List the following carbocations in order of decreasing stabilization energies:

I I, I I I, I, I V

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

I I I, I V, I I, I

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

I I I, I V, I, I I

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

I, I I, I V, I I I

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

Open in App

Solution

The correct option is B $I I I, I V, I I, I$
$I$ has an $s p^{2}$ hybridized carbocation which is highly unstable. Hence, $I$ will be the least stable of all. $I I I$ will be the most stable as it is tertiary carbocation.
Among $I I$ and $I V$ , $I V$ has $4$ $a l p h a -$ hydrogens whereas $I I$ has only, $α -$ hydrogens. Also $I V$ is a secondary carbocation while $I I$ is a primary one. Hence $I V$ will be more stable than $I I$ .
Hence, the decreasing order of stability or stabilization energy is $I I I > I V > I I > I$

Suggest Corrections

Similar questions

Arrange the following carbocations in decreasing order of stability

C l_{3} C^{+}

C l_{2} C H^{+}

C l C H_{2}^{+}

C H_{3}^{+}

The correct order for decreasing stability of the following free radicals is

Order of decreasing acidity of the following is

(I) HCOOH

(II) ${C H}_{3} C O O H$

(III) ${C l}_{2} C H C O O H$

(IV) ${C F}_{3} C O O H$

Join BYJU'S Learning Program