The correct statement about $\mathbf{S}$ is :
Match the reaction in LIST-I with one or more products in LIST-II and choose the correct option.
For the identification of $\beta$-naphthol using dye test, it is necessary to use
The acidic hydrolysis of ether (X) shown below is fastest when
Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists :
The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is
Match the reactions in Column I with appropriate types of steps/reactive intermediate involved in these reactions as given in Column II :
| Column I | Column II | ||
|---|---|---|---|
| (A) |  |
(P) | Nucleophilic substitution |
| (B) |  |
(Q) | Electrophilic substitution |
| (C) |  |
(R) | Dehydration |
| (D) |  |
(S) | Nucleophilic |
| (T) | Carbanion |
In the reaction
the products are :
Compound H is formed by the reaction of
The structure of compound I is :
The structures of compound J, K and L, respectively, are:
Which one of the following reagents is used in the above reaction?
The electrophile in this reaction is :
The structure of the intermediate I is
When benzene sulphonic acid and $p$-nitrophenol are treated with $\mathrm{NaHCO}_3$, the gases released respectively are
$\mathrm{SO}_2, \mathrm{NO}_2$
$\mathrm{SO}_2, \mathrm{NO}$
$\mathrm{SO}_2, \mathrm{CO}_2$
$\mathrm{CO}_2, \mathrm{CO}_2$
Both benzene sulphonic acid and $p$-nitrophenol are strong acids whereas, $\mathrm{NaHCO}_3$ is alkaline in nature.
The reaction between benzene sulphonic acid and $\mathrm{NaHCO}_3$ yields sodium salt of benzene sulphonic acid along with water and carbon dioxide.
The products formed during the reaction between $p$-nitrophenol and $\mathrm{NaHCO}_3$ are sodium salt of $p$-nitrophenol, water and carbon dioxide.(I) 1,2-Dihydroxybenzene
(II) 1,3-Dihydroxybenzene
(III) 1,4-Dihydroxybenzene
(IV) Hydroxy benzene
The increasing order of boiling points of the above mentioned alcohols is
I $<$ II $<$ III $<$ IV
I $<$ II $<$ IV $<$ III
IV $<$ I $<$ II $<$ III
IV $<$ II $<$ I $<$ III
The 1, 2-dihydroxybenzene has second OH group attached at ortho position. The 1, 2-dihydroxybenzene will participate in intramolecular hydrogen bonding as both OH groups are attached at adjacent position as shown below.
Whereas 1, 3-dihydroxybenzene and 1, 4-dihydroxybenzene form intermolecular hydrogen bonding as shown below:
$ \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\text { Intermolecular } \mathrm{H} \text { bonding in 1, 4-dihydroxybenzene } $
The strength of intermolecular hydrogen bonding is higher in 1, 4-dihydroxybenzene as compared to that of 1,3-dihydroxybenzene and 1,2-dihydroxybenzene. Thestrengthofhydrogen bonding is higher in 1, 3-dihydroxybenzene than 1,2-dihydroxybenzene.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.
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)
The observed pattern of electrophilic substitution can be explained by
In the following reaction, the product/products formed is/are
The major product(s) of the following reaction is(are)
In the reaction
The intermediate(s) is(are)
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
$
Explanation:
So, the possible structures which are optically active and have phenolic group are as followed :
Therefore, total optically active isomers will be 6.
The number of hydroxyl group(s) in Q is ___________.
Explanation:
Therefore, the number of hydroxyl groups is 4.
The number of resonance structure for N is _________.
Explanation:
The possible resonance structures for the given compound on loss of proton are as follows:
Hence, the number of possible resonance structures is nine.
In the following reaction
Identify X and Y.
Explanation:
When (1-methylcyclopentyl) methanol is treated with H$^+$ at high temperature, it is converted to 1-methyl-1-cyclohexene.
Therefore, X is 1-methyl-1-cyclohexene.
The mechanism for the same is represented below.
The H$^+$ on heating with (1-methylcyclopentyl) methanol get attached to O atom of hydroxy group, producing a positive charge on O atom. The elimination of water molecule results in the formation of primary carbocation, which undergoes ring expansion followed by elimination of proton yields product X.
The 1-methyl-1-cyclohexene undergoes ozonolysis followed by treatment with Zn/ CH$_3$COOH, it gives 6-oxoheptanal.
Therefore, compound Y is 6-oxoheptanal
Y on further undergoes aldol condensation reaction to form $\alpha$, $\beta$ unsaturated carbonyl compounds in presence of NaOH followed by heat.
Final Answer
Hints :
Ozonolysis of an unsaturated compound results in replacing double bonds with carbonyl group. The Zn in acetic acid is a reducing agent.