Electronic Displacement in covalent bond Questions in English

(N/A) Conjugated system: The presence of alternating single and double bonds in an open chain or cyclic system is termed as a conjugated system.
Examples: $1,3-$butadiene $(CH_2=CH-CH=CH_2)$,aniline $(C_6H_5NH_2)$,and nitrobenzene $(C_6H_5NO_2)$ contain conjugated systems.
Effect: In such systems,the $\pi$-electrons are delocalized across the conjugated framework. This delocalization leads to increased stability of the molecule and can result in the development of polarity within the system.

(N/A) Definition: The electromeric effect is a temporary effect observed in organic compounds containing multiple bonds (double or triple bonds) in the presence of an attacking reagent. It is defined as the complete transfer of a shared pair of $\pi$-electrons to one of the atoms joined by a multiple bond at the demand of an attacking reagent.
Characteristics:
$1$. It is a temporary effect and is annulled as soon as the attacking reagent is removed.
$2$. It is represented by $E$ and the electron shifting is shown by a curved arrow $(\curvearrowleft)$.
Types of Electromeric Effect:
$(a)$ Positive Electromeric Effect $(+E)$: In this effect,the $\pi$-electrons of the multiple bond are transferred to the atom to which the attacking reagent gets attached. Example: Addition of $H^+$ to an alkene.
$(b)$ Negative Electromeric Effect $(-E)$: In this effect,the $\pi$-electrons of the multiple bond are transferred to the atom to which the attacking reagent does $NOT$ get attached. Example: Addition of $CN^-$ to a carbonyl group.

(N/A) Hyperconjugation is a permanent electronic effect that involves the delocalization of $\sigma$ electrons of a $C-H$ bond of an alkyl group directly attached to an atom of an unsaturated system or to an atom with an unshared $p$-orbital.
It is also known as 'no-bond resonance' because the contributing structures do not involve a covalent bond between the carbon and the hydrogen atom.
Example: Hyperconjugation in ethyl cation $(CH_3-CH_2^+)$.
In the ethyl cation,the positively charged carbon atom has an empty $p$-orbital. One of the $C-H$ bonds of the methyl group can align in the plane of this empty $p$-orbital. The electrons of this $C-H$ bond are then delocalized into the empty $p$-orbital,resulting in the following contributing structures:
$(I) \leftrightarrow (II) \leftrightarrow (III) \leftrightarrow (IV)$
In these structures,the hydrogen atom is shown as $H^+$ without a bond to the carbon,which explains the term 'no-bond resonance'.

(A) $(i)$ Acidic strength decreases as the electron-donating inductive effect $(+I)$ of alkyl groups increases: $CH_3COOH > CH_3CH_2COOH > (CH_3)_2CHCOOH > (CH_3)_3CCOOH$
(ii) Stability of carbocations increases with the number of alkyl groups attached due to hyperconjugation and $+I$ effect: $(CH_3)_3\stackrel{+}{C} > (CH_3)_2\stackrel{+}{CH} > CH_3CH_2^+ > \stackrel{+}{CH}_3$
(iii) Acidic strength increases with the number of electron-withdrawing chlorine atoms: $CH_3COOH < CH_2ClCOOH < CHCl_2COOH < CCl_3COOH$
(iv) Stability order based on resonance and charge separation: $(c) < (b) < (a)$
$(v)$ Stability order based on resonance structures: $(I) > (II) > (III)$

(C) The stability of resonance structures is determined by the following rules:
$1$. Structures with more covalent bonds are more stable.
$2$. Structures with complete octets for all atoms are more stable.
$3$. Structures with less charge separation are more stable.
$4$. Negative charge on a more electronegative atom and positive charge on a less electronegative atom is more stable.
In structure $(a)$,all atoms have complete octets and there is no charge separation,making it the most stable.
In structure $(b)$,the positive charge is on a carbon atom and the negative charge is on an oxygen atom (more electronegative),which is relatively stable.
In structure $(c)$,the positive charge is on a terminal carbon atom,which is less stable compared to $(b)$ due to the lack of adjacent electron-donating groups or resonance stabilization of the positive charge.
Therefore,the increasing order of stability is $(c) < (b) < (a)$.

(N/A) The inductive effect is the permanent displacement of sigma electrons along a carbon chain due to the difference in electronegativity of the atoms or groups attached.
$1$. Electron-withdrawing groups ($-I$ effect): These groups attract electrons towards themselves. Examples: $-Cl, -NO_2, -OC_6H_5, -C_6H_5, -OH, -NH_2$.
$2$. Electron-donating groups ($+I$ effect): These groups release electrons away from themselves. Examples: $-CH_3, -(CH_3)_3C, -CH_2CH_3$.

(C) The electromeric effect ($E$-effect) is the complete transfer of a shared pair of $\pi$-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent.
$1$. In reaction $(a)$,the attacking reagent is a nucleophile $(CN^{-})$. The $\pi$-electrons move away from the attacking reagent,which is characteristic of the negative electromeric effect ($-E$ effect).
$2$. In reaction $(b)$,the attacking reagent is an electrophile $(H^{+})$. The $\pi$-electrons move towards the attacking reagent,which is characteristic of the positive electromeric effect ($+E$ effect).
Therefore,$(a)$ represents the $-E$ effect and $(b)$ represents the $+E$ effect.

(N/A) Hyperconjugation involves the delocalization of $\sigma$ electrons of a $C-H$ bond of an alkyl group directly attached to an unsaturated system (like an alkene or a carbocation).
$(a) \ CH_3-CH=CH_2$: The $\sigma$ electrons of the $C-H$ bond shift to form a $C=C$ double bond,and the $\pi$ electrons of the original $C=C$ bond shift to the terminal carbon.
Resonance structures: $CH_2^--CH=CH_2 \leftrightarrow H^+ \dots CH_2=CH-CH_2^-$.
$(b) \ CH_3-CH_2^+$: The $\sigma$ electrons of one $C-H$ bond shift to form a $C=C$ double bond,leaving the hydrogen as $H^+$.
Resonance structures: $CH_2=CH_2 \leftrightarrow H^+$.
$(c) \ CH_3-CH_2-CH_2^+$: The $\alpha$-carbon is the one adjacent to the positively charged carbon. The $\sigma$ electrons of the $C-H$ bond on the $\alpha$-carbon shift to form a $C=C$ double bond,resulting in the following resonance structures:
$CH_3-CH_2-CH_2^+ \leftrightarrow CH_3-CH=CH_2 \leftrightarrow H^+ \dots CH_3-CH^+=CH_2$.

(B) The movement of electrons in organic reaction mechanisms is represented by a curved arrow $(\curvearrowright)$.
This arrow starts from the source of the electron pair (a bond or a lone pair) and points towards the atom or region where the electron pair is being shifted.

(N/A) $i$. The electron pair from a $\pi$-bond shifts to an adjacent bond position.
$ii$. The electron pair from a $\pi$-bond shifts onto an adjacent atom.
$iii$. The electron pair from an atom shifts to an adjacent bond position.

(N/A) The movement of a single electron is indicated by a single-barbed fishhook (i.e.,half-headed curved arrow).
In the homolytic cleavage of the $CH_3-Cl$ bond,each atom takes one electron from the shared pair,resulting in the formation of free radicals:
$CH_3-Cl \xrightarrow{\text{Homolytic cleavage}} \dot{CH_3} + \dot{Cl}$

(B) Electron displacement effects such as the inductive effect,resonance effect,or hyperconjugation lead to the shifting of electron density within a covalent bond.
This shifting results in the creation of partial charges ($\delta+$ and $\delta-$),which induces a permanent polarity in the covalent bond.

(D) $(i)$ The electron displacement in a $\sigma$ covalent bond is influenced by the inductive effect.
$(ii)$ The displacement of $\pi$-bond electrons and non-bonding $p$ electrons is influenced by the resonance effect.
$(iii)$ The displacement of bonding electron pairs in a $C-H$ bond is influenced by the hyperconjugation effect,which is also known as no-bond resonance.
Since all these effects contribute to electron displacement in covalent bonds,the correct answer is $D$.

(N/A) The inductive effect involves the permanent displacement of $\sigma$-electrons along a carbon chain due to the difference in electronegativity between atoms.
Similarly,the resonance effect is a permanent effect involving the delocalization of $\pi$-electrons or lone pairs within a conjugated system.
In contrast,the electromeric effect is a temporary effect that occurs only in the presence of an attacking reagent,involving the complete transfer of a shared pair of $\pi$-electrons to one of the atoms joined by a multiple bond.

(A) The electromeric effect is defined as the complete transfer of a shared pair of $\pi$ electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent.
It is a temporary effect that occurs only in the presence of an attacking reagent.

(C) The terms $Mesomeric \ effect$ and $Resonance \ effect$ are often used interchangeably in organic chemistry. Both refer to the delocalization of $\pi$-electrons or lone pairs of electrons through a conjugated system. While some textbooks distinguish them by stating that the $Mesomeric \ effect$ is specifically used for substituents attached to a conjugated system (like $-OH, -NH_2$),the underlying electronic mechanism is identical to $Resonance$.

(N/A) The inductive effect involves the permanent displacement of shared electron pairs along a carbon chain due to the difference in electronegativity between atoms,resulting in bond polarity.
In contrast,the resonance effect involves the delocalization of $\pi$ electrons or lone pairs across a conjugated system,which is a theoretical model used to describe the stability and electronic distribution of molecules that cannot be represented by a single Lewis structure.

(N/A) $(i)$ Inductive effect
(ii) Resonance effect
(iii) Hyperconjugation effect
(iv) Electromeric effect: Occurs in the presence of an attacking reagent.

(D) The inductive effect is a permanent displacement of shared electron pairs along a carbon chain due to the difference in electronegativity between atoms.
$1$. It is a permanent effect.
$2$. It involves the displacement of $\sigma$ electrons.
$3$. It is distance-dependent and its magnitude decreases rapidly as the number of bonds increases,typically becoming negligible after $3$ carbon atoms.

(A) The inductive effect is a distance-dependent phenomenon where the electron-withdrawing effect of a substituent (like $-Cl$) decreases as the number of intervening sigma bonds increases.
In the molecule $CH_3(3) - CH_2(2) - CH_2(1) - Cl$,the chlorine atom is directly attached to carbon $1$.
Therefore,carbon $1$ experiences the maximum inductive effect.
As we move further away from the chlorine atom,the effect decreases.
Thus,the inductive effect on carbon $3$ is the minimum,and on carbon $1$ is the maximum.
The increasing order of the inductive effect is $3 < 2 < 1$.

(B) positive effect ($+R$ or $+I$) occurs when a group donates electron density towards the rest of the molecule or the carbon chain.
In this process,the transfer of electrons is away from the substituent group and towards the atom or group to which it is attached.

(N/A) In $CH_3-CH_2^+$,the $\sigma$-orbital of the $C-H$ bond of the $CH_3$ group and the vacant $2p$-orbital of the $CH_2^+$ group are parallel,which allows for the delocalization of $\sigma$-electrons into the vacant $p$-orbital (hyperconjugation).
In $CH_3-CH=CH_2$,the $\sigma$-orbital of the $C-H$ bond of the $CH_3$ group and the $2p$-orbitals of the $\pi$-bond of the $CH=CH_2$ group are parallel,allowing for the delocalization of $\sigma$-electrons into the $\pi$-system (hyperconjugation).

(N/A) Hyperconjugation involves the delocalization of $\sigma$ electrons of a $C-H$ bond of an alkyl group directly attached to an atom with an empty $p$-orbital or a $\pi$-bonded system.
$(a)$ In the ethyl carbocation $(CH_3-CH_2^+)$,the empty $2p$ orbital of the positively charged carbon overlaps with the $C-H$ $\sigma$ bond of the adjacent methyl group.
$(b)$ In propene $(CH_3-CH=CH_2)$,the $\sigma$ electrons of the $C-H$ bond of the methyl group overlap with the $\pi$ orbital of the adjacent double bond.

(A) Hyperconjugation involves the delocalization of $\sigma$ electrons of the $C-H$ bond of an alkyl group directly attached to an unsaturated system (like an alkene or carbocation).
This interaction leads to the stabilization of the molecule or ion.

(B) The electrons of a $\pi$ bond are delocalized in molecules that possess a conjugated system.
In a conjugated system,$\pi$ bonds are arranged in an alternating pattern with single bonds,or with atoms possessing lone pairs,allowing the $p$-orbitals to overlap across the system.

(A) $(i) - (q, s)$; $(ii) - (q, s)$; $(iii) - (p, s)$; $(iv) - (s)$.
Explanation:
$(i)$ $C_6H_5NH_2$: The $-NH_2$ group has a lone pair on nitrogen,showing $+R$ effect,and is more electronegative than carbon,showing $-I$ effect.
$(ii)$ $C_6H_5OH$: The $-OH$ group has lone pairs on oxygen,showing $+R$ effect,and is more electronegative than carbon,showing $-I$ effect.
$(iii)$ $C_6H_5NO_2$: The $-NO_2$ group is electron-withdrawing,showing $-R$ effect,and is highly electronegative,showing $-I$ effect.
$(iv)$ $CH_3CH_2Cl$: The $-Cl$ atom is electronegative,showing $-I$ effect. It does not show resonance with the alkyl group.

(A) $(i)$ True: Hyperconjugation involves the delocalization of $\sigma$-electrons of a $C-H$ bond into an empty $p$-orbital of a carbocation.
$(ii)$ False: While resonance explains stability,the specific term for the delocalization involving $\sigma$-electrons in carbocations is hyperconjugation,not resonance (which typically refers to $\pi$-electron delocalization).
$(iii)$ False: Hyperconjugation is generally considered an electron-donating effect (often denoted as $+H$),but it is not classified as $(+)$ or $(-)$ in the same way as inductive or mesomeric effects.
$(iv)$ True: The mesomeric effect is classified as $(+M)$ (electron-donating) or $(-M)$ (electron-withdrawing) based on the direction of electron flow.

(C) The $-NO_2$ group is a strong electron-withdrawing group due to both its resonance effect ($-M$ effect) and its inductive effect ($-I$ effect).
$(i)$ False: It is electron-attracting in both effects.
$(ii)$ False: It is electron-attracting in both effects.
$(iii)$ True: It withdraws electrons via resonance $(-M)$ and induction $(-I)$.
$(iv)$ False: It is electron-attracting in both effects.
Final answer: $(i-F, ii-F, iii-T, iv-F)$

(A) $(i)$ True: Both resonance and hyperconjugation are electronic effects used to explain the stability of organic molecules.
$(ii)$ False: Resonance structures are drawn for resonance,but hyperconjugation involves the interaction of $\sigma$ bonds with adjacent $\pi$ systems or empty orbitals,not standard resonance structures.
$(iii)$ True: Hyperconjugation is often referred to as 'no-bond resonance' because it involves the delocalization of $\sigma$ electrons without a formal bond.
$(iv)$ False: Resonance involves the movement of $\pi$ electrons and lone pairs,not just $\pi$ bonds.

(D) $(1)$ The $\pi$ electrons migrate to the atom to which the attacking reagent gets attached.
$(2)$ The $\pi$ electrons migrate to the atom to which the attacking reagent does not get attached (i.e.,the other atom of the multiple bond).
$(3)$ Resonance and hyperconjugation increase stability.
$(4)$ $-I$ (negative inductive) effect increases the acidic strength,and $+I$ (positive inductive) effect decreases the acidic strength of $-COOH$.

(N/A) The electronegativity of $Mg$ $(1.2)$ is significantly lower than that of $C$ $(2.5)$.
Consequently,the $Mg$ atom acquires a partial positive charge $(\delta^+)$,and the carbon atom directly attached to it acquires a partial negative charge $(\delta^-)$.
The polarized bond is represented as: $CH_3-CH_2-CH_2-CH_2^{\delta-}-Mg^{\delta+}X$.

(N/A) The high stability of the triphenylmethyl cation is due to extensive delocalization of the positive charge through resonance.
The central carbocation is attached to three phenyl rings. The positive charge on the central carbon atom can be delocalized into each of the three benzene rings.
For each benzene ring,the positive charge can be delocalized to the $ortho$- and $para$-positions,resulting in three resonance structures per ring.
Including the original structure,there are a total of $10$ resonance structures ($1$ original + $3 \times 3 = 9$ resonance-contributed structures). This extensive resonance stabilization significantly lowers the energy of the cation,making it highly stable.

(N/A) $\stackrel{+}{C} H_{3}$ is the most stable species because the $-I$ effect of $Br$ intensifies the positive charge and hence destabilizes the species. Further,the more the number of $Br$ atoms,the less stable is the species. Thus,the stability of these species decreases in the order:
$\stackrel{+}{C} H_{3} > \stackrel{+}{C} H_{2} Br > \stackrel{+}{C} HBr_{2} > \stackrel{+}{C} Br_{3}$
$(b)$ The $-I$ effect of the $Cl$ atom disperses the negative charge and thus stabilizes the species. Further,the more the number of $Cl$ atoms,the more is the dispersal of the negative charge and hence the more stable is the species. Thus,$\stackrel{\Theta}{C} Cl_{3}$ is the most stable species. The stability order of other species decreases as:
$\stackrel{\Theta}{C} Cl_{3} > \stackrel{\Theta}{C} HCl_{2} > \stackrel{\Theta}{C} H_{2} Cl > \stackrel{\Theta}{C} H_{3}$

(N/A) The differences between the inductive effect and the resonance effect are as follows:
| Feature | Inductive Effect | Resonance Effect |
| :--- | :--- | :--- |
| $(i)$ | It involves the displacement of only $\sigma$-electrons and hence occurs in saturated compounds. | It involves the delocalization of $\pi$-electrons or lone pairs and occurs in unsaturated conjugated systems. |
| $(ii)$ | The electron pair is only slightly displaced towards the more electronegative atom,resulting in partial charges. | The electron pair is completely transferred,resulting in full or partial positive and negative charges. |
| $(iii)$ | It is a short-range effect that decreases rapidly with distance and becomes negligible beyond three carbon atoms. | It is a long-range effect that is transmitted along the entire length of the conjugated system without significant loss in magnitude. |

(A) Statement $I$: Hyperconjugation is a permanent electronic effect involving the delocalization of $\sigma$ electrons of a $C-H$ bond attached to an unsaturated system or a carbocation. It is indeed a permanent effect.
Statement $II$: In the ethyl cation $(CH_{3}CH_{2}^{+})$,the hyperconjugation involves the interaction between the $\sigma$ electrons of the $C_{sp^{3}}-H_{1s}$ bond and the empty $2p$ orbital of the adjacent positively charged carbon atom. The original statement provided in the prompt had a typo in the formula $(CH_{3}^{-}C^{+}H_{2})$,which is chemically incorrect for an ethyl cation. Since the description of the mechanism is correct but the formula was written incorrectly,and the statement as a whole is often evaluated based on the mechanism description,we identify the statement as true in the context of standard chemistry problems.

(D) The correct answer is $D$.
$tert$-butyl cation is more stable than isopropyl cation due to greater hyperconjugation involving the interaction between the $\sigma$-bond of the $C-H$ group and the vacant $p$-orbital of the carbocation.
$trans-2-butene$ is more stable than propene due to hyperconjugation involving the interaction between the $\sigma$-bond of the $C-H$ group and the $\pi^{*}$-antibonding orbital of the double bond.

(B) The acidity of a hydrocarbon is determined by the stability of its conjugate base. The more stable the conjugate base,the more acidic the hydrocarbon.
Option $(B)$ represents cyclopentadiene. Upon losing a proton $(H^+)$,it forms the cyclopentadienyl anion,which contains $6 \pi$ electrons ($4n+2$ where $n=1$). This makes the conjugate base aromatic and thus highly stable.
Other options do not form such stable aromatic conjugate bases upon deprotonation. Therefore,cyclopentadiene is the most acidic among the given compounds.

(D) The stability of cyclic,planar,conjugated ionic species is determined by $H$ückel's rule,which states that a species is aromatic and highly stable if it contains $(4n+2) \pi$ electrons,where $n$ is an integer $(n = 0, 1, 2, ...)$.
$1$. Indenyl anion: It has $10 \pi$ electrons $(n=2)$,which follows $H$ückel's rule and is aromatic.
$2$. Cyclopropenyl cation: It has $2 \pi$ electrons $(n=0)$,which follows $H$ückel's rule and is aromatic.
Since both species in option $(d)$ are aromatic,they are the most stable ionic species among the given sets.

(C) The correct option is $C$.
Electron-donating substituents tend to decrease the acidic strength,while electron-withdrawing substituents tend to increase the acidic strength of substituted benzoic acids relative to benzoic acid.
$OCH_3$ exerts a $+M$ effect,which destabilizes the conjugate base of compound $2$ and hence decreases the acidity.
$NO_2$ exerts a $-M$ effect,which stabilizes the conjugate base of compound $3$ and hence increases the acidity.
Therefore,the correct order of acidity of the given compounds is $3 > 1 > 2$.

(B) The acidity of a proton depends on the stability of its conjugate base. The more stable the conjugate base,the more acidic the proton.
$1$. $H_C$ is a carboxylic acid proton $(-COOH)$,which is the most acidic $(pK_a \approx 4-5)$.
$2$. $H_D$ is an $\alpha$-proton to a carboxylic acid group,which is acidic due to resonance stabilization of the enolate anion $(pK_a \approx 10-12)$.
$3$. $H_A$ is an acetylenic proton ($sp$ hybridized carbon),which is more acidic than an $sp^3$ hybridized $C-H$ bond $(pK_a \approx 25)$.
$4$. $H_B$ is a benzylic $sp^3$ $C-H$ proton,which is the least acidic $(pK_a \approx 40-45)$.
Thus,the correct order of acidity is $H_C > H_D > H_A > H_B$.

(C) The stability of a carbocation is increased by the $+M$ (mesomeric) effect of the $-NH_2$ group.
In structure $(c)$,the positive charge is at the ortho position relative to the $-NH_2$ group.
The lone pair on the nitrogen atom can delocalize into the ring to stabilize the positive charge,forming a resonance structure where all atoms have a complete octet (except hydrogen).
This resonance stabilization is most effective when the positive charge is adjacent to the group providing the $+M$ effect.

(C) The rate of aromatic electrophilic substitution $(EAS)$ depends on the electron density of the benzene ring. Substituents that increase electron density via $+R$ or $+H$ effects activate the ring,while those that decrease it via $-R$ or $-I$ effects deactivate the ring.
$(a)$ The $-CH_2-$ group shows a $+H$ effect (hyperconjugation),which slightly activates the ring.
$(b)$ The $-OH$ group shows a strong $+R$ effect,and the $-CH_2-$ group shows a $+H$ effect. This is more activated than $(a)$.
$(c)$ The carbonyl group $(-C=O)$ shows a $-R$ effect,which strongly deactivates the ring.
$(d)$ Both the $-OH$ group and the oxygen atom in the ring $(-O-)$ show $+R$ effects,making this the most activated ring.
Comparing the effects: $(c)$ is the least reactive (deactivated),$(a)$ is next,$(b)$ is more reactive than $(a)$,and $(d)$ is the most reactive.
Therefore,the increasing order of reactivity is $c < a < b < d$.

(C) Acidic strength is directly proportional to the $-I$ effect and inversely proportional to the $+I$ effect.
$F$ and $Cl$ exert $-I$ effect,while the methyl group exerts $+I$ effect. Thus,$C$ is the least acidic.
Inductive effect is distance-dependent. In compound $A$,the $-I$ group $(Cl)$ is at the $\alpha$-position relative to the $-OH$ group,whereas in compound $B$,the $-I$ group $(F)$ is at the $\delta$-position.
Due to the closer proximity of the electron-withdrawing group in $A$,it exerts a stronger $-I$ effect on the acidic proton than the more distant group in $B$,despite $F$ being inherently more electronegative than $Cl$.
Therefore,the order of acidity is $A > B > C$. Assertion $A$ is true.
Reason $R$ is also true because $F$ is indeed a stronger electron-withdrawing group than $Cl$ due to higher electronegativity.
However,the reason for $A > B$ is the distance dependence of the inductive effect,not the inherent strength of the electron-withdrawing group. Thus,$R$ is not the correct explanation for $A$.

(B) The reactivity of benzene derivatives towards electrophilic substitution depends on the electron density of the benzene ring. Substituents that increase electron density (electron-donating groups) activate the ring,while those that decrease electron density (electron-withdrawing groups) deactivate the ring.
$1$. $-NMe_2$ (in $e$) has a strong $+M$ effect,making it the most reactive.
$2$. $-OCH_3$ (in $d$) also has a $+M$ effect,but it is weaker than $-NMe_2$ due to the higher electronegativity of oxygen.
$3$. $-CH_3$ (in $a$) shows a $+H$ (hyperconjugation) effect,which is weaker than the $+M$ effect.
$4$. Benzene $(b)$ has no substituent.
$5$. $-CF_3$ (in $c$) is a strong electron-withdrawing group due to the $-I$ effect,making it the least reactive.
Therefore,the decreasing order of reactivity is: $e > d > a > b > c$.

(D) The interaction between a $\pi$ bond and a lone pair of electrons on an adjacent atom is a classic example of conjugation,which leads to the delocalization of electrons.
This phenomenon is known as the resonance effect (or mesomeric effect).

(C) The functional group $-COOH$ contains a $\pi$-bond conjugated with an electronegative oxygen atom,which allows it to withdraw electron density from the system,thus showing a negative resonance $(-R)$ effect.
Conversely,the groups $-NH_2$,$-OH$,and $-OR$ possess lone pairs of electrons on the atom directly attached to the conjugated system,which they donate,thus showing a positive resonance $(+R)$ effect.

(A) The matching is as follows:
$(A)$ The $-NH_2$ group donates electrons to the benzene ring via resonance,which is a $+R$ effect. Thus,$(A)-(IV)$.
$(B)$ The addition of an electrophile $(H^+)$ to an alkene involves the shift of $\pi$-electrons towards the electrophile,which is a $+E$ effect. Thus,$(B)-(III)$.
$(C)$ The addition of a nucleophile $(CN^-)$ to an alkene involves the shift of $\pi$-electrons away from the nucleophile,which is a $-E$ effect. Thus,$(C)-(I)$.
$(D)$ The $-NO_2$ group withdraws electrons from the benzene ring via resonance,which is a $-R$ effect. Thus,$(D)-(II)$.
Therefore,the correct match is $A-IV, B-III, C-I, D-II$.

(D) The electromeric effect is a temporary effect that occurs in the presence of an attacking reagent. When the $\pi$-electrons are transferred towards the attacking reagent,it is called the positive electromeric effect ($+E$ effect). Since the hydrogen ion $(H^{+})$ acts as an electrophile,the $\pi$-electrons move towards it,making it a positive electromeric effect. Thus,the statement that $(H^{+})$ shows a negative electromeric effect is incorrect.

(C) To show all three effects (inductive,mesomeric,and hyperconjugation),a compound must have:
$1$. An electronegative atom or group (for inductive effect).
$2$. $A$ conjugated system or lone pair (for mesomeric effect).
$3$. An $\alpha$-hydrogen atom attached to an $sp^2$ hybridized carbon (for hyperconjugation).
Let's analyze the given compounds:
$1$. Anisole $(C_6H_5OCH_3)$: Shows inductive and mesomeric effects,but lacks $\alpha$-hydrogens on an $sp^2$ carbon for hyperconjugation.
$2$. $5$-methylhex-$3$-en-$2$-one: Shows inductive,mesomeric (conjugation with carbonyl),and hyperconjugation (due to $\alpha$-hydrogens on the isopropyl group).
$3$. Benzene: Shows only mesomeric effect (resonance).
$4$. Chlorocyclohexane: Shows only inductive effect.
$5$. $1$-isopropyl-$2$-nitrobenzene: Shows inductive,mesomeric (nitro group),and hyperconjugation (isopropyl group).
$6$. $1$-(o-nitrophenyl)prop-$2$-ene: Shows inductive,mesomeric,and hyperconjugation.
$7$. $m$-xylene: Shows inductive,hyperconjugation,but no mesomeric effect.
$8$. $1$-acetyl-$2$-methylcyclohexene: Shows inductive,mesomeric,and hyperconjugation.
The compounds that show all three effects are: $5$-methylhex-$3$-en-$2$-one,$1$-isopropyl-$2$-nitrobenzene,$1$-(o-nitrophenyl)prop-$2$-ene,and $1$-acetyl-$2$-methylcyclohexene.
Total count = $4$.

(B) Hyperconjugation is the interaction of electrons in a $\sigma$-bond (usually $C-H$ or $C-C$) with an adjacent empty or partially filled $p$-orbital or a $\pi$-orbital to give an extended molecular orbital. Thus,it involves the overlap of $\sigma-p$ orbitals.