Haloalkanes and Haloarenes

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.

Among the given options $ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl} $ does not undergo $ \mathrm{S}_{\mathrm{N}} 1 $ mechanism. It is because the carbocation (vinylic)

$ \mathrm{CH}_{3}-\mathrm{CH}=\stackrel{\oplus}{\mathrm{C}} \mathrm{H} \text { is very unstable. } $

The reagent used to obtained good yield of product is $ \mathrm{KI}, 95 \% \mathrm{H}_{3} \mathrm{PO}_{4} $.

Alcohol can be converted to alkyl halides using KI in presence of

$ \mathrm{H}_{3} \mathrm{PO}_{4} $. $ R-\mathrm{OH}+\mathrm{HI} \xrightarrow[\mathrm{H}_{3} \mathrm{PO}_{4}]{\mathrm{KI}} R \mathrm{I}+\mathrm{H}_{2} \mathrm{O} $

The complete reaction is as follows,

$ \left(\mathrm{CH}_3\right)_3 \mathrm{CBr} \xrightarrow[\Delta]{\text { Alc. } \mathrm{KOH}} \underset{\substack{\text { (Product) } \\\\ X}}{\left(\mathrm{CH}_3\right)_2 \mathrm{C}_2 \mathrm{H}_2} \xrightarrow{\mathrm{HBr}} \underset{\substack{\text { (Product) } \\\\\left(\mathrm{Y}\right)}}{\left(\mathrm{CH}_3\right)_3 \mathrm{CBr}} $

The reaction mentioned is Friedel-Crafts reaction.

$ \begin{aligned} & \text { Ethyl benzene }=\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_3 \\ & \text { Total mass }=8 \times 12+10 \times 1=106 \mathrm{~g} \\ & \text { Percentage of carbon }=\frac{96}{106} \times 100 \\ & \text { = 90.6\% } \end{aligned} $

Allylic halides are the compounds in which the halogen group is attached to a hybridised carbon $\left(s p^3\right)$ atom next to the carbon which is already in double bond with another carbon.

This is an example of $S_{\mathrm{N}} 1$ reaction. The complete reaction is as follows.

The sequence of reagents required to convert ethylbromide to propanal is $\mathrm{CH}_3 \mathrm{COOAg}$, DIBAL-H and $\mathrm{H}_2 \mathrm{O}$.

Allylic halides are those compound 156 where the halogen group is attached to a $s p^3$ hybridised carbon atom and at $\alpha$- to the carbon-carbon double bond. 4-chlorobut-1-ene is not allylic halide.

$ \text { The product } C \text { formed will be as follows } $

In the presence of peroxide, addition of HBr to unsymmetrical alkenes takes place contrary to the Markownikoff rule. In case of 2-butene both double bonded carbons are identical. Therefore, it does not observe anti-Markownikoff addition of HBr .

Among the given reaction 1 -iodopropane can be prepared by reactions given are $B$ and $C$ only.

When alkyl chloride/bromide is heated in presence of fluoride of same heavy metal alkyl fluoride is obtained. This reaction is named as Swarts reaction.

When chlorobenzene $(X)$ is reacted with reagent NaOH $(A)$ it produce phenol.

Hence, $A=$ (i) $\mathrm{NaOH}, 623 \mathrm{~K}, 300 \mathrm{~atm}$ (ii) $\mathrm{H}^{+} B=$ (i) NaOH , 443 K (ii) $\mathrm{H}^{+}$

The correct match is A-III, B-I, C-II.

Sandmeyer reaction is a reaction used to synthesis aryl halides from aryl diazonium salt.

Finkelstien reaction is a halogen $(X)$ exchange method. It is used to prepare alkyl iodide by reaction of alkyl chlorides/bromides with NaI in dry acetone.

$ \underset{(x=\mathrm{Cl}, \mathrm{Br})}{R-X}+\mathrm{NaI} \xrightarrow{\text { Acetone }} R \mathrm{I}+\mathrm{NaX} $

Swartz reaction is used to get alkyl fluorides from alkyl chlorides.

$ R-\mathrm{Br}+\mathrm{AgF} \longrightarrow R-\mathrm{F}+\mathrm{AgBr} $

The ammonia gas released will turn red litmus to blue. Thus, the compound $X$ and $Y$ are $\mathrm{KCN} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$.