Number of stereoisomeric forms of the compound ${C H}_{3} C H = C H C H B r - {C H}_{3}$ is:

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

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

Open in App

Solution

The correct option is D 4
The structure has 1 chiral carbon and a double bond.
So total no of isomers $= 4$
So, option $D$ is correct

Suggest Corrections

Similar questions

C H_{3} - (C H_{3}) C H - C H_{2} - C H_{3} C l_{2}, h v - --- \to N (isomeric products)

C5H11Clfractional distillation−−−−−−−−−−−−→M(isomeric products)

[C o (N H_{3})_{4} C l_{2}]^{+}

C h_{3} C = C - C H_{3}

C H_{3} C H_{2} - C H_{2} - C H_{3}

C H_{3} C H_{2} C = C H

C H_{3} C H = C H_{2}

C H_{3} - C O - C H_{2} C H_{2} C H_{2} C H_{3}

Join BYJU'S Learning Program