The increasing order of basicity for the foll | vedclass

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(C) Basicity $\propto \frac{1}{	ext{Stability of conjugate base}}$.Stability order of the given carbanions:$(v)$ $CN^{-}$: The negative charge is on

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$(v) < (iii) < (ii) < (iv) < (i)$

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(C) Basicity $\propto \frac{1}{ ext{Stability of conjugate base}}$.Stability order of the given carbanions:$(v)$ $CN^{-}$: The negative charge is on the nitrogen atom (more electronegative) and it is also stabilized by the $-I$ effect of the triple bond. It is the most stable.$(iii)$ $HC \equiv C^{-}$: The negative charge is on an $sp$ hybridized carbon atom. It is more stable than $sp^2$ or $sp^3$ hybridized carbons.$(ii)$ $H_2C=CH-CH_2^{-}$: The negative charge is stabilized by resonance (delocalization).$(iv)$ $CH_3^{-}$: The negative charge is on an $sp^3$ hybridized carbon atom.$(i)$ $(CH_3)_3C^{-}$: The negative charge is on an $sp^3$ hybridized carbon atom,which is further destabilized by the $+I$ effect of three methyl groups. It is the least stable.Stability order: $(v) > (iii) > (ii) > (iv) > (i)$.Since Basicity $\propto \frac{1}{ ext{Stability}}$,the increasing order of basicity is: $(v) < (iii) < (ii) < (iv) < (i)$.

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

Similar Questions

The correct order of the relative stability of the following carbanions is:
$I$. $HC \equiv C^{-}$
$II$. $CH_2=CH-CH_2^{-}$
$III$. $H_2C=CH^{-}$
$IV$. $(C_6H_5)_2CH^{-}$

Using curved-arrow notation,show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage:
$(a)$ $CH_{3}-SCH_{3}$
$(b)$ $CH_{3}-CN$
$(c)$ $CH_{3}-Cu$

The decreasing order of stability of the given carbocations is

Which of the following carbocations is expected to be most stable?

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

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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)$ $CN^{-}$