Alcohols, Phenols and Ethers

Given mass of phenol = 4.7 g

Molar mass of phenol (C₆H₆O) = (6 × 12) + (6 × 1) + 16 = 94 g/mol

So,

$ \text{Moles of phenol} = \frac{4.7}{94.0} = 0.05~\text{mol} $

From the reaction, 1 mole of phenol produces 1 mole of benzene (product X) if the reaction is complete (100% yield).

Therefore, 0.05 mol of phenol will give 0.05 mol of benzene at 100% completion.

The actual yield is 60% of the theoretical yield, so:

$ \text{Moles of X formed} = 0.05 \times 0.60 = 0.03~\text{mol} $

$ \text{Moles of X} = 3 \times 10^{-2}~\text{mol} $

Final Answer: (3)

Step 1: Calculate the percentage of oxygen in phenol.

The molecular formula of phenol is $ \mathrm{C_6H_6O} $. Its molecular mass is:

$ (6 \times 12) + (6 \times 1) + 16 = 94~\mathrm{g/mol} $

Percentage of oxygen in phenol:

$ \frac{16}{94} \times 100 = 17.02\% $

Step 2: Calculate the percentage of oxygen in o-nitrophenol.

When phenol reacts with dilute $ \mathrm{HNO_3} $, the steam-volatile product formed is o-nitrophenol ($ \mathrm{C_6H_5NO_3} $).

Molecular mass of o-nitrophenol:

$ (6 \times 12) + (5 \times 1) + 14 + (3 \times 16) = 139~\mathrm{g/mol} $

Percentage of oxygen in o-nitrophenol:

$ \frac{48}{139} \times 100 = 34.53\% $

Step 3: Find the percentage increase in oxygen.

Increase in oxygen percentage:

$ 34.53 - 17.02 = 17.5 $

Step 4: Express the answer in the required form.

The increase in percentage of oxygen with respect to phenol is:

$ 17.5 \times 10^{-1} = 175 $

$\begin{aligned} &0.1 \text { mole of compound ( } \mathrm{S} \text { ) weight in gm }\\ &\begin{aligned} & =0.1 \times \text { molar mass of compound }(\mathrm{S}) \\ & =0.1 \times 130=13 \mathrm{gm} \end{aligned} \end{aligned}$

$ \text { Moles of } x=16 \mathrm{~mol} \text {. } $ 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 $

Total no. of oxygen in A and B = 7 + 7 = 14

So, number of hyperconjugation structures in most stable carbocation

$=6+1=7$

For completion of reaction, we need 2 moles of $\mathrm{MeMgBr}$ per mole of reactant

1 mole for nucleophilic addition and 1 mole for acid base reaction.

The ligand present in Fehling's reagent is tartrate, specifically potassium sodium tartrate (KNaC₄H₄O₆•4H₂O), which acts as a chelating agent. The denticity of tartrate is four, meaning that it can bond to a metal ion through four different atoms, in this case, two oxygen atoms from each of the two carboxylate groups and two oxygen atoms from the two hydroxyl groups. This chelation allows the copper ions in Fehling's A to remain in solution, preventing them from precipitating out as insoluble copper hydroxides. The chelation also helps to stabilize the intermediate copper complexes formed during the redox reaction with reducing sugars, allowing for the formation of the brick-red precipitate of copper(I) oxide (Cu₂O).

Let initial moles of reactant taken = n

Total moles obtained for benzene sulphonic acid (with % yield = 60%) = 0.6n


Moles of benzene sulphonic acid before reaction II = 0.6n

Moles obtained for phenol (with % yield = 50%) = 0.6 × 0.5n = 0.3n

So over all % yield of complete reaction $=\frac{0.3 \mathrm{n}}{\mathrm{n}} \times 100=30$

Out of which only two are chiral

2 moles of CH₃MgBr reagent will be used. Chemical reaction is as follows

mass of Salicylaldehyde = 12 × 7 + 6 × 1 + 16 × 2 = 122

mass of carbon = 12 × 7 = 84

$ \therefore $ mass % of C in P = ${{84} \over {122}}$ $ \times $ 100

= 68.85 % $ \simeq $ 69 %

Phenolic ($-$OH) group gives positive test with neutral FeCl₃ solution, means organic compound (C₈H₁₀O₂) also rotate plane porarised light means compound is an optically active and chiral carbon atom is present.

So, the possible structures which are optically active and have phenolic group are as followed :



Therefore, total optically active isomers will be 6.

Therefore, the number of hydroxyl groups is 4.

The possible resonance structures for the given compound on loss of proton are as follows:

Hence, the number of possible resonance structures is nine.

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.