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.
Explain why nucleophilic substitution reactions are not very common in phenols?
Resonance is an important factor in phenols. During resonance -OH group in phenol gives its electrons to the benzene ring. As a result of this, the electron density on benzene ring is very high. This increased electron density repels nucleophiles.
Therefore, nucleophiles cannot attack the benzene ring and phenols usually do not give nucleophilic substitution reaction.
Preparation of alcohols from alkenes involves the electrophilic attack on alkene carbon atom. Explain its mechanism.
Preparation of alcohols from alkene by the hydration of alkenes in presence of sulphuric acid.
$$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$$
This addition reaction takes place in accordance with Markownikoff's rule.
Mechanism The mechanism of the reaction involves the following three steps
Step 1 Protonation of alkene to form carbocation by electrophilic attack of $\mathrm{H}_3^{+} \mathrm{O}$.
Step 2 Attack of water molecule to the secondary carbocation.
Step 3 Loss of the hydrogen from the protonated alcohol.
Explain why is $O=\mathrm{C}=O$ non-polar while $\mathrm{R}-O-\mathrm{R}$ is polar?
$\mathrm{CO}_2$ is a linear molecule. The dipole moment of two $\mathrm{C}=\mathrm{O}$ bonds are equal and opposite and they cancel each other and hence the dipole moment of $\mathrm{CO}_2$ is zero and it is a non-polar molecule.
While for ethers, two dipoles are pointing in the same direction. These two dipoles do not cancel the effect of each other. Therefore, there is a finite resultant dipoles and hence $R-O-R$ is a polar molecule.
