Properties of alcohol Questions in English

(C) Step $1$: Dehydration of cyclohexylmethanol using $H_3PO_4$ at $120^{\circ}C$ proceeds via an $E1$ mechanism to form the more stable alkene,$1$-methylcyclohexene,as the major product $A$.
Step $2$: Hydroboration-oxidation $(HBO)$ of $1$-methylcyclohexene involves the anti-Markovnikov addition of water across the double bond.
The $OH$ group attaches to the less hindered carbon atom (the $CH$ group of the double bond),while the $H$ atom attaches to the more hindered carbon atom (the $C-CH_3$ group).
Therefore,the final major product $P$ is $2$-methylcyclohexanol.

(D) The reaction is an acid-catalyzed dehydration of an alcohol.
$1$. The hydroxyl group is protonated by $H_2SO_4$ to form a good leaving group $(-OH_2^+)$.
$2$. Water is eliminated to form a stable benzylic carbocation.
$3$. $A$ proton is removed from the adjacent carbon to form an alkene.
$4$. The resulting alkene can exist as two geometric isomers: $A$ (trans-isomer) and $B$ (cis-isomer).
$5$. The trans-isomer $(A)$ is more stable due to reduced steric hindrance compared to the cis-isomer $(B)$.
$6$. According to the thermodynamic control of the reaction,the more stable product is formed as the major product.
Therefore,compound $A$ is the major product.

(C) The reaction of alcohols with conc. $HCl$ and anhydrous $ZnCl_{2}$ is known as the Lucas test.
This reaction proceeds via an $S_{N}1$ mechanism,where the rate-determining step is the formation of a carbocation.
Therefore,the reactivity of alcohols follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Among the given options:
$A$. Butan$-1-$ol is a $1^{\circ}$ alcohol.
$B$. Butan$-2-$ol is a $2^{\circ}$ alcohol.
$C$. $2-$methylpropan$-2-$ol is a $3^{\circ}$ alcohol.
$D$. $2-$methylpropan$-1-$ol is a $1^{\circ}$ alcohol.
Since $3^{\circ}$ alcohols form the most stable carbocation,$2-$methylpropan$-2-$ol reacts fastest.

(B) Statement $I$ is correct because the Lucas test uses a mixture of conc. $HCl$ and anhydrous $ZnCl_{2}$ (Lucas Reagent) to distinguish between $1^{\circ}$,$2^{\circ}$,and $3^{\circ}$ alcohols based on their reactivity.
Statement $II$ is incorrect because $3^{\circ}$ alcohols are the most reactive and produce turbidity immediately at room temperature,whereas $1^{\circ}$ alcohols are the least reactive and do not produce turbidity at room temperature.
The reactivity order towards Lucas reagent is $3^{\circ} > 2^{\circ} > 1^{\circ}$ alcohol.

(C) The reaction proceeds via an acid-catalyzed dehydration of a secondary alcohol.
$1$. Protonation of the $-OH$ group occurs to form a good leaving group $(-OH_2^+)$.
$2$. Loss of a water molecule generates a secondary carbocation.
$3$. $A$ $1,2$-methyl shift occurs to rearrange the secondary carbocation into a more stable tertiary carbocation.
$4$. Finally,deprotonation (loss of $H^+$) from the adjacent carbon leads to the formation of the most stable,highly substituted alkene (Zaitsev product).
Thus,the major product is $1,2,5$-trimethylcyclohex-$1$-ene.

(A) The isomers of $C_{4}H_{10}O$ that are alcohols are:
$1$. $CH_{3}CH_{2}CH_{2}CH_{2}OH$ (butan-$1$-ol): Achiral.
$2$. $CH_{3}CH(OH)CH_{2}CH_{3}$ (butan-$2$-ol): Chiral,as the $C_{2}$ carbon is bonded to four different groups $(-H, -OH, -CH_{3}, -CH_{2}CH_{3})$.
$3$. $(CH_{3})_{2}CHCH_{2}OH$ ($2$-methylpropan-$1$-ol): Achiral.
$4$. $(CH_{3})_{3}COH$ ($2$-methylpropan-$2$-ol): Achiral.
Thus,there is only $1$ chiral alcohol with the molecular formula $C_{4}H_{10}O$.

(A) The reaction between a monohydric alcohol and methylmagnesium iodide is: $ROH + CH_3MgI \rightarrow ROMgI + CH_4(g)$.
According to the stoichiometry,$1 \ mole$ of $ROH$ produces $1 \ mole$ of $CH_4$ gas.
At $NTP$ conditions,$1 \ mole$ of gas occupies $22.4 \ L$.
Given volume of $CH_4 = 3.1 \ L$.
Moles of $CH_4 = \frac{3.1}{22.4} \approx 0.1384 \ mol$.
Since moles of $ROH = \text{moles of } CH_4$,we have $0.1384 = \frac{\text{mass}}{\text{Molar Mass}} = \frac{4.5}{M.M}$.
$M.M = \frac{4.5}{0.1384} \approx 32.51 \ g/mol$.
The nearest integer is $33$.

(C) The reaction proceeds through the following steps:
$1$. Protonation of the hydroxyl group by $H^+$ followed by the loss of a water molecule to form a tertiary carbocation.
$2$. The double bond in the chain attacks the carbocation,leading to a cyclization reaction to form a more stable five-membered ring carbocation.
$3$. The bromide ion $(Br^-)$ then attacks the carbocation to form the final product.
$4$. The structure corresponding to this mechanism is option $C$.

(B) The reaction sequence involves the conversion of an alkene to an epoxide followed by the reduction of the epoxide to an alcohol.
$1$. The conversion of an alkene to an epoxide (epoxidation) requires a peroxyacid. $C_6H_5COOH$ (perbenzoic acid) is a standard reagent for this transformation.
$2$. The reduction of an epoxide to an alcohol requires a strong reducing agent like $LiAlH_4$,which attacks the less hindered carbon of the epoxide ring to yield the corresponding alcohol.
Therefore,$X = C_6H_5COOH$ and $Y = LiAlH_4$.

(B) The reaction proceeds in two steps:
$1$. Reduction of cyclobutyl methyl ketone with $NaBH_4/MeOH$ yields $1-cyclobutyl-1-ethanol$.
$2$. Treatment with $HBr$ leads to the protonation of the alcohol,followed by the loss of water to form a secondary carbocation.
$3$. This secondary carbocation undergoes ring expansion to form a more stable cyclopentyl carbocation.
$4$. $A$ $1,2-hydride$ shift occurs to form a more stable tertiary carbocation.
$5$. Finally,the nucleophilic attack by $Br^-$ yields $1-bromo-1-methylcyclopentane$ as the major product.

(A) The reaction involves the acid-catalyzed dehydration of an alcohol.
$1$. Protonation of the hydroxyl group occurs,followed by the loss of a water molecule to form a secondary carbocation.
$2$. $A$ $1,2$-methyl shift occurs to convert the secondary carbocation into a more stable tertiary carbocation.
$3$. Finally,deprotonation occurs to form the most substituted alkene,which is the thermodynamically controlled product.
$4$. The resulting major product is $1,2,3$-trimethylcyclohex-$1$-ene.

(A) The reaction involves the acid-catalyzed dehydration of an alcohol.
$1$. Protonation of the hydroxyl group occurs to form a good leaving group $(-OH_2^+)$.
$2$. Loss of water generates a secondary carbocation.
$3$. $A$ ring expansion occurs because the five-membered ring with a carbocation adjacent to it is less stable than a six-membered ring. The bond from the ring shifts to the carbocation center,resulting in a more stable six-membered ring with a carbocation at the adjacent position.
$4$. Finally,deprotonation occurs to form the most stable alkene,which is the more substituted alkene (Saytzeff product).
Thus,the major product is $1,2$-dimethylcyclohexene.

(B) The rate of dehydration of alcohols is directly proportional to the stability of the carbocation formed as an intermediate.
$(a)$ Phenol: The carbocation formed would be highly unstable due to the $sp^2$ hybridized carbon atom.
$(b)$ Cyclohexa$-2,4-$dien$-1-$ol: Forms a resonance-stabilized carbocation (aromatic character can be achieved).
$(c)$ Cyclohex$-2-$en$-1-$ol: Forms a resonance-stabilized allylic carbocation.
$(d)$ Cyclohexanol: Forms a secondary carbocation.
Comparing the stability: $(b) > (c) > (d) > (a)$.
Therefore,the decreasing order of dehydration is $b > c > d > a$.

(B) The reaction proceeds via the following steps:
$1$. Protonation of the alcohol group: The $-OH$ group is protonated by $H_3O^+$ to form a good leaving group $(-OH_2^+)$.
$2$. Formation of carbocation: Loss of water molecule generates a secondary carbocation.
$3$. Rearrangement: $A$ $1,2-CH_3^-$ shift occurs to form a more stable tertiary carbocation.
$4$. Cyclization: The double bond attacks the carbocation to form a new ring.
$5$. Deprotonation: Loss of a proton $(H^+)$ from the adjacent carbon leads to the formation of the final stable alkene product,which corresponds to the structure in option $B$.

(A) Butan$-1-$ol $(CH_3CH_2CH_2CH_2OH)$ contains an $-OH$ group,which allows it to undergo extensive intermolecular hydrogen bonding.
Ethoxyethane $(CH_3CH_2-O-CH_2CH_3)$ lacks a hydrogen atom attached to an electronegative atom like $O, N,$ or $F$,and therefore cannot form hydrogen bonds.
Stronger intermolecular forces (hydrogen bonding) in butan$-1-$ol lead to greater molecular association,resulting in a higher boiling point compared to ethoxyethane.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(A) The reaction proceeds in three steps:
$1$. Reaction of butan$-2-$ol with $NaI$ and $H_3PO_4$ replaces the $-OH$ group with $-I$ to form $2-$iodobutane $(CH_3-CH_2-CH(I)-CH_3)$.
$2$. Reaction of $2-$iodobutane with $Mg$ in dry ether forms the Grignard reagent,sec-butylmagnesium iodide $(CH_3-CH_2-CH(MgI)-CH_3)$.
$3$. The Grignard reagent reacts with $D_2O$ (deuterium oxide) to replace the $-MgI$ group with a deuterium atom,resulting in $2-$deuteriobutane $(CH_3-CH_2-CH(D)-CH_3)$.

(B) Step $1$: Hydroboration-oxidation of propene $(CH_3-CH=CH_2)$ with $B_2H_6$ followed by $H_2O_2, NaOH$ gives propan$-1-$ol $(CH_3-CH_2-CH_2-OH)$.
Step $2$: Oxidation of propan$-1-$ol with $PCC$ (Pyridinium chlorochromate) gives propanal $(CH_3-CH_2-CHO)$.
Step $3$: Reaction of propanal with methylmagnesium bromide $(CH_3MgBr)$ followed by acidic workup gives butan$-2-$ol $(CH_3-CH_2-CH(OH)-CH_3)$.

(B) The reaction of $3\text{-methylbutan-2-ol}$ with $HBr$ proceeds via an $S_N1$ mechanism.
$1$. Protonation of the alcohol group followed by the loss of a water molecule generates a secondary carbocation: $CH_{3}-CH(CH_{3})-CH^+-CH_{3}$.
$2$. This secondary carbocation undergoes a $1,2\text{-hydride shift}$ to form a more stable tertiary carbocation: $CH_{3}-C^+(CH_{3})-CH_{2}-CH_{3}$.
$3$. Finally,the bromide ion $(Br^-)$ attacks the tertiary carbocation to form the major product: $2\text{-bromo-2-methylbutane}$,which is $CH_{3}-C(Br)(CH_{3})-CH_{2}-CH_{3}$.

(A) The dehydration of alcohols under acidic conditions proceeds via the formation of a carbocation intermediate. The rate of dehydration depends on the stability of the carbocation formed. Among the given options,the molecule that can form the most stable carbocation or has a structural feature facilitating elimination will dehydrate most readily. However,looking at the provided options and standard chemical behavior,$\beta$-hydroxy carbonyl compounds (aldols) are known to undergo dehydration very easily due to the formation of a conjugated $\alpha,\beta$-unsaturated system. If we re-examine the structures,the question likely intends to identify the compound that forms a stable conjugated system upon dehydration. Given the options provided,the most stable product is formed by the dehydration of $\beta$-hydroxy compounds. Based on standard competitive chemistry problems of this type,the correct answer is $A$.

(D) Step $1$: Hydroboration-oxidation of styrene $(Ph-CH=CH_2)$ gives anti-Markovnikov addition of water,resulting in $Ph-CH_2-CH_2-OH$.
Step $2$: Reaction with $HBr$ converts the alcohol to an alkyl bromide: $Ph-CH_2-CH_2-OH + HBr \rightarrow Ph-CH_2-CH_2-Br + H_2O$.
Step $3$: Formation of Grignard reagent and reaction with formaldehyde $(HCHO)$ followed by hydrolysis:
$Ph-CH_2-CH_2-Br + Mg \xrightarrow{\text{ether}} Ph-CH_2-CH_2-MgBr$
$Ph-CH_2-CH_2-MgBr + HCHO \rightarrow Ph-CH_2-CH_2-CH_2-OMgBr$
$Ph-CH_2-CH_2-CH_2-OMgBr \xrightarrow{H_3O^{+}} Ph-CH_2-CH_2-CH_2-OH$.
Thus,the final product $A$ is $Ph-CH_2-CH_2-CH_2-OH$.

(D) Step $1$: Dehydrohalogenation of $CH_3-CH_2-CH_2-Br$ with $KOH_{(alc)}$ gives propene $(A)$ as $CH_3-CH=CH_2$.
Step $2$: Electrophilic addition of $HBr$ to propene follows Markovnikov's rule to give $2-$bromopropane $(B)$ as $CH_3-CH(Br)-CH_3$.
Step $3$: Nucleophilic substitution of $2-$bromopropane with $KOH_{(aq)}$ gives propan$-2-$ol $(C)$ as $CH_3-CH(OH)-CH_3$.

(D) Alcohols act as nucleophiles because the oxygen atom has lone pairs of electrons.
Alcohols act as electrophiles when the $C-O$ bond is broken,typically after protonation of the oxygen atom.
Thus,Assertion $A$ is true.
Alcohols react with active metals like $Na$,$K$,and $Al$ to form alkoxides and release $H_2$ gas,which confirms the acidic nature of alcohols.
Thus,Reason $R$ is true.
However,the reaction with active metals explains the acidity of alcohols,not their dual nature as nucleophiles and electrophiles.
Therefore,$R$ is not the correct explanation of $A$.

(B) The boiling point of organic compounds depends on intermolecular forces.
$CH_3CH_2CH_2CH_2OH$ is an alcohol,which exhibits strong intermolecular hydrogen bonding.
$CH_3CH_2CH_2CH_3$ (butane) is a non-polar alkane with weak van der Waals forces.
$CH_3CH_2CH_2CHO$ (butanal) and $C_2H_5OC_2H_5$ (diethyl ether) exhibit dipole-dipole interactions,which are weaker than hydrogen bonding.
Therefore,the alcohol has the highest boiling point.

(B) $1$. The reaction of $CH_3-CH_2-CH_2-Br$ with $NaOH$ in $C_2H_5OH$ is a dehydrohalogenation reaction (elimination),which produces propene $(CH_3-CH=CH_2)$ as Product $A$.
$2$. The hydration of propene in the presence of acid $(H_2O/H^+)$ follows Markovnikov's rule,leading to the formation of $2\text{-Propanol}$ as Product $B$.
$3$. The hydroboration-oxidation of propene $(B_2H_6, H_2O_2/OH^-)$ follows anti-Markovnikov's rule,leading to the formation of $1\text{-Propanol}$ as Product $C$.
$4$. Therefore,$B = 2\text{-Propanol}$ and $C = 1\text{-Propanol}$.

(A) The reaction of $1\text{-methylcyclohexanol}$ with $\text{conc. } H_2SO_4$ at high temperature involves acid-catalyzed dehydration. This proceeds via an $E1$ mechanism,where the hydroxyl group is protonated and leaves as water to form a tertiary carbocation. Subsequent loss of a proton from the adjacent carbon leads to the formation of the more stable,more substituted alkene,which is $1\text{-methylcyclohexene}$ $(A)$.
The reaction of $1\text{-methylcyclohexanol}$ with $CH_3COCl$ in the presence of pyridine is an acetylation reaction. The alcohol acts as a nucleophile,attacking the acetyl chloride to form an ester,$1\text{-methylcyclohexyl acetate}$ $(B)$.

(C) The Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$.
Alcohols react with the Lucas reagent to form alkyl chlorides,which appear as turbidity in the solution.
The reactivity order of alcohols towards the Lucas reagent is: $3^\circ > 2^\circ > 1^\circ$.
Tertiary $(3^\circ)$ alcohols react instantaneously with the Lucas reagent to produce immediate turbidity because they form stable carbocations.
Among the given options,$(CH_3)_3C-OH$ is a tertiary alcohol,hence it reacts instantaneously.

(D) Statement $(I)$ is true: The boiling point of alcohols and phenols increases with an increase in the number of carbon atoms due to an increase in van der Waals forces of attraction as the molecular weight increases $(B.P. \propto M.W.)$.
Statement $(II)$ is true: Alcohols and phenols exhibit intermolecular hydrogen bonding,which is significantly stronger than the dipole-dipole interactions found in ethers and haloalkanes of comparable molecular mass. Therefore,they have higher boiling points.

(C) The reaction sequence is as follows:
$1$. $A$ is Methanol $(CH_3OH)$. In acidic medium,it forms $CH_3-OH_2^+$,which reacts with $NaCN$ followed by hydrolysis to yield Acetic acid $(CH_3COOH)$ as $B$. Acetic acid is a common food preservative (vinegar).
$2$. $B$ (Acetic acid) is reduced by diborane $(B_2H_6)$ to form Ethanol $(C_2H_5OH)$ as $C$. Ethanol is used as a fuel additive.
$3$. $C$ (Ethanol) reacts with oleum $(H_2SO_4 + SO_3)$ at $140^{\circ} C$ to undergo intermolecular dehydration,yielding Diethyl ether $(C_2H_5-O-C_2H_5)$ as $D$,which is an inhalable anesthetic.
Therefore,the correct sequence is Methanol,Acetic acid,Ethanol,Diethyl ether.

(A) The Lucas test is used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction with Lucas reagent (a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$).
$1^{\circ}$ alcohols do not produce turbidity at room temperature.
$2^{\circ}$ alcohols produce turbidity within $5-10$ minutes.
$3^{\circ}$ alcohols produce turbidity immediately.
Among the given options:
$(1)$ $2-$Butanol is a $2^{\circ}$ alcohol.
$(2)$ $t-$Butyl alcohol is a $3^{\circ}$ alcohol.
$(3)$ Isobutyl alcohol is a $1^{\circ}$ alcohol.
$(4)$ Diphenyl carbinol is a $2^{\circ}$ alcohol,but due to the presence of two bulky phenyl groups,its reactivity is different from simple aliphatic $2^{\circ}$ alcohols.
Thus,$2-$Butanol is the standard example of a $2^{\circ}$ alcohol that produces turbidity within $5$ minutes.

(A) The reaction proceeds via the formation of a carbocation intermediate.
$1$. Protonation of the $-OH$ group by $HBr$ followed by the loss of $H_2O$ leads to the formation of a secondary $(2^{\circ})$ carbocation: $CH_3-CH(CH_3)-CH^+-CH_3$.
$2$. This $2^{\circ}$ carbocation undergoes a $1,2-H^-$ shift to form a more stable tertiary $(3^{\circ})$ carbocation: $CH_3-C^+(CH_3)-CH_2-CH_3$.
$3$. Finally,the bromide ion $(Br^-)$ attacks the $3^{\circ}$ carbocation to form the major product: $CH_3-C(Br)(CH_3)-CH_2-CH_3$.

(C) When vapours of primary $(1^{\circ})$ alcohols are passed over heated $Cu$ at $573 \ K$ $(300^{\circ} C)$,they undergo dehydrogenation to form aldehydes.
When vapours of secondary $(2^{\circ})$ alcohols are passed over heated $Cu$ at $573 \ K$ $(300^{\circ} C)$,they undergo dehydrogenation to form ketones.
When vapours of tertiary $(3^{\circ})$ alcohols are passed over heated $Cu$ at $573 \ K$ $(300^{\circ} C)$,they undergo dehydration to form alkenes.
Among the given options,tert-Butyl alcohol is a tertiary $(3^{\circ})$ alcohol,which undergoes dehydration to form an alkene (isobutylene) under these conditions.

(C) Primary $(1^{\circ})$ alcohols react with $Cu / \Delta$ to form aldehydes.
Secondary $(2^{\circ})$ alcohols react with $Cu / \Delta$ to form ketones.
Tertiary $(3^{\circ})$ alcohols do not undergo dehydrogenation in the presence of $Cu / \Delta$ because they lack an $\alpha$-hydrogen atom. Instead,they undergo dehydration to form alkenes.
Among the given options,$CH_3-C(CH_3)(CH_3)-OH$ is a tertiary alcohol and will not undergo dehydrogenation.

(B) The molecular formula $C_7H_8O$ corresponds to an aromatic compound.
$1$. $Na$ metal test: $A$ positive test indicates the presence of an acidic hydrogen (like in $-OH$ group).
$2$. Neutral $FeCl_3$ test: $A$ negative test indicates the absence of a phenolic $-OH$ group.
$3$. Lucas reagent test: $A$ positive test indicates the presence of an alcohol group (specifically,primary,secondary,or tertiary alcohol).
Analyzing the options:
- $C_6H_5OCH_3$ (Anisole) does not have an $-OH$ group,so it will not react with $Na$ metal.
- $C_6H_5CH_2OH$ (Benzyl alcohol) has an alcoholic $-OH$ group. It reacts with $Na$ metal (positive),does not give a color with neutral $FeCl_3$ (negative,as it is not a phenol),and reacts with Lucas reagent (positive,as it is a primary alcohol).
- $p-CH_3C_6H_4OH$ and $m-CH_3C_6H_4OH$ are cresols (phenols). They will give a positive test with neutral $FeCl_3$.
Therefore,the compound $A$ is $C_6H_5CH_2OH$.

(A) The boiling point of alcohols depends on the molecular mass and the extent of branching.
$1$. As the molecular mass increases,the boiling point increases. The order of molecular mass is: $\text{pentan-1-ol} (C_5H_{12}O) > \text{butan-1-ol} = \text{butan-2-ol} (C_4H_{10}O) > \text{propan-1-ol} (C_3H_8O)$.
$2$. For isomers with the same molecular mass,the boiling point decreases with an increase in branching because the surface area decreases,leading to weaker van der Waals forces. Thus,$\text{butan-1-ol}$ (straight chain) has a higher boiling point than $\text{butan-2-ol}$ (branched).
$3$. Combining these factors,the order is: $\text{pentan-1-ol} (a) > \text{butan-1-ol} (b) > \text{butan-2-ol} (c) > \text{propan-1-ol} (d)$.

(C) Statement-$1$ is correct: Lucas reagent $(conc. \ HCl + ZnCl_2)$ reacts with tertiary alcohols immediately to form turbidity,while secondary alcohols take about $5-10$ minutes to form turbidity. Thus,they can be distinguished.
Statement-$2$ is incorrect: Dichromate test (oxidation with $K_2Cr_2O_7/H^+$) is used to distinguish primary,secondary,and tertiary alcohols based on the products formed (aldehyde/acid,ketone,and no reaction/alkene respectively). However,it is not a specific test to distinguish between secondary and tertiary alcohols alone as effectively as the Lucas test or Victor Meyer's test,and specifically,both secondary and tertiary alcohols do not show the same color change pattern as primary alcohols. Therefore,Statement-$2$ is considered incorrect in the context of specific distinction.

(B) The reaction of a Grignard reagent $(R-MgBr)$ with formaldehyde $(HCHO)$ followed by hydrolysis with $D_2O$ proceeds as follows:
$1$. The nucleophilic cyclohexyl group $(C_6H_{11}^-)$ from the Grignard reagent attacks the electrophilic carbonyl carbon of formaldehyde $(HCHO)$:
$C_6H_{11}-MgBr + HCHO \rightarrow C_6H_{11}-CH_2-OMgBr$
$2$. Subsequent hydrolysis with $D_2O$ replaces the $MgBr$ group with a deuterium atom $(D)$ on the oxygen,forming a deuterated alcohol:
$C_6H_{11}-CH_2-OMgBr + D_2O \rightarrow C_6H_{11}-CH_2-OD + MgBr(OD)$
Thus,the product $A$ is cyclohexylmethanol where the hydroxyl hydrogen is replaced by deuterium,which corresponds to the structure in option $B$.

(C) The $Lucas$ test is used to distinguish between primary,secondary,and tertiary alcohols.
It involves the use of $Lucas$ reagent,which is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$.
$1.$ Tertiary alcohols react immediately with $Lucas$ reagent to form turbidity.
$2.$ Secondary alcohols react within $5$-$10$ minutes to form turbidity.
$3.$ Primary alcohols do not react at room temperature and show no turbidity.

(D) The 'Lucas' reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$.
The reaction of alcohols with the Lucas reagent proceeds via the formation of a carbocation intermediate,which is characteristic of the $S_N1$ mechanism.
Since the stability of carbocations follows the order: $3^\circ > 2^\circ > 1^\circ$,the tertiary $(3^\circ)$ alcohol forms the most stable carbocation and therefore reacts the fastest.
Thus,the tertiary alcohol reacts fastest via the $S_N1$ mechanism.

(A) The dehydration of alcohol in the presence of $H_3PO_4$ and heat proceeds via an $E1$ mechanism involving the formation of a carbocation intermediate.
$1$. Protonation of the $-OH$ group occurs,followed by the loss of a water molecule to form a secondary carbocation at the $C_2$ position.
$2$. The carbocation is adjacent to a phenyl group $(Ph)$ at $C_1$,which provides resonance stabilization.
$3$. Elimination of a proton $(H^+)$ can occur from either $C_1$ or $C_3$.
$4$. Elimination from $C_1$ leads to a double bond between $C_1$ and $C_2$,which is conjugated with the phenyl ring. This forms the most stable product,$1$-phenyl$-3-$methylcyclohex$-1-$ene.
Thus,the major product is $1-$phenyl$-3-$methylcyclohex$-1-$ene.

(A) In Victor Meyer's test,the alcohol is first converted to an alkyl iodide using $P/I_2$,then to a nitroalkane using $AgNO_2$,and finally treated with nitrous acid $(HNO_2)$.
For primary alcohols $(RCH_2OH)$,the resulting nitrolic acid forms a blood-red colour with alkali $(NaOH)$.
For secondary alcohols $(R_2CHOH)$,the pseudonitrol formed gives a blue colour with alkali.
Tertiary alcohols $(R_3COH)$ do not react with nitrous acid and remain colourless.

(D) The reaction of alcohols with concentrated $HCl$ in the presence of anhydrous $ZnCl_2$ is known as the Lucas test.
This test is used to distinguish between primary,secondary,and tertiary alcohols.
$3^{\circ}$ (tertiary) alcohols react immediately at room temperature to form a cloudy precipitate of alkyl chloride due to the formation of a stable carbocation.
Among the given options,$CH_3-CH_2-C(CH_3)_2-OH$ is a tertiary alcohol,hence it will react at once.

(A) Acid-catalyzed hydration involves the formation of a carbocation intermediate. The reactivity depends on the stability of the carbocation formed after the loss of water or the protonation of the double bond.
In the case of $1,2,3,4$-tetrahydronaphthalen-$1$-ol,the dehydration leads to the formation of a carbocation that is stabilized by resonance with the benzene ring,eventually leading to the formation of naphthalene,which is a highly stable aromatic system.
Therefore,the compound that can form the most stable carbocation intermediate or lead to the most stable product is the most reactive. Among the given options,the structure in option $A$ is the most reactive because its dehydration leads to the formation of the stable aromatic compound naphthalene.

(B) The general formula for a saturated acyclic alcohol is $C_nH_{2n+2}O$.
For $n = 5$,the formula is $C_5H_{12}O$.
Option $A$ is cyclopentanol,which has the formula $C_5H_{10}O$.
Option $B$ is pentan$-1-$ol,which is a straight-chain saturated alcohol with the formula $C_5H_{12}O$.
Option $C$ is pentan$-3-$ol,which is also a saturated alcohol with the formula $C_5H_{12}O$.
Option $D$ is cyclopent$-2-$en$-1-$ol,which has the formula $C_5H_8O$.
Both $B$ and $C$ are valid structures for $C_5H_{12}O$. However,in standard multiple-choice questions of this type,pentan$-1-$ol is a primary representative structure.

(B) An allylic alcohol is one in which the $-OH$ group is attached to a carbon atom next to a carbon-carbon double bond $(C=C)$.
$1$. $A$ primary $(1^{\circ})$ alcohol has the $-OH$ group attached to a carbon atom that is bonded to only one other carbon atom.
$2$. Let's analyze the options:
- $A$: $H_2C=CH-CH(CH_3)-OH$ is a secondary $(2^{\circ})$ allylic alcohol.
- $B$: $H_2C=CH-CH_2-OH$ is a primary $(1^{\circ})$ allylic alcohol because the $-OH$ group is attached to a $CH_2$ group.
- $C$: $H_2C=CH-C(CH_3)_2-OH$ is a tertiary $(3^{\circ})$ allylic alcohol.
- $D$: $CH_3-CH=CH-CH(CH_3)-OH$ is a secondary $(2^{\circ})$ allylic alcohol.
Therefore,the correct option is $B$.

(D) The boiling point of alcohols depends on the extent of intermolecular hydrogen bonding and the surface area of the molecule.
For isomeric alcohols,the boiling point decreases as the degree of branching increases because branching reduces the surface area,which in turn decreases the magnitude of van der Waals forces.
Comparing the given structures:
$A$: $n$-Butanol (primary alcohol,straight chain)
$B$: Isobutanol (primary alcohol,branched)
$C$: sec-Butanol (secondary alcohol,branched)
$D$: tert-Butanol (tertiary alcohol,highly branched)
Among these isomers,tert-Butanol has the most branching,which results in the smallest surface area and the weakest van der Waals forces. Therefore,it has the lowest boiling point.

(A) The boiling point of isomeric alcohols depends on the extent of branching.
As the branching increases,the surface area of the molecule decreases,which leads to a decrease in the magnitude of van der Waals forces of attraction.
Consequently,the boiling point decreases with an increase in branching.
The given compounds are:
$A$: Butan$-1-$ol (primary alcohol,straight chain)
$B$: $2-$Methylpropan$-1-$ol (primary alcohol,branched)
$C$: Butan$-2-$ol (secondary alcohol,branched)
$D$: $2-$Methylpropan$-2-$ol (tertiary alcohol,highly branched)
Among these,$A$ (Butan$-1-$ol) is a straight-chain primary alcohol with the largest surface area,resulting in the strongest intermolecular hydrogen bonding and van der Waals forces.
Therefore,Butan$-1-$ol has the highest boiling point.

(D) The boiling point of alcohols increases with an increase in the molecular mass due to the increase in the magnitude of van der Waals forces of attraction.
Among the given alcohols,$CH_3OH$ (Methanol) has the lowest molecular mass.
Therefore,$CH_3OH$ has the lowest boiling point.

(B) An allylic alcohol is one in which the $-OH$ group is attached to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C)$.
$A$: $CH_2=CH-CH_2-OH$ is a primary allylic alcohol because the $-OH$ group is attached to a primary carbon.
$B$: $CH_2=CH-CH(OH)-CH_3$ is a secondary allylic alcohol because the $-OH$ group is attached to a secondary carbon which is adjacent to a $C=C$ double bond.
$C$: $CH_3-CH=CH-CH_2-OH$ is a primary allylic alcohol.
$D$: $CH_2=CH-C(CH_3)_2-OH$ is a tertiary allylic alcohol because the $-OH$ group is attached to a tertiary carbon.
Therefore,the correct option is $B$.

(D) The boiling point of alcohols depends on the surface area of the molecule.
As the branching in the carbon chain increases,the surface area of the molecule decreases,which leads to weaker van der Waals forces of attraction.
Among the isomers of $C_4H_9OH$,$tert-$Butyl alcohol has the most compact structure due to maximum branching.
Therefore,$tert-$Butyl alcohol has the lowest boiling point.