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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:
Ethers are readily cleaved by HI or HBr at 373K to form an alcohol and an alkyl halide.
$\text{R}-\text{O}-\text{R}+\text{HX}\xrightarrow{373\text{K}}\text{R}-\text{X}+\text{R}-\text{OH}$
$\text{R}-\text{OH}+\text{HX}\xrightarrow{373\text{K}}\text{R}-\text{X}+\text{H}_2\text{O}$
Mixed ether, containing primary or secondary alkyl group, when heated with hydrogen halide, the lower alkyl group forms halide and higher will form an alcohol. Tertiary alkyl ether when heated with hydrogen halide gives tertiary alkyl halide.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
  1. $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{O}- \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 $
  2. $\text{CH}_3-\text{CH}-\text{CH}_2-\text{O}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
  3. $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{O}-\text{CH}_3$
  4. $\text{CH}_3-\text{CH}_2-\text{CH}-\text{O}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
  1. When $CH_2 = CH - O - CH_2 - CH_3$ reacts with one mole of HI, one of the products formed is:
  1. Ethane.
  2. Ethanol.
  3. Iodoethene.
  4. Ethanal.
  1. $(CH_3)_3COCH_3$ and $CH_3OC_2H_5$​​​​​​​ are treated with hydroiodic acid. The fragments obtained after reactions are respectively:
  1. $(CH_3)_3CI + CH_3OH; CH_3I + C_2H_5OH$
  2. $(CH_3)_3CI + CH_3OH; CH_3OH + C_2H_5I$
  3. $(CH_3)_3COH + CH_3I; CH_3OH + C_2H_5I$
  4. $CH_3I + (CH_3)_3COH; CH_3I + C_2H_5OH$
  1. Which of the following ether is unlikely to be cleaved by hot cone. HBr?
✓

Answer

  1. (a) $\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{O}- \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 $
Explanation:
$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{O}\text{CH}_3+\text{HI}\xrightarrow{\text{S}_\text{N}1}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \text{(Hot andconc.)}$
  1. (b)

Explanation:
  1. (d) Ethanal.
Explanation:
  1. (a) $(CH_3)_3CI + CH_3OH; CH_3I + C_2H_5OH$
Explanation:
When mixed ethers are used, the formation of alkyl iodide depends on the nature of alkyl groups. Methyl iodide is formed when one group is methyl and the other a primary or secondary alkyl group. Here reaction follows $S_N2$ mechanism and because of the steric effect of the larger group, $I^-$ attacks the smaller (Me) group.
$\text{CH}_3\text{OC}_2\text{H}_5+\text{HI}\rightarrow\text{CH}_3\text{I}+\text{C}_2\text{H}_5\text{OH}$
When the substrate is a methyl t-alkyl ether, the products are t-RI and MeOH. Here, reaction follows $S_N1$ mechanism and formation of products is controlled by the stability of carbocation. Since carbocation stability order is:
$3^\circ>2^\circ>1^\circ>\stackrel{+}{\hbox{ CH}}_3,$ therefore alkyl halide is always derived from tert-alkyl group.
$\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \\\text{CH}_3-\text{C}-\text{O}-\text{CH}_3+\text{HI}\xrightarrow[\text{S}_\text{N}1]{373\text{K}}\text{CH}_3-\text{C}-\text{I}+\text{CH}_3\text{OH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\ _\text{rert-Butyl methyl ether}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _\text{tert-Butyl iodide} $
  1. (a)

Explanation:
Diphenyl ethers are not cleaved by HBr or HI.

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The following questions are multiple choice questions. Choose the most appropriate answer:
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  1. 1-Butanol
  2. 2-Butanol
  3. 2-Methyl-1-propanol
  4. 2-Methyl-2-propanol
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  1. A substance $C_4H_{10}O (X)$ yields on oxidation a compound $C_4H_8O$ which gives an oxime and a positive iodoform test. The substance X on treatment with cone. $H_2SO_4$ gives $C_4H_8$. The structure of the compound (X) is:
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  1. In the oxidation of by acidified $K_2Cr_2O_7​​​​​​​$, the products are:
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  4. None of these.
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$\Delta\text{T}_\text{b}\propto\text{m}$
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  4. $(CH_3)_3N > CH_3NH_2 > (CH_3)_2NH$
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  3. Methylamine is a stronger base than ammonia.
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The half-life of a reaction is the time required for the concentration of reactant to decrease by half, i.e.,

$[\text{A}]_\text{t}=\frac{1}{2}[\text{A}]$
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this means $\text{t}\frac{1}{2}$ is independent of initial concentration. Figure shows that typical variation of concentration of reactant exhibiting first order kinetics. It may be noted that though the major portion of the first order kinetics may be over in a finite time, but the reaction will never cease as the concentration of reactant will be zero only at infinite time.
The following questions are multiple choice questions. Choose the most appropriate answer:
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  1. 2.303s
  2. 23.03s
  3. 230.3s
  4. 2303s
  1. The rate constant for a first order reaction is $7.0 \times 10^{-4} s^{-1}$. If initial concentration ofreactant is 0.080 M, what is the half life of reaction?
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  3. 12375s
  4. $10.10 \times 10^{-4}s$
  1. For the half-life period of a first order reaction, which one of the following statements is generally false?
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  2. It is independent of temperature.
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  4. None of these.
  1. The rate of a first order reaction is 0.04 mol $L^{-1} s^{-1}​​​​​​​$​​​​​​​ at 10 minutes and 0.03 mol $L^{-1} s^{-1}​​​​​​​$ at 20 minutes after initiation. The half-life of the reaction is:
  1. 4.408 min
  2. 44.086 min
  3. 24.086 min
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  1. The plot of $\text{t}_\frac{1}{2}$ vs initial concentration $[A]_0​​​​​​​$​​​​​​​ for a first order reaction is given by:
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The molecular compounds which are formed from the combination of two or more simple stable compounds and retain their identity in the solid as well as in the dissolved state are called coordination compounds. Their properties are completely different from the constituents. ln coordination compounds, the central metal atom or ion is linked to a number ofions or neutral molecules, called ligands, by coordinate bonds. For example, Dimethylglyoxime (dmg) is a bid en date ligand chelating large amounts of metals. When dimethyl glyoxime is added to alcoholic solution of $NiCl_2$ and ammonium hydroxide is slowly added to it, a rosy red precipitate of a complex is formed.
The following questions are multiple choice questions. Choose the most appropriate answer:
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  1. Oxidation number of Ni in the given complex is:
  1. $+3$
  2. $+1$
  3. $+2$
  4. $Zero$
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  1. $Sp^3$​​​​​​​, tetrahedral.
  2. $dsp^2$​​​​​​​, square planar.
  3. $Sp^3$​​​​​​​, square planar.
  4. $Sp^3d$, trigonal bipyramidal.
  1. Which of the following is true about this complex?
  1. It is paramagnetic, containing $2$ unpaired electrons.
  2. It is paramagnetic, containing $1$ unpaired electron.
  3. It is paramagnetic, containing $4$ unpaired electrons.
  4. It is diamagnetic with no unpaired electron.
  1. Which one will give test for $Fe^{3+}$​​​​​​​ ions in the solution?
  1. $[Fe(CN)_6]^{3-}$
  2. $[Fe(CN)_6]^{2-}$
  3. $(NH_4)_2SO_4·FeSO_4·6H_2O$
  4. $Fe_2(S0_4)_3$​​​​​​​
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Read the passage given below and answer the following questions:
When an aldehyde with no et-hydrogen reacts with concentrated aqueous $NaOH$, half the aldehyde is converted to carboxylic acid salt and other half is converted to an alcohol. In other words, half of the reactant is oxidized and other half is reduced. This reaction is known as Cannizzaro reaction.

The following questions are multiple choice questions. Choose the most appropriate answer:
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  1. Benzyl alcohol and sodium formate.
  2. Sodium benzoate and methyl alcohol.
  3. Sodium benzoate and sodium formate.
  4. Benzyl alcohol and methyl alcohol.
  1. Which of the following compounds will undergo Cannizzaro reaction?
  1. $CH_3CHO$
  2. $CH_3COCH_3$
  3. $C_6H_5CHO$
  4. $C_6H_5CH_2CHO$
  1. Trichloroacetaldehyde is subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate ion and another compound. The other compounds is:
  1. 2, 2, 2-trichloroethanol.
  2. Trichloromethanol.
  3. 2, 2, 2-trichloropropanol.
  4. Chloroform.
  1. In Cannizzaro reaction given below:
$2\text{PhCHO}\xrightarrow{\stackrel{-}{\hbox{ OH}}}\text{PhCH}_2+\text{OH}+\text{PhCO}_2^-$ the slowest step is:
  1. The attack $^-OH$ at the carboxyl group.
  2. The transfer of hydride to the carbonyl group.
  3. The abstraction of proton from the carboxylic group.
  4. The deprotonation of $PhCH_2OH$.
  1. Which of the following reaction will not result in the formation of carbon-carbon bonds?
  1. Cannizzaro reaction.
  2. Wurtz reaction.
  3. Reimer-Tiemann reaction.
  4. Friedel-Crafts' acylation.
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Read the passage given below and answer the following questions:
Arrangement of ligands in order of their ability to cause splitting $(\Delta)$ is called spectrochemical series. Ligands which cause large splitting (large $\Delta$) are called strong field ligands while those which cause small splitting (small $\Delta$) are called weak field ligands. When strong field ligands approach metal atom/ ion, the value of $\Delta_0$ is large, so that electrons are forced to get paired up in lower energy $t_{2g}$ orbitals. Hence, a low-spin complex is resulted from strong field ligand. When weak field ligands approach metal atom/ ion, the value of $\Delta_0$ is small, so that electrons enter high energy $e_g$ orbitals rather than pairing in low energy $t_{2g}$ orbitals. Hence, a high-spin complex is resulted from weak field ligands. Strong field ligands have tendency to form inner orbital complexes by forcing the electrons to pair up. Whereas weak field ligands have tendency to form outer orbital complex because inner electrons generally do not pair up.
In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Assertion: In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting.
Reason: Spectrochemical series is based on the absorption of light by complexes with different ligands.
  1. Assertion: In high spin situation, configuration of $d^5$ ions will be $\text{t}^3_{2\text{g}}\text{e}^2_\text{g}.$
Reason: In high spin situation, pairing energy is less than crystal field energy.
  1. Assertion: $F^-$ ion is a weak field ligand and fonns outer orbital complex.
Reason: $F^-$ ion cannot force the electrons of $d_{z^2}$ and $d_{x^2-y^2}$ orbitals of the inner shell to occupy $d_{xy}, d_{yz}$ and $d_{zx}$ orbitals of the same shell.
  1. Assertion: The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds.
Reason: In spectrochemical series, ligands are arranged in a series of increasing field strength.
  1. Assertion: $NF_3$ is a weaker ligand than $N(CH_3)_3$.
Reason: $NF_3$ ionizes to give $F^-$ ions in aqueous solution.
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