Haloalkanes and Haloarenes

In presence of peroxide, addition of HBr to an alkene follows free radical mechanism (peroxide effect).

For styrene $\mathrm{Ph{-}CH{=}CH_2}$:

Major product (with peroxide)

Br• adds in such a way that the more stable radical is formed.

If Br• adds to terminal carbon ($\mathrm{CH_2}$), the radical comes on benzylic carbon:

$\mathrm{Ph{-}\dot{C}H{-}CH_2Br}$

This is benzylic radical (resonance-stabilised), hence preferred. Then it abstracts H from HBr:

$\mathrm{Ph{-}CH_2{-}CH_2Br}$

So $1$-bromo-$2$-phenylethane is the major product.

Checking each statement

A. True

The reaction proceeds via the more stable radical intermediate (benzylic radical).

B. False

Peroxide does not generate $ \mathrm{H^\bullet}$.

It generates radicals that ultimately produce $ \mathrm{Br^\bullet}$ (the chain-carrying radical), e.g. by abstraction of H from HBr.

C. True

Benzoyl peroxide can form phenyl radicals ($\mathrm{Ph^\bullet}$), which can abstract H from HBr:

$\mathrm{Ph^\bullet + HBr \rightarrow PhH + Br^\bullet}$

So benzene ($\mathrm{PhH}$) can be formed as a byproduct.

D. False

$1$-bromo-$2$-phenylethane ($\mathrm{Ph{-}CH_2{-}CH_2Br}$) is the major, not minor, product in presence of peroxide.

E. True

In absence of peroxide, HBr adds by electrophilic addition via a carbocation intermediate (here benzylic carbocation), giving Markovnikov product.

Correct option: Option C (A, C & E Only)

Vinyl, allyl, benzyl and alkyl chlorides are identified from the position of $Cl$:

Step 1: Identify each example in List-II

1. III. $CH_2 = CHCl$

   Here $Cl$ is directly attached to a double-bond carbon $\Rightarrow$ vinyl chloride.

2. I. $CH_2 = CH-CH_2Cl$

   Here $Cl$ is on the carbon next to $C=C$ (allylic position) $\Rightarrow$ allyl chloride.

3. II. $CH_3-CH(Cl)CH_3$

   This is a saturated alkyl group with $Cl$ on an $sp^3$ carbon $\Rightarrow$ alkyl chloride (2-chloropropane).

4. IV. (structure shown is $C_6H_5-CH_2Cl$)

   $Cl$ is on the carbon next to benzene ring (benzylic position) $\Rightarrow$ benzyl chloride.

Step 2: Match List-I with List-II

• A. Vinyl chloride $\rightarrow$ III

• B. Benzyl chloride $\rightarrow$ IV

• C. Alkyl chloride $\rightarrow$ II

• D. Allyl chloride $\rightarrow$ I

So the correct option is:

$\boxed{\text{Option C: A-III, B-IV, C-II, D-I}}$

In $\mathrm{Ph}-\mathrm{OMe}$, the $-\mathrm{OMe}$ group is an electron-donating group and shows $+\mathrm{M}$ (plus mesomeric) effect. It donates electron density to the benzene ring, increases the electron density on the ring, and therefore activates the ring towards electrophilic substitution such as nitration.

In $\mathrm{Ph}-\mathrm{NO}_2$, the $-\mathrm{NO}_2$ group is a strong electron-withdrawing group and shows $-\mathrm{M}$ (minus mesomeric) effect. It pulls electron density away from the benzene ring and therefore strongly deactivates the ring towards nitration.

In $\mathrm{Ph}-\mathrm{Cl}$, the $-\mathrm{Cl}$ group is overall an electron-withdrawing group (due to its $-I$ or inductive effect). Because of this electron-withdrawing nature, it decreases the reactivity of the benzene ring towards nitration, but this effect is weaker than that of the $-\mathrm{NO}_2$ group.

Therefore, the increasing order of reactivity towards nitration is: $\mathrm{Ph}-\mathrm{NO}_2 < \mathrm{Ph}-\mathrm{Cl} < \mathrm{Ph}-\mathrm{OMe}$.

In chlorocyclohexane, chlorine shows only the $-\mathrm{I}$ (electron-withdrawing inductive) effect.

So, the $\mathrm{C}-\mathrm{Cl}$ bond remains a normal single bond and is more polar (more electron density is pulled towards chlorine).

In chlorobenzene, chlorine shows two effects: $-\mathrm{I}$ and $+\mathrm{M}$ (lone pair donation by resonance into the benzene ring).

Because $-\mathrm{I}>+\mathrm{M}$, the bond is still polar, but the $+\mathrm{M}$ effect reduces the polarity compared to chlorocyclohexane.

The $+\mathrm{M}$ effect also causes resonance between chlorine lone pairs and the aromatic ring, which gives the $\mathrm{C}-\mathrm{Cl}$ bond partial double bond character.

Rate of $\mathrm{S}_{\mathrm{N}} 1 \propto$ Stability of $\mathrm{C}^{\oplus}$ formed

(I) and (II) are unstable due to Bredt's rule, (I) has more + I effect.

$ \text { (II) }<\text { (I) }<\text { (III) }<\text { (IV) } $

$ \begin{aligned} &\text { In methanol, } \mathrm{CH}_3\mathrm{Br} \text{ reacts by } S_N2 \text{ mechanism, so the rate depends mainly on the nucleophilicity of the nucleophile.} \\[4pt] &\text { In a protic solvent like methanol, smaller anions like } \mathrm{F}^{\ominus} \text{ are strongly solvated (surrounded by solvent molecules).} \\ &\text { Because of this strong solvation, } \mathrm{F}^{\ominus} \text{ becomes less available to attack the carbon, so it is the weakest nucleophile.} \\[4pt] &\text { Larger anions like } \mathrm{I}^{\ominus} \text{ are less solvated in methanol, so they can attack more easily. Hence, } \mathrm{I}^{\ominus} \text{ is the strongest nucleophile.} \\[4pt] &\text { Among } \mathrm{C}_2\mathrm{H}_5\mathrm{O}^{\ominus} \text{ and } \mathrm{PhO}^{\ominus}, \text{ phenoxide } (\mathrm{PhO}^{\ominus}) \text{ is less nucleophilic.} \\ &\text { This is because its negative charge is delocalised (resonance) into the benzene ring, reducing its tendency to attack.} \\ &\text { Ethoxide } (\mathrm{C}_2\mathrm{H}_5\mathrm{O}^{\ominus}) \text{ has a more localised negative charge, so it is a better nucleophile.} \\[6pt] &\text { Order of nucleophilicity : }\\ &\mathrm{I}^{\ominus}>\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{\ominus}>\mathrm{PhO}^{\ominus}>\mathrm{F}^{\ominus} \end{aligned} $

For $CH_3Br$ (a methyl halide), the reaction proceeds by $S_N2$, so the rate increases with nucleophilicity of $Nu^-$.

For oxygen nucleophiles in the same row, nucleophilicity roughly follows basicity, and any resonance stabilization of the anion reduces nucleophilicity:

• $OH^-$: strong base, charge localized $\Rightarrow$ best nucleophile

• $PhO^-$: resonance-stabilized $\Rightarrow$ less nucleophilic than $OH^-$

• $CH_3COO^-$: even more resonance delocalization over two oxygens $\Rightarrow$ weaker nucleophile

• $ClO_4^-$: highly resonance-stabilized, essentially non-nucleophilic

$ OH^- \;>\; PhO^- \;>\; CH_3COO^- \;\gg\; ClO_4^- $

Correct option: B.

Type of reaction: This reaction happens by the $S_N1$ mechanism.

What affects the rate: The speed of an $S_N1$ reaction depends on how stable the carbocation (positively charged ion) formed in the process is.

Both reactions are electrophilic substitution reaction, $\mathrm{I}^{\text {st }}$ is sulphonation and $\mathrm{II}^{\text {nd }}$ is halogenation :

A nucleophilic substitution reaction is a chemical reaction in which an electron rich molecule (nucleophile) replaces a functional group in an electron deficient molecule (electrophile). Carbocation formation occurs in SN$^1$ mechanism (unimolecular nucleophilic substitution).

Carbocation is formed in the first step when the bond between carbon and the leaving group breaks. The nucleophile attack on the carbocation occurs in the second step.

The general stability order of carbocation is

Tertiory > Secondary > Primary

The stability order including allyl carbocation, benzyl carbocation, phenyl carbocation and triphenyl methyl $\left( {{{(Ph)}_3}{C^ + }} \right)$ carbocation is

Phenyl < Allyl < Benzyl < Triphenyl methyl

(1)

Carbocation formed from this halide is

Allylic carbocation stability is due to the resonance delocalization of the positive change across the adjacent double bond.

(2)

Stable due to the delocalization of positive change across all three phenyl rings.

(3)

Benzyl carbocation stability is because of the positive charge delocalization across the aromatic ring. So, it is mole stable than allylic carbocation but less stable than Ph$_3$C+ carbocation.

(4)

Delocalization is possible for the adjacent double bond. So, most stable carbocation is

The answer is option (2)

$\mathrm{\mathop {C{H_3} - C{H_2} - C{H_2} - N{O_2}}\limits_{(A)} \buildrel {Ag - N{O_2}} \over \longrightarrow C{H_3} - C{H_2} - C{H_2} - Br\buildrel {AgCN} \over \longrightarrow \mathop {C{H_3} - C{H_2} - C{H_2} - NC}\limits_{(B)}}$

Transition state (less charge density)

$\Rightarrow$ This reaction will proceed faster in less polar medium which will not increase the activation energy value.

Transition state (Higher charge density)

$\Rightarrow$ This reaction will proceed faster in more polar medium which will decrease the activation energy value.

Addition of $\mathrm{H}-\mathrm{Br}_{(\mathrm{aq})}$ to alkene follows electrophilic addition mechanism. In the rate determining step a carbocation intermediate is formed. Among P, Q, R & S compound Q will form most stable carbocation intermediate since it is resonance stabilized.

A is phenol and B is para benzoquinone.

Solvolysis or $\mathrm{S}_{\mathrm{N}} 1 \propto$ stability of carboccation Stability order

$\mathrm{RBr} \xrightarrow[\text { dry Ether }]{\mathrm{Mg}} \mathrm{RMgBr} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{R}-\mathrm{H}$

Hence, RBr Can be

Total 4 Structural isomers.

$\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_2-\mathrm{Cl}$ will undergo $\mathrm{S}_{\mathrm{N}} 1$ mechanism because $\mathrm{CH}_3-\mathrm{O}-\stackrel{+}{\mathrm{C}} \mathrm{H}_2$ is highly stable.

To determine the correct answer, we need to analyze both the Assertion (A) and the Reason (R) independently, as well as see if the Reason (R) provides the correct explanation for the Assertion (A).

Assertion (A) Analysis:

Assertion (A) states that $\mathrm{S}_{\mathrm{N}} 2$ reaction of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$ occurs more readily than the $\mathrm{S}_{\mathrm{N}} 2$ reaction of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$. In an $\mathrm{S}_{\mathrm{N}} 2$ reaction, the rate depends on the nucleophile as well as the substrate. Specifically, $\mathrm{S}_{\mathrm{N}} 2$ reactions proceed through a backside attack mechanism, leading to the formation of a transition state where the nucleophile and the leaving group are simultaneously bonded to the carbon atom. The benzyl group $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2-$ is more stable for $\mathrm{S}_{\mathrm{N}} 2$ reactions compared to the ethyl group $\mathrm{CH}_3 \mathrm{CH}_2-$ because the benzyl carbon's partial positive charge in the transition state is stabilized by the resonance of the phenyl ring.

Reason (R) Analysis:

Reason (R) explains that the partially bonded unhybridized p-orbital that develops in the trigonal bipyramidal transition state is stabilized by conjugation with the phenyl ring. This means that the presence of the phenyl ring can delocalize and stabilize the charge in the transition state, facilitating the $\mathrm{S}_{\mathrm{N}} 2$ reaction.

Combining Assertion (A) and Reason (R):

Given that the phenyl ring in $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$ can indeed stabilize the transition state through conjugation more effectively than an ethyl group, both the Assertion (A) and Reason (R) are correct. Furthermore, Reason (R) provides a valid explanation for Assertion (A), reinforcing why the $\mathrm{S}_{\mathrm{N}} 2$ reaction occurs more readily for $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br}$.

Thus, the most appropriate answer is:

Option B: Both (A) and (R) are correct and (R) is the correct explanation of (A)

Rate of $\mathrm{S}_{\mathrm{N} 2}$ reaction depends on steric hindrance, less the steric hindrance more will be rate of reaction , it will undergo S_N2 reaction fastest.

More stable double bond will be formed.

In , double bond is in resonance with the ring, thus, more stable.

Product '$\mathrm{A}$' is $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$

$\therefore$ Option (3) is correct.

Statement - I: Rate of $\mathrm{S}_{\mathrm{N}} 2 \propto[\mathrm{R}-\mathrm{X}]\left[\mathrm{Nu}^{-}\right]$

$\mathrm{S}_{\mathrm{N}} 2$ reaction is favoured by high concentration of nucleophile $(\mathrm{Nu}^{-})$ & less crowding in the substrate molecule.

Statement - II: Solvolysis follows $\mathrm{S}_{\mathrm{N}} 1$ path.

Both are correct Statements.

Vinyl carbon is $\mathrm{sp}^2$ hybridized aliphatic carbon

Covalent character of $\mathrm{AgCN}$.

Assertion (A): Given statement is correct because in phenol hydroxyl group cannot be replaced by halogen atom.

Reason (R) :

Given reason is false.

Hence Assertion (A) is correct but Reason (R) is false.

Since $\mathrm{CH}_3-\mathrm{CH}=\stackrel{+}{\mathrm{C}} \mathrm{H}$ is very unstable, $\mathrm{CH}_3-\mathrm{CH}= \mathrm{CH}-\mathrm{Cl}$ cannot give $\mathrm{S}_{\mathrm{N}^1}$ reaction.

The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is Option D. Let's break down why this is the case by understanding both the $S_N 1$ and $S_N 2$ reactions and their effects on the chirality of the alkyl halide.

$S_N 2$ Reaction:

The $S_N 2$ (Substitution Nucleophilic Bimolecular) reaction is a one-step process where the bond formation (nucleophile attacking the substrate) and the bond-breaking (leaving group leaving the substrate) processes occur simultaneously. This reaction type is associated with an inversion of configuration at the chiral center. This means that if the reactant molecule is chiral and the nucleophile attacks from the side opposite to the leaving group, the product will have the opposite configuration to the reactant. This phenomenon is known as Walden inversion. Therefore, it does not lead to racemisation but rather to inversion of the stereochemistry at the chiral center.

$S_N 1$ Reaction:

The $S_N 1$ (Substitution Nucleophilic Unimolecular) reaction is a two-step process. The first step involves the formation of a carbocation intermediate by the departure of the leaving group. This intermediate is planar, and hence, the nucleophile can attack from either side of the plane with equal probability. This results in the formation of products with both configurations - one retaining the original configuration and the other being its mirror image (inverted configuration). As both enantiomers are formed, the process leads to racemisation, especially in situations where the carbocation can freely rotate and is attacked with equal probability from both sides. Over time, if this reaction goes to completion and there's no bias in nucleophile attack direction, you'd expect a racemic mixture of products.

Thus, Option D is correct: Racemisation occurs in $S_N 1$ reaction, and inversion occurs in the $S_N 2$ reaction.