HI,the product formed is a)2,4-diiodobutane b)1,4-butanediol c)2-iodotetrahydrofuran d)4-iodo-1- butanol? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about When tetra hydra furan is treated with excess HI,the product formed is a)2,4-diiodobutane b)1,4-butanediol c)2-iodotetrahydrofuran d)4-iodo-1- butanol? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When tetra hydra furan is treated with excess HI,the product formed is a)2,4-diiodobutane b)1,4-butanediol c)2-iodotetrahydrofuran d)4-iodo-1- butanol?.
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