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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 $\mathrm{CH}_3 \mathrm{I}$.

59
Subjective

Match the items of Column I with items of Column II.

Column I Column II
A. Antifreeze used in car engine 1. Neutral ferric chloride
B. Solvent used in perfumes 2. Glycerol
C. Starting material for picric acid 3. Methanol
D. Wood spirit 4. Phenol
E. Reagent used for detection of phenolic group 5. Ethylene glycol
F. By product of soap industry used in cosmetics 6. Ethanol

Explanation

A. $\rightarrow(5)$ B. $\rightarrow(6)$ C. $\rightarrow(4)$ D. $\rightarrow(3)$ E. $\rightarrow$ (1) F. $\rightarrow$ (2)

A. IUPAC name of ethylene glycol is ethane $-1,2$ - diol. It is primarily used as raw material in the manufacturing of polyester fibers and fabric industry. A small percentage of it is used in antifreeze formulations.

B. Ethanol is a good solvent for fatty and waxy substances. Fats and waxes provide odour to the perfumes. Apart from being a good solvent, it is less irritating to the skin. So, it is used in perfumes.

C. Phenol is converted into picric acid (2, 4, 6-trinitro-phenol) by the reaction of phenol with conc. $\mathrm{HNO}_3$.

D. Methanol, $\mathrm{CH}_3 \mathrm{OH}$ is also known as 'wood spirit' as it was produced by the destructive distillation of wood.

E. Neutral ferric chloride give purple/red colour when treated with phenols. It is the reagent used for detection of phenolic group.

F. Soaps are prepared by the reactions of fatty acid with NaOH .

This glycerol (propan-1,2,3-triol) is the by product of soap industry and used in cosmetics.

60
Subjective

Match the items of Column I with items of Column II.

Column I Column II
A. Methanol 1. Conversion of phenol of o-hydroxysalicylic acid
B. Kolbe's reaction 2. Ethyl alcohol
C. Williamson's synthesis 3. Conversion of phenol to salicylaldehyde
D. Conversion of 2$^\circ$ alcohol to ketone 4. Wood spirit
E. Reimer-Tiemann reaction 5. Heated copper at 573K
F. Fermentation 6. Reaction of alkyl halide with sodium alkoxide

Explanation

A. $\rightarrow(4)$ B. $\rightarrow$ (1) C. $\rightarrow(6)$ D. $\rightarrow$ (5) E. $\rightarrow$ (3) F. $\rightarrow$ (2)

A. Methanol is also known as 'wood spirit' as it was produced by the destructive distillation of wood.

B. In Kolbe's reaction, 2 - hydroxy benzoic acid (salicylic acid) is prepared by the reaction of phenol with $\mathrm{CO}_2$ gas.

C. Williamson synthesis is an important method for the preparation of ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.

$$R-X+R-\mathrm{ONa} \longrightarrow \mathrm{ROR}+\mathrm{NaX}$$

D. When a $2^{\circ}$ alcohol is allowed to pass over heated copper at 573 K , dehydrogenation takes place and an ketone is formed.

E. On treating phenol with chloroform in the presence of NaOH , an aldehydic group is introduced at ortho position of benzene ring

F. Ethanol is prepared by the fermentation of sugars.

$$\begin{aligned} & \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Invertase }} \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \\ & \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \xrightarrow{\text { Zymase }} 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+2 \mathrm{CO}_2 \end{aligned}$$