Properties of alcohol Questions in English

(C) Viscosity is directly proportional to the extent of intermolecular hydrogen bonding.
Among the given options,$Glycerol$ $(CH_2OH-CHOH-CH_2OH)$ has the highest viscosity because it contains $3$ $-OH$ groups,allowing for extensive hydrogen bonding.
In comparison,$Ethylene \ glycol$ has $2$ $-OH$ groups,while $Ethanol$ and $Methanol$ each have only $1$ $-OH$ group.

(B) The reaction of a Grignard reagent with a carbonyl compound followed by hydrolysis produces an alcohol.
$Z$ reacts with conc. $HCl$ at room temperature to give immediate turbidity,which is the characteristic test for a tertiary $(3^{\circ})$ alcohol (Lucas test).
Among the options,the reaction of $CH_3MgBr$ $(X)$ with $CH_3COCH_3$ $(Y)$ produces $tert$-butyl alcohol,which is a tertiary alcohol.
The reaction is: $CH_3MgBr + CH_3COCH_3$ $\rightarrow (CH_3)_3COMgBr$ $\xrightarrow{H_3O^+} (CH_3)_3COH$.
Thus,the correct option is $B$.

(B) The acidic strength of alcohols depends on the stability of the conjugate base (alkoxide ion) formed after the release of a proton $(H^+)$.
Alkyl groups are electron-donating ($+I$ effect). As the number of alkyl groups attached to the carbon bearing the $-OH$ group increases,the electron density on the oxygen atom increases due to the $+I$ effect,which destabilizes the alkoxide ion.
Therefore,the stability of the alkoxide ion follows the order: Primary $(1^{\circ})$ > Secondary $(2^{\circ})$ > Tertiary $(3^{\circ})$.
Consequently,the acidic strength follows the same order: Primary $(I)$ > Secondary $(II)$ > Tertiary $(III)$.
Thus,the correct order is $(I) > (II) > (III)$.

(C) For the molecular formula $C_5H_{12}O$,the possible alcohols are:
$1^{\circ}$ Alcohols: $CH_3CH_2CH_2CH_2CH_2OH$ (pentan$-1-$ol),$CH_3CH_2CH_2CH(CH_3)OH$ (is incorrect,this is $2^{\circ}$),$CH_3CH(CH_3)CH_2CH_2OH$ ($3$-methylbutan$-1-$ol),$CH_3CH_2CH(CH_3)CH_2OH$ ($2$-methylbutan$-1-$ol),and $(CH_3)_3CCH_2OH$ ($2$,$2$-dimethylpropan$-1-$ol). Total $1^{\circ} = 4$.
$2^{\circ}$ Alcohols: $CH_3CH_2CH_2CH(OH)CH_3$ (pentan$-2-$ol),$CH_3CH_2CH(OH)CH_2CH_3$ (pentan$-3-$ol),and $CH_3CH(CH_3)CH(OH)CH_3$ ($3$-methylbutan$-2-$ol). Total $2^{\circ} = 3$.
$3^{\circ}$ Alcohols: $CH_3CH_2C(CH_3)(OH)CH_3$ ($2$-methylbutan$-2-$ol). Total $3^{\circ} = 1$.
Thus,the count is $4, 3, 1$.

(A) The reaction of $CH_3-CHO$ (acetaldehyde) with $C_2H_5MgBr$ (ethylmagnesium bromide) followed by hydrolysis yields $CH_3-CH(OH)-C_2H_5$ (butan$-2-$ol).
$CH_3-CHO + C_2H_5MgBr$ $\rightarrow CH_3-CH(OMgBr)-C_2H_5$ $\xrightarrow{H_2O} CH_3-CH(OH)-C_2H_5$.
Butan$-2-$ol is a secondary alcohol.
$(A)$ Secondary alcohols undergo dehydration with $20 \% H_3PO_4$ at $358 \ K$ to form alkenes. This is a correct statement.
$(B)$ Oxidation of secondary alcohols with $CrO_3$ gives a ketone (butanone). This is also a correct statement.
$(C)$ Butan$-2-$ol contains the $CH_3-CH(OH)-$ group,so it gives a positive iodoform test. Thus,this statement is incorrect.
$(D)$ It is not a vinylic alcohol.
Note: In many competitive contexts,$(A)$ is considered the primary characteristic reaction for this specific alcohol under these conditions.

(C) The reaction of alcohol with thionyl chloride $(SOCl_2)$ is the preferred method for preparing alkyl chlorides because the by-products formed ($SO_2$ and $HCl$) are gases.
Since these by-products escape into the atmosphere,the resulting alkyl chloride is obtained in a pure state.
The reaction is: $R-OH + SOCl_2 \rightarrow R-Cl + SO_2(g) + HCl(g)$.

(B) $1$. Dehydration of $X$ $(C_4H_{10}O)$ gives $C_4H_8$ (but$-2-$ene as major product). This implies $X$ is butan$-2-$ol.
$2$. Bromination of but$-2-$ene gives $2,3-$dibromobutane.
$3$. Treatment of $2,3-$dibromobutane with $Y$ $(NaNH_2)$ causes double dehydrohalogenation to form but$-2-$yne $(CH_3-C \equiv C-CH_3)$.
$4$. But$-2-$yne is an internal alkyne and does not have an acidic hydrogen,so it does not react with sodium metal.
$5$. Thus,$X$ is butan$-2-$ol and $Y$ is $(i) alc. KOH, (ii) NaNH_2$.

(B) $1$. Alcohol $X$ $(C_5H_{12}O)$ undergoes dehydration to form alkene $Y$ (major product).
$2$. $Y$ reacts with $HBr$ to form $Z$ ($C_5H_{11}Br$,major product) via Markovnikov addition.
$3$. $Z$ undergoes nucleophilic substitution in two steps,which is characteristic of an $S_N1$ mechanism. This implies $Z$ is a tertiary alkyl halide.
$4$. Among the options,$2$-methylbutan-$2$-ol is a tertiary alcohol. Its dehydration gives $2$-methylbut-$2$-ene as the major product (Saytzeff product).
$5$. Reaction of $2$-methylbut-$2$-ene with $HBr$ gives $2$-bromo-$2$-methylbutane $(Z)$ as the major product,which is a tertiary alkyl halide and follows the $S_N1$ mechanism (two steps).
$6$. Thus,$X$ is $2$-methylbutan-$2$-ol and $Y$ is $2$-methylbut-$2$-ene.

(B) The alcohol $X$ $(C_4H_{10}O)$ reacts with concentrated $HCl$ at room temperature to form $Y$ $(C_4H_9Cl)$. This indicates that $X$ is a tertiary alcohol,specifically $2$-methylpropan-$2$-ol (tert-butyl alcohol),because tertiary alcohols react rapidly with $HCl$ at room temperature (Lucas test).
The reaction is:
$(CH_3)_3C-OH + HCl \rightarrow (CH_3)_3C-Cl + H_2O$
When $X$ ($2$-methylpropan-$2$-ol) is passed over heated copper at $573 \ K$,it undergoes dehydration to form an alkene,$2$-methylpropene $(Z)$:
$(CH_3)_3C-OH \xrightarrow{Cu, 573 \ K} CH_3-C(CH_3)=CH_2 + H_2O$
Therefore,$Z$ is $2$-methylpropene.

(A) The conversion of propene $(CH_3-CH=CH_2)$ to $1$-chloropropane $(CH_3-CH_2-CH_2-Cl)$ requires an anti-Markovnikov hydration followed by the conversion of the primary alcohol to an alkyl chloride.
Step $1$: Hydroboration-oxidation of propene using $(i) (BH_3)_2$ and $(ii) H_2O_2 / OH^-$ yields propan-$1$-ol $(CH_3-CH_2-CH_2-OH)$.
Step $2$: The conversion of propan-$1$-ol to $1$-chloropropane is typically achieved using $HCl$ in the presence of a catalyst like $ZnCl_2$ (Lucas reagent) or other chlorinating agents. Although primary alcohols react slowly with Lucas reagent,the sequence provided in option $A$ represents the standard synthetic route for this transformation.

(A) The Lucas test uses conc. $HCl$ and $ZnCl_2$ to distinguish between $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ alcohols.
$3^{\circ}$ alcohols give immediate turbidity,$2^{\circ}$ alcohols give turbidity within $5-10$ minutes,and $1^{\circ}$ alcohols do not give turbidity at room temperature.
Since alcohol $X$ $(C_4H_{10}O)$ does not give turbidity at room temperature,it must be a $1^{\circ}$ alcohol.
Among the options,$n$-butanol is a $1^{\circ}$ alcohol.
$n$-butanol reacts with $PCC$ (Pyridinium chlorochromate) to form butanal $(Z)$.
Therefore,$X = n$-butanol,$Y = PCC$,and $Z = \text{butanal}$.

(B) The dehydration of alcohols to alkenes requires specific conditions depending on the degree of substitution of the alcohol.
Primary $(1^{\circ})$ alcohols,like butan$-1-$ol in option $(B)$,are the least reactive towards dehydration and typically require concentrated acid (e.g.,$95\% \ H_2SO_4$) and high temperatures (around $443 \ K$) to proceed.
Option $(B)$ suggests the dehydration of a primary alcohol at a very low temperature $(300 \ K)$ with a lower concentration of acid $(75\% \ H_2SO_4)$,which is insufficient to drive the elimination reaction.
Secondary and tertiary alcohols are more reactive and can undergo dehydration under milder conditions,making options $(C)$ and $(D)$ feasible.

(A) The reaction of ethanol with $H_2SO_4$ at $413 \ K$ leads to the formation of ethoxyethane (diethyl ether).
This process is an intermolecular dehydration reaction.
The mechanism involves the protonation of one ethanol molecule followed by the nucleophilic attack of another ethanol molecule on the protonated species,which follows the $S_N2$ pathway.

(D) The reaction sequence is as follows:
$1$. $1-$phenylethanol reacts with $HBr$ to form $1-$bromo$-1-$phenylethane,which then undergoes dehydrohalogenation with alcoholic $KOH$ to yield styrene $(C_6H_5CH=CH_2)$.
$2$. Styrene undergoes hydroboration-oxidation (using $B_2H_6$ followed by $NaOH/H_2O_2$) to produce $2-$phenylethanol $(C_6H_5CH_2CH_2OH)$ as the major anti-Markovnikov product.
Therefore,the major product '$B$' is $2-$phenylethanol.

(B) When tertiary butanol ($3^{\circ}$ alcohol) is passed over heated $Cu$ at $573 \ K$,isobutene is formed through an elimination reaction.
Since a water molecule is eliminated from the alcohol,this process is specifically classified as a dehydration reaction.
In contrast,$1^{\circ}$ alcohols undergo dehydrogenation to form aldehydes,and $2^{\circ}$ alcohols undergo dehydrogenation to form ketones under the same conditions.

(D) Lucas test is used to distinguish between primary,secondary,and tertiary alcohols.
Lucas reagent is a solution of anhydrous zinc chloride $(ZnCl_2)$ in concentrated hydrochloric acid $(HCl)$.
Lucas reagent converts alcohols into alkyl chlorides.
Tertiary alcohols react immediately with Lucas reagent,secondary alcohols react within $5 \ \text{minutes}$,while primary alcohols do not react at room temperature.

(B) Secondary alcohols like $CH_3-CH(OH)-CH_2-CH_3$ (butan$-2-$ol) on oxidation with strong oxidizing agents like alkaline $KMnO_4$ undergo oxidation to form ketones.
Specifically,butan$-2-$ol is oxidized to butanone $(CH_3-CO-CH_2-CH_3)$.
Primary alcohols like butan$-1-$ol $(CH_3-CH_2-CH_2-CH_2OH)$ are oxidized to carboxylic acids (butanoic acid) under these conditions.
Hence,the correct option is $(B)$.

(A) The dehydration of alcohols depends on the degree of substitution of the alcohol:
$1$. Primary $(1^{\circ})$ alcohols like ethanol undergo dehydration using concentrated $H_2SO_4$ at $443 \ K$.
$2$. Secondary $(2^{\circ})$ alcohols like propan$-2-$ol undergo dehydration using $85 \% \ H_3PO_4$ at $440 \ K$.
$3$. Tertiary $(3^{\circ})$ alcohols like tert-butyl alcohol undergo dehydration using $20 \% \ H_3PO_4$ at $358 \ K$.
Therefore,$X = H_2SO_4, 443 \ K$; $Y = 85 \% \ H_3PO_4, 440 \ K$; $Z = 20 \% \ H_3PO_4, 358 \ K$.

(C) When the vapours of a primary alcohol are passed over heated copper at $573 \ K$,dehydrogenation takes place and an aldehyde is formed: $R-CH_2OH \xrightarrow{Cu/573 \ K} R-CHO + H_2$.
In the case of tertiary alcohols,dehydration takes place to form an alkene: $(CH_3)_3C-OH \xrightarrow{Cu/573 \ K} (CH_3)_2C=CH_2 + H_2O$.
Thus,$X$ is an aldehyde $(RCHO)$ and $Y$ is an alkene $((CH_3)_2C=CH_2)$.

(B) In the reaction of alcohols with $PCl_5$,the hydroxyl group $(-OH)$ is replaced by a chlorine atom to form an alkyl chloride $(R-Cl)$.
The side products are phosphorus oxychloride $(POCl_3)$ and hydrogen chloride $(HCl)$.
The balanced chemical equation is: $R-OH + PCl_5 \rightarrow R-Cl + POCl_3 + HCl$.
Thus,$X = R-Cl$ and $Y = POCl_3$.

(B) The given reaction is the iodoform test for ethanol.
Ethanol $(C_2H_5OH)$ reacts with iodine $(I_2)$ in the presence of a base like sodium carbonate $(Na_2CO_3)$ to form iodoform $(CHI_3)$,which is also known as triiodomethane.
The balanced chemical equation is:
$C_2H_5OH + 4I_2 + 3Na_2CO_3 \longrightarrow CHI_3 + HCOONa + 5NaI + 3CO_2 + 2H_2O$
Therefore,'$X$' is $CHI_3$ (triiodomethane).

(A) In step-$1$,$CH_3 CH_2 OH$ is converted to $CH_3 CHO$ by $Cl_2$. This is an oxidation reaction where the primary alcohol group $(-CH_2 OH)$ is oxidized to an aldehyde group $(-CHO)$.
In step-$2$,$CH_3 CHO$ reacts with $3 Cl_2$ to form $Cl_3 C CHO$. This is a chlorination reaction where the hydrogen atoms of the methyl group are substituted by chlorine atoms.

(D) The reagent used for the conversion of allyl alcohol into propenal is $C_5H_5NH^{+} CrO_3Cl^{-}$ [$PCC$ (pyridinium chlorochromate)].
$CH_2=CH-CH_2-OH \xrightarrow{PCC} CH_2=CH-CHO$
(Allyl alcohol) (Propenal)
Here,the reagent used must selectively oxidize the primary alcohol group $(-CH_2OH)$ to an aldehyde $(-CHO)$ without affecting the carbon-carbon double bond. $PCC$ is a mild oxidizing agent that performs this specific transformation.

(C) positive iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
These groups are oxidized to the $CH_3CO-$ group in the presence of iodine and sodium hydroxide,which then reacts to form a pale yellow precipitate of iodoform $(CHI_3)$.
In the provided structures,compounds $(i)$ and $(ii)$ contain the $CH_3CH(OH)-$ moiety,which is capable of undergoing the iodoform reaction.
Therefore,these compounds will give a positive iodoform test.

(A) In alcohols,the oxygen atom is $sp^3$ hybridized.
Due to the presence of two lone pairs on the oxygen atom,the bond angle is slightly less than the ideal tetrahedral angle of $109.5^{\circ}$ $(109^{\circ} 28')$.
Therefore,the bond angle in alcohols is close to $109^{\circ}$.

(D) The general formula of a Grignard reagent is $R-Mg-X$,where $R$ is an alkyl or aryl group and $X$ is a halogen.
It acts as a nucleophile due to the partial negative charge on the carbon atom bonded to magnesium.
It is used to form new carbon-carbon bonds by reacting with carbonyl compounds.
It is an organomagnesium compound,not an organomanganese compound.
Therefore,option $D$ is incorrect.

(A) The reaction of alcohol $(R-OH)$ with phosphorus pentachloride $(PCl_5)$ is a standard method for the preparation of haloalkanes.
In this reaction,the hydroxyl group of the alcohol is replaced by a chlorine atom.
The balanced chemical equation is:
$R-OH + PCl_5 \longrightarrow R-Cl + HCl + POCl_3$
Comparing this with the given reaction $R-OH + PCl_5 \longrightarrow X + Y + Z$,we identify:
$X = R-Cl$
$Y = HCl$
$Z = POCl_3$
Therefore,the correct option is $A$.

(A) From the given reactions:
$1$. $X + HCl \xrightarrow{AlCl_3} C_2H_5Cl$ implies $X$ is ethene $(C_2H_4)$.
$2$. $Y + HCl \xrightarrow{ZnCl_2} C_2H_5Cl$ implies $Y$ is ethanol $(C_2H_5OH)$ (Lucas test).
$3$. The conversion of $Y$ (ethanol) to $X$ (ethene) is a dehydration reaction.
$4$. Ethanol undergoes dehydration to ethene when heated with $Al_2O_3$ at $350^{\circ}C$.
Reaction: $\underset{(Y)}{C_2H_5OH} \xrightarrow{Al_2O_3, 350^{\circ}C} \underset{(X)}{C_2H_4} + H_2O$.

(D) The boiling points depend on intermolecular forces.
$1$. $II$ ($n$-butanol) and $III$ (isobutanol) are alcohols and exhibit strong intermolecular hydrogen bonding,leading to higher boiling points compared to ethers and alkanes.
$2$. Between $II$ and $III$,$II$ is a straight-chain alcohol with a larger surface area,leading to stronger van der Waals forces than the branched $III$. Thus,$II > III$.
$3$. $I$ (diethyl ether) is polar due to the $C-O-C$ bond,giving it a higher boiling point than the non-polar alkane $IV$ ($n$-pentane).
$4$. Therefore,the order is $II > III > I > IV$.

(D) Rectified spirit ($95.6\%$ ethanol and $4.4\%$ water) cannot be concentrated further by simple distillation because it forms a constant boiling mixture (azeotrope).
To obtain absolute alcohol ($100\%$ ethanol),the water is removed by adding a dehydrating agent like quick lime $(CaO)$ or by distilling it over magnesium ethoxide,$Mg(OC_2H_5)_2$.

(D) Step $1$: Reaction of $CH_3CHO$ with $CH_3MgBr$ followed by hydrolysis gives $(A)$,which is propan$-2-$ol $(CH_3CH(OH)CH_3)$.
Step $2$: Dehydration of propan$-2-$ol with $H_2SO_4$ and heat gives $(B)$,which is propene $(CH_3CH=CH_2)$.
Step $3$: Hydroboration-oxidation of propene $(CH_3CH=CH_2)$ gives $(C)$,which is propan$-1-$ol $(CH_3CH_2CH_2OH)$.
Step $4$: $(A)$ is propan$-2-$ol and $(C)$ is propan$-1-$ol. These are position isomers because the position of the $-OH$ group differs.

(D) The Lucas test uses conc. $HCl$ and anhydrous $ZnCl_2$ to distinguish between primary,secondary,and tertiary alcohols.
Tertiary alcohols react instantly to produce turbidity due to the formation of an insoluble alkyl chloride.
Given the molecular formula $C_5H_{12}O$,the alcohol $X$ must be a tertiary alcohol,which is $2-$methyl$-2-$butanol.
Acid-catalysed hydration of $2-$methyl$-1-$butene follows Markovnikov's rule to yield $2-$methyl$-2-$butanol,which is a tertiary alcohol.
Therefore,option $D$ is the correct answer.

(B) The reaction involves the conversion of an alcohol to an alkyl iodide.
Using $H_2SO_4$ with $KI$ is not ideal because $H_2SO_4$ is an oxidizing agent that can oxidize $HI$ to $I_2$.
Phosphoric acid $(H_3PO_4)$ is a non-oxidizing acid and is preferred for the in-situ generation of $HI$ from $KI$.
The reaction is: $3 KI + H_3PO_4 \rightarrow 3 HI + K_3PO_4$.
This $HI$ then reacts with the alcohol via an $S_N2$ mechanism to give the alkyl iodide with a good yield.

(C) $1-$Propanol is a primary $(1^{\circ})$ alcohol, while $2-$propanol is a secondary $(2^{\circ})$ alcohol.
These can be distinguished by the Lucas test, which uses a mixture of anhydrous $ZnCl_2$ and concentrated $HCl$.
$2^{\circ}$ alcohols react with the Lucas reagent to produce turbidity within $5-10$ minutes at room temperature.
$1^{\circ}$ alcohols do not produce turbidity at room temperature.

(D) The chemical reaction in which alcohols and acids react with each other to form esters is called esterification.
In this reaction,the alcohol acts as a nucleophile and attacks the carbonyl carbon of the carboxylic acid.
The reactivity of alcohols towards esterification is primarily governed by steric hindrance.
As the bulkiness of the alkyl group attached to the carbon bearing the $-OH$ group increases,the steric hindrance increases,making it more difficult for the alcohol to attack the acid,thus decreasing the reactivity.
The given compounds are:
$(I) CH_3CH_2OH$ (Ethanol)
$(II) (CH_3)_2CHCH_2OH$ ($2-$methylpropan$-1-$ol)
$(III) (CH_3)_2CHCH(CH_3)CH_2OH$ ($2,3-$dimethylbutan$-1-$ol)
$(IV) CH_3CH_2CH_2OH$ (Propan$-1-$ol)
Comparing the steric hindrance,the order of reactivity is $(I) > (IV) > (II) > (III)$.
Therefore,the correct option is $D$.

(D) The rate of acid-catalyzed dehydration of alcohols depends on the stability of the carbocation intermediate formed during the reaction. The stability order of carbocations is $3^\circ > 2^\circ > 1^\circ$.
Let us analyze the carbocations formed from each alcohol:
$(I)$ $CH_3CH_2CH_2CH_2OH$ forms a $1^\circ$ carbocation: $CH_3CH_2CH_2CH_2^+$.
$(II)$ $CH_3-CH(CH_3)-CH_2OH$ forms a $1^\circ$ carbocation: $CH_3-CH(CH_3)-CH_2^+$,which can rearrange to a more stable $2^\circ$ carbocation.
$(III)$ $CH_3-C(CH_3)_2-OH$ forms a $3^\circ$ carbocation: $CH_3-C^+(CH_3)_2$.
$(IV)$ $CH_3-CH_2-CH(OH)-CH_3$ forms a $2^\circ$ carbocation: $CH_3-CH_2-CH^+-CH_3$.
Comparing the stability: The $3^\circ$ carbocation $(III)$ is the most stable,followed by the $2^\circ$ carbocation $(IV)$. Between the two $1^\circ$ carbocations,$(II)$ can undergo rearrangement to form a more stable carbocation,making it more reactive than the primary carbocation $(I)$.
Thus,the order of reactivity is $III > IV > II > I$.

(D) The reactant is propan-$2$-ol,which is a secondary alcohol.
When vapours of a secondary alcohol are passed over heated $Cu$ at $573 \ K$,dehydrogenation occurs,resulting in the loss of $H_2$ to form a ketone.
Therefore,propan-$2$-ol undergoes dehydrogenation to form propanone $(CH_3COCH_3)$.

(C) Assertion $(A)$ is true because tertiary alcohols react very rapidly with Lucas reagent to form alkyl chlorides,which appear as turbidity immediately.
Reason $(R)$ is false because Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$,not $HNO_3$ and $ZnCl_2$.

(C) The Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$. It is used to distinguish between primary,secondary,and tertiary alcohols based on the rate of formation of alkyl chlorides,which appear as white turbidity.
$1$. Tertiary alcohols $(3^\circ)$ react immediately to form white turbidity.
$2$. Secondary alcohols $(2^\circ)$ react within $5-10 \ min$ to form white turbidity.
$3$. Primary alcohols $(1^\circ)$ do not react at room temperature or react very slowly (after $30 \ min$ upon heating).
Analysis of the given compounds:
$(i)$ $n-$Butanol $(CH_3CH_2CH_2CH_2OH)$ is a primary alcohol.
$(ii)$ $tert-$Butanol $((CH_3)_3COH)$ is a tertiary alcohol.
$(iii)$ Benzyl alcohol $(C_6H_5CH_2OH)$ is a primary alcohol,but it reacts rapidly due to the stability of the carbocation formed (benzylic carbocation).
$(iv)$ Allyl alcohol $(CH_2=CH-CH_2OH)$ is a primary alcohol,but it also reacts rapidly due to the stability of the carbocation formed (allylic carbocation).
Therefore,$(ii)$,$(iii)$,and $(iv)$ give white turbidity almost immediately.

(D) Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$. The reaction of alcohols with Lucas reagent proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate. The reactivity of alcohols towards Lucas reagent follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
$(A)$ $CH_3-CH_2-CH_2-CH_2-OH$ is a $1^{\circ}$ alcohol.
$(B)$ $CH_3-CH_2-CH(OH)-CH_3$ is a $2^{\circ}$ alcohol.
$(C)$ $(CH_3)_2CH-CH_2-OH$ is a $1^{\circ}$ alcohol.
$(D)$ $(CH_3)_3C-OH$ is a $3^{\circ}$ alcohol.
Since $3^{\circ}$ alcohols form the most stable carbocation,they react fastest with Lucas reagent,often at room temperature.

(A) The compound $3-$methylpent$-1-$en$-3-$ol is an allylic alcohol. In the presence of an acid,it undergoes protonation of the $-OH$ group followed by the loss of water to form a resonance-stabilized carbocation: $CH_3-CH_2-C^+(CH_3)-CH=CH_2 \leftrightarrow CH_3-CH_2-C(CH_3)=CH-CH_2^+$.
This carbocation can lose a proton to form a more stable,conjugated diene or a substituted alkene. Specifically,the loss of optical activity occurs because the chiral center at $C-3$ is destroyed upon the formation of the planar carbocation intermediate. The major product formed is $3-$methylpenta$-1,3-$diene or a related stable alkene isomer depending on the specific rearrangement,but among the given options,the most stable alkene formed via rearrangement is $3-$methylpent$-2-$ene $(CH_3-CH_2-C(CH_3)=CH-CH_3)$.

(D) The reaction of hydrogen halides $(HX)$ with ethyl alcohol $(C_2H_5OH)$ involves the cleavage of the $C-O$ bond.
The reactivity depends on the strength of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the bond dissociation energy of the $H-X$ bond decreases,making the bond easier to break.
Therefore,the reactivity order is $HI > HBr > HCl > HF$.