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The product formed during the following reaction are
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- C
- D
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Verified Answer
The correct answer is:
The reaction involved is
This is because the reaction occurs by \( S_{N} I \) mechanism. The formation of products is governed by the stability of the
carbocation formed from the cleavage of \( \mathrm{C}-\mathrm{O} \) bond in the protonated ether (oxonium ion). Since tert-butyl carbocation
\( \left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}\right] \)is more stable than the methyl carbocation \( \mathrm{CH}_{3}^{+} \)
Therefore, cleavage of \( \mathrm{C}-\mathrm{O} \) bond gives a more stable carbocation \( \left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}\right] \)and methanol as products.
Then iodide ion \( \left(I^{-}\right) \)attacks tert-butyl carbocation to formtert-butyl iodide.
This is because the reaction occurs by \( S_{N} I \) mechanism. The formation of products is governed by the stability of the
carbocation formed from the cleavage of \( \mathrm{C}-\mathrm{O} \) bond in the protonated ether (oxonium ion). Since tert-butyl carbocation
\( \left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}\right] \)is more stable than the methyl carbocation \( \mathrm{CH}_{3}^{+} \)
Therefore, cleavage of \( \mathrm{C}-\mathrm{O} \) bond gives a more stable carbocation \( \left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}\right] \)and methanol as products.
Then iodide ion \( \left(I^{-}\right) \)attacks tert-butyl carbocation to formtert-butyl iodide.
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