Properties of alcohol Questions in English

(D) An allylic alcohol is one where the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond $(C=C)$.
$A$ tertiary alcohol is one where the carbon atom bonded to the $-OH$ group is attached to three other carbon atoms.
Let us analyze the structures:
$A$) $CH_2=CH-CH_2OH$ (Primary allylic alcohol)
$B$) $CH_2=CH-CH(OH)-CH_3$ (Secondary allylic alcohol)
$C$) $CH_2=C(CH_3)-CH_2OH$ (Primary allylic alcohol)
$D$) $CH_2=CH-C(CH_3)(OH)-CH_3$ (Tertiary allylic alcohol)
In option $D$,the carbon atom attached to the $-OH$ group is bonded to a vinyl group $(-CH=CH_2)$ and two methyl groups $(-CH_3)$,making it a tertiary allylic alcohol.

(A) The boiling point of isomeric alcohols depends on the surface area of the molecule.
As the branching in the carbon chain increases,the surface area decreases,which leads to weaker van der Waals forces of attraction between the molecules.
Therefore,the boiling point decreases with an increase in branching.
Among the given isomers,$n$-Butyl alcohol is a straight-chain alcohol with no branching,while the others have varying degrees of branching.
Thus,$n$-Butyl alcohol has the highest boiling point.

(C) In a tertiary benzylic alcohol,the $-OH$ group is bonded to a $sp^3$ hybridized tertiary carbon atom,which is directly attached to an aromatic ring.
In option $C$,the structure is $C_6H_5-C(CH_3)_2-OH$. Here,the carbon atom attached to the $-OH$ group is bonded to one phenyl group and two methyl groups,making it a tertiary carbon atom attached to an aromatic ring. Thus,it is a tertiary benzylic alcohol.

(C) Lower members of alcohols (like methanol and ethanol) are highly soluble in water due to hydrogen bonding and are also soluble in organic solvents. Therefore,the statement that they are insoluble in water and organic solvents is incorrect. Thus,option $C$ is the correct answer.

(A) Alcohols are polar molecules due to the presence of the $-OH$ group,which creates a dipole moment. Therefore,the statement that they are non-polar molecules is incorrect. Thus,option $(A)$ is the correct answer.

(A) The melting point of a compound depends on its molecular symmetry and packing in the crystal lattice.
$tert-$Butyl alcohol ($2-$methylpropan$-2-$ol) has a highly symmetrical structure compared to the other isomers.
Due to this high symmetry,it packs more efficiently into the crystal lattice,leading to stronger intermolecular forces and a significantly higher melting point.

(C) The boiling point of isomeric alcohols depends on the extent of branching in the carbon chain.
As the branching increases,the surface area of the molecule decreases,which reduces the magnitude of van der Waals forces of attraction.
Consequently,the boiling point decreases with an increase in branching.
Among the given isomers:
$n$-Butyl alcohol $(CH_3CH_2CH_2CH_2OH)$ has no branching.
Isobutyl alcohol $((CH_3)_2CHCH_2OH)$ has one branch.
sec-Butyl alcohol $(CH_3CH_2CH(OH)CH_3)$ has one branch.
tert-Butyl alcohol $((CH_3)_3COH)$ has the maximum branching (three methyl groups attached to the central carbon).
Therefore,tert-Butyl alcohol has the lowest boiling point.

(B) $(i)$ Boiling point $\propto$ Molar mass. Since all are isomers of $C_4H_9OH$,molar mass is the same.
$(ii)$ Boiling point $\propto \frac{1}{\text{Branching}}$.
$(iii)$ $tert-Butyl$ alcohol has the maximum branching,which reduces the surface area and decreases the van der Waals forces of attraction.
$(iv)$ Therefore,$tert-Butyl$ alcohol has the lowest boiling point.

(C) The acidity of alcohols depends on the stability of the alkoxide ion formed after the release of a proton $(H^+)$.
Alkyl groups are electron-donating ($+I$ effect) in nature.
As the number of alkyl groups attached to the carbon atom bearing the $-OH$ group increases,the electron density on the oxygen atom increases due to the $+I$ effect.
This destabilizes the alkoxide ion,thereby decreasing the acidic strength.
The order of acidity is: $CH_3OH > CH_3CH_2OH > (CH_3)_2CHOH > (CH_3)_3COH$.
Therefore,$CH_3OH$ is the most acidic.

(A) The solubility of alcohols in water depends on the balance between the hydrophilic hydroxyl $(-OH)$ group and the hydrophobic alkyl chain.
As the size of the hydrophobic alkyl chain increases,the solubility of the alcohol in water decreases.
All the given alcohols are isomers with the molecular formula $C_5H_{12}O$.
However,the solubility also depends on the branching of the alkyl chain.
Increased branching reduces the surface area of the hydrophobic part,which generally increases solubility.
Comparing the structures:
$1$. Pentan-$1$-ol (straight chain,$5$ carbons)
$2$. $2$-methyl butan-$2$-ol (branched,$5$ carbons)
$3$. Pentan-$2$-ol (branched,$5$ carbons)
$4$. $2,2$-Dimethyl propan-$1$-ol (highly branched,$5$ carbons)
Among these,Pentan-$1$-ol has the longest straight alkyl chain,which makes it the least soluble in water compared to the more branched isomers.

(C) The reaction of alcohols with Lucas reagent $(conc. \ HCl + ZnCl_2)$ follows the $S_N1$ mechanism,which depends on the stability of the carbocation intermediate.
Tertiary carbocations are the most stable,followed by secondary and then primary carbocations.
Therefore,tertiary alcohols react fastest with Lucas reagent at room temperature,resulting in immediate turbidity.
$2$-Methylpropan-$2$-ol is a tertiary alcohol,so it reacts immediately with Lucas reagent.

(C) For isomeric alcohols,the boiling point decreases as the extent of branching increases.
$n$-butyl alcohol $(CH_3CH_2CH_2CH_2OH)$ is a straight-chain alcohol with the highest surface area and strongest intermolecular hydrogen bonding.
tert-butyl alcohol $((CH_3)_3COH)$ is the most branched isomer,which reduces the surface area available for van der Waals interactions,resulting in the lowest boiling point.

(A) The boiling point of alcohols depends on the extent of intermolecular hydrogen bonding and the surface area of the molecule.
As the branching increases from primary $(1^{\circ})$ to secondary $(2^{\circ})$ to tertiary $(3^{\circ})$ alcohols,the surface area of the molecule decreases.
$A$ smaller surface area leads to weaker van der Waals forces of attraction.
Additionally,steric hindrance in tertiary alcohols reduces the effectiveness of intermolecular hydrogen bonding.
Therefore,the boiling point decreases in the order: $1^{\circ} > 2^{\circ} > 3^{\circ}$.

(D) The dehydration of alcohols with concentrated $H_2SO_4$ follows $E1$ mechanism,which involves the formation of a carbocation intermediate.
For $2-$Methylbutan$-2-$ol $(CH_3-CH_2-C(CH_3)(OH)-CH_3)$,the protonation of the $-OH$ group followed by the loss of water molecule generates a tertiary carbocation: $CH_3-CH_2-C^+(CH_3)-CH_3$.
This carbocation can lose a proton from the adjacent carbon atoms to form alkenes.
Loss of a proton from the $C-3$ position yields $2-$Methylbut$-2-$ene $(CH_3-CH=C(CH_3)-CH_3)$,which is the most stable alkene according to Saytzeff's rule.

(A) The reactivity of alcohols with halo acids depends on the stability of the carbocation intermediate formed during the reaction.
The order of stability of carbocations is $3^\circ > 2^\circ > 1^\circ > \text{methyl}$.
Therefore,the order of reactivity of alcohols is $3^\circ > 2^\circ > 1^\circ > \text{methyl}$.
$(I)$ $CH_3OH$ is a methyl alcohol.
$(II)$ $CH_3CH_2OH$ is a $1^\circ$ alcohol.
$(III)$ $(CH_3)_2CHOH$ is a $2^\circ$ alcohol.
$(IV)$ $(CH_3)_3COH$ is a $3^\circ$ alcohol.
Thus,the decreasing order of reactivity is $(IV) > (III) > (II) > (I)$.

(C) Grignard reagents $(CH_3MgX)$ are strong bases and react with compounds containing active hydrogen atoms (like the hydroxyl group in alcohols) to form alkanes.
In this reaction,the methyl group $(CH_3^-)$ from the Grignard reagent abstracts the acidic proton $(H^+)$ from the methanol $(CH_3OH)$ molecule.
This results in the formation of methane $(CH_4)$ as the product $P$ and the corresponding magnesium salt $MgX(OCH_3)$.
The reaction is: $CH_3OH + CH_3MgX \longrightarrow CH_4 + MgX(OCH_3)$.

(D) Step $1$: Dehydration of $Propan-1-ol$ with $Al_2O_3$ at $623 \ K$ gives $Propene$ $(A)$ as the product.
$CH_3CH_2CH_2OH \xrightarrow[623 \ K]{Al_2O_3} CH_3CH=CH_2$ $(A)$
Step $2$: $Propene$ $(A)$ reacts with $Conc. \ H_2SO_4$ followed by hydrolysis $(H_2O)$ to undergo hydration according to $Markovnikov's$ rule,yielding $Propan-2-ol$ $(B)$ as the major product.
$CH_3CH=CH_2 \xrightarrow[ii) \ H_2O]{i) \ Conc. \ H_2SO_4} CH_3CH(OH)CH_3$ $(B)$
Therefore,the product $B$ is $Propan-2-ol$.

(A) The Lucas reagent is a mixture of concentrated hydrochloric acid $(HCl)$ and anhydrous zinc chloride $(ZnCl_2)$.
It is used to differentiate between primary,secondary,and tertiary alcohols based on their reactivity.
$Tertiary$ alcohols react almost immediately with the Lucas reagent to form alkyl chlorides.
$Secondary$ alcohols react within $5-10$ minutes.
$Primary$ alcohols react very slowly or not at all at room temperature.

(C) In the reaction of alcohols with phosphorus trichloride $(PCl_3)$,the $C-O$ bond of the alcohol is broken to form an alkyl chloride.
The chemical equation is: $3 R-OH + PCl_3 \longrightarrow 3 R-Cl + H_3PO_3$.
In contrast,reactions with carboxylic acids,acid anhydrides,and acid chlorides typically involve the cleavage of the $O-H$ bond of the alcohol.

(B) When vapours of tertiary alcohols like $2-$methylpropan$-2-$ol are passed over hot copper at $573 \ K$,they undergo dehydration rather than dehydrogenation.
This reaction results in the formation of an alkene.
The reaction is: $(CH_3)_3C-OH \xrightarrow{Cu, 573 \ K} CH_3-C(CH_3)=CH_2 + H_2O$.
The product formed is $2-$methylpropene.

(D) The reactivity of alcohols with haloacids $(HX)$ follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
This is because the reaction proceeds via the formation of a carbocation intermediate,and the stability of the carbocation follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
Analyzing the given structures:
$(A)$ $CH_3CH_2C(CH_3)(OH)CH_3$ is a $3^{\circ}$ alcohol.
$(B)$ $CH_3CH_2CH(OH)CH_2CH_3$ is a $2^{\circ}$ alcohol.
$(C)$ $CH_3CH_2CH_2CH(OH)CH_3$ is a $2^{\circ}$ alcohol.
$(D)$ $CH_3CH_2C(OH)(CH_3)CH_2CH_3$ is a $3^{\circ}$ alcohol.
Both $(A)$ and $(D)$ are tertiary alcohols. However,$(D)$ is a more substituted tertiary alcohol ($3$-methylpentan$-3-$ol) compared to $(A)$ ($2$-methylbutan$-2-$ol),making the carbocation formed from $(D)$ slightly more stable due to increased hyperconjugation and inductive effects. Thus,$(D)$ reacts fastest.

(B) The Lucas reagent is a solution of anhydrous zinc chloride $(ZnCl_2)$ in concentrated hydrochloric acid $(HCl)$.
It is used to classify alcohols based on their reactivity.

(A) The dehydration of secondary alcohols like $Butan-2-ol$ $(CH_3-CH(OH)-CH_2-CH_3)$ typically requires $60 \% \ H_2SO_4$ at a temperature of $373 \ K$.
This reaction follows the $E1$ mechanism,where the major product is $But-2-ene$ (Saytzeff product) and the minor product is $But-1-ene$ (Hofmann product).

(B) Step $1$: $CH_{3}-CH_{2}-CH_{2}-OH + PCl_{3} \rightarrow CH_{3}-CH_{2}-CH_{2}-Cl (X) + H_{3}PO_{3}$.
Step $2$: $CH_{3}-CH_{2}-CH_{2}-Cl + alco.KOH \xrightarrow{\Delta} CH_{3}-CH=CH_{2} (Y) + KCl + H_{2}O$ (Dehydrohalogenation).
Step $3$: $CH_{3}-CH=CH_{2} + H_{2}O \xrightarrow{conc.H_{2}SO_{4}} CH_{3}-CH(OH)-CH_{3} (Z)$ (Acid-catalyzed hydration following Markovnikov's rule).
Thus,$Z$ is propan-$2$-ol,which is $(CH_{3})_{2}CH-OH$.

(A) The reaction of ethanol with $HCl$ to form ethyl chloride is known as the Grooves process.
In this reaction,anhydrous $ZnCl_{2}$ acts as a catalyst (Lewis acid) to facilitate the cleavage of the $C-O$ bond.
The reaction is: $C_{2}H_{5}OH + HCl \xrightarrow{\text{anhydrous } ZnCl_{2}} C_{2}H_{5}Cl + H_{2}O$.

(D) The correct answer is $D$ ($2-$Methylpropan$-2-$ol).
Primary and secondary alcohols can be oxidized to aldehydes/ketones using milder oxidizing agents like $PCC$ or $CrO_3$.
However,tertiary alcohols like $2-$methylpropan$-2-$ol are resistant to oxidation under normal conditions.
They require strong oxidizing agents such as acidic $KMnO_4$ and extreme conditions (high temperature) to undergo oxidation,which typically involves the cleavage of $C-C$ bonds to form smaller ketones or carboxylic acids.

(A) The final product is propanenitrile,which is $CH_3-CH_2-CN$.
This compound contains $3$ carbon atoms.
The reaction $B \xrightarrow{KCN, \Delta} CH_3-CH_2-CN$ indicates that $B$ must be an alkyl halide with $2$ carbon atoms,which is chloroethane $(CH_3-CH_2-Cl)$.
The reaction $A + SOCl_2 \xrightarrow{\text{pyridine}} B$ is the Darzens process for converting an alcohol into an alkyl chloride.
Therefore,$A$ must be ethanol $(CH_3-CH_2-OH)$.
Thus,$A$ is ethanol.

(D) The reaction of alcohol with Lucas reagent ($HCl +$ anhydrous $ZnCl_2$) proceeds via an $SN^1$ mechanism.
The rate of reaction with Lucas reagent follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$ alcohol.
$3^{\circ}$ alcohols react immediately with Lucas reagent to form alkyl chlorides,which appear as a cloudy layer (turbidity) in the solution.
Among the given options,$(CH_3)_3C-OH$ is a $3^{\circ}$ alcohol,therefore it reacts immediately.

(D) The reaction of alcohols with Lucas reagent (conc. $HCl +$ anhydrous $ZnCl_2$) is used to distinguish between primary,secondary,and tertiary alcohols.
Tertiary alcohols react most rapidly and give immediate turbidity at room temperature due to the formation of stable carbocations.
Among the given options,$2-$methylpropan$-2-$ol is a tertiary alcohol,while the others are primary or secondary alcohols.
Therefore,$2-$methylpropan$-2-$ol gives immediate turbidity.

(A) The general reaction of an alcohol $(R-OH)$ with sodium $(Na)$ is given by:
$2 R-OH + 2 Na \longrightarrow 2 R-ONa + H_2$
From the balanced chemical equation,$2 \ mol$ of alcohol produces $1 \ mol$ of hydrogen gas $(H_2)$.
Therefore,$1 \ mol$ of alcohol produces $\frac{1}{2} \ mol$ of $H_2$.
The molar mass of $H_2$ is $2 \ g/mol$.
Weight of $H_2 = \text{moles} \times \text{molar mass} = \frac{1}{2} \ mol \times 2 \ g/mol = 1 \ g$.
Thus,$1 \ g$ of hydrogen is produced.

(C) The mixture of anhydrous $ZnCl_2$ and conc. $HCl$ is known as Lucas reagent.
Lucas test is used to distinguish between primary,secondary,and tertiary alcohols.
Tertiary alcohols react immediately with Lucas reagent to produce turbidity due to the formation of alkyl chlorides.
Secondary alcohols give turbidity within $5-10 \ \text{min}$,while primary alcohols do not produce turbidity at room temperature.
Among the given options,$2-\text{hydroxy}-2-\text{methylpropane}$ is a $3^{\circ}$ alcohol,which is the most reactive and produces turbidity immediately.

(C) When $Ethyl \ alcohol$ $(C_{2}H_{5}OH)$ is treated with $H_{2}SO_{4}$ at $413 \ K$,the reaction proceeds via an $S_{N}2$ mechanism to form $Diethyl \ ether$ $(C_{2}H_{5}OC_{2}H_{5})$.
Step $1$: Protonation of $Ethyl \ alcohol$ to form protonated $Ethyl \ alcohol$.
Step $2$: Nucleophilic attack of another molecule of $Ethyl \ alcohol$ on the protonated $Ethyl \ alcohol$ with the loss of a water molecule.
Step $3$: Loss of a proton to form $Diethyl \ ether$.

(C) An allylic alcohol is one in which the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond $(C=C-C-OH)$.
$A$. $H_2C=CH-CH_2-OH$: Primary allylic alcohol ($-OH$ is on a primary carbon).
$B$. $CH_3-CH=CH-CH_2-OH$: Primary allylic alcohol ($-OH$ is on a primary carbon).
$C$. $H_2C=CH-CH(OH)-CH_3$: Secondary allylic alcohol ($-OH$ is on a secondary carbon).
$D$. $H_2C=CH-C(OH)(CH_3)_2$: Tertiary allylic alcohol ($-OH$ is on a tertiary carbon).
Therefore,the correct option is $C$.

(D) The chemical structure of propylene glycol (propane-$1,2$-diol) is $CH_3-CH(OH)-CH_2OH$.
As shown in the structure,there are two hydroxyl $(-OH)$ groups attached to the carbon chain.
Therefore,the number of hydroxyl groups present in propylene glycol is $2$.

(C) Ethylene glycol $(HO-CH_2-CH_2-OH)$ is used as an antifreeze agent in automobile radiators because it lowers the freezing point of water,preventing it from freezing in cold weather.

(C) The term 'anhydride' in this context refers to the compound formed by the removal of a water molecule $(H_2O)$ from a diol.
$2,3-$butanediol has the structure $CH_3-CH(OH)-CH(OH)-CH_3$.
Upon dehydration,it loses a water molecule to form an epoxide (oxirane derivative).
The reaction is: $CH_3-CH(OH)-CH(OH)-CH_3 \xrightarrow{-H_2O} CH_3-CH(O)CH-CH_3$.
The product formed is $2,3-$epoxybutane.
Therefore,$2,3-$epoxybutane is the anhydride of $2,3-$butanediol.

(B) The reduction of aldehydes gives primary alcohols,and the reduction of ketones gives secondary alcohols.
$1$. Butan$-1-$ol (primary alcohol) can be obtained by the reduction of butanal.
$2$. Butan$-2-$ol (secondary alcohol) can be obtained by the reduction of butan$-2-$one.
$3$. $2-$methyl propan$-1-$ol (primary alcohol) can be obtained by the reduction of $2-$methyl propanal.
$4$. $2-$methyl propan$-2-$ol is a tertiary alcohol. Tertiary alcohols cannot be prepared by the reduction of carbonyl compounds (aldehydes or ketones) as they require the addition of a Grignard reagent to a ketone.

(B) Solubility in water depends on the ability of the compound to form hydrogen bonds with water molecules.
$tert$-Butyl alcohol ($CH_3)_3COH$ has a small hydrophobic alkyl group relative to the hydrophilic hydroxyl group,allowing it to be miscible in water.
Phenol has a large hydrophobic benzene ring,which reduces its solubility compared to lower alcohols.
$o$-Nitrophenol exhibits intramolecular hydrogen bonding,which decreases its solubility in water.
$p$-Nitrophenol has intermolecular hydrogen bonding,but the hydrophobic benzene ring still limits its solubility compared to $tert$-butyl alcohol.
Therefore,$tert$-butyl alcohol has the highest solubility among the given options.

(A) The reaction of an organic compound '$A$' with $PCl_3$ to form an alkyl chloride $(C_3H_7Cl)$ suggests that '$A$' is a primary alcohol,$CH_3-CH_2-CH_2-OH$ (Propan$-1-$ol).
Reaction $1$: $CH_3-CH_2-CH_2-OH + PCl_3 \rightarrow CH_3-CH_2-CH_2-Cl + H_3PO_3$.
Reaction $2$: Oxidation of a primary alcohol with $PCC$ (Pyridinium chlorochromate) yields an aldehyde.
$CH_3-CH_2-CH_2-OH \xrightarrow{PCC} CH_3-CH_2-CHO$ (Propanal,$C_3H_6O$).
Thus,'$A$' is Propan$-1-$ol.

(B) The given reaction involves the dehydrogenation of a secondary alcohol $(R-CH(OH)-R)$ in the presence of copper $(Cu)$ catalyst at $573 \ K$.
Secondary alcohols undergo dehydrogenation to form ketones as the product $P$.
The reaction is: $R-CH(OH)-R \xrightarrow{Cu, 573 \ K} R-CO-R + H_2$.
Therefore,the product $P$ is $R-CO-R$.

(A) The solubility of organic compounds in water depends on their ability to form hydrogen bonds with water molecules.
$1$. Alcohols $(R-OH)$ can form strong hydrogen bonds with water due to the highly electronegative oxygen atom.
$2$. Amines $(R-NH_2)$ can also form hydrogen bonds with water, but the $N-H \cdots O$ hydrogen bond is weaker than the $O-H \cdots O$ hydrogen bond because nitrogen is less electronegative than oxygen.
$3$. Alkanes are non-polar hydrocarbons and cannot form hydrogen bonds with water, making them essentially insoluble.
Therefore, the correct decreasing order of solubility is $Alcohols > Amines > Alkanes$.

(C) Step $1$: Reaction of isopropylamine with $HNO_2$ gives propan-$2$-ol $(A)$.
$CH_3-CH(NH_2)-CH_3 + HNO_2 \rightarrow CH_3-CH(OH)-CH_3 + N_2 + H_2O$
Step $2$: Oxidation of propan-$2$-ol $(A)$ gives acetone $(B)$.
$CH_3-CH(OH)-CH_3 \xrightarrow{[O]} CH_3-CO-CH_3$
Step $3$: Reaction of acetone $(B)$ with methylmagnesium iodide $(CH_3MgI)$ followed by hydrolysis gives $2$-methylpropan-$2$-ol $(C)$.
$CH_3-CO-CH_3 + CH_3MgI$ $\rightarrow (CH_3)_3C-OMgI$ $\xrightarrow{H_2O/H^+} (CH_3)_3C-OH$
Thus,the final product $C$ is $2$-methylpropan-$2$-ol.

(A) The boiling point depends on the strength of intermolecular forces.
$CH_3(CH_2)_2CH_2OH$ is a primary alcohol,which exhibits strong intermolecular hydrogen bonding due to the highly electronegative oxygen atom.
$CH_3(CH_2)_2CH_2NH_2$ and $(C_2H_5)_2NH$ are amines,which also exhibit hydrogen bonding,but the $N-H$ bond is less polar than the $O-H$ bond,resulting in weaker hydrogen bonding.
$C_2H_5CH(CH_3)_2$ is an alkane,which only has weak London dispersion forces.
Therefore,the alcohol has the highest boiling point.

(B) An allylic alcohol is one where the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond $(C=C-C-OH)$.
$A$ $1^{\circ}$ allylic alcohol has the $-OH$ group attached to a primary carbon ($CH_2$ group) which is adjacent to a double bond.
$1.$ $CH_2=CH-CH(CH_3)-OH$: The $-OH$ is on a secondary carbon,so it is a $2^{\circ}$ allylic alcohol.
$2.$ $CH_2=CH-CH_2-OH$: The $-OH$ is on a primary carbon adjacent to the double bond,so it is a $1^{\circ}$ allylic alcohol.
$3.$ $CH_2=CH-C(CH_3)_2-OH$: The $-OH$ is on a tertiary carbon,so it is a $3^{\circ}$ allylic alcohol.
$4.$ $CH_3-CH=CH-CH_2-CH_2-OH$: The $-OH$ is on a carbon that is not adjacent to the double bond,so it is not an allylic alcohol.
Therefore,the correct option is $B$.

(C) trihydric alcohol is an alcohol that contains three hydroxyl $(-OH)$ groups in its molecule.
$1$. Quinol (Benzene$-1,4-$diol) has two $-OH$ groups (dihydric).
$2$. Catechol (Benzene$-1,2-$diol) has two $-OH$ groups (dihydric).
$3$. Glycerol $(CH_2(OH)-CH(OH)-CH_2(OH))$ has three $-OH$ groups (trihydric).
$4$. Resorcinol (Benzene$-1,3-$diol) has two $-OH$ groups (dihydric).
Therefore,the correct answer is Glycerol.

(B) primary alcohol is one in which the hydroxyl group $(-OH)$ is attached to a primary carbon atom (a carbon atom attached to only one other carbon atom).
$1$. $2-$methylpropan$-2-$ol: The $-OH$ group is attached to a tertiary carbon atom. It is a tertiary alcohol.
$2$. $2-$methylpropan$-1-$ol: The structure is $(CH_3)_2CH-CH_2OH$. The $-OH$ group is attached to a primary carbon atom. It is a primary alcohol.
$3$. Cyclobutanol: The $-OH$ group is attached to a secondary carbon atom. It is a secondary alcohol.
$4$. butan$-2-$ol: The $-OH$ group is attached to a secondary carbon atom. It is a secondary alcohol.
Therefore,$2-$methylpropan$-1-$ol is a primary alcohol.

(B) secondary alcohol is an alcohol where the hydroxyl group $(-OH)$ is attached to a carbon atom that is connected to two other carbon atoms.
In the structure $(CH_3)_2CH-CH(OH)-CH_2-CH_2-CH_3$,the carbon bearing the $-OH$ group is connected to an isopropyl group and a propyl group,making it a secondary alcohol.

(B) An allylic alcohol is one where the $-OH$ group is attached to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond $(C=C)$.
$1$. $A$ secondary allylic alcohol is one where the carbon atom attached to the $-OH$ group is bonded to two other carbon atoms.
$2$. Let us analyze the options:
- Option $A$: $CH_2=CH-C(CH_3)_2-OH$ is a tertiary allylic alcohol.
- Option $B$: $CH_2=CH-CH(CH_3)-OH$ is a secondary allylic alcohol because the carbon attached to $-OH$ is bonded to one $-CH_3$ group and one $-CH=CH_2$ group.
- Option $C$: $CH_2=CH-CH_2-OH$ is a primary allylic alcohol.
- Option $D$: $CH_3-CH=CH-CH_2-OH$ is a primary allylic alcohol.
Therefore,the correct option is $B$.

(B) An allylic alcohol is one where the $-OH$ group is attached to a carbon atom adjacent to a carbon-carbon double bond.
For an allylic alcohol to be secondary,the carbon atom bearing the $-OH$ group must be attached to two other carbon atoms.
In $CH_2=CH-CH(OH)-CH_3$ (but$-3-$en$-2-$ol),the $-OH$ group is on the $C-2$ carbon,which is adjacent to the double bond $(C-3=C-4)$ and is attached to two other carbon atoms ($C-1$ and $C-3$).
Thus,but$-3-$en$-2-$ol is an allylic secondary alcohol.

(C) benzylic alcohol is one where the hydroxyl group $(-OH)$ is attached to a carbon atom that is directly bonded to a benzene ring.
In a tertiary $(3^{\circ})$ alcohol,the carbon atom bearing the $-OH$ group is attached to three other carbon atoms.
Let's analyze the structures:
$A$: Phenyl methanol $(C_6H_5CH_2OH)$ - The carbon attached to $-OH$ is bonded to one carbon (the benzene ring),making it a primary $(1^{\circ})$ alcohol.
$B$: $1$-phenyl ethanol $(C_6H_5CH(OH)CH_3)$ - The carbon attached to $-OH$ is bonded to two carbons,making it a secondary $(2^{\circ})$ alcohol.
$C$: $2$-phenyl propan-$2$-ol $(C_6H_5C(CH_3)_2OH)$ - The carbon attached to $-OH$ is bonded to three carbons (two methyl groups and one phenyl group),making it a tertiary $(3^{\circ})$ alcohol.
$D$: $1$-phenyl propan-$2$-ol $(C_6H_5CH_2CH(OH)CH_3)$ - This is not a benzylic alcohol because the $-OH$ is not on the carbon directly attached to the benzene ring.
Therefore,the correct option is $C$.