Alcohol, Phenols and Ethers

tert-butoxide reacts with methyl bromide to form ether product, $t$-butyl methyl ether. This reaction proceeds via $\mathrm{S}_{\mathrm{N}} 2$ mechanism.

Methoxide reacts with $t$-butyl bromide to give alkene product. This reaction proceeds via $\mathrm{E}_2$ mechanism.

Decarboxylation of carboxylic acids takes place when they react with soda lime.

Phenol changes into benzene on heating with zinc dust. When phenolic acid is reacted with aqueous bromine, Br is substituted at all ortho and para positions in the same molecule.

Statement (A) and (C) are correct while (B) and (D) are incorrect. Their correct forms are as follows.

Phenol is used in the manufacturing of bakelite not melamine plastic.

Phenol when heated with acetyl chloride gives phenyl ethanoate not phenetole.

$ \text { In reaction } A \text {, } $

$ \text { In reaction } B \text {, } $

Step (i) and (ii) The reaction of cumene with oxygen results in the formation of peroxide which on acidic treatment gives phenol and acetone.

Step (iii) Phenol is $0,-p$-directing group. At low temperature and in the presence of non-polar solvent like $\mathrm{CHCl}_3$, phenoxide ion is not generated. The ring doesn't get highly activated for trisubstitution. So the phenol takes part in the electrophilic substitution reaction and gives a mixture of ortho and para bromophenol.

The para-bromophenol is the major product.

The alcohol acts as nucleophile and attacks the carbonyl carbon of acid in esterification.

As the bulkiness of the nucleophilic group increases the reactivity towards esterification decreases.

The nucleophilicity order is

$ \mathrm{I}>\mathrm{IV}>\mathrm{II}>\mathrm{III} $

Therefore, reactivity towards esterification decreases in same order.

$\mathrm{CO}+\mathrm{H}_2 \xrightarrow[\text { High pressure }(673 \mathrm{~K}, 300 \mathrm{~atm}) \mathrm{ZnO}-\mathrm{Cr}_2 \mathrm{O}_3]{\text { Migh temperature }} \underset{\text { Methanol }}{\mathrm{CH}_3 \mathrm{OH}}$

Methanol is industrially prepared when mixture of carbon monoxide and hydrogen is subjected to high temperature and pressure in presence of $\mathrm{ZnO}-\mathrm{Cr}_2 \mathrm{O}_3$ as catalyst.

Acidity $\propto$ stability of the anion formed. The anions formed are

Chlorine is an ortho-para directing group. Chlorine is electronegative species but it has lone pair of electrons which has tendency to added in resonance structure of benzene ring which develop negative charge on ortho and para position and minimum electron density at meta position.

Alkyl bromide in presence of $\mathrm{Mg} / \mathrm{ether}$ gives $R \mathrm{Mg}^{+} \mathrm{Br}^{-}$. Here, alkyl group is nucleophilic and attacks $\mathrm{CO}_2$ to form carboxylic acid.

$\mathrm{SOCl}_2$ is used to replace $-\mathrm{OH}$ group to $-\mathrm{Cl}$ group and evolved $\mathrm{SO}_2$ and $\mathrm{HCl}$ gas.

Cyclohexanol on reaction with H₂SO₄ and heating gives cyclohexene as follows

Methyl alcohol can be distinguished from ethyl alcohol by iodoform test as follows by using sodium hydroxide and iodine.

${C_2}{H_5}OH\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{Iodoform\,test}^{NaOH/{I_2}}} CH{I_3}$ (yellow ppt.)

$C{H_3}OH\buildrel {NaOH/{I_2}} \over \longrightarrow $ No. ppt.

$ \text { The given chain reaction give the final product as follows : } $

(i) with KI , we get $\mathrm{CH}_3 \mathrm{I}$, which on treatment with Mg , dry ether give ethane $\left(\mathrm{CH}_3-\mathrm{CH}_3\right)$.

(ii) on treatment with

followed by $\mathrm{H}_2 \mathrm{O}$ and $20 \% \mathrm{H}_3 \mathrm{PO}_4$ at 358 K , we get the final product. Hence, option (a) is the correct answer.

So, reactions I, III and IV produce $n$-butanol.

$ \text { ∵ In } $

$ \text { In first reaction, } $

$ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl} \xrightarrow[-\mathrm{Cl}]{ } \mathrm{CH}_3 \mathrm{CH}_2 \stackrel{\oplus}{\mathrm{C}} \mathrm{H}_2 $

$ \text { In second reaction, } $

$ \text { Friedel-craft alkylation } $

$ \text { After Friedel Craft acylation, } $

∵ (i) and (ii) gives Kolbe reaction and (iii), (iv) and (v) gives the final product as:

$ \text { Hence, option (a) is the correct answer. } $

$ \text { The major product in the following is } $

From, the above reaction (i) and (iv) gives alcohol and phenol respectively.

$ \text { In this reaction } P \text { and } Q \text { are } $

Phenol on reaction with chloroform an KOH gives salicylaldehyde, which with 50% KOH solution undergoes Cannizzaro's reaction.