Alcohol, Phenol and Ether MCQ Questions & Answers in Organic Chemistry | Chemistry

\[\underset{t\text{-butyl alcohol}}{\mathop{{{H}_{3}}C\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C-}}}\,OH}}\,\xrightarrow[ZnC{{l}_{2}}\left( \text{anhy}\text{.} \right)]{\text{conc}\text{.}\,HCl}\]      \[\underset{t\text{-butylchloride}}{\mathop{{{H}_{3}}C\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,\,C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{-C-}}}\,Cl}}\,\]

The reaction of phenol with $$NaOH$$  and $$C{O_2}$$  is known as Kolbe-Schmidt or Kolbe's reaction. The product formed is salicylic acid.

Key Concept When intermediate carbocation is stable, no rearrangement takes place in carbocation.

Weakest acid has the strongest conjugate base. Since $$ROH$$  is the weakest acid, therefore $$R{O^ - }$$  is the strongest base.

\[\underset{\text{2, 3-Dimethyl-2, 3-butanediol}}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ \,\,\,\,C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} \,\,\,\,OH \\ | \end{smallmatrix}}{\mathop{-C-}}}\,\underset{\begin{smallmatrix} |\,\,\,\, \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} OH \\ |\,\,\,\, \end{smallmatrix}}{\mathop{C-}}}\,C{{H}_{3}}}}\,\]
It has two tertiary alcoholic groups.

The acidic behaviour of phenols may be explained on the basis of two following reasons.
(a) Due to resonance (which is not possible in alcohols), the oxygen atom of the $$- OH$$  group acquires a positive charge which helps in the release of a proton.

(b) In the dissociation of phenol to phenoxide ion and a proton the equilibrium lies mainly towards the right hand side as the resulting phenoxide ion is more stabilised by resonance as compared to phenol.

The acidic strength of phenols depends on the nature of substituents present in the benzene nucleus.
Electron withdrawing groups like $$ - N{O_2}, - CN, - CHO, - COOH,$$       etc, when present at the ortho and $$para$$  - positions with respect to phenolic group increases the acidity of phenol due to greater stabilisation of phenoxide ion. While the presence of electron releasing group like $$ - N{H_2}, - C{H_3},$$    etc, decrease the acidity of phenols. This explains the following order of acidity
$$p$$ - nitrophenol > phenol > $$p$$ - cresol.

TIPS/Formulae :
Compounds having \[\underset{\begin{smallmatrix} \parallel \\ O \end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{3}}\]   groups show positive
iodoform test.
Hence, \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}\underset{\begin{smallmatrix} \parallel \\ O \end{smallmatrix}}{\mathop{-C-}}\,C{{H}_{3}}\]
( pentanone - 2 ) gives this test.

\[\begin{align} & 3C{{H}_{3}}\,CH=C{{H}_{2}}\xrightarrow{{{B}_{2}}{{H}_{6}}}{{\left( C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}} \right)}_{3}}B \\ & \xrightarrow[{}]{\frac{{{H}_{2}}{{O}_{2}}}{O{{H}^{-}}}}C{{H}_{3}}-\underset{\text{Propan-}1-ol}{\mathop{C{{H}_{2}}-C{{H}_{2}}}}\,-OH \\ \end{align}\]
Here, half mol of $$\left( {{B_2}{H_6}} \right)$$  diborane react with propane by Markownikoff's addition it gives tripropyl borane called hydroboration. In presence of $${H_2}{O_2}$$  in basic medium tripropyl borane gives alcohol. Remember that product is Anti Markownikoff’s rule that is 1-propanol. Reaction is called hydroboration oxidation.

$$Victor{\text{ }}Meyer's\,\,test:$$     The various steps involved are

\[\left( \text{iii} \right){{R}_{3}}COH\xrightarrow{HI}{{R}_{3}}CI\xrightarrow{AgN{{O}_{2}}}\]       \[{{R}_{3}}CN{{O}_{2}}\xrightarrow{HN{{O}_{2}}}\text{No}\,\,\text{reaction}\]       \[\xrightarrow{KOH}\text{Colourless}\]