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Carbocations is considered one the most difficult concept.
48 Questions around this concept.
The shape of carbocation is
The total number of optically active compounds formed in the following reaction is :

The order of stability of the following carbocations:
is :
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Which of the following carbocations is most stable?
The correct stability order of carbocations is
The correct order of stability of the carbocations is
Which is the most stable carbocation
Which one of the following reactions is NOT possible?
The stability of carbocations
${ }_{(1)}\left(\mathrm{CH}_3\right)_3 \mathrm{C}^{\oplus}$ (II) $\left(\mathrm{CH}_3\right)_2 \mathrm{C}^{\oplus}\left(\mathrm{OCH}_3\right)$
(III) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C}^{\oplus} \mathrm{H}_2$ (Iv) $\mathrm{CH}_3 \mathrm{C}^{\oplus} \mathrm{HCH}_2 \mathrm{CH}_3$
follows the order
Which one of the carbocations from the following
is most stable ?
A species having a carbon atom possessing a sextet (6 in a group) of electrons and a positive charge is called a carbocation (earlier called carbonium ion). The +CH3 ion is known as a methyl cation or methyl carbonium ion. Carbocations are classified as primary, secondary or tertiary depending on whether one, two or three carbons are directly attached to the positively charged carbon. Carbocations are highly unstable and reactive species. Alkyl groups directly attached to the positively charged carbon stabilise the carbocations due to inductive and hyperconjugation effects. The observed order of carbocation stability is: +CH3 < CH3C+H2 < (CH3)2C+H < (CH3)3C+. These carbocations have a trigonal planar shape with positively charged carbon being sp2 hybridised. Thus, the shape of +CH3 may be considered as being derived from the overlap of three equivalent C(sp2) hybridised orbitals with 1s orbital of each of the three hydrogen atoms. Each bond may be represented as C(sp2)–H(1s) sigma bond. The remaining carbon orbital is perpendicular to the molecular plane and contains no electrons. The shape of methyl carbocation is given as below:

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