Phenol is less acidic than $$o$$ - nitrophenol as electron withdrawing $$\left( { - N{O_2}} \right)$$ group increases the acidity of phenols while electron donating groups $$\left( { - C{H_3}, - OC{H_3}} \right)$$ decrease the acidity of phenols. Phenols are more acidic than alcohols.
213.
Conversion of ethyl alcohol into acetaldehyde is an example of
\[C{{H}_{3}}-C{{H}_{2}}-OH\xrightarrow[KMn{{O}_{4}}]{\left[ O \right]}\] \[C{{H}_{3}}-CHO\left( \text{Oxidation} \right)\]
214.
Identify $$X, Y$$ and $$Z$$ in the following sequence of reactions \[C{{H}_{3}}CHBrC{{H}_{3}}\xrightarrow{\text{alc}\text{.}\,KOH}X\] \[\xrightarrow[\text{Peroxide}]{HBr}Y\xrightarrow{C{{H}_{3}}ONa}Z\]
215.
Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether due to the presence of
Alcohols have higher boiling points as compared to other organic compounds of similar molecular masses such as ethers. This is due to the presence of intermolecular hydrogen bonding in alcohols which is absent in ethers. Because of hydrogen bonding in alcohols, these exist as associated molecules rather than discrete molecules. Consequently, a large amount of energy is required to break these bonds and therefore, their boiling points are high.
216.
Which one of the following substituents at $$para$$ - position is most effective in stabilizing the phenoxide ion ?
Electron withdrawing group stabilises the benzene ring due to delocalisation of charge.
$$ - C{H_3}$$ and $$ - C{H_2}OH$$ are electron donating group and hence decrease the stability of benzene ring $$ - OC{H_3}$$ is weaker electron withdrawing group than $$ - COC{H_3}.$$ Hence $$ - COC{H_3}$$ group more stabilize the phenoxide ion at $$p$$ - position.
217.
Arrange the following compounds in order of decreasing
acidity :
Electron withdrawing substituents like $$ - N{O_2},\,C{\text{l}}$$ increase the acidity of phenol while electron releasing substituents like $$ - C{H_3}, - OC{H_3}$$ decreases acidity.
hence the correct order of acidity will be
Further $$\left( { - I} \right)N{O_2} > \left( { - I} \right)Cl\,\,{\text{and}}\,\left( { + I} \right)C{H_3} > \left( { + I} \right)OC{H_3}$$
218.
The electrophile involved in the above reaction is
A
trichloromethyl anion \[\left( \overset{\Theta }{\mathop{C}}\,C{{l}_{\text{3}}} \right)\]
B
formyl cation \[\left( \overset{\Theta }{\mathop{C}}\,HO \right)\]
C
dichloromethyl cation \[\left( \overset{\Theta }{\mathop{C}}\,HC{{l}_{2}} \right)\]
NOTE: This is Riemer - Tiemann reaction and the electrophile is dichlorocarbene.
219.
\[\xrightarrow[\text{Pyridine}]{SOC{{l}_{2}}}\left( A \right)\xrightarrow[\left( {{H}_{2}}O \right)]{{{O}_{3}}/Zn}\underset{{{C}_{5}}{{H}_{9}}ClO}{\mathop{\left( B \right)}}\,\xrightarrow{NaB{{H}_{4}}}\left( C \right)\] Compound $$(C)$$ is