Arrange the following compounds in increasing order of acidity and give a suitable explanation.
Phenol, o-nitrophenol, o-cresol
Nitro group shows $-I$ and $-R$ effect as follows
Due to this $-I$ and $-R$ effect of o-nitrophenol, it is a stronger acid than phenol. On the other hand, $-\mathrm{CH}_3$ group produces $+I$ effect $-I$ and $-R$ effect increases the acidic strength by increasing the polarity of -OH bond while $+I$ effect decreases the polarity due to increase in electron density on -OH bond. So, o-cresol is a weaker acid than phenol. Thus, the correct order is o-cresol < phenol
Alcohols react with active metals e.g., $\mathrm{Na}, \mathrm{K}$ etc., to give corresponding alkoxides. Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols.
An electron releasing group produces $+I$ effect so it increases the electron density on oxygen and decreases the polarity of O - H bond. As the number of alkyl group increases from $1^{\circ}$ to $3^{\circ}$ alcohols, the electron density on the O - H bond increases. It will finally decrease the polarity and increase the strength of $\mathrm{O}-\mathrm{H}$ bond in going from $1^{\circ}$ to $3^{\circ}$ alcohols. Thus, acidic strength decreases in the following order
As we know that, sodium metal is basic in nature and alcohols are acidic in nature. Thus, reactivity of alcohol with sodium metal decreases with decrease in acidic strength. Therefore, the correct order is $1^{\circ}>2^{\circ}>3^{\circ}.$
What happens when benzene diazonium chloride is heated with water?
When benzene diazonium chloride is heated with water then phenol is formed.
Arrange the following compounds in decreasing order of acidity.
$$\mathrm{H}_2 \mathrm{O}, \mathrm{ROH}, \mathrm{HC} \equiv \mathrm{CH}$$
A stronger acid displaces a weaker acid from its salt. When $R \mathrm{ONa}$ is treated with $\mathrm{H}_2 \mathrm{O}$, it forms ROH . So, water is a stronger acid than ROH .
$$\underset{\text { Stronger acid }}{\mathrm{HOH}}+\mathrm{RONa} \longrightarrow \mathrm{NaOH} \underset{\text { Weaker acid }}{+\mathrm{ROH}}$$
Similarly, when sodium ethynide is treated with water and alcohol, then acetylene is obtained.
$\underset{\substack{\text { Stronger } \\ \text { acid }}}{\mathrm{HOH}}+\mathrm{CH} \equiv \mathrm{CNa} \longrightarrow \underset{\text { Weak acid }}{\mathrm{HC} \equiv \mathrm{CH}}+\mathrm{NaOH}$
$\underset{\substack{\text { Stronger } \\ \text { acid }}}{\mathrm{ROH}}+\mathrm{HC} \equiv \mathrm{CNa} \longrightarrow \underset{\text { Weak acid }}{\mathrm{HC} \equiv \mathrm{CH}}+\mathrm{RONa}$
Thus, water and alcohol are stronger acid than ethyne and the decreasing order of acidity of given compounds are
$\mathrm{HOH > ROH > HC \equiv CH}$
Name the enzymes and write the reactions involved in the preparation of ethanol from sucrose by fermentation.
Sucrose is converted to glucose and fructose in the presence of an enzyme, invertase or sucrase. Glucose and fructose undergo fermentation in the presence of another enzyme, zymase. Both these enzymes are present in yeast.
$\underset{\text { Sucrose }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Invertase/sucrose }} \underset{\text { Glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { Fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}$
$\underset{\substack{\text { Glucose /Fructose }}}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6} \xrightarrow{\text { Zymase }} \underset{\substack{\text { Ethanol }}}{2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}}+2 \mathrm{CO}_2$