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Alcohols, Phenols, and Ethers — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryAlcohols, Phenols, and Ethers4 Marks
Question
Read the passage given below and answer the following questions:
A compound (X) containing C, H and O is unreactive towards sodium. It also does not react with Schiff s reagent. On refluxing with an excess of hydroiodic acid, (X) yields only one organic product ( Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorus and iodine. The compound (Z) on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. The compound (X) is an:
  1. Acid.
  2. Aldehyde.
  3. Alcohol.
  4. Ether.
  1. The IUPAC name of the acid formed is:
  1. Methanoic acid.
  2. Ethanoic acid.
  3. Propanoic acid.
  4. Butanoic acid.
  1. Compound (Y) is:
  1. Ethyl iodide.
  2. Methyl iodide.
  3. Propyl iodide.
  4. Mixture of (a) and (b).
  1. Compound (Z) is:
  1. Methanol.
  2. Ethanol.
  3. Propanol.
  4. Butanol.
  1. Compound (X) on treatment with excess of Cl2 in presence of tight gives:
  1. $\propto-$ Chlorodiethyl ether.
  2. $\propto,\propto'-$ Dichlorodiethyl ether.
  3. Perchlorodiethyl ether.
  4. None of these.
✓

Answer

  1. (d) Ether.

Explanation:

Since the compound X is unreactive towards sodium so it is neither an acid nor an alcohol. Since the compound X is unreactive towards Schiff's base so it is not an aldehyde.

The compound X forms only one product on reaction with excess HI, indicates that the compound X may be ether.

  1. (b) Ethanoic acid.

​​​​​​​​​​​​​​Explanation:

The reactions can be written as:

Since the equivalent weight of carboxylic acid is 60. So, it must be CH3COOH i.e., ethanoic acid.

  1. (a) Ethyl iodide.

​​​​​​​​​​​​​​​​​​​​​Explanation:

The alcohol Z in that case should be C2H5OH and the compound Y should be ethyl iodide. X is therefore diethyl ether (C2H5 — O — C2H5)

  1. (b) Ethanol.
  2. (c) Perchlorodiethyl ether.

​​​​​​​​​​​​​​​​​​​​​​​​​​​​Explanation:

In the presence of light and excess of chlorine, all the hydrogen atoms of diethyl ether are substituted to give perchlorodiethyl ether.

$\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3+10\text{Cl}_2\xrightarrow{\text{hu}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(excess)}$

$\text{CCl}_3\text{CC}_2-\text{O}-\text{CCl}_2-\text{CCl}_3+\text{10HCI}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Perchlorodierhyl ether }$​​​​​​​

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Molar conductance $(\wedge_\text{m})$ and equivalent conductance $(\wedge_\text{e})$ of an electrolyte solution are calculated as

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where, M = molarity of solution and N is normality of solution. Molar conductance of strong electrolyte depends on the concentration.

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Read the passage given below and answer the following questions:

Aldehydes and ketones are reduced to primary and secondary alcohols respectively by NaBH4 or LiAlH4 as well as catalytic hydrogenation. The carbonyl group of aldehydes and ketones is reduced to  group on treatment with Zn-Hg and cone. HCl (Clemmensen reduction) or with hydrazine followed by NaOH or KOH in highly boiling solvent such as ethylene glycol (Wolff-Kishner reduction).Aldehydes differ from ketones in their oxidation reactions. Aldehydes are easily oxidised to carboxylic acids on treatment with HNO3, KMnO4, K2Cr2O7 etc. Even mild oxidising agents mainlyTollens' reagent and Fehling's solution also oxidise aldehydes. Ketones are generally oxidised under vigorous conditions i.e., strong oxidising agents and at elevated temperatures, to give mixture of carboxylic acids having lesser number of C-atoms than the parent ketone.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which of the following cannot be made by reduction of ketone or aldehyde with NaBH4 in methanol?
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  4. Benzophenone
  1. A substance C4H10O (X) yields on oxidation a compound C4H8O which gives an oxime and a positive iodoform test. The substance X on treatment with cone. H2SO4 gives C4H8. The structure of the compound (X) is:
  1. CH3CH2CH2CH2OH
  2. CH3CH(OH)CH2CH3
  3. (CH3)3COH
  4. CH3CH2 - O - CH2CH3
  1. In the oxidation of  by acidified K2Cr2O7, the products are:

  1. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\ ^\text{14}\text{C}-\text{OH}$ and $\text{CH}_3\text{CH}_2\text{COOH}$

  2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3(\text{CH}_2)_2\text{COOH}-\text{C}-\text{OH}$ and $ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{COOH}$

  3. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 14\\\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}+\text{HCOOH}$

  4. None of these.

  1. The appropriate reagent for the following transformation is:

  1. $\text{Na}_2\text{NH}_2,^-\text{OH}$

  2. $\text{NaBH}_4$

  3. $\frac{\text{H}_2}{\text{Ni}}$

  4. $\text{AICl}_3$

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