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

Vedclass · Accepted Answer

$3 > 2 > 5 > 4 > 1$

Vedclass · Answer

(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 increasing order of basicity for the following intermediates is (from weak to strong):
$(i)$ $(CH_3)_3C^{-}$
$(ii)$ $H_2C=CH-CH_2^{-}$
$(iii)$ $HC \equiv C^{-}$
$(iv)$ $CH_3^{-}$
$(v)$ $CN^{-}$

Arrange the following cations in decreasing order of their stability:
$(a)$ $CH_3^+$
$(b)$ $F-CH_2-CH_2^+$
$(c)$ $Br-CH_2-CH_2^+$
$(d)$ $CH_3-CH_2-CH_2^+$

Which of the following cannot exist at room temperature?

State True or False for the following statements:
$(i)$ Carbon possesses $sp^2$ hybridization in a carbocation.
$(ii)$ Carbon possesses $sp^3$ hybridization in a carbocation.
$(iii)$ $A$ carbocation is formed by homolytic fission of a bond.
$(iv)$ Carbocations are very stable.

Which of the following is the most stable carbocation?

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