Properties of alcohol Questions in English

(D) The dehydration of alcohols to form alkenes or ethers requires a dehydrating agent.
$H_2SO_4$ (concentrated),$Al_2O_3$ (heated),and $H_3PO_4$ (concentrated) are all well-known dehydrating agents used in organic chemistry to remove a water molecule from alcohols.
Therefore,all of these can act as a dehydrating agent.

(B) Glycerol $(CH_2OH-CHOH-CH_2OH)$ reacts with excess $HI$ to form $1,2,3$-triiodopropane.
This intermediate is unstable and undergoes elimination of $I_2$ to form allyl iodide $(CH_2=CH-CH_2I)$.

(D) The ease of dehydration of alcohols follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$ alcohol. This is because the rate-determining step involves the formation of a carbocation intermediate. $3^{\circ}$ carbocations are the most stable. Among the given options,$2-$methylpropan$-2-$ol and $2-$methylbutan$-2-$ol are both $3^{\circ}$ alcohols. However,$2-$methylbutan$-2-$ol forms a slightly more stable carbocation due to the presence of more alkyl groups providing hyperconjugation and inductive effects,making it the most reactive towards dehydration.

(B) The reduction of a ketone with $LiAlH_4$ yields a secondary alcohol.
In option $(B)$,$C_6H_5COCH_3$ (acetophenone) is a ketone.
Upon reaction with $LiAlH_4$ followed by $H^{+}$,it is reduced to $C_6H_5-CH(OH)-CH_3$,which is a secondary $(2^o)$ alcohol.
Option $(A)$ produces a tertiary alcohol,option $(C)$ produces a primary alcohol,and option $(D)$ is a halogenation reaction.

(B) The conversion of cyclohexanol to cyclohexene is an acid-catalyzed dehydration reaction.
Concentrated $H_3PO_4$ (phosphoric acid) is preferred over other mineral acids like $H_2SO_4$ or $HCl$ because it is a non-oxidizing dehydrating agent,which minimizes side reactions like oxidation or substitution.
The reaction is as follows:
$C_6H_{11}OH \xrightarrow{Conc. H_3PO_4, \Delta} C_6H_{10} + H_2O$
Thus,option $(B)$ is the correct answer.

(D) Acidity is determined by the stability of the conjugate base formed after the loss of a proton $(H^+)$.
In $CH_4$,$C_2H_6$,and $CH \equiv CH$,the negative charge on the conjugate base resides on carbon atoms with $sp^3$,$sp^3$,and $sp$ hybridization,respectively.
In $C_2H_5OH$,the negative charge resides on an oxygen atom,which is much more electronegative than carbon.
The order of acidity is: $C_2H_5OH > CH \equiv CH > C_2H_6 > CH_4$.
Therefore,$C_2H_5OH$ is the most acidic compound.

(C) $C_2H_5OH$ contains a $CH_3CH(OH)-$ group,which gives a positive iodoform test with $I_2$ and $NaOH$ to form yellow precipitate of $CHI_3$.
$CH_3OH$ does not contain this group and does not give the iodoform test.
$C_2H_5OH + 4I_2 + 6NaOH \rightarrow CHI_3 + HCOONa + 5NaI + 5H_2O$
$CH_3OH + I_2 + NaOH \rightarrow \text{No reaction}$

(A) The conversion of tertiary butyl alcohol to tertiary butyl chloride is a nucleophilic substitution reaction ($S_N1$ mechanism).
Tertiary alcohols react readily with concentrated $HCl$ in the presence of anhydrous $ZnCl_2$ (Lucas reagent) to form alkyl chlorides.
The reaction is: $(CH_3)_3C-OH + HCl \xrightarrow{anhydrous ZnCl_2} (CH_3)_3C-Cl + H_2O$.

(A) $Glycerol$ reacts with concentrated $HNO_3$ in the presence of $H_2SO_4$ to form glyceryl trinitrate (nitroglycerin).
Nitroglycerin is an explosive liquid,which is absorbed on an inert material like $Kieselguhr$ (a porous siliceous earth) to form $Dynamite$.

(B) Methyl alcohol $(CH_3OH)$ is toxic because it is oxidized in the liver to formaldehyde and then to formic acid. Formic acid causes metabolic acidosis and can lead to blindness or death. The statement provided in the option is scientifically incorrect as methanol does not react with nitrogen to form $CN^-$ in the lungs. However,based on the provided options,$B$ is the intended answer.

(D) Glycerol is a versatile compound used in various industries.
$1$. It acts as a sweetening agent in food and beverages.
$2$. It is a key ingredient in the manufacture of high-quality soaps.
$3$. It is used as a raw material in the production of nitro-glycerine,which is an explosive.
Therefore,all the given options are correct.

(B) Liquor poisoning,often referred to as methanol poisoning,occurs due to the presence of $CH_3OH$ (methyl alcohol) in alcoholic beverages. $CH_3OH$ is toxic to the human body as it is oxidized to formaldehyde and then to formic acid in the liver,which can lead to blindness and death.

(C) The process of making ethanol unfit for drinking by adding toxic substances like methanol,pyridine,or copper sulfate is known as denaturing. The resulting mixture is called $Denatured \ spirit$.

(C) Denatured spirit is ethanol that has been rendered unfit for drinking by the addition of small amounts of poisonous substances like methanol,pyridine,or copper sulfate. Due to its properties as an alcohol,it is primarily used as a solvent in the preparation of varnishes,paints,and lacquers.

(B) Dynamite is an explosive material consisting of a mixture of glyceryl trinitrate (nitroglycerine) and glyceryl dinitrate,which is absorbed on an inert absorbent material like kieselguhr.

(C) Alcoholic beverages,such as wine,beer,or distilled spirits,contain ethyl alcohol,also known as ethanol,with the chemical formula $C_2H_5OH$ (or $CH_3CH_2OH$),as the primary intoxicating agent.

(C) Tonics generally contain $Ethanol$ $(C_2H_5OH)$,which acts as a solvent for various medicinal ingredients.

(C) Widespread deaths due to liquor poisoning occur due to the presence of methyl alcohol $(CH_3OH)$ in liquor.
Methyl alcohol is toxic and,upon ingestion,is oxidized to formaldehyde and formic acid in the liver,which causes blindness and death.

(A) $(A)$. Ethylene glycol is commonly used as an antifreeze in automobile radiators.
It works by lowering the freezing point of water, which prevents the coolant from freezing in cold climates.

(C) The reaction of ethylene oxide (oxirane) with a Grignard reagent $(RMgI)$ involves the nucleophilic attack of the alkyl group $(R^-)$ on one of the carbon atoms of the epoxide ring,leading to ring opening.
The reaction proceeds as follows:
$C_2H_4O + RMgI \to R - CH_2 - CH_2 - OMgI$
Subsequent hydrolysis of the intermediate yields the primary alcohol:
$R - CH_2 - CH_2 - OMgI + H_2O \to R - CH_2 - CH_2 - OH + MgI(OH)$
Thus,the product $A$ is $R - CH_2 - CH_2 - OH$.

(A) The correct order is ${1^o} > {2^o} > {3^o}$.
As the number of branches in the alkyl group increases,the surface area of the molecule decreases.
This leads to a decrease in the magnitude of van der Waals forces of attraction between the molecules.
Consequently,the boiling point decreases from primary to tertiary alcohols.

(C) The dehydration of alcohols proceeds via the formation of a carbocation intermediate.
The stability of carbocations follows the order: $3^o > 2^o > 1^o$.
Therefore,the rate of dehydration of alcohols follows the same order: $3^o > 2^o > 1^o$.

(D) . The distinction between primary,secondary,and tertiary alcohols can be performed using all three methods: oxidation,the Victor Meyer test,and the Lucas test.

(A) The correct answer is $A$. In the dehydration of alcohols using concentrated $H_2SO_4$,the reaction proceeds via the formation of a carbonium ion (carbocation) intermediate.
Step $1$: Protonation of the alcohol group: $R-CH_2-CH_2-OH + H^+ \rightarrow R-CH_2-CH_2-OH_2^+$
Step $2$: Loss of water molecule to form a carbonium ion: $R-CH_2-CH_2-OH_2^+ \rightarrow R-CH_2-CH_2^+ + H_2O$
Step $3$: The carbonium ion may undergo $1, 2-$ hydride or $1, 2-$ methyl shifts to form a more stable carbonium ion before losing a proton to form an alkene.

(A) Isopropyl alcohol is a secondary alcohol. On oxidation,secondary alcohols are oxidized to ketones.
The reaction is as follows:
$CH_3CH(OH)CH_3 \xrightarrow{[O]} CH_3COCH_3 + H_2O$
Here,$CH_3CH(OH)CH_3$ is isopropyl alcohol (propan$-2-$ol) and $CH_3COCH_3$ is acetone (propanone).

(B) Hot concentrated $HNO_3$ is a strong oxidizing agent.
It oxidizes secondary alcohols like $2-propanol$ $(CH_3CH(OH)CH_3)$ to the corresponding ketone,which is $propanone$ $(CH_3COCH_3)$.
The reaction is: $CH_3CH(OH)CH_3 + [O] \xrightarrow{\text{conc. } HNO_3} CH_3COCH_3 + H_2O$.

(C) When $Glycerol$ is heated with $KHSO_4$,it undergoes dehydration by the loss of two molecules of water $(H_2O)$.
This reaction results in the formation of an $\alpha, \beta$-unsaturated aldehyde known as $Acryl \ aldehyde$ (also called $Acrolein$).
The reaction is represented as:
$CH_2OH-CHOH-CH_2OH \xrightarrow{KHSO_4, \Delta} CH_2=CH-CHO + 2H_2O$

(A) Chromic acid in aqueous acetone at $5 - 10^{\circ}C$ is known as the $Jones$ reagent.
It is a selective oxidizing agent that oxidizes secondary alcohols to ketones and primary alcohols to carboxylic acids.
$A$ key feature of the $Jones$ reagent is that it does not oxidize or affect carbon-carbon double or triple bonds.
In option $(A)$,a secondary alcohol containing a triple bond is correctly oxidized to a ketone while the triple bond remains intact.
Options $(B)$ and $(D)$ show primary alcohols being oxidized to aldehydes,which is incorrect as the $Jones$ reagent typically oxidizes primary alcohols to carboxylic acids.
Option $(C)$ requires much stronger oxidizing conditions.

(C) The iodoform test is positive for compounds containing a methyl ketone group $(CH_3-CO-)$ or a methyl carbinol group $(CH_3-CH(OH)-)$.
Among the given options,$Butan-2-ol$ $(CH_3-CH(OH)-CH_2-CH_3)$ contains the $CH_3-CH(OH)-$ group and thus gives a positive iodoform test.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$A$ Isopropyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group and gives a positive test.
$B$ Ethanol $(CH_3CH_2OH)$ contains the $CH_3CH(OH)-$ group and gives a positive test.
$C$ Ethanal $(CH_3CHO)$ contains the $CH_3CO-$ group and gives a positive test.
$D$ Benzyl alcohol $(C_6H_5CH_2OH)$ does not contain either the $CH_3CO-$ or the $CH_3CH(OH)-$ group,therefore it will not give the iodoform test.

(B) The oxidation of primary alcohols like benzyl alcohol $(C_6H_5CH_2OH)$ with a strong oxidizing agent like $KMnO_4$ leads to the formation of a carboxylic acid.
The reaction is: $C_6H_5CH_2OH + 2[O] \xrightarrow{KMnO_4} C_6H_5COOH + H_2O$.
Therefore,the product obtained is benzoic acid.

(D) Tertiary alcohols $(3^{\circ})$ are resistant to oxidation under mild conditions. However,under drastic conditions (e.g.,using concentrated $HNO_3$ or $KMnO_4$ at high temperatures),they undergo carbon-carbon bond cleavage.
First,the tertiary alcohol is oxidized to a ketone by losing one carbon atom.
Then,the resulting ketone undergoes further oxidation,which involves the cleavage of the carbon-carbon bond,resulting in the loss of more carbon atoms to form carboxylic acids.
Specifically,for a tertiary alcohol with $n$ carbon atoms,the oxidation process involves the loss of carbon atoms at each step of the cleavage,leading to products with fewer carbon atoms than the parent alcohol. The reaction follows the general pathway: $R_3C-OH$ $\xrightarrow{[O]} R_2C=O$ $\xrightarrow{[O]} R-COOH$.
Thus,the resulting carboxylic acid has fewer carbon atoms than the original tertiary alcohol.

(C) . $C_{10}H_{17}OH$ (Geraniol),a liquid terpene alcohol,reacts with $HCOOH$ to form an ester that possesses a rose-like odour.

(C) Viscosity is directly related to the extent of intermolecular hydrogen bonding.
$CH_2OH-CH_2OH$ (Ethylene glycol) contains two $-OH$ groups per molecule,which allows for extensive intermolecular hydrogen bonding compared to $CH_3OH$ and $C_2H_5OH$,which have only one $-OH$ group.
Therefore,ethylene glycol has the highest viscosity.

(B) Methanol $(CH_3OH)$ contains an oxygen atom bonded to a hydrogen atom,which allows for the formation of strong intermolecular hydrogen bonds between its molecules.
Methyl thiol $(CH_3SH)$ contains a sulfur atom,which is less electronegative than oxygen and has a larger atomic size,preventing it from forming significant hydrogen bonds.
Due to the presence of intermolecular hydrogen bonding in methanol,it requires more energy to overcome these forces,resulting in a higher boiling point compared to methyl thiol.

(B) Methanol and ethanol are capable of forming $H$-bonds with water molecules.
Due to this intermolecular $H$-bonding,they are soluble in water.

(A) Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom like $O$,$N$,or $F$.
Among the given options,ethanol $(C_2H_5OH)$ contains an $-OH$ group,which allows for strong intermolecular hydrogen bonding.
Diethyl ether $(C_2H_5-O-C_2H_5)$ lacks a hydrogen atom bonded to oxygen,so it cannot form $H$-bonds.
Ethyl chloride $(C_2H_5Cl)$ does not have hydrogen bonded to a highly electronegative atom.
Triethylamine $((C_2H_5)_3N)$ has nitrogen,but no hydrogen atom is directly attached to it,preventing $H$-bonding.
Therefore,ethanol exhibits the maximum $H$-bonding.

(A) Lucas reagent is a mixture of anhydrous $ZnCl_2$ and concentrated $HCl$. It is used to distinguish between primary,secondary,and tertiary alcohols.

(A) The iodoform test is given by compounds containing the $CH_3CH(OH)-$ group (like $2$-alkanols) or the $CH_3CO-$ group (methyl ketones).
Among the given options,$C_2H_5OH$ (ethanol) contains the $CH_3CH(OH)-$ group,which reacts with $I_2$ and $NaOH$ to form iodoform $(CHI_3)$.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
Among the given options,$CH_3CH_2OH$ (ethanol) contains the $CH_3CH(OH)-$ group.
When $CH_3CH_2OH$ reacts with $I_2$ and $NaOH$,it forms $CHI_3$ (iodoform),which is a yellow precipitate.

(C) Neopentyl alcohol ($CH_3)_3CCH_2OH$ reacts with $HCl$ in the presence of $ZnCl_2$ to form a primary carbocation. This primary carbocation undergoes a $1,2$-methyl shift to form a more stable tertiary carbocation ($t$-pentyl carbocation). The chloride ion then attacks this tertiary carbocation to yield $t$-pentyl chloride as the major product.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$A$: $CH_3CH_2CH(OH)CH_2CH_3$ is a secondary alcohol but does not have a $CH_3CH(OH)-$ group.
$B$: $C_6H_5CH_2CH_2OH$ is a primary alcohol and does not have the required group.
$C$: $CH_3CH(CH_3)CH_2OH$ is a primary alcohol and does not have the required group.
$D$: $PhCH(OH)CH_3$ contains the $CH_3CH(OH)-$ group attached to a phenyl ring,which satisfies the condition for a positive iodoform test.
Therefore,the correct option is $D$.

(B) The reaction of alcohols with halogen acids proceeds via the formation of a carbocation intermediate.
The reactivity of alcohols towards halogen acids follows the order: $3^\circ > 2^\circ > 1^\circ$ alcohol.
This is because the stability of the carbocation formed is $3^\circ > 2^\circ > 1^\circ$.
$(1)$ is a primary $(1^\circ)$ alcohol: $CH_3CH_2CH_2OH$.
$(2)$ is a secondary $(2^\circ)$ alcohol: $CH_3CH_2CH(CH_3)OH$.
$(3)$ is a tertiary $(3^\circ)$ alcohol: $CH_3CH_2C(CH_3)_2OH$.
Therefore,the order of reactivity is $(3) > (2) > (1)$.

(B) Statement $A$ is incorrect because alkyl halides do not react with alkenes to form products in this manner; rather,they are typically formed from alkenes via hydrohalogenation.
Statement $B$ is correct: $CH_3COCH_3 + CH_3MgBr$ $\rightarrow (CH_3)_3COMgBr$ $\xrightarrow{H_2O} (CH_3)_3COH$ (tert-butyl alcohol),but wait,the question asks for butan-$2$-ol. Actually,the reaction of $CH_3MgBr$ with acetaldehyde gives propan-$2$-ol,and with acetone gives tert-butyl alcohol. Thus,statement $B$ is also chemically incorrect as stated.
However,in the context of standard multiple-choice questions,statement $A$ is fundamentally flawed regarding the reaction type. Let us re-evaluate: $C_2H_4Cl_2$ has two isomers: $1,1$-dichloroethane and $1,2$-dichloroethane. This is correct.
Reduction of ethyl chloride $(C_2H_5Cl)$ with $Zn-Cu/C_2H_5OH$ gives ethane $(C_2H_6)$. This is correct.
Therefore,statement $B$ is the intended incorrect statement as it describes the wrong product for the reaction of acetone with methyl magnesium bromide.

(A) The reaction of ethyl alcohol $(CH_3CH_2OH)$ with bleaching powder $(CaOCl_2)$ involves the formation of chloroform $(CHCl_3)$.
Step $1$: Bleaching powder reacts with water to produce chlorine: $CaOCl_2 + H_2O \rightarrow Ca(OH)_2 + Cl_2$.
Step $2$: Chlorine oxidizes ethyl alcohol to acetaldehyde: $CH_3CH_2OH + Cl_2 \rightarrow CH_3CHO + 2HCl$.
Step $3$: Acetaldehyde is chlorinated to form chloral $(CCl_3CHO)$,which then reacts with calcium hydroxide to yield chloroform: $2CH_3CHO + 6Cl_2 \rightarrow 2CCl_3CHO + 6HCl$ and $2CCl_3CHO + Ca(OH)_2 \rightarrow 2CHCl_3 + (HCOO)_2Ca$ (or calcium acetate depending on conditions).
The final major organic product is chloroform $(CHCl_3)$.

(D) The reactivity of alcohols towards $H-X$ depends on the stability of the carbocation intermediate formed after the loss of water.
$(I)$ Vinyl alcohol $(CH_2=CH-OH)$ forms a vinyl carbocation $(CH_2=CH^+)$,which is highly unstable.
$(II)$ Allyl alcohol $(CH_2=CH-CH_2OH)$ forms an allyl carbocation $(CH_2=CH-CH_2^+)$,which is resonance stabilized.
$(III)$ Ethanol $(CH_3-CH_2OH)$ forms a primary carbocation $(CH_3-CH_2^+)$.
$(IV)$ Isopropyl alcohol $(CH_3-CH(OH)-CH_3)$ forms a secondary carbocation $(CH_3-CH^+-CH_3)$,which is more stable than a primary carbocation due to inductive effect and hyperconjugation.
Stability order of carbocations: $Allyl > Secondary > Primary > Vinyl$.
Therefore,the reactivity order is: $II > IV > III > I$.

(A) Lucas reagent is a mixture of concentrated hydrochloric acid $(HCl)$ and anhydrous zinc chloride $(ZnCl_2)$.
It is used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction,which results in the formation of alkyl chlorides (turbidity).
Therefore,the correct composition is concentrated $HCl$ and anhydrous $ZnCl_2$.

(B) The reaction of alcohols with hydrogen halides $(HX)$ in the presence of $ZnX_2$ (Lucas reagent) follows the mechanism where the bond strength of $H-X$ determines the reactivity.
As the size of the halogen atom increases,the bond dissociation energy of the $H-X$ bond decreases,making it easier to break.
The bond dissociation energy order is $HI < HBr < HCl$.
Therefore,the reactivity order for the reaction is $HI > HBr > HCl$.

(C) The reaction of ethanol with chlorine occurs in two main steps:
$1$. Ethanol is oxidized to acetaldehyde by chlorine: $CH_3CH_2OH + Cl_2 \to CH_3CHO + 2HCl$
$2$. Acetaldehyde then reacts with excess chlorine to form chloral: $CH_3CHO + 3Cl_2 \to CCl_3CHO + 3HCl$
Therefore,the final product formed is chloral.

(A) Grignard reagents $(RMgX)$ act as strong bases when they react with compounds containing active hydrogen atoms (like alcohols,$R-OH$).
In this reaction,$PhMgBr$ acts as a base and abstracts the acidic proton from the hydroxyl group of the alcohol (propan$-2-$ol).
The reaction is: $PhMgBr + (CH_3)_2CH-OH \to Ph-H + (CH_3)_2CH-OMgBr$.
The product $A$ is benzene $(Ph-H)$.