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54
Subjective

Explain why alcohols and ethers of comparable molecular mass have different boiling points?

Explanation

Boiling point depends upon the strength of intermolecular forces of attraction. Higher these forces of attraction, more will be the boiling point. Alcohols undergo intermolecular hydrogen bonding. So, the molecules of alcohols are held together by strong intermolecular forces of attraction.

But in ethers no hydrogen atom is bonded to oxygen. Therefore, ethers are held together by weak dipole-dipole forces, not by strong hydrogen bond.

Since, lesser amount of energy is required than to break weak dipole-dipole forces in ethers than to break strong hydrogen bonds in alcohol.

55
Subjective

The carbon-oxygen bond in phenol is slightly stronger than that in methanol. Why ?

Explanation

Positive resonance effect is observed in phenols. Due to this $+R$ effect, lone pair of electrons of $-$OH group are in conjugation with $\pi$ electrons of the ring and the following resonance hybrid are obtained.

From the above resonating structure, it is very clear that $\mathrm{C}-\mathrm{O}$ bond of phenol acquires some partial double bond character while the $\mathrm{C}-\mathrm{O}$ bond of methanol is purely single bond. Therefore, the carbon-oxygen bond in phenol is slightly stronger than that in methanol.

56
Subjective

Arrange water, ethanol and phenol in increasing order of acidity and give reason for your answer.

Explanation

The phenoxide ion obtained after the removal of a proton is resonance stabilised as follows

Whereas, the ethoxide ion obtained after the removal of a proton is not stabilised but destabilised due to $+I$ effect of $-\mathrm{C}_2 \mathrm{H}_5$ group. Therefore, phenol is a stronger acid than ethanol. $+I$ effect of $\mathrm{CH}_3-\mathrm{CH}_2$ group increase the electron density on $\mathrm{O}-\mathrm{H}$ bond in ethanol. As the electron density on the O - H bond of ethanol is more than that of water. So, ethanol is a weaker acid than water. Thus, the increasing order of acidity is

ethanol < water < phenol

57
Subjective

Match the structures of the compounds given in Column I with the name of the compounds given in Column II.

Column I Column II
A. 1. Hydroquinone
B. 2. Phenetole
C. 3. Catechol
D. 4. o-cresol
E. 5. Quinone
F. 6. Resorcinol
7. Anisole

Explanation

A. $\rightarrow(4)$ B. $\rightarrow$ (3) C. $\rightarrow(6)$ D. $\rightarrow$ (1) E. $\rightarrow$ (7) F. $\rightarrow$ (2)

A. Cresols are organic compounds which are methyl phenols. There are three forms of cresol-o-cresol, $p$-cresol and $m$-cresol.

B. Catechol is also known as pyrocatechol. Its IUPAC name is 1,2-dihydrobenzene. It is used in the production of pesticides, perfumes and pharmaceuticals.

C. Its IUPAC name is 1,3-dihydroxybenzene. Resorcinol is used to treat acne, seborrheic dermatitis and other skin disorder.

D. Hydroquinone is also known as quinol. Its IUPAC name is 1, 4- dihydroxybenzene. It is a white granular solid. It is a good reducing agent.

E. Anisole or methoxy benzene, is a colourless liquid with a smell reminiscent of anise seed.

F. Phenetole is an organic compound. It is also known as ethylphenyl ether. It is volatile in nature and its vapours are explosive in nature.

58
Subjective

Match the starting material given in Column I with the products formed by these (Column II) in the reaction with HI.

Column I Column II
A. 1.
B. 2.
C. 3.
D. 4.
5.
6.
7.

Explanation

A. $\rightarrow(4)$ B. $\rightarrow(5)$ C. $\rightarrow$ (2) D. $\rightarrow$ (1)

A. $\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_3$ is a symmetrical ether so the products are $\mathrm{CH}_3 \mathrm{I}$ and $\mathrm{CH}_3 \mathrm{OH}$.

B. $\operatorname{In}\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{O}-\mathrm{CH}_3$ unsymmetrical ether, one alkyl group is primary while another is secondary. So, it follows $\mathrm{S}_{\mathrm{N}}{ }^2$ mechanism. Thus, the halide ion attacks the smaller alkyl group and the products are

C. In this case, one of the alkyl group is tertiary and the other is primary. It follows $S_N{ }^1$ mechanism and halide ion attacks the tertiary alkyl group and the products are $\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{I}$ and $\mathrm{CH}_3 \mathrm{OH}$.

D. Here, the unsymmetrical ether is alkyl aryl ether. In this ether $\mathrm{O}-\mathrm{CH}_3$ bond is weaker than $\mathrm{O}-\mathrm{C}_6 \mathrm{H}_5$ bond which has partial double bond character due to resonance. So, the halide ion attacks on alkyl group and the products are $\mathrm{C}_6 \mathrm{H}_5-\mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{I}$.