Properties of alcohol Questions in English

(A) In step-$1$: $CH_3 CH_2 OH + Cl_2 \rightarrow CH_3 CHO + 2HCl$. Here,$Cl_2$ acts as an oxidizing agent,converting the primary alcohol $(-CH_2OH)$ into an aldehyde $(-CHO)$.
In step-$2$: $CH_3 CHO + 3Cl_2 \rightarrow CCl_3 CHO + 3HCl$. Here,$Cl_2$ replaces the hydrogen atoms of the methyl group with chlorine atoms,which is a chlorination reaction (specifically,a substitution reaction).

(A) Ethyl alcohol $(C_2H_5OH)$ contains an active hydrogen atom attached to an oxygen atom. When it reacts with a Grignard reagent like methyl magnesium iodide $(CH_3MgI)$,the active hydrogen is replaced by the methyl group to form methane gas $(CH_4)$.
The reaction is as follows:
$C_2H_5OH + CH_3MgI \rightarrow CH_4 \uparrow + C_2H_5OMgI$

(D) The balanced chemical equation for the reaction of ethanol with phosphorus tribromide is:
$3 CH_3CH_2OH + PBr_3 \rightarrow 3 CH_3CH_2Br + H_3PO_3$
In this reaction,$3$ moles of ethanol react with $1$ mole of $PBr_3$ to produce $3$ moles of bromoethane and $1$ mole of phosphorous acid $(H_3PO_3)$.
Therefore,$X$ is $H_3PO_3$.

(C) The reaction of ethanol with $PCl_5$ is as follows:
$C_2H_5OH + PCl_5 \rightarrow C_2H_5Cl (A) + POCl_3 + HCl$
Thus,$A$ is $C_2H_5Cl$ (ethyl chloride).
Next,the reaction of ethyl chloride $(A)$ with silver nitrite $(AgNO_2)$ is:
$C_2H_5Cl + AgNO_2 \rightarrow C_2H_5NO_2 (B) + AgCl$
Thus,$B$ is $C_2H_5NO_2$ (nitroethane).
Therefore,$A$ and $B$ are $C_2H_5Cl$ and $C_2H_5NO_2$ respectively.

(A) The dehydration of alcohols in the presence of $H_3PO_4$ proceeds via the formation of a carbocation intermediate. The rate of reaction depends on the stability of the carbocation formed.
$I$: Phenol (or cyclohexadienol derivative) forms a highly stable aromatic carbocation (benzene ring system).
$II$: Cyclohexadienol forms a conjugated diene system upon dehydration.
$III$: Cyclohexenol forms a simple alkene.
Comparing the stability of the carbocations formed:
$1$. Compound $II$ (cyclohexa$-1,4-$dien$-1-$ol) leads to the formation of benzene,which is highly stable due to aromaticity.
$2$. Compound $III$ (cyclohex$-2-$en$-1-$ol) leads to the formation of cyclohexadiene.
$3$. Compound $I$ (phenol) is not an alcohol in the aliphatic sense and does not undergo dehydration to an alkene under these conditions; however,based on the provided structures,$II$ is the most reactive due to the formation of aromatic benzene,followed by $III$,and $I$ is the least reactive.
Thus,the correct order of reactivity is $II > III > I$.

(C) The reaction of ethanol $(C_2H_5OH)$ with chlorine $(Cl_2)$ proceeds as follows:
$1$. Ethanol is oxidized by chlorine to form acetaldehyde $(CH_3CHO)$,which is $X$.
$CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl$
$2$. Acetaldehyde then reacts with further chlorine to form chloral $(CCl_3CHO)$,which is $Y$.
$CH_3CHO + 3Cl_2 \rightarrow CCl_3CHO + 3HCl$
Thus,$X$ is $CH_3CHO$ and $Y$ is $CCl_3CHO$.

(C) $1$. The reaction of propanone $(CH_3)_2CO$ with methylmagnesium bromide $(CH_3MgBr)$ in the presence of ether forms an addition product $A$ (an alkoxide complex).
$2$. Acidic hydrolysis $(H_3O^+)$ of $A$ yields $B$,which is $2-$methylpropan$-2-$ol $(CH_3)_3C-OH$ (a tertiary alcohol).
$3$. When tertiary alcohols are passed over heated copper $(Cu)$ at $573 \ K$,they undergo dehydration to form alkenes.
$4$. Therefore,$2-$methylpropan$-2-$ol undergoes dehydration to form $2-$methylprop$-1-$ene $(CH_3)_2C=CH_2$ as the final product $C$.

(B) The reaction in $(A)$ is not feasible because the $C-O$ bond in phenol has partial double bond character due to resonance,making it resistant to nucleophilic substitution by $SOCl_2$.
The reaction in $(B)$ is feasible. $PCC$ (Pyridinium chlorochromate) is a selective oxidizing agent that oxidizes primary allylic alcohols to aldehydes without affecting the double bond.
The reaction in $(C)$ is not feasible because $KMnO_4/H^{+}$ is a strong oxidizing agent that will oxidize the primary alcohol directly to a carboxylic acid $(CH_3CH_2COOH)$,not an aldehyde.
The reaction in $(D)$ is not feasible because $20\%\ H_3PO_4$ is a dilute acid and is not a strong enough dehydrating agent to convert an alcohol to an alkene; concentrated $H_3PO_4$ or $H_2SO_4$ is required.

(C) secondary $(2^{\circ})$ alcohol is one in which the hydroxyl group $(-OH)$ is attached to a carbon atom that is bonded to two other carbon atoms.
$1$. $2$-methyl-$1$-propanol: $(CH_3)_2CH-CH_2OH$ (Primary alcohol)
$2$. $2$-methyl-$2$-propanol: $(CH_3)_3C-OH$ (Tertiary alcohol)
$3$. $2$-butanol: $CH_3-CH(OH)-CH_2-CH_3$ (Secondary alcohol)
$4$. $1$-butanol: $CH_3-CH_2-CH_2-CH_2OH$ (Primary alcohol)
Thus,$2$-butanol is a secondary alcohol.

(D) secondary $(2^{\circ})$ alcohol is one in which the hydroxyl group $(-OH)$ is attached to a carbon atom that is bonded to two other carbon atoms.
$1$. $2$-methyl-$2$-propanol: $(CH_3)_3COH$ is a tertiary $(3^{\circ})$ alcohol.
$2$. $1$-propanol: $CH_3CH_2CH_2OH$ is a primary $(1^{\circ})$ alcohol.
$3$. $1$-butanol: $CH_3CH_2CH_2CH_2OH$ is a primary $(1^{\circ})$ alcohol.
$4$. $2$-pentanol: $CH_3CH(OH)CH_2CH_2CH_3$ is a secondary $(2^{\circ})$ alcohol,as the $-OH$ group is attached to the $C2$ atom,which is bonded to $C1$ and $C3$.

(B) $PBr_3$ converts ethanol into ethyl bromide. The compound $X$ is $CH_3CH_2Br$.
Alcoholic $KOH$ causes dehydrohalogenation of ethyl bromide to form ethene. The compound $Y$ is $CH_2=CH_2$.
Ethene reacts with cold concentrated $H_2SO_4$ followed by hydrolysis to form ethanol. The compound $Z$ is $CH_3CH_2OH$.
Reaction sequence:
$CH_3CH_2OH$ $\xrightarrow{PBr_3} (X) CH_3CH_2Br$ $\xrightarrow{alc. KOH} (Y) CH_2=CH_2$ $\xrightarrow[{(ii) H_2O, \text{heat}}]{{(i) H_2SO_4, \text{RT}}} (Z) CH_3CH_2OH$
Therefore,the compound $Z$ is $CH_3CH_2OH$.

(B) Step $(i)$: Benzene reacts with $Br_2$ in the presence of anhydrous $FeBr_3$ to form bromobenzene via electrophilic aromatic substitution.
Step (ii): Bromobenzene undergoes Friedel-Crafts alkylation with $CH_3Cl$ and anhydrous $AlCl_3$. Since the $-Br$ group is ortho/para directing,the major product is $p$-bromotoluene.
Step (iii): $p$-Bromotoluene reacts with $Mg$ in dry ether to form the Grignard reagent,$p$-tolylmagnesium bromide $(CH_3-C_6H_4-MgBr)$.
Step (iv) and $(v)$: The Grignard reagent reacts with acetaldehyde $(CH_3CHO)$ followed by hydrolysis $(H_2O)$ to form a secondary alcohol. The nucleophilic carbon of the Grignard reagent attacks the carbonyl carbon of acetaldehyde,resulting in $1-(4-methylphenyl)ethanol$.

(D) The reaction of cyclopentanol with $NaH$ forms a sodium alkoxide intermediate.
This alkoxide then reacts with $CS_{2}$ to form a sodium xanthate salt.
Finally,the reaction with $CH_{3}I$ (methyl iodide) results in the formation of a methyl xanthate ester.
This sequence is known as the Chugaev elimination precursor synthesis.
Therefore,the product is a xanthate.
Thus,option $(d)$ is the correct answer.

(C) The oxidation of allyl alcohol $(CH_2=CH-CH_2OH)$ with a peracid $(RCO_3H)$ is an epoxidation reaction.
This reaction yields glycidol,which has the molecular formula $C_3H_6O_2$.
The structure of glycidol is $CH_2(O)CH-CH_2OH$ (or $2,3-$epoxypropan$-1-$ol).
In this molecule,the carbon atom at the $C-2$ position is bonded to four different groups ($-H$,$-CH_2OH$,and the two carbons of the epoxide ring),making it an asymmetric (chiral) carbon atom.
Therefore,the correct structure is $CH_2(O)CH-CH_2OH$.

(B) The given sequence of chemical reactions is:
$C_{4}H_{10}O$ $\xrightarrow{K_{2}Cr_{2}O_{7} / H_{2}SO_{4}} C_{4}H_{8}O$ $\xrightarrow{I_{2} / NaOH, \text{Warm}} CHI_{3}$
$(N)$
Step $1$: The compound $N$ is oxidized by $K_{2}Cr_{2}O_{7} / H_{2}SO_{4}$ to a ketone $(C_{4}H_{8}O)$. This indicates that $N$ is a secondary alcohol.
Step $2$: The ketone formed undergoes the iodoform reaction with $I_{2} / NaOH$ to produce iodoform $(CHI_{3})$. This confirms that the ketone must have a methyl group attached to the carbonyl carbon (i.e.,it is a methyl ketone).
Butan-$2$-ol $(CH_{3}CH(OH)CH_{2}CH_{3})$ is a secondary alcohol. Upon oxidation,it yields butan-$2$-one $(CH_{3}COCH_{2}CH_{3})$,which is a methyl ketone and gives a positive iodoform test.
Therefore,$N$ is butan-$2$-ol.

(C) The Lucas reagent test is used to distinguish between primary,secondary,and tertiary alcohols.
Tertiary alcohols react immediately at room temperature to form turbidity.
Secondary alcohols react within $5-10$ minutes.
Primary alcohols do not react with Lucas reagent at room temperature.
Among the given options,$CH_{3}CH_{2}CH_{2}OH$ is a primary alcohol,so it does not react at room temperature.

(B) The reaction between an alcohol $(R-OH)$ and a Grignard reagent $(R'-MgX)$ is an acid-base reaction.
Alcohols contain a weakly acidic hydrogen atom attached to an oxygen atom.
Grignard reagents $(R'-MgX)$ contain a highly nucleophilic/basic alkyl group $(R'^- )$.
The $R'^- $ group abstracts the acidic proton from the alcohol to form an alkane $(R'-H)$ and an alkoxide salt $(R-OMgX)$.
The reaction is: $R-OH + R'-MgX \rightarrow R-OMgX + R'-H$ (Alkane).

(C) The boiling point depends on the strength of intermolecular forces.
$1$. Ethoxyethane $(II)$ is an ether,which exhibits only weak dipole-dipole interactions.
$2$. $N$-ethylethanamine $(I)$ is a secondary amine,which exhibits hydrogen bonding,but it is weaker than the hydrogen bonding in alcohols due to the lower electronegativity of $N$ compared to $O$.
$3$. Butan$-2-$ol $(III)$ is an alcohol,which exhibits strong intermolecular hydrogen bonding.
Therefore,the order of boiling points is $II < I < III$.

(B) The boiling point depends on the strength of intermolecular forces. The order of strength of intermolecular forces is: $\text{Hydrogen bonding} > \text{Dipole-dipole interaction} > \text{Van der Waals forces}$.
$1$. $1$-butanol $(CH_3CH_2CH_2CH_2OH)$ exhibits intermolecular hydrogen bonding,which is the strongest force among the given compounds,resulting in the highest boiling point.
$2$. Butanone $(CH_3COCH_2CH_3)$ is a polar molecule and exhibits dipole-dipole interactions,which are stronger than Van der Waals forces but weaker than hydrogen bonding.
$3$. $n$-pentane and isopentane are non-polar alkanes and exhibit only Van der Waals forces. For alkanes,the boiling point decreases with branching because branching reduces the surface area of contact,leading to weaker Van der Waals forces. Thus,$n$-pentane has a higher boiling point than isopentane.
Therefore,the correct order of boiling points is: $\text{isopentane} < n\text{-pentane} < \text{butanone} < 1\text{-butanol}$.

(B, C) $n-$butanol is a straight-chain alcohol,while $t-$butanol is a branched-chain alcohol.
Branching decreases the surface area of the molecule,which reduces the strength of intermolecular van der Waals forces,leading to a lower boiling point for $t-$butanol compared to $n-$butanol.
Additionally,branching increases the solubility of alcohols in water because it reduces the hydrophobic character of the alkyl group.
Therefore,$t-$butanol is more soluble in water and has a lower boiling point than $n-$butanol.
Thus,both statements $B$ and $C$ are correct.

(B) secondary alcohol is one where the $-OH$ group is attached to a carbon atom that is bonded to two other carbon atoms.
Analyzing the structures:
$(I)$ Contains two primary alcohols $(-CH_2OH)$.
$(II)$ Contains a secondary alcohol (the $-OH$ is on a $CH$ group bonded to two carbons).
$(III)$ Contains a tertiary alcohol (the $-OH$ is on a carbon bonded to three other carbons).
$(IV)$ Contains secondary alcohols (the $-OH$ groups are on $CH$ groups bonded to two carbons).
$(V)$ Contains a secondary alcohol (the $-OH$ is on a $CH$ group bonded to two carbons).
$(VI)$ Contains tertiary alcohols (the $-OH$ groups are on carbons bonded to three other carbons).
Thus,compounds $(II)$,$(IV)$,and $(V)$ contain at least one secondary alcohol.
There are $3$ such compounds.

(D) During acetylation,each $-OH$ group is replaced by an $-OCOCH_3$ group.
This results in the replacement of one hydrogen atom (mass $1 \ u$) with an acetyl group ($-COCH_3$,mass $43 \ u$).
Therefore,the net increase in molar mass for each $-OH$ group is $43 - 1 = 42 \ g \ mol^{-1}$.
Let $n$ be the number of $-OH$ groups.
The increase in molar mass is $290 - 122 = 168 \ g \ mol^{-1}$.
Thus,$n \times 42 = 168$.
$n = \frac{168}{42} = 4$.
So,there are $4$ hydroxyl groups present in compound $(X)$.

(A) The reaction of ethanol with concentrated $H_2SO_4$ at $413 \ K$ is an example of intermolecular dehydration of alcohols to form ethers.
This reaction proceeds via an $S_N2$ mechanism.
In the first step,the alcohol is protonated by the acid.
In the second step,a second molecule of ethanol acts as a nucleophile and attacks the protonated alcohol,displacing a water molecule to form diethyl ether.
Since the rate-determining step involves the collision of two molecules,it is a bimolecular nucleophilic substitution reaction $(S_N2)$.

(C) The reaction shows the oxidation of a primary alcohol (cyclohexylmethanol) to an aldehyde (cyclohexanecarbaldehyde).
Pyridinium chlorochromate $(C_5H_5NH^+CrO_3Cl^-)$,commonly known as $PCC$,is a mild oxidizing agent.
It selectively oxidizes primary alcohols to aldehydes and prevents further oxidation to carboxylic acids.
Other reagents like $KMnO_4$ are strong oxidizing agents that would oxidize the primary alcohol directly to a carboxylic acid.

(B) The dehydration of ethanol $(\text{CH}_3\text{CH}_2\text{OH})$ with concentrated $\text{H}_2\text{SO}_4$ depends on the reaction conditions.
$1$. At $413 \text{ K}$ $(140^{\circ}\text{C})$,the reaction proceeds via an $S_N2$ mechanism to form diethyl ether $(\text{C}_2\text{H}_5\text{OC}_2\text{H}_5)$.
$2$. At $443 \text{ K}$ $(170^{\circ}\text{C})$,the reaction proceeds via an $E1$ mechanism to form ethene $(\text{CH}_2 = \text{CH}_2)$.
$3$. Ethyl hydrogen sulfate $(\text{C}_2\text{H}_5\text{HSO}_4)$ is formed as an intermediate in both processes.
$4$. Acetylene $(\text{CH} \equiv \text{CH})$ is not a product of this reaction.