As it is given that compound X on reacting with C₆H₅SO₂Cl give a solid product which is insoluble in alkali, this shows that there is no hydrogen attached to N in the product.
Thus, the compound X is a secondary amine. The reaction involved is as follows

(a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}\)
Mechanism: HBr \(\longrightarrow\) H⁺+ Br^-

It fIlows S_N2 mechanism.
(b) Reimer Tiemann reaction:

The Lucas test differentiates between primary, secondary, and tertiary alcohols.
It is based on the difference in reactivity of the alcohols with HCl in an S_N1 reaction:
ROH + HCl→RCl+H₂O
It works because secondary carbocations are more stable and form faster than primary carbocations, and tertiary carbocations are so stable that the reaction takes place almost immediately.

\(\begin{equation} \text { (i) } \mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}+\mathrm{HI} \longrightarrow \mathrm{CH}_{3}-\mathrm{OH}+\mathrm{CH}_{3}-\mathrm{I} \end{equation}\)
(ii) (a) Protonation of alkene to form carbocation by elecrophilic attack of H₃O⁺.

(b) Nucleophilic attack of water on carbocation.

(c) Deprotonation to form an alcohol.

Iodide ion is a powerful nucleophile and hence reacts rapidly with RCl to form RI.
KI ——— K⁺ + I^-; R — Cl + I^- ——— R — I + Cl^-
Further I^- is a better leaving group than CI^- ion, therefore, RI is more rapidly hydrolysed than RCl to form ROH.
HO^- + R — I ——— R — OH + I^-

(i) \({ CH }_{ 3 }CH={ CH }_{ 2 }+{ H }_{ 2 }O\overset { { H }^{ + } }{ \rightleftharpoons } { CH }_{ 3 }-\underset { \overset { | }{ OH } }{ C } -{ CH }_{ 3 }\)

(i) \({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -O-{ CH }_{ 3 }+HI\longrightarrow { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -I+\underset { Methanol }{ { CH }_{ 3 }OH } \)
(ii) \({ CH }_{ 3 }-CH={ CH }_{ 2 }\overset { (i){ B }_{ 2 }{ H }_{ 6 } }{ \underset { (ii)3{ H }_{ 2 }{ O }_{ 2 }/{ H }^{ - } }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }OH\)
 

\((a)\quad { CH }_{ 3 }\underset { Butanal }{ { CH }_{ 2 }{ CH }_{ 2 } } CHO\overset { { H }_{ 2 }/Ni }{ \longrightarrow } { CH }_{ 3 }\underset { Butanal-1-ol }{ { CH }_{ 2 }{ CH }_{ 2 } } OH\\ (b)\quad { CH }_{ 3 }\underset { Propene }{ -CH } ={ CH }_{ 2 }+{ H }_{ 2 }O\overset { { H }_{ 2 }/{ SO }_{ 4 } }{ \longrightarrow } { CH }_{ 3 }-\underset { \overset { | }{ \underset { Prppane-2-ol }{ OH } } }{ CH } -{ CH }_{ 3 }\\ (c)\quad { CH }_{ 3 }-\overset { \underset { \parallel }{ O } }{ C } -{ CH }_{ 3 }+{ CH }_{ 3 }MgBr\longrightarrow { CH }_{ 3 }-\overset { \underset { | }{ OMgBr } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -{ CH }_{ 3 }\overset { { H }_{ 2 }/O{ H }^{ + } }{ \longrightarrow } { CH }_{ 3 }-\overset { \underset { | }{ OMgBr } }{ \underset { \overset { | }{ \quad \quad { CH }_{ 3 }\quad } }{ C } } -{ CH }_{ 3 }\)
Propane               Methyl magnesium                         Tert.butoxy                           2-Methylpropan-2-ol
                                  bromide                               magnesium bromide

Reaction of HI with methoxymethane

(a) Cumene (2-phenyl propane) is used in the industrial preparation phenol.

(ii) In non-aqueous medium

(c) In the case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to highly activating effect of -OH group attached to benzene ring. Br₂ + H₂O \(\longrightarrow \) HOBr.
Secondly, HOBr breaks into OH^- and Br⁺: \(\mathrm{HOBr} \longrightarrow \mathrm{O} \mathrm{H}+\mathrm{Br}^{+}\) .

(i) In aryl alkyl ethers, +R-effect of the alkoxy group (-OR) increases the electron density on the benzene ring thereby, activating the benzene ring towards electrophilic substitution reactions.

(ii) Alkoxy group (-OR) is ortho, para directing and activates the aromatic ring towards electrophilic substitution.
It directs the incoming electrophilic at o- and p-positions. This is because of the presence of the negative charges at o- and p-positions indicate the more electron density at these positions.

(i) 2, 5-Dimethylphenol
(ii) 2,6-Dimethylphenol
(iii) 1-Methoxy-2-methylpropane.

(i) 2, 2, 4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butane -2, 3-diol.

(i) It is due to higher molecular weight, more surface area more vav deer waals, forces of Attraction in C₂H₂Oh.
(ii) Phenol is the strongest acid than alcohol because phenoxide ion is more stable due to resonance than ethoxide ion.
(iii) It is because ethers are not associated with inter molecular H- bonding, whereas ethanol.

Benzene is heated with conc. sulphuric acid to form benzene sulphonic acid. It is then heated with sodium hydroxide to give sodium phenoixde. Hydrolysis with sulphuric acid gives phenol. The reaction sequence is given above.

(i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.

(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.
 
(iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol on hydrolysis.
 
(iv) When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.