Properties of alcohol Questions in English

(C) $n$-Propyl alcohol $(CH_3CH_2CH_2OH)$ is a primary $(1^{\circ})$ alcohol,while isopropyl alcohol $(CH_3CH(OH)CH_3)$ is a secondary $(2^{\circ})$ alcohol.
When oxidized with potassium dichromate $(K_2Cr_2O_7)$ in acidic medium,primary alcohols are oxidized to aldehydes and then to carboxylic acids with the same number of carbon atoms.
Secondary alcohols are oxidized to ketones with the same number of carbon atoms,which are further oxidized to carboxylic acids with fewer carbon atoms.
Since the products formed are different,they can be distinguished by this method.

(B) $1$. The starting material is isopropyl alcohol $(CH_3)_2CHOH$.
$2$. Oxidation of a secondary alcohol with dilute oxidizing agents (like $KMnO_4$ or $CrO_3$) yields a ketone,which is acetone $(CH_3COCH_3)$. Thus,$X = CH_3COCH_3$.
$3$. The reaction of acetone with methylmagnesium bromide $(CH_3MgBr)$ followed by hydrolysis is a Grignard reaction.
$4$. $CH_3COCH_3 + CH_3MgBr \rightarrow (CH_3)_3COMgBr$.
$5$. Subsequent hydrolysis of $(CH_3)_3COMgBr$ yields tertiary butyl alcohol $(CH_3)_3COH$. Thus,$Y = (CH_3)_3COH$.

(D) Step $1$: $(CH_3)_2CH-OH + PBr_3 \rightarrow (CH_3)_2CH-Br$ ($X$ is isopropyl bromide).
Step $2$: $(CH_3)_2CH-Br + alc. KOH \rightarrow CH_3-CH=CH_2$ ($Y$ is propene).
Step $3$: $CH_3-CH=CH_2 + (i) H_2SO_4 / (ii) H_3O^+ \rightarrow CH_3-CH(OH)-CH_3$ (Isopropyl alcohol or propan-$2$-ol).
Since $CH_3-CH(OH)-CH_3$ is not among the given options,the correct answer is $D$.

(C) The given product is ethylene glycol diacetate,which has the structure $CH_2(OCOCH_3)-CH_2(OCOCH_3)$.
This compound is formed by the acetylation of ethylene glycol $(HO-CH_2-CH_2-OH)$ with an acetylating agent like acetic anhydride $((CH_3CO)_2O)$ or acetyl chloride $(CH_3COCl)$.
Looking at the options,option $C$ mentions glycol and $CH_3COCl$,which is a valid pathway for the synthesis of this ester.
Note: The structure provided in the question represents ethylene glycol diacetate,not a glycerol derivative.

(A) Isopropyl alcohol is a secondary alcohol with the structure $(CH_3)_2CHOH$.
Upon oxidation,secondary alcohols are converted into ketones.
The oxidation of isopropyl alcohol $(CH_3)_2CHOH$ using an oxidizing agent like $K_2Cr_2O_7/H^+$ yields acetone $(CH_3)_2C=O$.

(A) Ethylene glycol $(HOCH_2-CH_2OH)$ is a vicinal diol. When it is treated with a strong oxidizing agent like acidic $KMnO_4$,the primary alcohol groups are oxidized to carboxylic acid groups. The reaction proceeds as follows:
$HOCH_2-CH_2OH + 2[O] \xrightarrow{H^+, KMnO_4} HOOC-COOH + 2H_2O$.
The product formed is oxalic acid.

(C) The iodoform test is used to distinguish between alcohols containing the $CH_3CH(OH)-$ group and those that do not.
$C_2H_5OH$ (ethanol) contains the $CH_3CH(OH)-$ group,so it reacts with $I_2$ and $NaOH$ to form a yellow precipitate of iodoform $(CHI_3)$.
$CH_3OH$ (methanol) does not contain this group and does not give the iodoform test.
Therefore,the iodoform test is the correct method to distinguish between them.

(B) The reaction of alcohols with halogen acids $(HX)$ involves the cleavage of the $C-O$ bond.
The reactivity of halogen acids depends on the strength of the $H-X$ bond.
As the size of the halogen atom increases from $Cl$ to $I$,the bond dissociation energy of the $H-X$ bond decreases,making it easier to break.
Therefore,the order of reactivity is $HI > HBr > HCl$.

(A) Dehydrogenation of alcohols involves the removal of hydrogen atoms from the carbon atom bearing the hydroxyl group and the oxygen atom of the hydroxyl group.
Primary alcohols $(R-CH_2OH)$ undergo dehydrogenation to form aldehydes $(R-CHO)$.
Secondary alcohols $(R_2CHOH)$ undergo dehydrogenation to form ketones $(R_2C=O)$.
$Propan-2-ol$ $(CH_3-CH(OH)-CH_3)$ is a secondary alcohol.
Upon dehydrogenation,it loses $H_2$ to form $Propanone$ $(CH_3-CO-CH_3)$.

(B) Glycerol $(CH_2OH-CHOH-CH_2OH)$ reacts with excess $HI$ to undergo substitution and reduction reactions.
First,the hydroxyl groups are replaced by iodine to form $1,2,3-triiodopropane$ $(CH_2I-CHI-CH_2I)$.
This compound is unstable and loses $I_2$ to form allyl iodide $(CH_2=CH-CH_2I)$.
Allyl iodide can further react with $HI$ to form $1-iodopropane$ or $2-iodopropane$ $(CH_3-CHI-CH_3)$ via addition reactions.
However,the compound $CH_2OH-CHI-CH_2OH$ is not formed because the reaction proceeds through the complete substitution of all $-OH$ groups.

(B) The mechanism for the acid-catalyzed dehydration of alcohols to alkenes involves three steps:
$1$. Protonation of the alcohol molecule: The oxygen atom of the alcohol group gets protonated by the acid $(H_2SO_4)$ to form an alkyloxonium ion $(R-OH_2^+)$.
$2$. Formation of carbocation: The alkyloxonium ion loses a water molecule to form a carbocation.
$3$. Formation of alkene: The carbocation loses a proton to form an alkene.

(A) The rate of acid-catalyzed esterification of alcohols depends on the steric hindrance around the hydroxyl group.
As the degree of branching at the carbon atom attached to the $-OH$ group increases,the steric hindrance increases,which makes the attack of the carboxylic acid on the alcohol more difficult.
Therefore,the reactivity order is: Primary $(1^{\circ})$ > Secondary $(2^{\circ})$ > Tertiary $(3^{\circ})$.
In the given options:
$CH_3(CH_2)_2OH$ is a primary alcohol $(1^{\circ})$,
$CH_3CH_2CH(OH)CH_3$ is a secondary alcohol $(2^{\circ})$,
$(CH_3)_3C-OH$ is a tertiary alcohol $(3^{\circ})$.
Thus,the correct decreasing order is $CH_3(CH_2)_2OH > CH_3CH_2CH(OH)CH_3 > (CH_3)_3C-OH$.

(C) $1-$propanol is a primary alcohol $(CH_3CH_2CH_2OH)$,which on oxidation with hot $Cu$ gives an aldehyde (propanal,$CH_3CH_2CHO$).
$2-$propanol is a secondary alcohol $(CH_3CH(OH)CH_3)$,which on oxidation with hot $Cu$ gives a ketone (acetone,$CH_3COCH_3$).
Aldehydes give a positive test with Fehling's solution (red precipitate),whereas ketones do not.
Therefore,heating with $Cu$ followed by treatment with Fehling's solution distinguishes the two.

(C) The dehydration of alcohols with concentrated ${H_2SO_4}$ proceeds via an $E1$ mechanism,which involves the formation of a carbocation intermediate.
The rate of dehydration depends on the stability of the carbocation formed.
In the given options,all are secondary benzylic alcohols. The stability of the resulting benzylic carbocation is influenced by the substituent present on the benzene ring.
Electron-donating groups $(EDG)$ stabilize the carbocation,while electron-withdrawing groups $(EWG)$ destabilize it.
The methoxy group $(-OCH_3)$ is a strong electron-donating group due to its $+M$ effect,which significantly stabilizes the carbocation.
The nitro group $(-NO_2)$ and chlorine $(-Cl)$ are electron-withdrawing groups,which destabilize the carbocation.
Therefore,$p-CH_3O.C_6H_4.CH(OH)CH_3$ forms the most stable carbocation and undergoes dehydration most easily.

(A) The reaction of alcohols with $HBr$ proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate.
The rate of reaction depends on the stability of the carbocation formed.
$2$-Methylpropan-$2$-ol is a tertiary $(3^{\circ})$ alcohol,which forms a stable tertiary carbocation.
Propan-$2$-ol is a secondary $(2^{\circ})$ alcohol,while $2$-Methylpropan-$1$-ol and Propan-$1$-ol are primary $(1^{\circ})$ alcohols.
Since the stability order of carbocations is $3^{\circ} > 2^{\circ} > 1^{\circ}$,the tertiary alcohol ($2$-Methylpropan-$2$-ol) will react the fastest.

(B) The dehydration of cyclohexanol to cyclohexene is an elimination reaction ($E1$ mechanism).
Concentrated phosphoric acid $(H_3PO_4)$ is the preferred reagent for this dehydration because it is a non-oxidizing acid and minimizes side reactions like oxidation or substitution compared to concentrated sulfuric acid or hydrohalic acids.
Thus,heating cyclohexanol with concentrated $H_3PO_4$ yields cyclohexene efficiently.

(B) Phenylmagnesium bromide $(C_6H_5MgBr)$ is a Grignard reagent,which acts as a strong base.
Methanol $(CH_3OH)$ contains an acidic proton attached to the oxygen atom.
The reaction proceeds as an acid-base reaction where the Grignard reagent abstracts the acidic proton from methanol:
$C_6H_5MgBr + CH_3OH \rightarrow C_6H_6 + Mg(OCH_3)Br$
Here,$C_6H_6$ is benzene and $Mg(OCH_3)Br$ is bromomagnesium methoxide.

(C) The Assertion is true because $CaCl_2$ reacts with alcohols and $NH_3$ to form addition compounds (e.g.,$CaCl_2 \cdot 4C_2H_5OH$ and $CaCl_2 \cdot 8NH_3$),making it unsuitable as a drying agent for these substances.
The Reason is false because $CaCl_2$ is actually a very effective and commonly used desiccant for many other gases and solvents.

(B) In the petroleum industry,zeolites are used to convert alcohols directly into hydrocarbons by dehydration.
Zeolites are indeed porous catalysts,which allows them to act as shape-selective catalysts.
However,the dehydration of alcohols to hydrocarbons is specifically due to the acidic nature of the zeolites,not just their porosity.
Therefore,both statements are correct,but the Reason is not the correct explanation for the Assertion.

(D) The reaction involves an aqueous solution of $KOH$,which acts as a source of $OH^-$ ions.
In the presence of $aq. KOH$,the nucleophilic substitution reaction takes place where the chloride ion $(-Cl)$ is replaced by the hydroxyl group $(-OH)$.
Therefore,the reaction is: $CH_3CH(Cl)CH_2CH_2OH + KOH(aq) \rightarrow CH_3CH(OH)CH_2CH_2OH + KCl$.
The major product is $CH_3CH(OH)CH_2CH_2OH$.

(B) $1$. $CH_3CH_2CH_2Br$ reacts with aqueous $NaOH$ (nucleophilic substitution) to form propan-$1$-ol $(X)$: $CH_3CH_2CH_2Br + NaOH(aq) \rightarrow CH_3CH_2CH_2OH + NaBr$.
$2$. Propan-$1$-ol $(X)$ undergoes dehydration with $Al_2O_3$ at high temperature to form propene $(Y)$: $CH_3CH_2CH_2OH \xrightarrow{Al_2O_3, \Delta} CH_3CH=CH_2 + H_2O$.
$3$. Propene $(Y)$ reacts with $Cl_2/H_2O$ (hypochlorous acid,$HOCl$) to form chlorohydrin $(Z)$: $CH_3CH=CH_2 + Cl_2 + H_2O \rightarrow CH_3CH(OH)CH_2Cl + HCl$. The hydroxyl group attaches to the more substituted carbon (Markovnikov addition).

(C) The formation of diethyl ether from ethanol $(C_2H_5OH)$ in the presence of concentrated $H_2SO_4$ at $413 \ K$ $(140^\circ C)$ is an example of an intermolecular dehydration reaction.
$2 \ C_2H_5OH \xrightarrow{H_2SO_4, 413 \ K} C_2H_5-O-C_2H_5 + H_2O$

(A) $1$. The molecular formula $C_4H_{10}O$ corresponds to an alcohol.
$2$. Lucas reagent $(conc. HCl + ZnCl_2)$ reacts with tertiary alcohols instantaneously at room temperature to form alkyl chlorides.
$3$. Compound $B$ on heating with alcoholic $KOH$ (dehydrohalogenation) gives isobutene $(CH_3-C(CH_3)=CH_2)$.
$4$. The structure of isobutene indicates that the precursor alkyl chloride $(B)$ must be $2-$chloro$-2-$methylpropane.
$5$. Therefore,compound $A$ must be $2-$methylpropan$-2-$ol ($tert-$butyl alcohol).
$6$. The reaction is: $(CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O$.
$7$. Thus,$A$ is $2-$methylpropan$-2-$ol and $B$ is $2-$chloro$-2-$methylpropane.

(D) The dehydration of alcohols to form ethers using a trace of sulphuric acid follows an $S_N2$ mechanism.
Primary alcohols are the least sterically hindered and therefore undergo $S_N2$ substitution most readily to form ethers.
Secondary and tertiary alcohols are more prone to elimination reactions (forming alkenes) due to steric hindrance and the stability of the carbocation intermediate.
Among the given options,$1-$Pentanol is a primary alcohol,while the others are secondary or tertiary alcohols.
Therefore,$1-$Pentanol gives the best yield of dialkyl ether.

(D) The acidity of an alcohol depends on the stability of the conjugate base (alkoxide ion) formed after the loss of a proton $(H^+)$.
In $ClCH_2CH_2OH$,the chlorine atom exerts a $-I$ (negative inductive) effect,which is electron-withdrawing.
This $-I$ effect helps in dispersing the negative charge on the oxygen atom of the alkoxide ion $(ClCH_2CH_2O^-)$,thereby increasing its stability compared to the ethoxide ion $(CH_3CH_2O^-)$.
Since the conjugate base of $ClCH_2CH_2OH$ is more stable,it is a stronger acid than $CH_3CH_2OH$.

(A) In the Victor-Meyer's test,the alcohols are converted into nitroalkanes,which are then treated with nitrous acid $(HNO_2)$ and finally made alkaline with $NaOH$.
For $1^o$ alcohols,the product is a nitrolic acid,which gives a blood-red colour in alkaline solution.
For $2^o$ alcohols,the product is a pseudonitrol,which gives a blue colour in alkaline solution.
For $3^o$ alcohols,the product does not react with nitrous acid and remains colourless in alkaline solution.
Therefore,the colours are red,blue,and colourless respectively.

(A) The given order of dehydration is correct due to the following reasons:
$(i)$ Alcohols that lead to the formation of conjugated alkenes are dehydrated more readily because the resulting product is more stable.
$(ii)$ $2-Cyclohexenol$ is dehydrated more easily than $3-cyclohexenol$ because the carbocation formed from the former is stabilized by resonance with the double bond (allylic carbocation),making it more stable than the carbocation from the latter.
$(iii)$ Phenol is not dehydrated under standard conditions due to the high stability of the $C-O$ bond resulting from resonance,which gives it partial double-bond character.

(B) Tertiary alkyl halides like $CH_3-C(CH_3)_2-Cl$ undergo elimination $(E2)$ when treated with $NaCN$ (which acts as a base) to form isobutylene $(CH_2=C(CH_3)_2)$.
Subsequent hydration with $dil. H_2SO_4$ follows Markovnikov's rule to yield tert-butyl alcohol $(CH_3-C(CH_3)_2-OH)$.
Reaction:
$CH_3-C(CH_3)_2-Cl$ $\xrightarrow{NaCN} CH_2=C(CH_3)_2 (A)$ $\xrightarrow{dil. H_2SO_4} CH_3-C(CH_3)_2-OH (B)$

(C) When vapours of a secondary alcohol are passed over heated copper at $573 \; K$,dehydrogenation occurs to form a ketone.
Reaction: $CH_3-CH(OH)-CH_3 \xrightarrow{Cu / 573 \; K} CH_3-CO-CH_3 + H_2$
Thus,a $2^o$ alcohol yields a ketone.

(C) The haloform reaction is given by compounds containing the $CH_3-CO-$ group or the $CH_3-CH(OH)-$ group.
Compound $A$ with molecular formula $C_8H_{10}O$ that contains the $CH_3-CH(OH)-$ group is $1-phenylethanol$ $(C_6H_5-CH(OH)-CH_3)$.
$NaOI$ is prepared by the reaction of $I_2$ with $NaOH$.
Therefore,$A$ is $1-phenylethanol$ and $Y$ is $I_2$.

(D) The reaction sequence is as follows:
$1$. Compound $A$ is ethanol $(C_{2}H_{5}OH)$.
$2$. When $A$ $(C_{2}H_{5}OH)$ reacts with $Na$,it forms sodium ethoxide $(B = C_{2}H_{5}ONa)$.
$3$. When $A$ $(C_{2}H_{5}OH)$ reacts with $PCl_{5}$,it forms ethyl chloride $(C = C_{2}H_{5}Cl)$.
$4$. $B$ $(C_{2}H_{5}ONa)$ and $C$ $(C_{2}H_{5}Cl)$ react via Williamson's synthesis to produce diethyl ether $(C_{2}H_{5}-O-C_{2}H_{5})$.
Therefore,the correct order is $A = C_{2}H_{5}OH, B = C_{2}H_{5}ONa, C = C_{2}H_{5}Cl$.

(A) The boiling point of a compound depends on the intermolecular forces present.
Alcohols possess $-OH$ groups,which allow for extensive intermolecular hydrogen bonding,leading to significantly higher boiling points compared to ethers,which lack such hydrogen bonding.
Compound $A$ is $1,3,5-\text{tris(hydroxymethyl)cyclohexane}$,which contains three $-OH$ groups,facilitating strong intermolecular hydrogen bonding.
Compound $B$ is $1,3,5-\text{trimethoxycyclohexane}$,which is an ether and lacks hydrogen bonding.
Therefore,$A$ has a higher boiling point than $B$.

(N/A)

Structures of Alcohols	$IUPAC$ Name
$(i) \ CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}OH$	Pentan$-1-$ol
$(ii) \ CH_{3}CH(OH)CH_{2}CH_{2}CH_{3}$	Pentan$-2-$ol
$(iii) \ CH_{3}CH_{2}CH(OH)CH_{2}CH_{3}$	Pentan$-3-$ol
$(iv) \ CH_{3}CH(CH_{3})CH_{2}CH_{2}OH$	$3-$Methylbutan$-1-$ol
$(v) \ CH_{3}CH_{2}CH(CH_{3})CH_{2}OH$	$2-$Methylbutan$-1-$ol
$(vi) \ CH_{3}C(OH)(CH_{3})CH_{2}CH_{3}$	$2-$Methylbutan$-2-$ol
$(vii) \ (CH_{3})_{3}CCH_{2}OH$	$2,2-$Dimethylpropan$-1-$ol
$(viii) \ CH_{3}CH(CH_{3})CH(OH)CH_{3}$	$3-$Methylbutan$-2-$ol

(N/A) $(i)$ Phenol $(C_{6}H_{5}OH)$ contains a polar $-OH$ group and a non-polar phenyl ring. It is partially soluble in water.
$(ii)$ Toluene $(C_{6}H_{5}CH_{3})$ is a non-polar hydrocarbon. It is insoluble in water.
$(iii)$ Formic acid $(HCOOH)$ can form strong hydrogen bonds with water. It is highly soluble in water.
$(iv)$ Ethylene glycol $(HOCH_{2}CH_{2}OH)$ has two polar $-OH$ groups and forms extensive hydrogen bonds. It is highly soluble in water.
$(v)$ Chloroform $(CHCl_{3})$ is a non-polar organic solvent. It is insoluble in water.
$(vi)$ Pentanol $(C_{5}H_{11}OH)$ has a polar $-OH$ group but a long hydrophobic alkyl chain. It is partially soluble in water.

(N/A) In the presence of sulphuric acid $(H_2SO_4)$,$KI$ produces $HI$.
$2 KI + H_2SO_4 \longrightarrow 2 KHSO_4 + 2 HI$
Since $H_2SO_4$ is a strong oxidizing agent,it oxidizes the $HI$ produced in the reaction to $I_2$.
$2 HI + H_2SO_4 \longrightarrow I_2 + SO_2 + 2 H_2O$
As a result,the $HI$ required for the conversion of alcohol to alkyl iodide is consumed,and the reaction cannot proceed.
Therefore,sulphuric acid is not used; instead,a non-oxidizing acid such as phosphoric acid $(H_3PO_4)$ is used.

(N/A) $(i)$ $CH_{3}CH_{2}CH_{2}CH_{2}OH$ (Butan$-1-$ol) $\xrightarrow{HCl + ZnCl_{2}}$ No reaction at room temperature. Primary alcohols do not react with Lucas' reagent at room temperature.
$(ii)$ $CH_{3}CH_{2}C(OH)(CH_{3})_{2}$ ($2$-Methylbutan$-2-$ol,$3^{\circ}$) $\xrightarrow{HCl + ZnCl_{2}}$ $CH_{3}CH_{2}C(Cl)(CH_{3})_{2}$ ($2$-Chloro$-2-$methylbutane) + $H_{2}O$. Tertiary alcohols react immediately with Lucas' reagent to form white turbidity.
$(b)$
$(i)$ $CH_{3}CH_{2}CH_{2}CH_{2}OH + HBr \to CH_{3}CH_{2}CH_{2}CH_{2}Br$ ($1$-Bromobutane) + $H_{2}O$.
$(ii)$ $CH_{3}CH_{2}C(OH)(CH_{3})_{2} + HBr \to CH_{3}CH_{2}C(Br)(CH_{3})_{2}$ ($2$-Bromo$-2-$methylbutane) + $H_{2}O$.
$(c)$
$(i)$ $CH_{3}CH_{2}CH_{2}CH_{2}OH + SOCl_{2} \to CH_{3}CH_{2}CH_{2}CH_{2}Cl$ ($1$-Chlorobutane) + $SO_{2} + HCl$.
$(ii)$ $CH_{3}CH_{2}C(OH)(CH_{3})_{2} + SOCl_{2} \to CH_{3}CH_{2}C(Cl)(CH_{3})_{2}$ ($2$-Chloro$-2-$methylbutane) + $SO_{2} + HCl$.

(N/A) $(i)$ Acid-catalyzed dehydration of $1$-methylcyclohexanol proceeds via the formation of a tertiary carbocation. The loss of a proton from the adjacent carbon leads to the formation of $1$-methylcyclohexene as the major product.
$(ii)$ Acid-catalyzed dehydration of butan-$1$-ol proceeds via the formation of a primary carbocation,which undergoes rearrangement to a more stable secondary carbocation. Subsequent loss of a proton yields but-$2$-ene as the major product.
Reaction for $(i)$:
$1$-methylcyclohexanol $\xrightarrow{H^+}$ $1$-methylcyclohexene $+ H_2O$
Reaction for $(ii)$:
$CH_3CH_2CH_2CH_2OH \xrightarrow{H^+} CH_3CH=CHCH_3$ (but-$2$-ene,major product) $+ H_2O$

(N/A) Propanol $(CH_3CH_2CH_2OH)$ undergoes intermolecular hydrogen bonding due to the presence of the polar $-OH$ group.
On the other hand,butane $(C_4H_{10})$ is a non-polar hydrocarbon and only exhibits weak van der Waals forces.
Because of the strong intermolecular hydrogen bonds in propanol,extra energy is required to overcome these forces during the phase transition.
Therefore,propanol has a significantly higher boiling point than butane.

(N/A) Alcohols are capable of forming intermolecular hydrogen bonds with water molecules due to the presence of the polar $-OH$ group.
In contrast,hydrocarbons are non-polar and cannot form hydrogen bonds with water.
As a result,alcohols are significantly more soluble in water than hydrocarbons of comparable molecular masses.

(N/A) The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:
Step $1$: Protonation of ethanol to form ethyl oxonium ion:
$CH_3-CH_2-OH + H^{+} \rightleftharpoons CH_3-CH_2-OH_2^{+}$
Step $2$: Formation of carbocation (rate-determining step):
$CH_3-CH_2-OH_2^{+} \rightarrow CH_3-CH_2^{+} + H_2O$
Step $3$: Elimination of a proton to form ethene:
$CH_3-CH_2^{+} \rightarrow CH_2=CH_2 + H^{+}$
The acid consumed in step $1$ is released in step $3$. After the formation of ethene,it is removed to shift the equilibrium in a forward direction.

(N/A) $(i)$ Acidified potassium permanganate $(KMnO_4/H^+)$ or acidified potassium dichromate $(K_2Cr_2O_7/H^+)$.
$(ii)$ Pyridinium chlorochromate $(PCC)$ in dichloromethane $(CH_2Cl_2)$.
$(iii)$ Bromine water $(Br_2/H_2O)$.
$(iv)$ Acidified potassium permanganate $(KMnO_4/H^+)$.
$(v)$ Concentrated phosphoric acid $(H_3PO_4)$ or concentrated sulfuric acid $(H_2SO_4)$ at $440 \ K$.
$(vi)$ Sodium borohydride $(NaBH_4)$ or Lithium aluminium hydride $(LiAlH_4)$.

(N/A) Ethanol $(C_2H_5OH)$ undergoes intermolecular hydrogen bonding due to the presence of the polar $-OH$ group,which leads to the association of molecules.
Extra energy is required to break these hydrogen bonds during the phase transition from liquid to gas.
On the other hand,methoxymethane $(CH_3OCH_3)$ is an ether and does not possess an $-OH$ group,so it does not undergo intermolecular hydrogen bonding.
Hence,the boiling point of ethanol is significantly higher than that of methoxymethane.

(A) The molecular masses of the given compounds are in the range $44$ to $46 \ g/mol$.
$CH_3CH_2OH$ (ethanol) exhibits extensive intermolecular $H$-bonding,leading to molecular association,which results in the highest boiling point.
$CH_3CHO$ (acetaldehyde) is more polar than $CH_3OCH_3$ (methoxy methane),resulting in stronger dipole-dipole interactions.
$CH_3CH_2CH_3$ (propane) is a non-polar alkane with only weak van der Waals forces.
Therefore,the increasing order of boiling points is: $CH_3CH_2CH_3 < CH_3OCH_3 < CH_3CHO < CH_3CH_2OH$.

(N/A) $1$. In alcohols and phenols,the oxygen atom of the $-OH$ group is highly electronegative,which creates a partial negative charge $(\delta^-)$ on the oxygen and a partial positive charge $(\delta^+)$ on the hydrogen atom.
$2$. This polarity leads to the formation of intermolecular hydrogen bonds between the hydrogen atom of one molecule and the oxygen atom of another molecule.
$3$. Effect on Physical Properties: Due to the presence of these intermolecular hydrogen bonds,alcohols and phenols exhibit higher boiling points compared to alkanes,ethers,or haloalkanes of comparable molecular masses.
$4$. Additionally,these hydrogen bonds allow alcohols and phenols to form hydrogen bonds with water molecules,which explains their solubility in water.

(N/A) The boiling points of alcohols and phenols increase with an increase in the number of carbon atoms due to an increase in the strength of van der Waals forces.
For example,the boiling points follow the order: $CH_3OH < C_2H_5OH < C_3H_7OH < C_4H_9OH$.
In alcohols,as the branching in the carbon chain increases,the boiling point decreases because the surface area decreases,which leads to a decrease in the strength of van der Waals forces.
For example,the boiling points follow the order: $CH_3CH_2CH_2CH_2OH > (CH_3)_2CHCH_2OH > (CH_3)_3COH$.

(N/A) The solubility of alcohols and phenols in water is due to their ability to form hydrogen bonds with water molecules.
The solubility of alcohols and phenols decreases as the size of the alkyl or aryl group increases.
Lower molecular weight alcohols are miscible with water in all proportions. The solubility increases with an increase in the number of $-OH$ groups.

(N/A) Alcohol compounds exhibit three main types of reactions:
$1$. Reactions involving the cleavage of $O-H$ bond: In these reactions,alcohols act as nucleophiles. For example,the reaction of alcohols with carboxylic acids,acid chlorides,or acid anhydrides to form esters.
$R-OH + R'COOH \xrightarrow{H^+} R-COOR' + H_2O$
$2$. Reactions involving the cleavage of $C-O$ bond: In these reactions,the $C-O$ bond breaks,and the alcohol acts as an electrophile. For example,the reaction with hydrogen halides $(HX)$ to form alkyl halides.
$R-OH + HX \rightarrow R-X + H_2O$
$3$. Reactions involving both alkyl and hydroxyl groups (Oxidation and Dehydrogenation): These reactions involve the cleavage of both $O-H$ and $C-H$ bonds. For example,the oxidation of primary alcohols to aldehydes and then to carboxylic acids.
$R-CH_2OH$ $\xrightarrow{[O]} R-CHO$ $\xrightarrow{[O]} R-COOH$

(N/A) Reaction of alcohols with active metals: Like water,alcohols react with active metals such as $Li, Na, K, Al$ to form corresponding alkoxides with the evolution of hydrogen gas. Examples are as follows:
$2 ROH + 2 Na \longrightarrow 2 R-O-Na + H_2(g)$
$2 CH_3OH + 2 Na \longrightarrow 2 CH_3ONa + H_2(g)$
$2 CH_3CH_2OH + Mg \longrightarrow (CH_3CH_2O)_2Mg + H_2(g)$
$6 (CH_3)_3C-OH + 2 Al \longrightarrow 2 [(CH_3)_3C-O]_3Al + 3 H_2(g)$
These reactions involve the cleavage of the $O-H$ bond.
$(b)$ Acidic nature of alcohols: The above reactions show that alcohols are acidic in nature. In fact,alcohols are $Br\ddot{o}nsted$ acids and can donate a proton to a strong base $(B:)$.
$B: + H-\ddot{O}-R \longrightarrow B-H + :\ddot{O}-R^-$