What is the decreasing order of stability (mo | vedclass

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(B) The stability of carbocations is determined by the number of $\alpha$-hydrogens (hyperconjugation) and inductive effects. $1$. Carbocation $3$ is

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$3 > 2 > 5 > 4 > 1$

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(B) The stability of carbocations is determined by the number of $\alpha$-hydrogens (hyperconjugation) and inductive effects. $1$. Carbocation $3$ is a tertiary carbocation with $9 \alpha$-hydrogens,making it the most stable. $2$. Carbocation $2$ is a secondary carbocation with $6 \alpha$-hydrogens. $3$. Carbocation $5$ is a secondary carbocation with $4 \alpha$-hydrogens. $4$. Carbocation $4$ is a primary carbocation with $2 \alpha$-hydrogens and an additional $+I$ effect from the methyl group. $5$. Carbocation $1$ is a primary carbocation with $2 \alpha$-hydrogens. Comparing these,the order is $3 > 2 > 5 > 4 > 1$.

What is the decreasing order of stability (most stable $\to$ least stable) of the following carbocations?

Similar Questions

The correct stability order of the following carbocations is:
$I$. $H_2C^+-CH=CH-CH_3$
$II$. $H_2C^+-CH=CH-BMe_2$
$III$. $H_2C^+-CH=CH-NMe_2$
$IV$. $H_2C^+-CH=CH-OMe$

The hybridization of the positively charged carbon and the negatively charged carbon in the following structures are respectively:

The stability of the carbocations follows the order:

Consider the following carbocations. What is the relative stability order of these carbocations?
$(1)$ $CH_3O-C_6H_4-CH_2^+$
$(2)$ $C_6H_5-CH_2^+$
$(3)$ $CH_3-C_6H_4-CH_2^+$

In the potential energy diagram shown,the point $X$ represents:

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