Arrange the following carbocations in decreasing order of stability

Question

Arrange the following carbocations in decreasing order of stability (A) IV > III > I > V > II (B) IV > I > III > II > V (C) V > IV > I > III > II (D) V > IV > III > I > II

Tardigrade Admin · Accepted Answer

Cyclopropylmethyl carbocation is especially stable because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p -orbitals of the cationic carbon. Thus, <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9c4c2c79055a76a545f2dafb49d38bb5-.png" />

Q. Arrange the following carbocations in decreasing order of stability

$I V > III > I > V > II$

$I V > I > III > II > V$

$V > I V > I > III > II$

$V > I V > III > I > II$

Cyclopropylmethyl carbocation is especially stable because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant $p$ -orbitals of the cationic carbon. Thus,

Q. Arrange the following carbocations in decreasing order of stability

IV>III>I>V>II

IV>I>III>II>V

V>IV>I>III>II

V>IV>III>I>II

Cyclopropylmethyl carbocation is especially stable because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p -orbitals of the cationic carbon. Thus,

$I V > III > I > V > II$

$I V > I > III > II > V$

$V > I V > I > III > II$

$V > I V > III > I > II$

Cyclopropylmethyl carbocation is especially stable because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant $p$ -orbitals of the cationic carbon. Thus,