Reactive Intermediates Questions in English

(A) The bond dissociation energy required to form the benzyl radical $(C_6H_5CH_2^{\bullet})$ from toluene $(C_6H_5CH_3)$ is less than that required to form the methyl radical $(CH_3^{\bullet})$ from methane $(CH_4)$.
This is because the benzyl radical is stabilized by resonance with the benzene ring,which lowers the energy of the radical intermediate.
In contrast,the methyl radical lacks such resonance stabilization.

(B) The homolytic fission of the $C-C$ bond in ethane $(CH_3-CH_3)$ results in the formation of two methyl free radicals $(CH_3^{\bullet})$.
$CH_3-CH_3 \xrightarrow{\text{Homolytic fission}} 2CH_3^{\bullet}$
In the methyl free radical,the central carbon atom is bonded to three hydrogen atoms and possesses one unpaired electron in an unhybridized $p$-orbital.
Therefore,the carbon atom is $sp^2$-hybridized with a trigonal planar geometry.

(A) The stability of carbocations is determined by the extent of resonance and inductive effects.
$(A)$ $Ph_3C^+$ (Triphenylmethyl cation) is the most stable because the positive charge is delocalized over three benzene rings through resonance.
$(B)$ $Ph_2CH^+$ has resonance with two benzene rings.
$(C)$ $PhCH_2^+$ has resonance with one benzene ring.
$(D)$ $CH_3^+$ has no resonance stabilization.
Therefore,$Ph_3C^+$ is the most stable.

(B) The correct answer is $(B)$ $CH_3-C^+(CH_3)_2$.
Carbocation stability is determined by the inductive effect $(+I)$ and hyperconjugation.
$CH_3-C^+(CH_3)_2$ is a $3^\circ$ carbocation,which is stabilized by the electron-donating effect of three methyl groups.
$CH_3-CH^+(CH_3)$ is a $2^\circ$ carbocation,$CH_3-CH_2^+$ is a $1^\circ$ carbocation,and $CH_3^+$ is a methyl carbocation.
Stability order: $3^\circ > 2^\circ > 1^\circ > \text{methyl}$.

(D) The correct order of stability is $(ii) > (i) > (iii)$.
$(ii)$ $CH_3-CH^{+}-OCH_3$ is the most stable because the lone pair on the oxygen atom provides strong resonance stabilization ($+M$ effect) to the carbocation.
$(i)$ $CH_3-CH^{+}-CH_3$ is a secondary carbocation stabilized by hyperconjugation and the inductive effect $(+I)$ of two methyl groups.
$(iii)$ $CH_3-CH^{+}-COCH_3$ is the least stable because the carbonyl group $(-COCH_3)$ exerts a strong electron-withdrawing effect ($-I$ and $-M$ effects),which destabilizes the positive charge on the carbon atom.

(B) The stability of carbanions is determined by the inductive effect and resonance effect.
$1.$ $C_6H_5CH_2^-$ (Benzyl carbanion) is the most stable due to resonance stabilization by the phenyl ring.
$2.$ Among the alkyl carbanions,stability decreases as the number of electron-donating alkyl groups increases due to the $+I$ effect.
$3.$ The order of stability for alkyl carbanions is: primary $(CH_3CH_2^-)$ > secondary $((CH_3)_2CH^-)$ > tertiary $((CH_3)_3C^-)$.
$4.$ Combining these,the overall order of decreasing stability is: $C_6H_5CH_2^- > CH_3CH_2^- > (CH_3)_2CH^- > (CH_3)_3C^-$,which corresponds to $4 > 3 > 2 > 1$.

(B) Dehydrogenation of alcohols typically involves the removal of hydrogen to form a carbocation intermediate or carbonyl compound.
Among the given options,$CH_3-C(CH_3)_2-OH$ is a tertiary $(3^{\circ})$ alcohol.
Upon dehydration or related processes,it forms a tertiary carbocation,$(CH_3)_3C^+$,which is the most stable due to the $+I$ effect and hyperconjugation of nine $\alpha$-hydrogens.

(C) species is paramagnetic if it contains at least one unpaired electron.
$1$. $A$ $Carbanion$ $(R_3C^-)$ has a lone pair of electrons on the carbon atom,meaning all electrons are paired (diamagnetic).
$2$. $A$ $Carbonium$ $ion$ $(R_3C^+)$ has an empty $p$-orbital and all electrons are paired (diamagnetic).
$3$. $A$ $Free$ $radical$ $(R_3C^.)$ contains an unpaired electron in its orbital,which makes it paramagnetic.
Therefore,the correct option is $C$.

(A) The correct answer is $A$.
$A$ carbanion is a species containing a carbon atom with a negative charge and an unshared pair of electrons.

$1$. \text{Species}	$2$. \text{Charge on central } $C$ \text{-atom}
$A$. \text{Carbanion}	\text{Negative}
$B$. \text{Carbonium ion (Carbocation)}	\text{Positive}
$C$. \text{Free radical}	\text{Neutral}

(D) The stability order of free radicals is $Tertiary > Secondary > Primary > Methyl$.
Greater the number of alkyl groups attached to the carbon atom carrying the odd electron,the greater is the delocalization of the odd electron through hyperconjugation and inductive effects,making the free radical more stable.

(A) carbocation $(R_3C^+)$ has $3$ bond pairs around the central carbon atom,totaling $6$ electrons.
$A$ carbanion $(R_3C^-)$ has $3$ bond pairs and $1$ lone pair,totaling $8$ electrons.
$A$ free radical $(R_3C\cdot)$ has $3$ bond pairs and $1$ odd electron,totaling $7$ electrons.
Therefore,the carbocation contains three pairs of electrons around the central carbon atom.

(A) The stability of carbanions is determined by the inductive effect of alkyl groups. Alkyl groups are electron-donating ($+I$ effect),which increases the electron density on the negatively charged carbon atom,thereby destabilizing the carbanion.
Therefore,the stability order is: $CH_3^- > 1^o > 2^o > 3^o$.
Thus,the methyl carbanion $(CH_3^-)$ is the most stable.

(B) The stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$.
$(CH_3)_3C^{+}$ is a $3^{\circ}$ (tert-butyl) carbocation,$(CH_3)_2CH^{+}$ is a $2^{\circ}$ (sec-butyl) carbocation,and $CH_3CH_2^+$ is a $1^{\circ}$ (n-butyl) carbocation.
Since the stability of carbocations increases with the number of alkyl groups attached to the positively charged carbon due to the inductive effect and hyperconjugation,the $3^{\circ}$ carbocation is the most stable.
Therefore,the correct option is $(b)$.

(B) In a $Carbanion$,the carbon atom is bonded to three atoms or groups (trivalent) and possesses a lone pair of electrons,resulting in a total of $8$ electrons (complete octet) in its valence shell.
It is an electron-rich species.
It is $sp^3$ hybridized,where one of the hybrid orbitals contains the lone pair of electrons,giving it a pyramidal geometry.

(B) The stability of carbocations is determined by the dispersal of the positive charge.
Alkyl groups stabilize carbocations through the inductive effect and hyperconjugation,with the order being: $3^\circ \text{alkyl} > 2^\circ \text{alkyl} > 1^\circ \text{alkyl} > \text{methyl}$.
However,resonance stabilization is significantly more effective than inductive effects. The triphenylmethyl carbocation is stabilized by resonance with three phenyl rings,which delocalize the positive charge over a large number of carbon atoms.
Therefore,the triphenylmethyl carbocation is the most stable among the given options.

(B) In an $S_N1$ reaction,the rate-determining step involves the formation of a carbocation intermediate.
Since the central carbon atom in the carbocation is $sp^2$ hybridized,it adopts a trigonal planar geometry.
Therefore,the correct geometry is planar.

(C) The stability of a carbanion is increased by electron-withdrawing groups $(EWG)$ and decreased by electron-donating groups $(EDG)$ due to the inductive and resonance effects.
$1.$ $CH_3^-$ is a simple alkyl carbanion.
$2.$ $CH_3CH_2^-$ is less stable than $CH_3^-$ due to the $+I$ effect of the methyl group.
$3.$ $p-O_2N-C_6H_4-CH_2^-$ contains a nitro group $(-NO_2)$,which is a strong electron-withdrawing group ($-I$ and $-M$ effect). This group effectively disperses the negative charge on the carbon atom,making it the most stable among the given options.
$4.$ $p-CH_3-C_6H_4-CH_2^-$ contains a methyl group $(-CH_3)$,which is an electron-donating group ($+I$ and hyperconjugation),thus destabilizing the carbanion.
Therefore,the correct option is $C$.

(C) The correct option is $(C)$.
In the benzyl carbocation $(C_6H_5CH_2^+)$,the positive charge on the carbon atom is delocalized over the benzene ring through resonance.
This resonance stabilization makes the benzyl carbocation significantly more stable than the primary alkyl carbocations ($CH_3CH_2^+$ and $CH_3CH_2CH_2^+$) and the methyl carbocation $(CH_3^+)$.

(B) The stability of carbocations follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ} > \text{methyl}$.
$(CH_3)_3C^{+}$ is a $3^{\circ}$ carbocation,which is stabilized by the inductive effect of three methyl groups and hyperconjugation.
Therefore,$(CH_3)_3C^{+}$ is the most stable cation among the given options.

(C) The correct answer is $C$. In the triphenylmethyl carbonium ion,the $\pi$ electrons of all three benzene rings are delocalized with the vacant $p$-orbital of the central carbon atom.
This leads to extensive resonance stabilization.
It is the most stable among the given carbonium ions.
The ion $(CH_3)_3C^+$ is stabilized by hyperconjugation,which is a weaker effect compared to the extensive resonance provided by three phenyl groups.

(B) The stability of a carbocation is increased by electron-donating groups $(EDG)$ and decreased by electron-withdrawing groups $(EWG)$.
$1$. The $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effect),which destabilizes the carbocation.
$2$. The $-OCH_3$ group is an electron-donating group ($+M$ effect),which significantly stabilizes the carbocation by donating electron density through resonance.
$3$. The $-Cl$ group is electron-withdrawing ($-I$ effect) but also has a weak $+M$ effect; overall,it is less destabilizing than $-NO_2$ but less stabilizing than $-OCH_3$.
$4$. The unsubstituted benzyl carbocation $(C_6H_5-CH_2^+)$ serves as the reference.
Therefore,the species with the $-OCH_3$ group is the most stable.

(B) The stability of free radicals is primarily determined by hyperconjugation and the inductive effect of alkyl groups.
As the number of alkyl groups attached to the carbon atom bearing the odd electron increases,the number of hyperconjugative structures increases,leading to greater stability.
Therefore,the order of stability is $3^\circ > 2^\circ > 1^\circ$.

(C) The stability of carbanions increases with an increase in the $s-$character of the hybrid orbitals of the carbon atom bearing the negative charge.
This is because a higher $s-$character makes the orbital more electronegative,which better stabilizes the negative charge.
The $s-$character in different hybridizations is: $sp$ $(50\%)$,$sp^2$ $(33.3\%)$,and $sp^3$ $(25\%)$.
Therefore,the correct order of stability is: $sp^3 < sp^2 < sp$.

(D) The stability of carbocations follows the order: $3^\circ > 2^\circ > 1^\circ > CH_3^+$.
This is due to the inductive effect ($+I$ effect) and hyperconjugation from the alkyl groups.
$CH_3-C^+(CH_3)-CH_3$ is a tertiary $(3^\circ)$ carbocation,which is the most stable among the given options.

(A) . $({C_6}{H_5})_2{CH}^+$ is the most stable carbonium ion because the positive charge is delocalized over two phenyl rings through resonance,providing greater stabilization compared to the other options.

(B) The stability of carbonium ions (carbocations) is determined by the $+I$ effect of alkyl groups and hyperconjugation.
Tertiary carbocations are more stable than secondary,which are more stable than primary,which are more stable than methyl carbocations.
The order is: $(CH_3)_3C^+ > (CH_3)_2CH^+ > CH_3CH_2^+ > CH_3^+$.

(B) migration of hydrogen with a pair of electrons is called a hydride shift.
In this process,the hydrogen atom moves along with its bonding pair of electrons $(H^-)$ to an adjacent carbon atom,typically to form a more stable carbocation.

(C) The stability of a carbocation is determined by the extent of resonance and inductive effects.
$1$. $(CH_3)_3C^+$ is a tertiary alkyl carbocation stabilized by hyperconjugation and inductive effects.
$2$. $C_6H_5CH_2^+$ is a benzyl carbocation stabilized by resonance with one phenyl ring.
$3$. $(C_6H_5)_2CH^+$ is a benzhydryl carbocation stabilized by resonance with two phenyl rings.
$4$. $(C_6H_5)_3C^+$ is a trityl carbocation stabilized by resonance with three phenyl rings.
Since $(C_6H_5)_3C^+$ has the maximum number of phenyl rings,it exhibits the greatest degree of resonance stabilization,making it the most stable carbocation among the given options.

(B) The stability of carbocations is primarily determined by the inductive effect and hyperconjugation.
$3^o$ carbocations are the most stable due to the maximum number of alkyl groups providing electron density through the $+I$ effect and hyperconjugation.
$2^o$ carbocations are less stable than $3^o$ but more stable than $1^o$.
$1^o$ carbocations are less stable than $2^o$.
$CH_3^+$ (methyl carbocation) is the least stable.
Therefore,the correct order of stability is $3^o C^+ > 2^o C^+ > 1^o C^+ > CH_3^+$.

(D) The stability of a carbocation is determined by the electronic effects of the substituents attached to the benzene ring.
Electron-donating groups $(EDG)$ increase the stability of a carbocation by donating electron density through resonance or inductive effects.
Electron-withdrawing groups $(EWG)$ decrease the stability of a carbocation by withdrawing electron density.
In the given options:
$1$. $-NO_2$ is a strong $EWG$ ($-M$ and $-I$ effect).
$2$. $-Cl$ is an $EWG$ ($-I$ effect,though it has $+M$ effect,the $-I$ effect dominates in this position).
$3$. $-H$ is the reference (no effect).
$4$. $-OCH_3$ is an $EDG$ ($+M$ effect).
Since $-OCH_3$ is an electron-donating group,it stabilizes the positive charge on the benzylic carbon most effectively.
Therefore,$p-CH_3O-C_6H_4-CH_2^+$ is the most stable carbocation.

(A) The stability of free radicals is determined by the number of alkyl groups attached to the carbon atom bearing the unpaired electron.
Alkyl groups provide stability through the inductive effect and hyperconjugation.
As the number of alkyl groups increases,the stability of the free radical increases.
Therefore,the order of stability is: $3^\circ$ (tertiary) > $2^\circ$ (secondary) > $1^\circ$ (primary) > $\dot{C}H_3$ (methyl radical).

(D) To determine the stability of carbocations,we analyze the electronic effects of the substituents attached to the positively charged carbon atom.
$(I) \ CH_3-C^+(CH_3)-CH_3$ is a tertiary $(3^{\circ})$ carbocation. It is stabilized by the $+I$ effect of three methyl groups and hyperconjugation.
$(II) \ CH_3-CH^+(OCH_3)$ is a secondary $(2^{\circ})$ carbocation. The oxygen atom in the $-OCH_3$ group has lone pairs,which provide strong $+M$ (mesomeric) stabilization to the carbocation,making it highly stable.
$(III) \ CH_3-CH^+(COCH_3)$ is a secondary $(2^{\circ})$ carbocation. The $-COCH_3$ group exerts a strong $-I$ (inductive) effect and a $-M$ (mesomeric) effect,both of which destabilize the carbocation.
Comparing these,$(II)$ is the most stable due to resonance stabilization ($+M$ effect). $(I)$ is more stable than $(III)$ because $(III)$ is destabilized by electron-withdrawing groups. Thus,the order of stability is $(II > I > III)$.

(B) The stability of a carbanion is increased by electron-withdrawing groups $(-EWG)$ and decreased by electron-donating groups $(-EDG)$ attached to the ring.
The $-NO_2$ group is a strong electron-withdrawing group due to both $-I$ and $-M$ effects.
The $-Cl$ group is electron-withdrawing due to the $-I$ effect but electron-donating due to the $+M$ effect (overall $-I$ dominates).
The $-OCH_3$ group is a strong electron-donating group due to the $+M$ effect.
The $-NO_2$ group stabilizes the negative charge on the carbanion most effectively through resonance and inductive effects.
Therefore,$p-NO_2-C_6H_4-CH_2^-$ is the most stable carbanion.

(D) The stability of a carbanion is directly proportional to the electron-withdrawing effect ($-I$ and $-M$ effects) and inversely proportional to the electron-donating effect ($+I$ and $+M$ effects).
In the given options,the $-NO_2$ group is a strong electron-withdrawing group ($-I$ and $-M$ effects),which stabilizes the negative charge on the carbanion by dispersing it through resonance and induction.
Therefore,$p-NO_2-C_6H_4-CH_2^-$ is the most stable carbanion.

(A) The stability of carbocations is determined by factors such as inductive effect,hyperconjugation,and resonance.
$1$. $(CH_3)_3C^+$ is a tertiary carbocation,which is highly stable due to the $+I$ effect of three methyl groups and hyperconjugation.
$2$. $(CH_3)_2CH^+$ is a secondary carbocation,which is moderately stable.
$3$. $CH_2=CH-CH_2^+$ is an allyl carbocation,which is stabilized by resonance.
$4$. $H_3C^+$ is a methyl carbocation,which has no alkyl groups to provide electron density through $+I$ effect or hyperconjugation.
Therefore,$H_3C^+$ is the least stable carbocation.

(C) Free radicals possess an unpaired electron,which makes them highly unstable and reactive. They have a strong tendency to pair their electron to complete their octet,making them chemically active.

(A) The stability of a free radical is determined by the number of resonance structures it can form.
In the structure $\phi - \dot{C}H - \phi$ (diphenylmethyl radical),the unpaired electron is delocalized over two phenyl rings,providing a large number of resonance structures.
In $\phi - \dot{C}H_2$ (benzyl radical),it is delocalized over one phenyl ring.
In $\phi - \dot{C}H - CH_3$,it is delocalized over one phenyl ring and stabilized by hyperconjugation.
Therefore,$\phi - \dot{C}H - \phi$ shows the maximum resonance.

(A) The stability of a carbanion is determined by the hybridization of the carbon atom bearing the negative charge.
$1$. $HC \equiv C^{\ominus}$ has $sp$ hybridization ($50\% \ s$-character).
$2$. $(CH_3)_2C = CH^{\ominus}$ has $sp^2$ hybridization ($33.3\% \ s$-character).
$3$. $C_6H_5^{\ominus}$ has $sp^2$ hybridization ($33.3\% \ s$-character).
$4$. $(CH_3)_3C - CH_2^{\ominus}$ has $sp^3$ hybridization ($25\% \ s$-character).
Greater $s$-character increases the electronegativity of the carbon atom,making it better at holding the negative charge.
Therefore,the $sp$ hybridized carbanion $HC \equiv C^{\ominus}$ is the most stable.

(A) The stability of carbocations is determined by the dispersal of positive charge.
$(i) \ CH_3^+$ is a methyl carbocation with no electron-donating groups.
$(ii) \ CH_3CH_2^+$ is a primary carbocation stabilized by the $+I$ effect of the methyl group.
$(iii) \ CH_3OCH_2^+$ is stabilized by the resonance effect ($+M$ effect) of the oxygen atom,where the lone pair on oxygen donates electron density to the vacant $p$-orbital of the carbocation,making it the most stable.
Therefore,the correct order of stability is $(iii) > (ii) > (i)$.

(A) The stability of carbanions is determined by the inductive effect ($-I$ and $+I$ effects) and resonance effect.
$1$. $CCl_3^-$: The three chlorine atoms exert a strong $-I$ effect,which stabilizes the negative charge significantly.
$2$. ${C_6H_5}CH_2^-$: The negative charge is stabilized by resonance with the phenyl ring.
$3$. ${(CH_3)_2}CH^-$: This is a secondary carbanion,which is destabilized by the $+I$ effect of two methyl groups.
$4$. ${(CH_3)_3}C^-$: This is a tertiary carbanion,which is most destabilized by the $+I$ effect of three methyl groups.
Therefore,the decreasing order of stability is $CCl_3^- > {C_6H_5}CH_2^- > {(CH_3)_2}CH^- > {(CH_3)_3}C^-$. The correct option is $A$.

(D) The stability of free radicals is determined by resonance and hyperconjugation.
$1$. Resonance stabilized radicals are more stable than alkyl radicals.
$2$. $(C_6H_5)_3\dot{C}$ (triphenylmethyl radical) is the most stable due to extensive resonance with three phenyl rings.
$3$. $(C_6H_5)_2\dot{C}H$ (diphenylmethyl radical) is next in stability due to resonance with two phenyl rings.
$4$. Among alkyl radicals,$(CH_3)_3\dot{C}$ (tert-butyl radical) is more stable than $(CH_3)_2\dot{C}H$ (isopropyl radical) due to more hyperconjugation.
Thus,the increasing order of stability is: $(CH_3)_2\dot{C}H < (CH_3)_3\dot{C} < (C_6H_5)_2\dot{C}H < (C_6H_5)_3\dot{C}$.

(B) $1$. The stability of carbanions is primarily governed by resonance and the inductive effect.
$2$. Resonance: $Q$ $(C_6H_5\bar{C}H_2)$ is stabilized by resonance with the benzene ring,and $S$ $(H_2\bar{C}^{-}CH=CH_2)$ is stabilized by resonance with the double bond. $Q$ is more stable than $S$ due to the greater delocalization of the negative charge over the aromatic ring.
$3$. Inductive Effect: Alkyl groups exhibit a $+I$ effect,which destabilizes carbanions. $P$ $(\bar{C}H_3)$ has no alkyl groups,while $R$ $((CH_3)_2\bar{C}H)$ has two electron-donating methyl groups. Thus,$P$ is more stable than $R$.
$4$. Combining these factors,the stability order is $R < P < S < Q$.

(D) The stability of carbocations is primarily determined by the inductive effect and hyperconjugation.
$3^{\circ}$ carbocations are the most stable due to the maximum number of alkyl groups providing electron density through the $+I$ effect and hyperconjugation.
$2^{\circ}$ carbocations are less stable than $3^{\circ}$ but more stable than $1^{\circ}$.
$1^{\circ}$ carbocations are less stable than $2^{\circ}$.
$\overset{\oplus}{C}H_3$ (methyl carbocation) is the least stable as it lacks any alkyl groups to stabilize the positive charge.
Therefore,the correct order of stability is $3^{\circ} > 2^{\circ} > 1^{\circ} > \overset{\oplus}{C}H_3$.

(B) The stability of carbocations is determined by inductive effect,hyperconjugation,and resonance.
$3^\circ$ carbocations are stabilized by hyperconjugation and inductive effects of three alkyl groups.
Benzyl carbocation $(C_6H_5CH_2^+)$ is stabilized by resonance with the benzene ring.
Resonance stabilization is generally more effective than the hyperconjugation and inductive effects present in $3^\circ$ alkyl carbocations.
Therefore,the order of stability is: $\text{benzyl} > 3^\circ > 2^\circ > 1^\circ$.

(D) The stability of a carbanion is determined by electronic effects such as inductive effect $(I)$,resonance,and hybridization.
$1$. The $sp^3$ hybridized carbon in $(CH_3)_3C^-$ has three electron-donating alkyl groups ($+I$ effect),which increase the electron density on the negatively charged carbon,destabilizing it.
$2$. The $sp^3$ hybridized carbon in $CH_3CH_2^-$ has one electron-donating alkyl group.
$3$. The $sp$ hybridized carbon in $HC \equiv C^-$ is more stable due to higher $s$-character.
$4$. The $(C_6H_5)_3C^-$ is highly stable due to extensive resonance with three phenyl rings.
Therefore,$(CH_3)_3C^-$ is the least stable carbanion due to the maximum $+I$ effect of three methyl groups.

(C) The stability of carbanions depends on the $s$-character of the carbon atom bearing the negative charge.
As the $s$-character increases,the electronegativity of the carbon atom increases,making it better at stabilizing the negative charge.
$(1)$ $CH_3-CH_2^-$: $sp^3$ hybridized ($25\% \ s$-character).
$(2)$ $CH_2=CH^-$: $sp^2$ hybridized ($33.3\% \ s$-character).
$(3)$ $HC \equiv C^-$: $sp$ hybridized ($50\% \ s$-character).
The order of stability is $sp > sp^2 > sp^3$,which corresponds to $(3) > (2) > (1)$.

(B) The stability of carbocations is determined by the electronic effects of the substituents attached to the positively charged carbon atom.
$1$. In $(Y)$,the $-OCH_3$ group exerts a strong $+M$ (mesomeric) effect,which significantly stabilizes the carbocation by donating electron density through resonance.
$2$. In $(X)$,the two $-CH_3$ groups provide stability through $+I$ (inductive) effect and hyperconjugation.
$3$. In $(Z)$,the $-COCH_3$ group is an electron-withdrawing group due to its $-I$ and $-M$ effects,which destabilizes the carbocation.
Therefore,the order of stability is $Y > X > Z$.

(C) The stability of carbon free radicals is determined by the number of alkyl groups attached to the carbon atom bearing the odd electron.
These alkyl groups provide stability through the inductive effect ($+I$ effect) and hyperconjugation.
As the number of alkyl groups increases,the stability of the radical increases.
Therefore,the order of stability is: ${3^\circ} > {2^\circ} > {1^\circ} > \mathop C\limits^\bullet {H_3}$.

(C) carbanion is a species containing a carbon atom with a negative charge and an unshared pair of electrons.
Due to the presence of a lone pair of electrons,it can donate these electrons to an electrophile,making it a $Nucleophile$.
Additionally,because it can accept a proton $(H^+)$ from an acid,it also acts as a $Base$.
Therefore,a carbanion acts as both a base and a nucleophile.