The compound that will react with $H B r$ most readily is:

{(C H_{3})}_{3} C O H

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

{(C H_{3})}_{2} C H O H

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

C H_{3} C H_{2} O H

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

C_{6} H_{5} O H

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

Open in App

Solution

The correct option is A ${(C H_{3})}_{3} C O H$
Reactivity order of alcohols with HBr is $3^{\circ} > 2^{\circ} > 1^{\circ}$ alcohol.
The option (A) is a $3^{\circ}$ alcohol and hence it will react most readikly with HBr.
Phenol (option D) does not react with $H B r$ . (Ar-O) bond is not broken unless the ring is activated by the presence of electron withdrawing group (e.g., $N O_{2}$ ) at o- and p-positions.

Suggest Corrections

Similar questions

C H_{3} C H_{2} O H 1

(C H_{3})_{2} C H O H 2

(C H_{3})_{3} C O H 3

N a O H

C H_{3} C H_{2} O H + H C l a n h . Z n C l_{2} - ----- \to

C H_{3} C H_{2} O H + H C l \to

(C H_{3})_{3} C O H + H C l \to

(C H_{3})_{2} C H O H + H C l a n h . Z n C l_{2} - ----- \to

C H_{3} C H_{2} O H,

Join BYJU'S Learning Program