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.
$\underset{\text { Sucrose }}{\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Invertase/sucrose }} \underset{\text { Glucose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}+\underset{\text { Fructose }}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6}$
$\underset{\substack{\text { Glucose /Fructose }}}{\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6} \xrightarrow{\text { Zymase }} \underset{\substack{\text { Ethanol }}}{2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}}+2 \mathrm{CO}_2$
How can propan-2-one be converted into tert-butyl alcohol?
Propan-2-one is a ketone. Its structural formula is 
ketones when treated with Grignard reagent give tertiary alcohols.
Write the structures of the isomers of alcohols with molecular formula $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$. Which of these exhibits optical activity?
Some compounds can rotate the plane polarised light, when it is passed through their solution. Such compounds are called optically active compounds. The structures of the isomers of alcohols with molecular formula $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$ are as follows
| (i) | $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$ | Butan-1-ol |
|---|---|---|
| (ii) |  |
Butan-2-ol |
| (iii) |  |
2-methylpropan-1-ol |
| (iv) |  |
2-methylpropan-2-ol |
The asymmetry of the molecule is responsible for the optical activity in a molecule. If all the four substituents attached to the carbon are different then the carbon is called asymmetric or chiral carbon and such a molecule is called asymmetric molecule. In the above explained structure, it is only butan-2-ol which contains a chiral carbon and hence it is optically active.
Explain why is OH group in phenols more strongly held as compared to OH group in alcohols?
In phenols, the electron pairs on oxygen atom of -OH group are in conjugation (or resonance) with $\pi$ electrons of the ring and the following resonating structures are possible
Out of these five resonating structures, II, III and IV structures contain a carbon-oxygen double bond character. In other words, carbon-oxygen bond in phenol acquires a partial double bond character due to resonance. But in alcohols carbon-oxygen bond in alcohols is purely single bond. Therefore, -OH group in phenols is more strongly held as compared to - OH group in alcohols.