Alcohols, Phenols and Ethers
The reactions which produce alcohol as the product are :
A. $\mathrm{CH}_4 + \mathrm{O}_2 \xrightarrow{\mathrm{Mo}_2\mathrm{O}_3,\ \Delta} $
B. $2\mathrm{CH}_3\mathrm{CH}_3 + 3\mathrm{O}_2 \xrightarrow{(\mathrm{CH}_3\mathrm{COO})_2\mathrm{Mn},\ \Delta} $
C. $(\mathrm{CH}_3)_3\mathrm{CH} \xrightarrow{\mathrm{KMnO}_4} $
D. $ 2\mathrm{CH}_4 + \mathrm{O}_2 \xrightarrow{\mathrm{Cu}/523\ \mathrm{K}/100 \ \mathrm{atm}.} $
E. $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \xrightarrow{\mathrm{KMnO}_4/\mathrm{H}^+} $
Choose the correct answer from the options given below :
B, D and E Only
A and D Only
A, C and E Only
C and D Only
$ \text { Consider the following reaction sequence } $
Compound $(\mathrm{y})$ develops characteristic colour with neutral $\mathrm{FeCl}_3$ solution.
Identify the INCORRECT statement from the following for the above sequence.
Compound y will dissolve in $\mathrm{NaHCO}_3$ and evolve a gas.
Compound x is more acidic than compound y .
Both compounds x and y will dissolve in NaOH .
Both compounds x and y will burn with sooty flame.
$ \text { From the following, how many compounds contain at least one secondary alcohol?} $
Choose the correct answer from the options given below :
Three
Two
Four
Five
A hydroxy compound $(\mathrm{X})$ with molar mass $122 \mathrm{~g} \mathrm{~mol}^{-1}$ is acetylated with acetic anhydride, using a large excess of the reagent ensuring complete acetylation of all hydroxyl groups. The product obtained has a molar mass of $290 \mathrm{~g} \mathrm{~mol}^{-1}$. The number of hydroxyl groups present in compound $(\mathrm{X})$ is:
2
4
3
5
A mixed ether $(\mathrm{P})$, when heated with excess of hot concentrated hydrogen iodide produces two different alkyl iodides which when treated with aq. NaOH give compounds $(\mathrm{Q})$ and $(\mathrm{R})$. Both $(\mathrm{Q})$ and $(\mathrm{R})$ give yellow precipitate with NaOI . Identify the mixed ether $(\mathrm{P})$ :
$ \text { 3, 3-Dimethyl-2-butanol cannot be prepared by : } $
$ \text { Choose the correct answer from the options given below : } $
B and E Only
B, C and E Only
B and C Only
B Only
Given below are two statements :
Statement I : Phenol on treatment with $\mathrm{CHCl}_3$ /aq. KOH under refluxing condition, followed by acidification produces $p$-hydroxy benzaldehyde as the major product and $o$-hydroxy benzaldehyde as the minor product.
Statement II : The mixture of $p$-hydroxybenzaldehyde and $o$ hydroxybenzaldehyde can be easily separated through steam distillation.
In the light of the above statements, choose the correct answer from the options given below
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is false but Statement II is true
Given below are two statements :
Statement I : Dimethyl ether is completely soluble in water. However, diethyl ether is soluble in water to a very small extent.
Statement II : Sodium metal can be used to dry diethyl ether and not ethyl alcohol.
In the light of given statements. choose the correct answer from the options given below
Benzene is treated with oleum to produce compound $(\mathrm{X})$ which when further heated with molten sodium hydroxide followed by acidification produces compound $(\mathrm{Y})$. The compound Y is treated with zinc metal to produce compound $(Z)$. Identify the structure of compound $(Z)$ from the following option.
Which one of the following, with HBr will give a phenol?
Given below are two statements :
Statement I : 
Statement II : In 
, intramolecular substitution takes place first by involving lone pair of electrons on nitrogen.
In the light of the above statements, choose the most appropriate answer from the options given below
Given below are two statements :
Statement (I) : The boiling points of alcohols and phenols increase with increase in the number of C -atoms.
Statement (II) : The boiling points of alcohols and phenols are higher in comparison to other class of compounds such as ethers, haloalkanes.
In the light of the above statements, choose the correct answer from the options given below :
What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol?
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}, \mathrm{H}, \mathrm{O}, \mathrm{Br}$ are $12,1,16,80$ respectively )
0.1 mole of compound ‘S’ will weigh ___________ g.
(Given molar mass in g mol−1 C : 12, H : 1, O : 16)
Explanation:
$\begin{aligned} &0.1 \text { mole of compound ( } \mathrm{S} \text { ) weight in gm }\\ &\begin{aligned} & =0.1 \times \text { molar mass of compound }(\mathrm{S}) \\ & =0.1 \times 130=13 \mathrm{gm} \end{aligned} \end{aligned}$
For the reaction sequence given below, the correct statement(s) is(are)
P is optically active.
S gives Bayer’s test
Q gives effervescence with aq. NaHCO3.
R is an alkyne.
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ______.
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80
Explanation:
First step is Fitting reaction (Wurtz-Fittig reaction) Fitting reaction: Aryl halides when treated with sodium in dry ether, two. aryl groups joined to gether to form another aromatic compounds.Here, 2 molecules of the reactant $X$ react with 2 Na in presence of dry ether and forms a coupling product with the elimination of 2 NaBr .
The reaction proceeds through a radical mechanism where the sodium metal abstracts a bromine atom. from the compound $x$, forming an aryl radical. These aryl radicals then combine to foem a biaryl (two acyl groups linked to gether)
$
\text { Overall reaction can be written as, }
$
Second step is the hydrolysis reaction. The acetal hydrolysis occurs and forms - CHO group.
2 moles of $x$ reacts to form 1 mol $P$. So, 1 mole of $X$ gives $\frac{1}{2}$ mol of $P$.
Given moles of $x$ is 16 mol.
So, the mole of $P=16 \mathrm{~mol} \times \frac{1}{2}$
$
=8 \mathrm{~mol}
$
$
\begin{aligned}
&\text { Ratio between } x \text { and } P \text {, }\\
&\begin{aligned}
& x: P \\
& 2 \mathrm{~mol}: 1 \mathrm{~mol} \\
& 1 \mathrm{~mol}: \frac{1}{2} \mathrm{~mol} \\
& 16 \mathrm{~mol}: 16 \times \frac{1}{2} \mathrm{~mol}=8 \mathrm{~mol}
\end{aligned}
\end{aligned}
$
$
\begin{aligned}
& P \xrightarrow[\text { 2) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {1) } \mathrm{NaOH}, \Delta} Q \\
& (100 \%) \quad(50 \%)
\end{aligned}
$It is Cannizzaro reaction: One-CHO group converts to
$-\mathrm{CH}_2 \mathrm{OH}$ and the other - CHO to - COO .
Moles of $P \equiv 8 \mathrm{~mol}$
The reaction gives $50 \% Q$ from $100 \% P$. So, the moles of $Q$ is half of moles of $P$. Moles of $Q=\frac{8}{2}$ $=4 \mathrm{~mol}$
$
Q \frac{1) \mathrm{NaOH}, \mathrm{CaO}}{2) \Delta} R
$$
(50 \%)\,\,\,\,\,\,(50 \%)
$
-COOH group is removed in this reaction. On heating with a mixture of NaOH and CaO (sodium hydroxide and calcium oxide), gives decarboxylation reaction. Cao acts as a drying agent and catalyst. NaOH is used to facilitate the removal of carbon dioxide $\left(\mathrm{CO}_2\right)$.$
\mathrm{NaOH}+ \mathrm{Cao} \text { mixture is known as soda lime. }
$$
\text { - } \mathrm{CH}_2 \mathrm{OH} \text { group has no change during this reaction. }
$
$
\text { Moles of } Q=4 \mathrm{~mol}
$So, theoretically moles of $R$ is 4 mol $(1: 1$ ratio. between $Q$ and $R$ ).
But, the yield given is $50 \%$. so the moles of $R$ is half of moles of $Q$.$
\text { Moles of } \quad R=\frac{4}{2}
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
=2 \mathrm{~mol}
$$
\begin{aligned}
& \mathrm{R} \xrightarrow{\mathrm{PBr}_3,\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O}} \mathrm{~T} \\
& 50 \%\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad 50 \%
\end{aligned}
$
When alcohol reacts with $\mathrm{PBr}_3$ in the presence of $\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O}$ (solvent), the reaction proceeds via $S_N 2$ mechanism and the hyphoxyl group is replaced by a bromine atom.$
\text { So, - } \mathrm{CH}_2 \mathrm{OH} \text { group becomes - } \mathrm{CH}_2 \mathrm{Br} \text {. }
$
Moles of $R=2 \mathrm{~mol}$.
The yield given is $50 \%$. So, the moles of is $T_{\text { }}$ half of moles of $R$.
Moles of $T=\frac{2}{2}$
$
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=1 \mathrm{~mol}
$
$
\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{O} \text { is the solvent }
$
reaction is known as Williamson ether synthesis. Ether compound is formed by reacting an alkoxide $\left(RO^-\right)^{}$with an alkyl halide $(R x)$. The reaction follows $S_{N 2}$ mechanism where the alkoxide acts as a nucleophile, attacking the alkylhalide's carbon atom and displacing the halide.
$
\text { Formula: }\left(\mathrm{C}_6 \mathrm{H}_5\right)_2\left(\mathrm{C}_6 \mathrm{H}_4\right)_2\left(\mathrm{CH}_2\right)_2 \mathrm{O}
\mathrm{C}_{26} \mathrm{H}_{22} \mathrm{O}$Moles of T. $=1 \mathrm{~mol}$
Moles of s is half of moles of $T(50 \%$ yeld)
So, moles of $S=\frac{1}{2} \mathrm{~mol}$
$
\text { Moles of } S \text { is determined. }
$
To calculate the amount, molecular weight must be known.
Molecular weight of $S \Rightarrow 26 \times 12 \mathrm{~g} / \mathrm{mol}+22 \times 1 \mathrm{~g} / \mathrm{mol}+1 \times$
$16 \mathrm{~g} / \mathrm{mol}$$
=312+22+16=350 \mathrm{~g} / \mathrm{mol}
$$
\begin{aligned}
\text { Moles } & =\frac{\text { mass }}{\text { molarmass }} \\
\text { So, mass } & =\text { moles } x \text { molar mass } \\
& =\frac{1}{2} \text { mot } \times 350 \text { g/mol } \\
& =175 . \mathrm{g}
\end{aligned}
$$
\text { Answer: } 175
$
Match List I with List II
Choose the correct answer from the options given below :
Which one the following compounds will readily react with dilute $\mathrm{NaOH}$ ?
Identify the major products A and B respectively in the following set of reactions.
The major products formed :
A and B respectively are :
In Reimer - Tiemann reaction, phenol is converted into salicylaldehyde through an intermediate. The structure of intermediate is __________.
Consider the above reaction sequence and identify the major product P.
Which one of the following reactions is NOT possible?
Given below are two statement :
Statements I : Bromination of phenol in solvent with low polarity such as $\mathrm{CHCl}_3$ or $\mathrm{CS}_2$ requires Lewis acid catalyst.
Statements II : The Lewis acid catalyst polarises the bromine to generate $\mathrm{Br}^{+}$.
In the light of the above statements, choose the correct answer from the options given below :
For the Compounds :
(A) $\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$
(B) $\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$
(C) 
(D) 
The increasing order of boiling point is :
Choose the correct answer from the options given below :
Common name of Benzene - 1,2 - diol is -
| List I (Reactants) | List II (Product) |
|---|---|
| (A) Phenol, Zn/Δ | (I) Salicylaldehyde |
| (B) Phenol, CHCl3, NaOH, HCl | (II) Salicylic acid |
| (C) Phenol, CO2, NaOH, HCl | (III) Benzene |
| (D) Phenol, Conc. HNO3 | (IV) Picric acid |
Choose the correct answer from the options given below :
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: Alcohols react both as nucleophiles and electrophiles.
Reason R: Alcohols react with active metals such as sodium, potassium and aluminum to yield corresponding alkoxides and liberate hydrogen.
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:
Assertion A: $\mathrm{pK}_{\mathrm{a}}$ value of phenol is 10.0 while that of ethanol is 15.9 .
Reason R: Ethanol is stronger acid than phenol.
In the light of the above statements, choose the correct answer from the options given below:
Phenol treated with chloroform in presence of sodium hydroxide, which further hydrolyzed in presence of an acid results
Match List I with List II
| List - I (Compound) |
List - II ($\mathrm{pK_a}$ value) |
||
|---|---|---|---|
| (A) | Ethanol | (I) | 10.0 |
| (B) | Phenol | (II) | 15.9 |
| (C) | m-Nitrophenol | (III) | 7.1 |
| (D) | p-Nitrophenol | (IV) | 8.3 |
Choose the correct answer from the options given below:
The major product(P) in the following reaction is
Major product formed in the following reaction is a mixture of :
Phenolic group can be identified by a positive:
Match List - I with List - II.
| List - I (Reaction) |
List - II (Reagent(s)) |
||
|---|---|---|---|
| (A) |  |
(I) | $\mathrm{Na_2Cr_2O_7,H_2SO_4}$ |
| (B) |  |
(II) | (i) $\mathrm(NaOH)$, (ii) $\mathrm{CH_3Cl}$ |
| (C) |  |
(III) | (i) $\mathrm{NaOH,CHCl_3}$, (ii) $\mathrm{NaOH}$, (iii) $\mathrm{HCl}$ |
| (D) |  |
(IV) | (i) $\mathrm{NaOH}$, (ii) $\mathrm{CO_2}$, (iii) $\mathrm{HCl}$ |
Choose the correct answer from the options given below :
Given below are two statements :
Statement (I) : p-nitrophenol is more acidic than m-nitrophenol and o-nitrophenol.
Statement (II) : Ethanol will give immediate turbidity with Lucas reagent.
In the light of the above statements, choose the correct answer from the options given below :
The ascending order of acidity of $-\mathrm{OH}$ group in the following compounds is :
Choose the correct answer from the options given below:
Consider the given chemical reaction sequence :
Total sum of oxygen atoms in Product A and Product B are ________.
Explanation:
Total no. of oxygen in A and B = 7 + 7 = 14
Reaction of iso-propylbenzene with $\mathrm{O}_2$ followed by the treatment with $\mathrm{H}_3 \mathrm{O}^{+}$forms phenol and a by-product $\mathbf{P}$. Reaction of $\mathbf{P}$ with 3 equivalents of $\mathrm{Cl}_2$ gives compound $\mathbf{Q}$. Treatment of $\mathbf{Q}$ with $\mathrm{Ca}(\mathrm{OH})_2$ produces compound $\mathbf{R}$ and calcium salt $\mathbf{S}$.
The correct statement(s) regarding $\mathbf{P}, \mathbf{Q}, \mathbf{R}$ and $\mathbf{S}$ is(are)
'A' formed in the above reaction is
The major product for the following reaction is:
Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Order of acidic nature of the following compounds is A > B > C.
Reason R : Fluoro is a stronger electron withdrawing group than Chloro group.
In the light of the above statements, choose the correct answer from the options given below:
2-Methyl propyl bromide reacts with $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{O}^{-}$ and gives 'A' whereas on reaction with $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are :
Correct statements for the given reaction are :
A. Compound '$\mathrm{B}$' is aromatic
B. The completion of above reaction is very slow
C. 'A' shows tautomerism
D. The bond lengths of C-C in compound $B$ are found to be same
Choose the correct answer from the options given below:
Compound 'B' is
Product [X] formed in the above reaction is :
Arrange the following compounds in increasing order of rate of aromatic electrophilic substitution reaction
The correct order for acidity of the following hydroxyl compound is :
A. $\mathrm{CH}_{3} \mathrm{OH}$
B. $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$
C. 
D. 
E. 
Choose the correct answer from the options given below:
Incorrect method of preparation for alcohols from the following is: