Haloalkanes and Haloarenes

$ \text { The complete reaction is as follows } $

Thus, the correct decreasing order will be

$ 3^{\circ}>2^{\circ}>1^{\circ} \text { or } Y>X>Z $

$ \text { The reaction sequence is as follows } $

$ \text { So, } Z \text { is propanone. } $

The complete reaction sequence is as follows.

The complete reaction mechanism is, as follows,

$\mathrm{C}_6 \mathrm{Cl}_6$ is the molecular formula of product D. Thus, its, empirical formula is CCl.

$ \text { The complete reaction sequence is } $

$\mathrm{S}_{\mathrm{N}} 1$ mechanism depends on carbocation stability. The more stable the carbocation formed after the leaving group departs, the faster is the $\mathrm{S}_{\mathrm{N}} 1$ reaction.

In

tertiary benzylic carbocation is stabilised by $+I$ effect of $\mathrm{CH}_3$ and 2 phenyl group, which is most stable among the given options.

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.

Swarts reaction is a halogen exchange reaction. It is used to get alkyl fluorides from alkyl chlorides or bromides.

$ R-\mathrm{Cl}+\mathrm{AgF} \longrightarrow R-\mathrm{F}+\mathrm{AgCl} $

Among the given options, chlorobenzene is least reactive towards nucleophilic substitution reaction due to the delocalisation of the electrons of $\mathrm{C}-\mathrm{Cl}$ bond and a partial double bond character developed in the bond which makes it difficult for nucleophile to deave $\mathrm{C}-\mathrm{Cl}$ bond.

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[\mathrm{Mg} / \text { Dry ether }]{\mathrm{D}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{D}$

For $\mathrm{S}_{\mathrm{N}} 1$ reaction, carbocation intermediate is formed and stability of carbocation defines feasibility of reaction. The order of stability of carbocation is tertiary > secondary > primary

Order of reactivity of alkyl halide is tertiary halide $>$ secondary halide $>$ primary halide.

Swart's reaction is used for synthesis of alkyl fluorides from alkyl bromides.

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \xrightarrow{\mathrm{CoF}_2} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{~F} $

The order of reactivity towards $\mathrm{S}_{\mathrm{N}} 2$ reaction for alkyl halide is methyl $>$ primary halide $\left(1^{\circ}\right)>$ secondary halide $\left(2^{\circ}\right)>$ tertiary halide $\left(3^{\circ}\right)$.

Thus, option (d) is least reactive towards $\mathrm{S}_{\mathrm{N}} 2$ reaction as it is a $3^{\circ}$ halide.

Primary halides undergo $\mathrm{S}_{\mathrm{N}} 2$ reaction with $\mathrm{aq} . \mathrm{KOH}$. The reaction forms a transition state where the attacking nucleophile and leaving group both are partially bonded to carbon.

$3^{\circ}$ alkyl halide undergoes nucleophilic substitution reaction via formation of a stable $3^{\circ}$ carbocation. Thus, it shows $\mathrm{S}_{\mathrm{N}} 1$ reaction where rate of reaction depends only on concentration of substrate.

$ \text { (C) } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{I} \xrightarrow{\text { Alc. } \mathrm{KOH}}\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 $

Primary halides prefer to undergo E2 elimination reaction.

The reaction path followed is E1 reaction.

∴ The correct option is $\mathrm{A} \rightarrow \mathrm{IV}, \mathrm{B} \rightarrow \mathrm{I}, \mathrm{C} \rightarrow \mathrm{II}$ and $\mathrm{D} \rightarrow \mathrm{III}$.

$\therefore X$ is alc. KOH $Y$ is NaOI

Due to presence of peroxide, addition of HBr takes place according to anti-Markownikoff rule in which negative part of addendum goes to carbon which has more number of H -atom attached to it.

Hydrolysis of cyano group will yield or produce.

(i) Firstly toluene reacts with bromine in the presence of light to give benzyl bromide. This reaction is free radical substitution reaction.

(ii) Benzyl bromide reacts with magnesium to give benzyl magnesium bromide (Grignard reagent).

(iii), (iv) In next step, benzyl magnesium bromide reacts with formaldehyde $\left(\mathrm{CH}_2 \mathrm{O}\right)$ followed by acidic hydrolysis to give 2-phenylethanol.

(v) In last step, alcohol is converted to alkyl halide in presence of $\mathrm{PBr}_3$.

Mechanism involved

Step (i) Bromination reaction Electrophilic substitution takes place.

Step (ii) Friedel crafts alkylation Electrophilic substitution takes place. As, Br is ortho-para directing group, so, alkylation occurs at para-position.

$ \text { Step (iii) Grignard reagent formation } $

Step (iv) Nucleophilic addition reaction of Grignard reagent

$ \text { Step (v) Hydrolysis } $

The $\mathrm{S}_{\mathrm{N}} 1$ reaction proceed by the formation of carbonium ion. The order of stability of carbonium ion is as follows

$ 3^{\circ}>2^{\circ}>1^{\circ} $

Also, more the number of carbon atoms, more will be the stability of carbonium ion. Therefore, the correct order of reactivity of $\mathrm{S}_{\mathrm{N}} 1$ reaction is as follows

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

Presence of $-\mathrm{NO}_2$ group will favour chlorobenzene towards nucleophilic substitution, $\operatorname{ArS}_{\mathrm{N}} 2$ (via addition elimination mechanism) $-R$ effect of $-\mathrm{NO}_2$ at ortho position (I) will favour the reaction most in comparison to $-I$ effect of $-\mathrm{NO}_2$ at meta position (III).

$ \text { So,the reactivity order : I }>\text { III }>\text { II } \text {. } $

Step 1 First of all cyclohexanone react with Et MgBr and $\mathrm{H}_2 \mathrm{O}$.

Step 2 In the presence of $20 \% \mathrm{H}_3 \mathrm{PO}_4$ elimination undergoes and water is removed resulting alkene is formed.

Step 3 HBr react with alkene forms haloderivative.