In which case the product is a mixture of geometrical isomers

P h C H O H C N / N a O H ⟶

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

P h C H O A g N O_{3} / N H_{4} O H ⟶

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

P h C H O A c_{2} O / A c O N a ⟶

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

P h C H O H C H O / O H^{(-)} ⟶

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

Open in App

Solution

The correct option is C $P h C H O A c_{2} O / A c O N a ⟶$
$P h C H O A c_{2} O / A c O N a ⟶ P h C H = C H C O O H$
(show geometrical isomerism.

Suggest Corrections

Similar questions

[P t (N H_{3}) (N O_{2}) P y (N H_{2} O H)]^{+}

C H_{3} C H = N - O H

C H_{3} - C H = C H - C H = C H - C_{2} H_{5}

In a complex MA₃X₃, unidentate ligands A and X may give:

[M (N H_{3})_{2} (p y)_{2} (N O_{2})_{2}] + N H_{3} - --- \to - N O_{2} [M (N H_{3})_{3} (p y)_{2} (N O_{2})]

Join BYJU'S Learning Program