You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction

II. CH 2 =CH...

Question

II. ${C H}_{2} = C H C O O H △ - \to {C H}_{2} = {C H}_{2}$

III. $C H_{3} C H (C O O H)_{2} △ - \to C H_{3} C H_{2} C O O H$

Consider the above given reactions,
In which of the cases, parent compound loses its functional group in preference :

I, I I

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

I, I I, I I I

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

I I, I I I

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

I, I I I

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

Open in App

Solution

The correct option is A $I, I I$

$C H_{2} = C H C O O H \to C H_{2} = C H_{2}$ :-looses its preference.

$C H_{3} - C H (C O O H)_{2} \to C H_{3} C H_{2} C O O H$ :- $- C O O H$ does not loses its preference.

Suggest Corrections

Similar questions

C H I \equiv ⊖ C, C H_{2} = I I ⊖ C H, C H_{3} - I I I ⊖ C H_{2}

^{\oplus} C H_{2} - C O C H_{3}

^{\oplus} C H_{2} - O C H_{3}

^{\oplus} C H_{2} - C H_{3}

E_{2}

^{'} R^{'}

R - C H = C H_{2}

^{'} R^{'}

M e -

E t -

M e_{2} C H -

M e_{3} C -

(i) C H_{3} - + C H - C H_{2} - O C H_{3}

(i i) C H_{3} - + C H - C H_{2} - C H_{3}

(i i i) C H_{3} - + C H - O | | C - C H_{3}

Join BYJU'S Learning Program