The correct statement about the given reaction is:

M e_{3} C - O M e

with anhydrous

H I

gives this reaction

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M e_{3} C - O M e

with concentrated

H I

gives this reaction

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both A and B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C Both A and B
In this reaction, $H I$ is added to an ether. One of the alkyl groups in this ether is a tert-alkyl group. Hence, tert-alkyl iodide and lower alcohol will be formed in this reaction.
Both anhydrous and conc. $H I$ can be used for this reaction.

Suggest Corrections

Similar questions

H I

M e_{3} C - O - C H_{2} C H_{3} + H I 1 mole Δ ⟶

M e_{3} C - O | | C - 18 O - C M e_{3} d i l H_{2} S O_{4} - ----- \to

((M e)_{3} C - O - C l)

M e_{3} C - O - C H_{2} C H_{3} + H I (1 mole) Δ - \to

Join BYJU'S Learning Program