Properties of alcohol Questions in English

(A) Although ethanol and dimethyl ether have the same molecular formula $C_2H_6O$,ethanol has a boiling point of $78.2 \ ^\circ C$ while dimethyl ether has a boiling point of $-23.6 \ ^\circ C$.
This difference is due to the presence of intermolecular hydrogen bonding in ethanol,which is absent in dimethyl ether.

(B) Methanol $(CH_3OH)$ and ethanol $(C_2H_5OH)$ contain a polar covalent $O-H$ bond.
Because of this,they are capable of forming hydrogen bonds with water molecules.
This ability to form hydrogen bonds makes them miscible in water.

(B) The boiling point depends on the strength of intermolecular forces.
Ethyl alcohol $(C_2H_5OH)$ exhibits intermolecular hydrogen bonding,which is a strong dipole-dipole interaction.
Acetone,diethyl ether,and chloroform exhibit weaker dipole-dipole or London dispersion forces.
Therefore,ethyl alcohol has the highest boiling point among the given options.

(D) Glycerol $(CH_2OH-CHOH-CH_2OH)$ contains three $-OH$ groups,which facilitate extensive intermolecular hydrogen bonding. This strong network of hydrogen bonds restricts the flow of molecules,making it highly viscous.

(C) The metal used to remove the last traces of water from alcohol is $Calcium$ $(Ca)$. $Calcium$ reacts with water to form $Calcium$ hydroxide $(Ca(OH)_2)$,which is a solid and can be easily separated,thereby drying the alcohol.

(C) Lead is resistant to corrosion by dilute $H_2SO_4$ and $HCl$ due to the formation of an insoluble protective layer of $PbSO_4$ or $PbCl_2$ on its surface.
However,lead is readily corroded by organic acids like acetic acid in the presence of oxygen (air).
The reaction is: $Pb + 2CH_3COOH + \frac{1}{2}O_2 \to Pb(CH_3COO)_2 + H_2O$.

(D) The heterolytic cleavage of a bond depends on the difference in electronegativity between the bonded atoms.
In $CH_3CH_2OH$,the $O-H$ bond is the most polar because oxygen is significantly more electronegative than hydrogen.
Therefore,the $O-H$ bond undergoes heterolytic cleavage most readily to form $CH_3CH_2O^-$ and $H^+$,or in some cases $CH_3CH_2^+$ and $OH^-$,depending on the reaction conditions.
Among the given options,the $O-H$ bond is the most susceptible to heterolytic cleavage due to the high electronegativity of the oxygen atom.

(C) primary alcohol is defined by the structure $R-CH_2-OH$. The isomers of $C_5H_{11}OH$ that are primary alcohols are:
$1$. $CH_3-CH_2-CH_2-CH_2-CH_2-OH$ (pentan-$1$-ol)
$2$. $CH_3-CH(CH_3)-CH_2-CH_2-OH$ ($3$-methylbutan-$1$-ol)
$3$. $CH_3-CH_2-CH(CH_3)-CH_2-OH$ ($2$-methylbutan-$1$-ol)
$4$. $CH_3-C(CH_3)_2-CH_2-OH$ ($2,2$-dimethylpropan-$1$-ol)
Thus,there are $4$ primary alcohol isomers.

(A) $1$. $CH_3-CH_2-CH_2-CH_2-OH$ (butan$-1-$ol)
$2$. $CH_3-CH(OH)-CH_2-CH_3$ (butan$-2-$ol)
$3$. $CH_3-CH(CH_3)-CH_2-OH$ ($2-$methylpropan$-1-$ol)
$4$. $CH_3-C(OH)(CH_3)-CH_3$ ($2-$methylpropan$-2-$ol)
There are a total of $4$ possible alcoholic isomers for the molecular formula $C_4H_{10}O$.

(A) The dehydration of $butan-2-ol$ $(CH_3CH_2CH(OH)CH_3)$ in the presence of $H^{+}$ involves the elimination of a water molecule to form an alkene $[F]$.
The possible alkenes formed are:
$1$. $But-1-ene$ $(CH_3CH_2CH=CH_2)$
$2$. $But-2-ene$ $(CH_3CH=CHCH_3)$
Both $But-1-ene$ and $But-2-ene$ are structural isomers of each other.
Therefore,there are $2$ possible structural isomers for $[F]$.

(A) The dehydration of alcohols follows the order: $3^\circ > 2^\circ > 1^\circ$ alcohol,because the rate-determining step involves the formation of a carbocation intermediate.
$CH_3-CH_2-C(OH)(CH_3)-CH_2-CH_3$ is a $3^\circ$ alcohol,which forms a stable $3^\circ$ carbocation upon dehydration.
$CH_3-CH_2-CH_2-CH(OH)-CH_3$ is a $2^\circ$ alcohol.
$CH_3-CH_2-CH_2-CH_2-CH_2-OH$ and $CH_3-CH_2-CH(CH_3)-CH_2-CH_2-OH$ are $1^\circ$ alcohols.
Since the $3^\circ$ carbocation is the most stable,the $3^\circ$ alcohol undergoes dehydration most easily.

(B) Dehydration of alcohols involves the formation of a carbocation intermediate.
$3^{\circ}$ alcohols form $3^{\circ}$ carbocations,which are the most stable due to the inductive effect and hyperconjugation.
In $(CH_3)_3COH$,the dehydration leads to the formation of the tert-butyl carbocation,$(CH_3)_3C^+$,which is a $3^{\circ}$ carbocation and highly stable.
Therefore,the correct option is $B$.

(A) Grignard reagents $(RMgX)$ are strong bases and react with compounds containing active hydrogen atoms (like alcohols,water,or amines) to form alkanes.
The reaction is as follows:
$C_2H_5OH + CH_3MgBr \rightarrow CH_4 + Mg(OC_2H_5)Br$
Here,the $CH_3^-$ group from the Grignard reagent abstracts the acidic proton $(H^+)$ from the ethyl alcohol $(C_2H_5OH)$ to form methane $(CH_4)$.

(A) Compounds containing active hydrogen atoms (such as those in $ROH, H_2O, RNH_2$,or $RCOOH$) react with Grignard reagents $(R'MgX)$ to form hydrocarbons (alkanes) via an acid-base reaction.
$CH_3CH_2OH + CH_3MgBr \rightarrow CH_4 + Mg(Br)(OCH_2CH_3)$
In this reaction,the ethanol $(CH_3CH_2OH)$ provides an active hydrogen atom attached to the oxygen,which reacts with the alkyl group of the Grignard reagent to produce methane $(CH_4)$,which is a hydrocarbon. Other options like aldehydes $(CH_3CHO)$ and ketones $(CH_3COCH_3)$ undergo nucleophilic addition reactions with Grignard reagents to form alcohols,not hydrocarbons.

(D) $CH_3MgI$ is a Grignard reagent,which acts as a strong base and a nucleophile.
It reacts with compounds containing active hydrogen atoms (like those attached to $O$,$N$,or $S$) to form alkanes.
$(a)$ $CH_3MgI + C_2H_5OH \to CH_4 + C_2H_5OMgI$
$(b)$ $CH_3MgI + CH_3-CH_2-NH_2 \to CH_4 + CH_3CH_2NHMgI$
Since both $C_2H_5OH$ and $CH_3-CH_2-NH_2$ contain active hydrogen atoms,both will produce methane.

(A) The dehydration of ethanol $(C_2H_5OH)$ with concentrated $H_2SO_4$ at $160-170\,^{\circ}C$ leads to the formation of ethylene $(CH_2=CH_2)$ through an elimination reaction.
The chemical equation is:
$CH_3-CH_2-OH \xrightarrow[160-170\,^{\circ}C]{Conc. H_2SO_4} CH_2=CH_2 + H_2O$

(C) The dehydration of $3, 3-$dimethyl$-2-$butanol with ${H_2SO_4}$ proceeds via the formation of a carbocation.
$1.$ Protonation of the alcohol group leads to the formation of a $2^o$ carbocation: $CH_3-CH^+-C(CH_3)_2-CH_3$.
$2.$ This $2^o$ carbocation undergoes a $1, 2-$methyl shift to form a more stable $3^o$ carbocation: $CH_3-CH(CH_3)-C^+(CH_3)-CH_3$.
$3.$ Loss of a proton from the $3^o$ carbocation yields the most stable alkene,which is $2, 3-$dimethyl$-2-$butene,according to Saytzeff's rule.

(C) $1$. To decolorize aqueous bromine,the compound must contain an unsaturated bond (alkene or alkyne).
$2$. To give white fumes of $HCl$ on reaction with $PCl_5$,the compound must contain an alcoholic group $(-OH)$.
$3$. Option $C$ $(CH_3CH = CHCH_2CH_2OH)$ contains both a double bond (for bromine decolorization) and an $-OH$ group (for $HCl$ fumes).
$4$. The reaction is: $CH_3CH = CHCH_2CH_2OH + PCl_5 \rightarrow CH_3CH = CHCH_2CH_2Cl + HCl + POCl_3$.

(C) When ethyl alcohol $(CH_3CH_2OH)$ is heated with concentrated $H_2SO_4$ at $443 \ K$,it undergoes dehydration to form ethene $(C_2H_4)$.
$CH_3-CH_2-OH \xrightarrow[443 \ K]{Conc. \ H_2SO_4} CH_2=CH_2 + H_2O$

(A) According to $Saytzeff's$ rule,in the dehydration of alcohols,the more stable (more substituted) alkene is the major product.
$CH_3-CH=CH-CH_3$ $(but-2-ene)$ is more substituted and stable than $CH_2=CH-CH_2-CH_3$ $(but-1-ene)$.
Therefore,$CH_3-CH=CH-CH_3$ is the major product.

(C) Grignard reagents $(R-MgX)$ act as strong bases and react with compounds containing active hydrogen atoms (such as alcohols,$R'-OH$) to form alkanes $(R-H)$.
In this reaction,the isobutyl group $(CH_3-CH(CH_3)-CH_2^-)$ from the Grignard reagent abstracts the acidic proton from the ethyl alcohol to form isobutane $(CH_3-CH(CH_3)-CH_3)$.
The reaction is:
$CH_3-CH(CH_3)-CH_2-MgBr + CH_3-CH_2-OH \rightarrow CH_3-CH(CH_3)-CH_3 + CH_3-CH_2-OMgBr$.

(D) The correct answer is $D$. When vapours of a tertiary alcohol are passed over heated copper $(Cu)$ at $573 \, K$,it undergoes dehydration to form an alkene because tertiary alcohols lack $\alpha$-hydrogen atoms required for dehydrogenation.
$CH_3-C(OH)(CH_3)-CH_3 \xrightarrow{Cu, 573 \, K} CH_3-C(CH_3)=CH_2 + H_2O$

(B) The reaction involves the removal of a water molecule $(H_2O)$ from an alcohol $(2\text{-methylpropan-}2\text{-ol})$ in the presence of a dehydrating agent like $H_2SO_4$ to form an alkene $(2\text{-methylpropene})$.
This process is known as a dehydration reaction.

(D) The reaction of ethanol with concentrated sulfuric acid at $110 \, ^\circ C$ leads to the formation of ethyl hydrogen sulphate.
$CH_3CH_2OH + H_2SO_4 \xrightarrow{110 \, ^\circ C} CH_3CH_2HSO_4 + H_2O$
Thus,ethyl hydrogen sulphate is obtained from ethanol.

(B) The reaction is an acid-catalyzed dehydration of ethanol.
When ethanol $(C_2H_5OH)$ is heated with concentrated $H_2SO_4$ at $165^{\circ}C - 170^{\circ}C$,it undergoes intermolecular dehydration to form ethene $(CH_2 = CH_2)$ and water.
The chemical equation is: $C_2H_5OH \xrightarrow{\text{Conc. } H_2SO_4, 170^{\circ}C} CH_2 = CH_2 + H_2O$.

(A) The reaction sequence is as follows:
$1$. Hydration of ethyne: $CH \equiv CH + H_2O \xrightarrow{HgSO_4/H_2SO_4} CH_3CHO$ (Ethanal)
$2$. Reaction with Grignard reagent: $CH_3CHO + CH_3MgBr \xrightarrow{H_2O} CH_3-CH(OH)-CH_3$ (Propan$-2-$ol)
$3$. Bromination: $CH_3-CH(OH)-CH_3 \xrightarrow{P/Br_2} CH_3-CH(Br)-CH_3$ ($2$-Bromopropane).

(B) The dehydration of $2-$butanol $(CH_3-CH(OH)-CH_2-CH_3)$ involves the removal of a water molecule to form an alkene.
According to the $Saytzeff$ rule,in elimination reactions,the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons) is the major product.
In this case,$2-$butene $(CH_3-CH=CH-CH_3)$ is more substituted than $1-$butene $(CH_2=CH-CH_2-CH_3)$,hence it is the major product.

(A) The reaction of ethyl alcohol with $HI$ in the presence of red phosphorus is a reduction reaction.
Red phosphorus acts as a reducing agent that reduces the alcohol to the corresponding alkane.
The chemical equation is:
$CH_3CH_2OH + 2HI \xrightarrow{\text{Red } P} CH_3CH_3 + H_2O + I_2$
Thus,ethyl alcohol is reduced to ethane $(C_2H_6)$.

(C) $CH_2=CH_2 \xrightarrow{HBr} CH_3-CH_2Br (X)$
$CH_3-CH_2Br \xrightarrow{Hydrolysis} CH_3-CH_2OH (Y)$
$CH_3-CH_2OH \xrightarrow[I_2 \ \text{excess}]{Na_2CO_3} CHI_3 (Z) + HCOONa + NaI + H_2O$
This is the iodoform test,where ethanol reacts with iodine in the presence of base to form a yellow precipitate of iodoform $(CHI_3)$.

(A) Lucas reagent is a solution of anhydrous $ZnCl_2$ in concentrated $HCl$.
This reagent is used to classify alcohols of low molecular weight. The reaction involves the substitution of the hydroxyl group $(-OH)$ with a chloride ion $(Cl^-)$.
The reaction proceeds via an $SN1$ mechanism,where the rate of reaction depends on the stability of the carbocation intermediate. Tertiary alcohols react fastest,followed by secondary,while primary alcohols react very slowly or not at all at room temperature.
Therefore,the correct composition is anhydrous $ZnCl_2$ and concentrated $HCl$.

(B) The reaction of an alcohol with thionyl chloride $(SOCl_2)$ in the presence of pyridine is a standard method for the preparation of alkyl chlorides.
This specific reaction is known as Darzen's procedure.
It is preferred because the by-products ($SO_2$ and $HCl$) are gases,which escape,leaving behind pure alkyl chloride.

(D) The reaction of ethyl alcohol with thionyl chloride in the presence of pyridine is known as the Darzens process.
The reaction is: $C_2H_5OH + SOCl_2 \xrightarrow{\text{Pyridine}} C_2H_5Cl + SO_2 + HCl$.
Pyridine is used to neutralize the $HCl$ produced in the reaction,which drives the reaction to completion.

(A) The reactivity of hydrogen halides $(HX)$ towards alcohols 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.
Therefore,the ease of cleavage of the $H-X$ bond increases in the order $HF < HCl < HBr < HI$.
Thus,the decreasing order of reactivity is $HI > HBr > HCl > HF$.

(A) The reaction $R-OH + HX \to R-X + H_2O$ proceeds via the formation of a carbocation intermediate.
The stability of the carbocation formed determines the reactivity of the alcohol.
The stability order of carbocations is $3^o > 2^o > 1^o$.
Therefore,the reactivity order of alcohols towards $HX$ is $3^o > 2^o > 1^o$ (Tertiary > Secondary > Primary).

(A) The reaction of ethyl alcohol $(C_2H_5OH)$ with $KI$ in the presence of $Na_2CO_3$ is an iodoform test.
Ethyl alcohol is oxidized to acetaldehyde,which then reacts with iodine (generated in situ) in the presence of base to form iodoform $(CHI_3)$.
$C_2H_5OH + 4I_2 + 6Na_2CO_3 \rightarrow CHI_3 + HCOONa + 5NaI + 5NaHCO_3$.
$CHI_3$ appears as yellow crystals.

(D) $CaOCl_2 + H_2O \to Ca(OH)_2 + Cl_2$
$CH_3-CH(OH)-CH_3 + Cl_2 \to CH_3-CO-CH_3 + 2HCl$
$CH_3-CO-CH_3 + 3Cl_2 \to CCl_3-CO-CH_3 + 3HCl$
$2CCl_3-CO-CH_3 + Ca(OH)_2 \to 2CHCl_3 + (CH_3COO)_2Ca$
Chloroform is obtained from Propanol-$2$ via the haloform reaction.

(A) Lucas reagent is a solution of anhydrous zinc chloride $(ZnCl_2)$ in concentrated hydrochloric acid $(HCl)$.
It is primarily used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction to form alkyl chlorides.

(C) $C_2H_5OH$ (ethanol) undergoes the haloform reaction with bleaching powder $(CaOCl_2)$.
$1. CaOCl_2 + H_2O \to Ca(OH)_2 + Cl_2$
$2. CH_3CH_2OH + Cl_2 \to CH_3CHO + 2HCl$
$3. CH_3CHO + 3Cl_2 \to CCl_3CHO + 3HCl$
$4. 2CCl_3CHO + Ca(OH)_2 \to 2CHCl_3 + (HCOO)_2Ca$
Thus,ethanol produces trichloromethane $(CHCl_3)$.

(A) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$C_2H_5OH$ (ethanol) contains the $CH_3CH(OH)-$ group.
When $C_2H_5OH$ is warmed with $I_2$ and $NaOH$,it undergoes the iodoform reaction to produce yellow crystals of iodoform $(CHI_3)$.
The reaction is: $C_2H_5OH + 4I_2 + 6NaOH \rightarrow CHI_3 + HCOONa + 5NaI + 5H_2O$.

(A) $CH_3-CH_2-CH_2-Br$ undergoes dehydrohalogenation with alcoholic $KOH$ to form propene $(A)$.
Propene then reacts with $HBr$ via Markovnikov addition to form $2$-bromopropane $(B)$.
Finally,$2$-bromopropane reacts with aqueous $KOH$ via nucleophilic substitution to form propan-$2$-ol $(C)$.
$CH_3-CH_2-CH_2-Br$ $\xrightarrow{KOH(alc)} CH_3-CH=CH_2 (A)$ $\xrightarrow{HBr} CH_3-CH(Br)-CH_3 (B)$ $\xrightarrow{KOH(aq.)} CH_3-CH(OH)-CH_3 (C)$

(A) Phenyl magnesium bromide $(PhMgBr)$ is a Grignard reagent,which acts as a strong base.
$t-$butanol $((CH_3)_3COH)$ contains an acidic proton on the hydroxyl group.
When they react,the Grignard reagent abstracts the acidic proton from the alcohol to form an alkane.
The reaction is: $(CH_3)_3COH + PhMgBr \to PhH + (CH_3)_3COMgBr$.
Thus,the product formed is benzene $(PhH)$.

(B) The structure of But$-2-$ol is $CH_3-CH(OH)-CH_2-CH_3$.
In this molecule,the hydroxyl group $(-OH)$ is attached to a carbon atom that is further bonded to two other carbon atoms.
Therefore,it is a secondary $(2^{\circ})$ alcohol.

(C) The structure of $pent-3-ol$ is $CH_3-CH_2-CH(OH)-CH_2-CH_3$.
In this molecule,the hydroxyl group $(-OH)$ is attached to a carbon atom that is bonded to two other carbon atoms.
Therefore,it is a secondary alcohol.

(B) The structural formula of Glycerine (Glycerol) is $CH_2(OH)-CH(OH)-CH_2(OH)$.
In this structure,the two terminal $-OH$ groups are attached to primary carbon atoms,and the middle $-OH$ group is attached to a secondary carbon atom.
Therefore,it contains two primary and one secondary $-OH$ groups.

(C) In a tertiary alcohol,the hydroxyl group $(-OH)$ is attached to a tertiary carbon atom (a carbon atom bonded to three other carbon atoms).
In $CH_3-C(CH_3)(OH)-CH_3$ ($2$-methylpropan-$2$-ol),the carbon atom bearing the $-OH$ group is bonded to three methyl groups,making it a tertiary alcohol.

(B) primary alcohol is one in which the $-OH$ group is attached to a primary carbon atom (a carbon atom bonded to only one other carbon atom).
$1$. Butan-$1$-ol $(CH_3CH_2CH_2CH_2OH)$: The $-OH$ group is attached to a terminal carbon,which is bonded to only one other carbon. Thus,it is a primary alcohol.
$2$. Butan-$2$-ol $(CH_3CH(OH)CH_2CH_3)$: The $-OH$ group is attached to a carbon bonded to two other carbons. Thus,it is a secondary alcohol.
$3$. Propan-$2$-ol (Isopropyl alcohol) $(CH_3CH(OH)CH_3)$: The $-OH$ group is attached to a carbon bonded to two other carbons. Thus,it is a secondary alcohol.
Therefore,butan-$1$-ol is the primary alcohol.

(B) Cyclohexanol is a secondary alcohol because the $-OH$ group is linked to a $2^o$ carbon atom in the cyclohexane ring.

(B) Methylated spirit is a mixture of $5-10 \%$ methanol and the remaining ethanol.
It is also known as denatured alcohol because the addition of methanol makes it toxic and unfit for human consumption.