Haloalkanes and Haloarenes

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 } $

Cl has $-I$ and $+R$-effect. Due to electron donating resonance effect it is ortho/para-directing and donates the electron density to benzene ring. Thus, the incoming electrophile -Br gets attached on para-position and para disubstituted product is obtained as major product and ortho as minor product.

Alkenyl bromide reacts with Mg in dry ether to form Grignard reagent.

Grignard reagent abstracts acidic proton. So, it takes $\mathrm{D}^{+}$ form $\mathrm{D}_2 \mathrm{O}$.

Finkelstein reaction is an $\mathrm{S}_{\mathrm{N}} 2$ reaction that involves the exchange of one halogen by iodine. In this reaction, alkyl chloride/bromides are treated with sodium iodide in acetone to form alkyl iodide and $\mathrm{NaCl} / \mathrm{NaBr}$. Acetone is used because $\mathrm{NaCl} / \mathrm{NaBr}$ are less soluble in acetone hence, they can precipitate out.

$R-\mathrm{Cl}+\mathrm{NaI} \xrightarrow{\text { Acetone }} R-\mathrm{I}+\mathrm{NaCl} \downarrow$

Density is directly proportional to mass. The molar masses of the compounds are as follows:

$ \begin{aligned} \mathrm{CCl}_4 & \rightarrow 12 + 142 = 154 \ \mathrm{g/mol} \\ n-\mathrm{C}_3\mathrm{H}_7\mathrm{I} & \rightarrow (12 \times 3) + (7 \times 1) + 127 = 170 \ \mathrm{g/mol} \\ n-\mathrm{C}_3\mathrm{H}_7\mathrm{Br} & \rightarrow (12 \times 3) + (7 \times 1) + 80 = 123 \ \mathrm{g/mol} \\ n-\mathrm{C}_3\mathrm{H}_7\mathrm{Cl} & \rightarrow (12 \times 3) + (7 \times 1) + 35.5 = 78.5 \ \mathrm{g/mol} \\ \end{aligned} $

Since the molar mass of $n-\mathrm{C}_3\mathrm{H}_7\mathrm{I}$ is the highest, it will also have the highest density.

Benzyl chloride reacts with Mg in presence of dry ether to form Grignard reagent (G.R.) and on treating with $\mathrm{CO}_2$, G.R. form carboxylate ion which on hydrolysis give carboxylic acid.

$X$ on treatment with $\mathrm{LiAlH}_4$ ether followed by acidic hydrolysis give alcohol as $Y$ product.

$\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$ is used to reduce carbonyl compounds to alkane. This reaction is called Clemmenson reduction. But in presence of basic groups like $-\mathrm{OH},-\mathrm{NH}_2$ etc. side reaction also takes place. In this case substitution of alcoholic group by $\mathrm{Cl}^{\ominus}$ will take place.

Amides reacts with Br$_2$ in presence of an alkali base (KOH, NaOH) to form amine. This reaction is known as Hofmann bromamide reaction. This reaction proceeds via formation of nitrene intermediate.

First step in the given reaction is "Lucas test" in which alcohols react with $\mathrm{HCl} / \mathrm{ZnCl}_2$ to give alkyl halides.

Second step proceed via nucleophilic substitution bimolecular $(\mathrm{S}_{\mathrm{N}} 2$) mechanism. Here, $\mathrm{I}^{-}$ replaces Cl to form alkyl iodide.

Third step is "Wurtz reaction" in which two alkyl iodide reacts with Na in presence of dry ether to form alkane with even number of carbon atoms.

1-chlorobutane reacts with aq. KOH to give 1-butanol by S_N2 reaction.

However, 1-chlorobutane reacts with alc. KOH to give 1-butene by elimination reaction.

For a compound to be aromatic, it must have (4n + 12) $\pi$-electrons.

Here, two double bonds and 4$\pi$-electrons are present in the compound. Therefore, it is anti-aromatic.

Homocyclic compounds are ring structures that consist only of carbon atoms with in the ring. Bromo group is attached to ring system as substituent, so $\mathrm{C_6H_6Br_2}$ is homocyclic but not aromatic.

Hence, option (c) is correct.

With mixture of conc. HNO$_3$ and conc. H$_2$SO$_4$, anisole gives ortho and para-nitro anisole.

Methoxy group is activating and ortho-para directing group.