Surprisingly, the reaction shown below goes through classical carbocation. What is the major product of this reaction ?

t r a n s - 1, 3 - d i b r o m o c y c l o h e x a n e

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

c i s - 1, 3 - d i b r o m o c y c l o h e x a n e

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

t r a n s - 1, 2 - d i b r o m o c y c l o h e x a n e

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

c i s - 1, 2 - d i b r o m o c y c l o h e x a n e

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

Open in App

Solution

The correct option is D $t r a n s - 1, 3 - d i b r o m o c y c l o h e x a n e$
As refer to the above image, $- B r$ attaches to the carbon which is more stable due to more hyperconjugated structures. So, the product formed is $t r a n s -$ $1, 3 - d i b r o m o - c y c l o - h e x a n e$ .

Suggest Corrections

Similar questions

K M n O_{4}

Which of the following reagents would distinguish cis-cyclopenta -1, 2- diol from the trans-isomer?

$B (O H)_{3} + N a O H ⇌ N a B O_{2} + N a [B (O H)_{4}]^{-} + H_{2} O$

How can the reaction above be made to proceed in the forward reaction?

3

1, 2

3

1

3

1

Join BYJU'S Learning Program