Properties of alcohol Questions in English

(N/A) Esterification: The reaction of alcohols and phenols with carboxylic acids,acid chlorides,and acid anhydrides to form esters is known as esterification.
$(i)$ Reaction of alcohol/phenol with acid $(R'COOH)$: This is an esterification reaction where the $O-H$ bond breaks.
$(ii)$ Reaction of alcohol/phenol with acid anhydride: In this reaction,the $O-H$ bond of the alcohol or phenol breaks to form $-OCOR'$. This reaction is carried out in the presence of a small amount of concentrated sulfuric acid and is reversible.
$(iii)$ Reaction of alcohol/phenol with acid chloride $(R'COCl)$: In this reaction,the $O-H$ bond breaks to form $-OCOR'$. This reaction is carried out in the presence of a base,pyridine,to neutralize the $HCl$ formed,which shifts the equilibrium to the right (product side).
$(b)$ Acetylation: Acetylation is a type of esterification where the $-OH$ group is converted into $-COCH_3$. It involves the introduction of an acetyl group $(CH_3CO-)$ into alcohols and phenols. Acetylation of salicylic acid produces aspirin,which is a drug with analgesic,anti-inflammatory,and antipyretic properties.

(N/A) Esterification: The reaction of alcohols and phenols with carboxylic acids,acid chlorides,and acid anhydrides to form esters is known as esterification.
$(i)$ Reaction of alcohols/phenols with carboxylic acids $(R'COOH)$: This reaction involves the cleavage of the $O-H$ bond of the alcohol or phenol.
$(ii)$ Reaction of alcohols/phenols with acid anhydrides: In this reaction,the $O-H$ bond of the alcohol or phenol breaks to form $-OCOR'$. This reaction is carried out in the presence of a small amount of concentrated $H_2SO_4$ and is reversible.
$(iii)$ Reaction of alcohols/phenols with acid chlorides $(R'COCl)$: In this reaction,the $O-H$ bond breaks to form $-OCOR'$. This reaction is carried out in the presence of a base like pyridine to neutralize the $HCl$ produced,which shifts the equilibrium to the right (product side).
$(b)$ Acetylation: Acetylation is a specific type of esterification where the $-OH$ group is converted into $-COCH_3$. It involves the introduction of an acetyl group $(CH_3CO-)$ into alcohols and phenols. Acetylation of salicylic acid produces aspirin,which is a well-known analgesic,anti-inflammatory,and antipyretic drug.

(N/A) Reactions involving the cleavage of the $C-O$ bond occur in alcohols. In phenols,the $C-O$ bond cleavage only occurs upon heating with zinc dust.
$(a)$ Reaction of alcohols with $HX$ (hydrogen halides) and Lucas test:
$(i)$ Alcohols react with hydrogen halides $(HX)$ to form alkyl halides.
$R-OH + HX \rightarrow R-X + H_2O$
In this reaction,the $C-O$ bond is broken,a $C-X$ bond is formed,and the $-OH$ group is removed. This reaction proceeds via nucleophilic substitution mechanisms ($S_N1$ and $S_N2$).
The order of reactivity of hydrogen halides $(HX)$ with alcohols is $3^{\circ} > 2^{\circ} > 1^{\circ}$.
$(ii)$ Alkyl bromides are formed by heating alcohols with $48\% \ HBr$.
$(iii)$ Reaction with $HCl$ (Lucas test):
The catalyst $ZnCl_2$ is required for the reaction of primary and secondary alcohols with $HCl$. The Lucas reagent is a mixture of concentrated $HCl$ and $ZnCl_2$. $1^{\circ}, 2^{\circ},$ and $3^{\circ}$ alcohols can be distinguished using the Lucas test.
Alcohols are soluble in the Lucas reagent,but the resulting alkyl halides are immiscible,causing turbidity in the solution.
Tertiary alcohols react immediately with the Lucas reagent to form alkyl halides,causing immediate turbidity and the formation of a separate layer.
Secondary alcohols react slowly with the Lucas reagent,producing turbidity after some time,followed by the formation of a separate layer.
Primary alcohols do not react with the Lucas reagent at room temperature,so no turbidity or separate layer is observed for a long time.
Thus,the reactivity order of alcohols is $3^{\circ} > 2^{\circ} > 1^{\circ}$,allowing for the identification of $3^{\circ}, 2^{\circ},$ and $1^{\circ}$ alcohols using the Lucas test.

(N/A) Reactions involving the cleavage of $C-O$ bond in alcohols:
$(a)$ Reaction with $HX$ and Lucas test:
$(i)$ Alcohols react with hydrogen halides $(HX)$ to form alkyl halides: $R-OH + HX \rightarrow R-X + H_2O$.
The reactivity order of $HX$ with alcohols is $3^{\circ} > 2^{\circ} > 1^{\circ}$.
$(ii)$ Lucas test: The reaction of alcohols with concentrated $HCl$ in the presence of anhydrous $ZnCl_2$ is known as the Lucas test.
$3^{\circ}$ alcohols react immediately to form turbidity.
$2^{\circ}$ alcohols form turbidity within $5-10$ minutes.
$1^{\circ}$ alcohols do not react at room temperature.
$(b)$ Reaction with $PX_3$:
Alcohols react with phosphorus trihalides ($PX_3$,where $X = Cl, Br$) to form alkyl halides: $3R-OH + PX_3 \rightarrow 3R-X + H_3PO_3$.
$(c)$ Dehydration reactions:
Alcohols undergo dehydration in the presence of acid catalysts (like conc. $H_2SO_4$ or $H_3PO_4$) to form alkenes.
The ease of dehydration follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.

(N/A) The dehydration of ethanol to ethene proceeds via a $\beta$-elimination mechanism in three steps:
Step-$1$: Protonation of ethanol to form ethyl oxonium ion.
$CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2OH_2^+$
Step-$2$: Formation of carbocation. This is the slow,rate-determining step where the $C-O$ bond breaks and water is eliminated.
$CH_3CH_2OH_2^+ \rightleftharpoons CH_3CH_2^+ + H_2O$
Step-$3$: Formation of ethene by elimination of a proton. The carbocation loses a $\beta$-proton to form ethene.
$CH_3CH_2^+ \rightleftharpoons CH_2=CH_2 + H^+$

(N/A) $(i)$ Dehydration of alcohols occurs by heating them in the presence of protic acids like concentrated $H_2SO_4$ or $H_3PO_4$ or catalysts like anhydrous $ZnCl_2$ or alumina to form alkenes.
$CH_3CH_2OH \xrightarrow{\text{conc. } H_2SO_4} CH_2=CH_2 + H_2O$
$(ii)$ The dehydration of ethanol proceeds in three steps via a $\beta$-elimination mechanism.
Step-$1$: Protonation of alcohol. The proton $(H^+)$ from the acid attacks the $-OH$ group of the alcohol to form an oxonium ion $(-OH_2^+)$.
Step-$2$: Formation of carbocation. The $C-O$ bond breaks,releasing $H_2O$ and forming a carbocation. This is the slow and rate-determining step.
Step-$3$: Formation of ethene. $A$ $\beta$-proton is eliminated from the carbocation to form ethene. The $H^+$ used in Step-$1$ is released,regenerating the acid catalyst.

(N/A) Oxidation: The oxidation of alcohols involves the breaking of $C-H$ and $O-H$ bonds to form a carbon-oxygen double bond $(C=O)$.
Oxidation/Dehydrogenation: In this process,$C-H$ and $O-H$ bonds are broken and dihydrogen $(H_2)$ is released,which is why this process is called dehydrogenation.
The products of oxidation depend on the oxidizing agents used and the type of alcohol.
$(b)$ Oxidation of primary alcohols: Depending on the oxidizing agent used,$1^{\circ}$ alcohols are oxidized to aldehydes,which are further oxidized to carboxylic acids.
$(i)$ $1^{\circ}$-Alcohol $\rightarrow$ Carboxylic acid: Strong oxidizing agents like potassium permanganate $(KMnO_4)$ are used to obtain carboxylic acids directly from $1^{\circ}$-alcohol compounds.

(N/A) Oxidation: The oxidation of alcohols involves the breaking of $O-H$ and $C-H$ bonds,leading to the formation of a carbon-oxygen double bond $(C=O)$.
Dehydrogenation: In this process,$C-H$ and $O-H$ bonds are broken,and dihydrogen $(H_2)$ is released; hence,these are called dehydrogenation reactions.
The products of oxidation depend on the oxidizing agents used and the type of alcohol.
$(b)$ Oxidation of primary alcohols: Depending on the oxidizing agent used,$1^{\circ}$ alcohols are oxidized to aldehydes,which are further oxidized to carboxylic acids.
$(i)$ $1^{\circ}$-Alcohol $\rightarrow$ Carboxylic acid: Strong oxidizing agents like potassium permanganate $(KMnO_4)$ are used to obtain carboxylic acids directly from $1^{\circ}$-alcohols.

(N/A) Oxidation: Oxidation of alcohols involves the breaking of $O-H$ and $C-H$ bonds to form a carbon-oxygen double bond $(C=O)$.
$(b)$ Dehydrogenation: In this process,$C-H$ and $O-H$ bonds break to release dihydrogen $(H_2)$,hence it is called a dehydrogenation reaction.
$(c)$ Oxidation of $1^o$ alcohols: $1^o$ alcohols are oxidized to aldehydes,which are further oxidized to carboxylic acids using strong oxidizing agents like $KMnO_4$.
$(d)$ Oxidation of $2^o$ alcohols: $2^o$ alcohols are oxidized to ketones using oxidizing agents like $CrO_3$.
$(e)$ Oxidation of $3^o$ alcohols: $3^o$ alcohols are resistant to oxidation under normal conditions but undergo dehydration to form alkenes in the presence of hot concentrated $H_2SO_4$ or by heating with copper at $573 K$.

(N/A) Denatured alcohol is ethanol $\left( C _{2} H _{5} OH \right)$ mixed with methanol $\left( CH _{3} OH \right)$.
Effects: When a person consumes denatured alcohol,methanol is oxidized in the body to methanal $\left( HCHO \right)$ and subsequently to methanoic acid $\left( HCOOH \right)$. This process causes severe toxicity,leading to blindness and potentially death.
Remedies: The treatment for methanol poisoning involves the intravenous administration of diluted ethanol. Ethanol acts as a competitive inhibitor for the enzyme responsible for the oxidation of methanol. This allows the body time to excrete the methanol through the kidneys,thereby preventing the formation of toxic methanoic acid and saving the patient.

(N/A) Methanol is $CH_3OH$,while ethanol is $C_2H_5OH$. Both undergo biological oxidation to form their corresponding aldehydes $(HCHO, CH_3CHO)$ and subsequently acids $(HCOOH, CH_3COOH)$.
Side Effects: Denatured alcohol is ethanol mixed with methanol. If an addict consumes denatured alcohol,methanol is first converted into methanal and then into methanoic acid. This can lead to blindness and death.
Remedies: $A$ patient suffering from the toxic effects of methanol consumption is treated by intravenous administration of dilute ethanol. The enzyme responsible for the oxidation of aldehyde $(HCHO)$ to acid is inhibited by ethanol,allowing the kidneys time to excrete methanol,thereby saving the patient.

(N/A) Acidic dehydration of alcohols involves the removal of a water molecule from an alcohol in the presence of an acid catalyst to form an alkene.
General Reaction:
$CH_3-CH_2-OH \xrightarrow[\Delta, 443 \ K]{Conc. \ H_2SO_4} CH_2=CH_2 + H_2O$
Mechanism and Explanation:
$1$. In this reaction,the $-OH$ group is removed from the $\alpha$-carbon,and a hydrogen atom is removed from the $\beta$-carbon.
$2$. This results in the formation of a $\pi$-bond between the $\alpha$ and $\beta$ carbons.
$3$. Because a hydrogen atom is eliminated from the $\beta$-position,this process is also known as a $\beta$-elimination reaction.
$4$. Common dehydrating agents used include concentrated $H_2SO_4$,$H_3PO_4$,or anhydrous $Al_2O_3$ at high temperatures.

(N/A) When alcohols are treated with concentrated $H_{2}SO_{4}$,they undergo dehydration to form alkenes,a process known as $\beta$-elimination reaction.
$(i)$ $CH_{3}CH_{2}CH_{2}OH \xrightarrow[\Delta, 443 \ K]{Conc. \ H_{2}SO_{4}} CH_{3}CH=CH_{2} + H_{2}O$ (Propene)
$(ii)$ $CH_{3}CH(OH)CH_{3} \xrightarrow[438-443 \ K]{H_{2}SO_{4}} CH_{3}CH=CH_{2} + H_{2}O$ (Propene)
$(iii)$ $C_{6}H_{11}OH \xrightarrow[373-383 \ K]{H_{2}SO_{4}/H_{3}PO_{4}} C_{6}H_{10} + H_{2}O$ (Cyclohexene)
$(iv)$ $(CH_{3})_{3}COH \xrightarrow[\Delta, 360 \ K]{H_{2}SO_{4}} CH_{3}-C(CH_{3})=CH_{2} + H_{2}O$ ($2$-Methylpropene)

(N/A) Production of methanol: Methanol is known as wood spirit,as it was formerly produced by the destructive distillation of wood. Currently,it is produced by the catalytic hydrogenation of carbon monoxide in the presence of $Cr_2O_3-ZnO$ catalyst at high temperature $(573-673 \ K)$ and high pressure $(200-300 \ atm)$.
$CO + 2H_2 \longrightarrow CH_3OH$
$(b)$ Physical properties: Methanol is a colorless liquid and is highly poisonous in nature. Its boiling point is $337 \ K$.
$(c)$ Physiological effects: Consumption of even small quantities of methanol can cause blindness,and large quantities can lead to death.
$(d)$ Uses: It is used as a solvent in paints and varnishes,and primarily in the manufacture of formaldehyde $(HCHO)$.

(N/A) Production of Ethanol: The production of ethanol is described as follows:
$(i)$ Fermentation of Sugars: This is the oldest method. Sugars from molasses,sugarcane,or fruits like grapes are fermented in the presence of the enzyme invertase to produce glucose and fructose.
$C_{12}H_{22}O_{11} + H_{2}O \xrightarrow{\text{invertase}} C_{6}H_{12}O_{6} + C_{6}H_{12}O_{6}$
Glucose and fructose have the same molecular formula $C_{6}H_{12}O_{6}$ but different structures. The enzyme zymase is obtained from yeast. Fermentation of glucose and fructose by zymase produces ethanol.
$C_{6}H_{12}O_{6} \longrightarrow 2C_{2}H_{5}OH + 2CO_{2}$
$(ii)$ Ethanol from Grapes: When grapes ripen,the quantity of sugar increases,and yeast grows on the surface of the ripe grapes. When the grape sugar and the zymase in the yeast come into contact,fermentation begins. Fermentation occurs under anaerobic conditions (absence of oxygen),releasing ethanol and carbon dioxide.
If the concentration of alcohol produced exceeds $14\%$,the action of zymase is inhibited. If air enters the fermentation mixture,oxygen in the air oxidizes ethanol to ethanoic acid,which spoils the taste of alcoholic beverages.
$(b)$ Physical Properties: Ethanol is a colorless liquid. Its boiling point is $351\,K$.
$(c)$ Uses: It is used as a solvent in the paint industry and for the synthesis of many organic compounds.
$(d)$ Denaturation of Alcohol: Industrial alcohol is made unfit for drinking by adding small amounts of copper sulfate to give it color and pyridine to make it foul-smelling. This process is called denaturation of alcohol.

(N/A) The solubility of ethers in water is comparable to that of alcohols of similar molecular mass.
For example,ethoxyethane and butan-$1$-ol both have the same molecular mass of $74 \ g \ mol^{-1}$ and their solubility in water is quite similar. However,alcohols are generally more soluble than ethers because they can form stronger hydrogen bonds with water molecules.

Property	Ethoxyethane	Butan-$1$-ol
Group	Ether	Alcohol
Molecular Mass	$74 \ g \ mol^{-1}$	$74 \ g \ mol^{-1}$
Formula	$C_2H_5-O-C_2H_5$	$CH_3CH_2CH_2CH_2OH$
Miscibility	$7.5 \ g$ per $100 \ mL$ water	$9 \ g$ per $100 \ mL$ water

(N/A) Methoxymethane $(CH_3-O-CH_3)$ and ethanol $(C_2H_5OH)$ are isomers with the same molecular formula $C_2H_6O$ and the same molar mass of $46 \ g \ mol^{-1}$. Despite this,their boiling points are different.
In methoxymethane,there is an ether group,and there is no polar $O-H$ bond. Therefore,there is no intermolecular hydrogen bonding in ethers.
In ethanol,the $O-H$ bond is polar,and there is intermolecular hydrogen bonding between different ethanol molecules. Let $C_2H_5 = R$,then the intermolecular $H$-bonding is as shown in the figure.
Due to these intermolecular hydrogen bonding forces,ethanol has a higher boiling point compared to methoxymethane.

(N/A) The reaction sequence proceeds as follows:
$1$. $CH_3-CH_2-CH(OH)-CH_3 + PBr_3 \rightarrow CH_3-CH_2-CH(Br)-CH_3 (A)$ (Substitution of $-OH$ with $-Br$)
$2$. $CH_3-CH_2-CH(Br)-CH_3 + Mg \xrightarrow{\text{ether}} CH_3-CH_2-CH(MgBr)-CH_3 (B)$ (Formation of Grignard reagent)
$3$. $CH_3-CH_2-CH(MgBr)-CH_3 + CH_3CHO \rightarrow CH_3-CH_2-CH(CH_3)-CH(OMgBr)-CH_3 (C)$ (Nucleophilic addition to acetaldehyde)
$4$. $CH_3-CH_2-CH(CH_3)-CH(OMgBr)-CH_3 + H_3O^+ \rightarrow CH_3-CH_2-CH(CH_3)-CH(OH)-CH_3 (D)$ (Hydrolysis to form secondary alcohol)

(A) $1$. The starting material is butan$-2-$ol,a secondary alcohol. Dehydrogenation of secondary alcohols with $Cu$ at $520 \ K$ yields a ketone. Thus,$(A)$ is butan$-2-$one: $CH_3-CH_2-CO-CH_3$.
$2$. The reaction of butan$-2-$one with ethylmagnesium bromide $(C_2H_5MgBr)$ is a Grignard reaction. The nucleophilic ethyl group attacks the carbonyl carbon,forming an intermediate alkoxide: $(B)$ is $CH_3-CH_2-C(OMgBr)(CH_3)-C_2H_5$.
$3$. Acidic hydrolysis of the intermediate $(B)$ with $H_3O^+$ yields the corresponding tertiary alcohol: $(C)$ is $CH_3-CH_2-C(OH)(CH_3)-C_2H_5$ ($3$-methylpentan$-3-$ol).

(N/A) The conversion of methanol $(CH_3OH)$ to higher alcohols involves the formation of aldehydes followed by Grignard reaction:
$1$. Conversion to Ethanol:
$CH_3OH$ $\xrightarrow{Cu, \Delta} HCHO$ $\xrightarrow{CH_3MgBr} CH_3CH_2OMgBr$ $\xrightarrow{H_3O^+} CH_3CH_2OH$
$2$. Conversion to Propan-$2$-ol:
$CH_3CH_2OH$ $\xrightarrow{Cu, \Delta} CH_3CHO$ $\xrightarrow{CH_3MgBr} CH_3CH(OMgBr)CH_3$ $\xrightarrow{H_3O^+} CH_3CH(OH)CH_3$
$3$. Conversion to $2$-methylpropan-$2$-ol:
$CH_3CH(OH)CH_3$ $\xrightarrow{Cu, \Delta} CH_3COCH_3$ $\xrightarrow{CH_3MgBr} (CH_3)_3COMgBr$ $\xrightarrow{H_3O^+} (CH_3)_3COH$

(A) The reaction sequence is as follows:
$1$. Cyclohexanol is oxidized by $CrO_3$ in $H_2SO_4$ (Jones reagent) to form Cyclohexanone $(A)$.
$2$. Cyclohexanone reacts with $CH_3MgI$ followed by acid hydrolysis to form $1-$methylcyclohexanol $(B)$.
$3$. $1-$methylcyclohexanol undergoes acid-catalyzed dehydration with $H_2SO_4$ and heat to form $1-$methylcyclohexene $(C)$.
$4$. $1-$methylcyclohexene undergoes hydroboration-oxidation with $(BH_3)_2$ followed by $H_2O_2/NaOH$ to form $2-$methylcyclohexanol $(D)$.

(N/A) The dehydration of $2$-methylbutan-$1$-ol with concentrated $H_2SO_4$ at $443 \ K$ proceeds via the formation of a carbocation intermediate.
$1$. Protonation of the alcohol: The hydroxyl group is protonated to form a good leaving group $(-OH_2^+)$.
$2$. Formation of carbocation: Loss of water molecule generates a $1^{\circ}$ carbocation.
$3$. Rearrangement: The $1^{\circ}$ carbocation undergoes a $1,2$-hydride shift to form a more stable $3^{\circ}$ carbocation.
$4$. Elimination: Loss of a proton from the $3^{\circ}$ carbocation yields $2$-methylbut-$2$-ene as the major product,following Saytzeff's rule.

(A) The reaction of ethanol with concentrated $H_2SO_4$ at $413 \ K$ leads to the formation of diethyl ether via intermolecular dehydration.
At $443 \ K$,the reaction leads to the formation of ethene.
Therefore,the correct condition is $Conc. H_2SO_4, 413 \ K$.

(N/A) The solubility of alcohols in water is primarily due to the presence of the polar $-OH$ group,which allows alcohols to form intermolecular $H$-bonding with water molecules.
$R-OH \cdots O(H)-H \cdots O(H)-R$

(B) The process of adding substances like pyridine,methanol,or copper sulphate to ethanol to make it unfit for human consumption is known as denaturation. The resulting mixture is called denatured alcohol.

(PCC) The conversion involves the oxidation of a secondary allylic alcohol to an $\alpha,\beta$-unsaturated ketone.
Pyridinium chlorochromate $(PCC)$ is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes and secondary alcohols to ketones without affecting carbon-carbon double bonds.
Therefore,the required reagent is $PCC$.

(N/A) $2$-chloroethanol is more acidic than ethanol.
This is because the chlorine atom exerts a strong $-I$ effect,which pulls the electron density away from the oxygen atom.
This weakens the $O-H$ bond,facilitating the release of the proton $(H^+)$ and stabilizing the resulting alkoxide ion.

(A) The conversion of a primary alcohol like ethanol $(CH_3CH_2OH)$ to an aldehyde like ethanal $(CH_3CHO)$ requires a mild oxidizing agent that stops the oxidation at the aldehyde stage without further oxidizing it to a carboxylic acid. Pyridinium chlorochromate $(PCC)$ is the ideal reagent for this transformation.

(C) Acidified $KMnO_4$ or acidified $K_2Cr_2O_7$ acts as a strong oxidizing agent and oxidizes ethanol to ethanoic acid.
$CH_3-CH_2-OH$ $\xrightarrow{[O]} CH_3-CHO$ $\xrightarrow{[O]} CH_3-COOH$

(N/A) The order of reactivity of alcohols with sodium metal is:
Primary alcohol $>$ Secondary alcohol $>$ Tertiary alcohol
The $+I$ effect of the alkyl groups increases the electron density on the oxygen atom. As a result,the electropositive hydrogen of the $-OH$ group is more tightly held by the molecule,which makes the release of $H^{+}$ difficult.
The effect is more pronounced if there are more alkyl groups in the molecule. Thus,tertiary alcohol is the least acidic and shows the least reactivity towards sodium metal.

(N/A) The isomers of alcohols with molecular formula $C_4H_{10}O$ are:
$1.$ $Butan-1-ol$: $CH_3-CH_2-CH_2-CH_2-OH$
$2.$ $Butan-2-ol$: $CH_3-CH_2-CH(OH)-CH_3$
$3.$ $2-Methylpropan-1-ol$: $(CH_3)_2CH-CH_2-OH$
$4.$ $2-Methylpropan-2-ol$: $(CH_3)_3C-OH$
Among these,$Butan-2-ol$ exhibits optical activity because the $C_2$ carbon atom is chiral,as it is attached to four different groups: $-H$,$-OH$,$-CH_3$,and $-C_2H_5$.

(N/A) The reaction of alcohols with Lucas reagent proceeds via the formation of a carbocation intermediate,which is the rate-determining step.
The stability of carbocations follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Tertiary $(3^{\circ})$ alcohols form the most stable carbocation,making the cleavage of the $C-OH$ bond easiest.
Consequently,the reactivity order is $3^{\circ} \text{ alcohol} > 2^{\circ} \text{ alcohol} > 1^{\circ} \text{ alcohol}$.

(N/A) In alcohols of low molecular mass,the size of the non-polar alkyl group is small,which exerts less steric hindrance. As a result,intermolecular $H$-bonds with water molecules are easily formed by the hydroxyl group of the alcohol. Hence,alcohols of low molecular mass are readily soluble in water.

(N/A) Alcohols contain a polar $-OH$ group,which allows for the formation of intermolecular $H$-bonds between alcohol molecules. This association requires more energy to break during vaporization.
In contrast,ethers do not contain a hydrogen atom bonded directly to an oxygen atom,so they cannot form intermolecular $H$-bonds. Their intermolecular forces are primarily dipole-dipole interactions,which are weaker than $H$-bonds.
Therefore,for comparable molecular masses,alcohols have significantly higher boiling points than ethers.

(N/A) For reaction $1$,the dehydrohalogenation of $2$-bromo-$3$-methylbutane with alcoholic $KOH$ follows $Saytzeff$ rule. The more substituted alkene,$2$-methyl-$2$-butene,is the major product $(80\%)$,while $3$-methyl-$1$-butene is the minor product $(20\%)$.
For reaction $2$,the acid-catalyzed dehydration of $3$-methylbutan-$2$-ol follows $Saytzeff$ rule. The more substituted alkene,$2$-methyl-$2$-butene,is the major product,while $3$-methyl-$1$-butene is the minor product.

(D) The reaction of cyclohexanol with alcoholic $KOH$ under heating conditions leads to the dehydration of the alcohol.
This is an elimination reaction where a water molecule $(-H_2O)$ is removed from the cyclohexanol molecule.
The hydroxyl group $(-OH)$ is removed from the carbon atom,and a hydrogen atom is removed from an adjacent carbon atom,resulting in the formation of a double bond.
Thus,the main product $X$ is cyclohexene.

(A) The normal boiling points of the given substances are as follows:
Ether: $308 \ K$
Carbon tetrachloride $(CCl_4)$: $349 \ K$
Ethanol $(C_2H_5OH)$: $351.3 \ K$
Water $(H_2O)$: $373 \ K$
Therefore,the increasing order is: $\text{Ether} < \text{Carbon tetrachloride} < \text{Ethanol} < \text{Water}$.

(A-III, B-IV, C-II, D-I) The reactions are as follows:
$A$. $ROH + SOCl_2 \rightarrow RCl + SO_2 + HCl$ (Darbzens process) $(A \rightarrow iii)$
$B$. $ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl$ $(B \rightarrow iv)$
$C$. $3ROH + PX_3 \rightarrow 3RX + H_3PO_3$ $(C \rightarrow ii)$
$D$. $ROH + NaBr + H_2SO_4 \rightarrow RBr + NaHSO_4 + H_2O$ $(D \rightarrow i)$
Therefore,the correct matching is $A-iii, B-iv, C-ii, D-i$.

(A) $A \rightarrow ii$ (or $iv$),$B \rightarrow iii$,$C \rightarrow iv$ (or $ii$),$D \rightarrow i$.
$1$. $Butan-1-ol$ reacts with $HCl + ZnCl_2$ (Lucas reagent) to form $1-chlorobutane$ $(CH_3CH_2CH_2CH_2Cl)$.
$2$. $Butan-1-ol$ reacts with $HBr$ to form $1-bromobutane$ $(CH_3CH_2CH_2CH_2Br)$.
$3$. $Butan-1-ol$ reacts with $SOCl_2$ (Thionyl chloride) to form $1-chlorobutane$ $(CH_3CH_2CH_2CH_2Cl)$.
$4$. $Butan-1-ol$ reacts with $H_2SO_4$ upon heating (dehydration) to form $but-1-ene$ (which isomerizes to $but-2-ene$,$CH_3CH=CHCH_3$).

(A-III, B-II, C-I, D-IV) The dehydration of alcohols follows the $E1$ or $E2$ mechanism,typically yielding the most stable alkene (Saytzeff's rule).
$A$. $1$-Methylcyclohexanol dehydrates to form $1$-methylcyclohexene $(iii)$.
$B$. Butan-$1$-ol dehydrates to form but-$1$-ene $(ii)$.
$C$. tert-Butyl alcohol ($2$-methylpropan$-2-$ol) dehydrates to form isobutylene $(i)$.
$D$. Butan-$2$-ol dehydrates to form but-$2$-ene as the major product $(iv)$.
Therefore,the correct matching is: $A-iii, B-ii, C-i, D-iv$.

(D) $1$. The reaction of the ether $CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3$ with $HI$ proceeds via an $S_N2$ mechanism.
$2$. The iodide ion $(I^-)$ attacks the less sterically hindered carbon atom,which is the ethyl group $(CH_2-CH_3)$,resulting in the formation of $CH_3-CH_2-I$ and the alcohol $[A]$,which is $2-methylbutan-1-ol$ $(CH_3-CH_2-CH(CH_3)-CH_2OH)$.
$3$. The dehydration of $2-methylbutan-1-ol$ with concentrated $H_2SO_4$ at high temperature involves the formation of a carbocation intermediate followed by rearrangement to a more stable carbocation.
$4$. The primary carbocation $CH_3-CH_2-CH(CH_3)-CH_2^+$ rearranges to the more stable tertiary carbocation $CH_3-CH_2-C^+(CH_3)_2$ via a hydride shift.
$5$. Elimination of a proton from this tertiary carbocation yields the most stable alkene,$2-methylbut-2-ene$ $(CH_3-CH=C(CH_3)_2)$,as the major product $[B]$.

(A) The organic compound $(A)$ with molecular formula $C_{6}H_{12}O_{2}$ is an ester.
Upon hydrolysis with dil. $H_{2}SO_{4}$,it yields a carboxylic acid $(B)$ and an alcohol $(C)$.
The alcohol $(C)$ gives immediate white turbidity with Lucas reagent (anhydrous $ZnCl_{2}$ and conc. $HCl$),which is a characteristic test for a tertiary alcohol.
Among the given options,the ester that produces a tertiary alcohol (tert-butyl alcohol) upon hydrolysis is tert-butyl acetate.
The structure shown in option $(A)$ corresponds to tert-butyl acetate,which hydrolyzes to form acetic acid and tert-butyl alcohol.
Thus,the correct compound is $(A)$.

(A) The reaction of methylenecyclohexane with $H_3O^+$ proceeds via acid-catalyzed hydration,which follows Markownikoff's rule. The carbocation formed at the tertiary carbon is more stable,leading to the formation of $1$-methylcyclohexanol $(Y)$.
The reaction with $(i) B_2H_6/THF$ and $(ii) H_2O_2/OH^-$ is hydroboration-oxidation,which follows anti-Markownikoff's rule. The hydroxyl group adds to the less substituted carbon,resulting in the formation of cyclohexylmethanol $(X)$.

(A) The reaction sequence is as follows:
$1$. $CH_2=CH_2$ reacts with $HOCl$ (hypochlorous acid) to form ethylene chlorohydrin $(X)$: $CH_2=CH_2 + HOCl \rightarrow CH_2(OH)-CH_2(Cl)$.
$2$. $CH_2(OH)-CH_2(Cl)$ (ethylene chlorohydrin) reacts with a base like $NaHCO_3$ or $NaOH$ to undergo hydrolysis or epoxide formation followed by ring opening to yield ethylene glycol $(CH_2(OH)-CH_2(OH))$.
$3$. Thus,$X$ is $CH_2(OH)-CH_2(Cl)$ and $Y$ is $NaHCO_3$.

(C) $1$. The molecular formula of the reactant is $C_{8}H_{8}O$. Given the reaction with $C_{2}H_{5}MgBr$ (Grignard reagent) and subsequent hydrolysis,it is likely an aldehyde or ketone. Acetophenone $(C_{6}H_{5}COCH_{3})$ fits the formula $C_{8}H_{8}O$.
$2$. Reaction: $C_{6}H_{5}COCH_{3} + C_{2}H_{5}MgBr \rightarrow C_{6}H_{5}C(OH)(CH_{3})(C_{2}H_{5})$. This product $A$ is $2$-phenylbutan-$2$-ol.
$3$. Compound $A$ is a tertiary alcohol. Tertiary alcohols react instantly with Lucas reagent $(conc. \ HCl + ZnCl_{2})$ to form alkyl chlorides.
$4$. The reaction of $2$-phenylbutan-$2$-ol with $HCl$ replaces the $-OH$ group with $-Cl$ to form $2$-chloro-$2$-phenylbutane $(C_{10}H_{13}Cl)$.
$5$. Comparing this with the given options,the structure corresponding to $2$-chloro-$2$-phenylbutane is option $C$.

(A) The reaction of a secondary alcohol with $HCl$ proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.
$1$. Protonation of the $-OH$ group by $HCl$ converts it into a good leaving group $(-OH_2^+)$.
$2$. Loss of water molecule generates a secondary carbocation at the carbon atom attached to the cyclohexane ring.
$3$. This secondary carbocation undergoes a $1,2$-hydride shift to form a more stable tertiary carbocation at the ring carbon.
$4$. Finally,the nucleophilic chloride ion $(Cl^-)$ attacks the tertiary carbocation to form the major product,$1$-ethyl-$1$-chloro-$2$-methylcyclohexane.

(B) The reaction involves the treatment of a polyfunctional compound with $SOCl_2$.
$SOCl_2$ (thionyl chloride) is a reagent commonly used to convert alcohols into alkyl chlorides.
In the given substrate,there are three hydroxyl groups:
$1$. $A$ phenolic $-OH$ group.
$2$. $A$ secondary aliphatic $-OH$ group on the saturated ring.
$3$. $A$ primary benzylic $-CH_2OH$ group.
$SOCl_2$ reacts readily with aliphatic alcohols (both primary and secondary) to form alkyl chlorides.
However,it does not typically convert phenolic $-OH$ groups into aryl chlorides under standard conditions.
Therefore,both the secondary $-OH$ and the primary benzylic $-CH_2OH$ groups will be converted into $-Cl$ and $-CH_2Cl$ respectively.
The resulting major product $A$ will have a phenolic $-OH$ group intact,a secondary $-Cl$ group,and a $-CH_2Cl$ group.
This corresponds to the structure shown in option $B$.

(A) Step $1$: Hydration of propene $(C_{3}H_{6})$ with $H^{+}/H_{2}O$ follows Markovnikov's rule to form propan$-2-$ol $(A)$:
$CH_{3}-CH=CH_{2} + H_{2}O \xrightarrow{H^{+}} CH_{3}-CH(OH)-CH_{3} (A)$
Step $2$: Propan$-2-$ol $(A)$ is a secondary alcohol with a methyl group attached to the carbinol carbon,which undergoes the iodoform reaction with $KIO/dil. KOH$ to form iodoform $(B)$ and potassium acetate $(C)$:
$CH_{3}-CH(OH)-CH_{3} + 4KIO \rightarrow CHI_{3} (B) + CH_{3}COOK (C) + 3KOH + H_{2}O$
Thus,$B$ is $CHI_{3}$ and $C$ is $CH_{3}COOK$.

(D) The reaction of acetone $(CH_3COCH_3)$ with ethylmagnesium bromide $(C_2H_5MgBr)$ in the presence of dry ether is a nucleophilic addition reaction.
The nucleophilic ethyl group $(C_2H_5^-)$ attacks the electrophilic carbonyl carbon of acetone to form an intermediate magnesium alkoxide.
Subsequent hydrolysis with $H_3O^+$ yields the final alcohol product.
The reaction sequence is:
$CH_3COCH_3 + C_2H_5MgBr \rightarrow CH_3-C(OMgBr)(CH_3)-C_2H_5$
$CH_3-C(OMgBr)(CH_3)-C_2H_5 + H_2O/H^+ \rightarrow CH_3-C(OH)(CH_3)-CH_2CH_3 + Mg(OH)Br$
The structure of the product is $CH_3-C(OH)(CH_3)-CH_2-CH_3$.
Selecting the longest carbon chain containing the $-OH$ group gives a butane chain. Numbering from the end closest to the $-OH$ group,the $-OH$ is at position $2$ and a methyl group is also at position $2$.
Thus,the $IUPAC$ name is $2-$methylbutan$-2-$ol.

(B) The reaction of $CH_3MgBr$ (excess) with phenyl acetate $(C_6H_5OCOCH_3)$ proceeds as follows:
$1$. The first equivalent of $CH_3MgBr$ attacks the carbonyl carbon of the ester,followed by the elimination of the phenoxide ion $(C_6H_5O^-)$ to form acetophenone $(C_6H_5COCH_3)$.
$2$. The second equivalent of $CH_3MgBr$ attacks the carbonyl carbon of acetophenone to form a tertiary alkoxide intermediate.
$3$. Upon hydrolysis,this intermediate yields $2$-phenylpropan-$2$-ol,which is a tertiary alcohol.
Therefore,phenyl acetate is the most appropriate compound to yield a tertiary alcohol as the primary organic product.