Properties of alcohol Questions in English

(A) Lucas test is used to distinguish between primary,secondary,and tertiary alcohols.
Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$.
The reaction proceeds via an $SN^1$ mechanism,where a carbocation is formed as an intermediate.
The rate of the reaction depends on the stability of the carbocation formed.
The stability order of carbocations is tertiary $>$ secondary $>$ primary.
Therefore,the reactivity order of alcohols with Lucas reagent is tertiary $>$ secondary $>$ primary.
$(CH_3)_3COH$ is a tertiary alcohol,so it reacts most readily with the Lucas reagent.

(D) The reaction sequence is as follows:
$1$. Compound '$A$' reacts with $PCl_5$ to form '$B$'. Since the final product is propanoic acid $(CH_3CH_2COOH)$,which contains $3$ carbon atoms,'$B$' must be ethyl chloride $(C_2H_5Cl)$.
$2$. Therefore,'$A$' is ethyl alcohol $(C_2H_5OH)$.
$3$. The reaction steps are:
$C_2H_5OH + PCl_5 \to C_2H_5Cl + POCl_3 + HCl$
$C_2H_5Cl + KCN \to C_2H_5CN + KCl$
$C_2H_5CN + 2H_2O + H^+ \to CH_3CH_2COOH + NH_4^+$

(B) The dehydration of alcohols follows the $E1$ mechanism,which involves the formation of a carbocation intermediate. The rate of dehydration depends on the stability of the carbocation formed.
For $2$-phenyl-$2$-butanol,the loss of the hydroxyl group results in a carbocation at the $C2$ position. This carbocation is both tertiary and benzylic,making it exceptionally stable due to resonance with the phenyl ring and inductive effects from alkyl groups.
In contrast,the other options form secondary or primary carbocations,which are significantly less stable than the tertiary benzylic carbocation formed by $2$-phenyl-$2$-butanol. Therefore,$2$-phenyl-$2$-butanol undergoes dehydration most readily.

(D) The reaction of an organic compound with acidified $K_2Cr_2O_7$ (an oxidizing agent) suggests that $X$ is a secondary alcohol.
Secondary alcohols are oxidized to ketones $(Y)$.
Ketones containing the $CH_3CO-$ group undergo the iodoform test with $I_2$ and $NaOH$ (or $Na_2CO_3$) to form tri-iodomethane ($CHI_3$,iodoform).
$CH_3CH(OH)CH_3 \xrightarrow{[O]} CH_3COCH_3$ ($Y$ is acetone).
$CH_3COCH_3 + 3I_2 + 4NaOH \rightarrow CHI_3 + CH_3COONa + 3NaI + 3H_2O$.
Thus,$X$ is $CH_3CH(OH)CH_3$ (propan$-2-$ol).

(D)
The oxidation of primary alcohols like ethanol $(C_2H_5OH)$ with strong oxidizing agents such as chromic acid ($CrO_3$ in $H_2SO_4$) results in the formation of carboxylic acids.
The reaction is:
$C_2H_5OH \xrightarrow{CrO_3} CH_3COOH$
Thus,ethanol is oxidized to ethanoic acid (acetic acid).

(A) Secondary alcohols undergo dehydrogenation when passed over heated copper at $300\,^{\circ}C$ to form ketones.
$CH_3-CH(OH)-CH_3 \xrightarrow[Cu]{300\,^{\circ}C} CH_3-CO-CH_3 + H_2$
Thus,isopropyl alcohol (propan$-2-$ol) forms acetone (propanone).

(A) Dehydrogenation of a secondary alcohol like isopropyl alcohol $(CH_3-CH(OH)-CH_3)$ involves the removal of hydrogen in the presence of a catalyst like copper at $573 \ K$.
The reaction is: $CH_3-CH(OH)-CH_3 \xrightarrow{Cu, 573 \ K} CH_3-CO-CH_3 + H_2$.
This results in the formation of a ketone,which is acetone $(CH_3-CO-CH_3)$.

(B) The reaction of alcohols with Lucas reagent (conc. $HCl + ZnCl_2$) is used to distinguish between primary,secondary,and tertiary alcohols.
$(i) \ CH_3CH_2CH_2OH$ is a primary $(1^{\circ})$ alcohol,which does not react at room temperature.
$(ii) \ CH_3-CH(OH)-CH_3$ is a secondary $(2^{\circ})$ alcohol,which reacts in about $5-10$ minutes to form turbidity.
$(iii) \ CH_3-C(CH_3)(OH)-CH_3$ is a tertiary $(3^{\circ})$ alcohol,which reacts immediately to form turbidity.
Therefore,$(iii)$ reacts immediately,$(ii)$ reacts in about $5$ minutes,and $(i)$ does not react at room temperature.

(A) The dehydration of alcohols in the presence of a concentrated acid catalyst like $H_2SO_4$ at high temperatures $(160-170 \ ^oC)$ results in the formation of an alkene with the same number of carbon atoms.
Specifically,the dehydration of ethanol $(C_2H_5OH)$ proceeds as follows:
$CH_3CH_2OH \xrightarrow{conc. H_2SO_4, 160-170 \ ^oC} CH_2=CH_2 + H_2O$
Thus,ethylene $(CH_2=CH_2)$ is obtained from ethanol $(C_2H_5OH)$.

(D) The oxidation of ethyl alcohol $(C_2H_5OH)$ proceeds in two steps:
$1$. First,it is oxidized to acetaldehyde $(CH_3CHO)$.
$2$. Further oxidation of acetaldehyde yields acetic acid $(CH_3COOH)$ as the final product.
$C_2H_5OH + [O] \to CH_3CHO + H_2O$
$CH_3CHO + [O] \to CH_3COOH$

(A) When alcohols are passed over heated copper at $300 \ ^oC$,they undergo dehydrogenation or dehydration.
$1^\circ$ alcohols are dehydrogenated to aldehydes.
$2^\circ$ alcohols are dehydrogenated to ketones.
$3^\circ$ alcohols undergo dehydration to form alkenes (olefins).
Phenol does not undergo this reaction to form benzyl alcohol,as the structure of phenol $(C_6H_5OH)$ cannot be converted to benzyl alcohol $(C_6H_5CH_2OH)$ through this process.

(D) Traces of water from ethanol are removed by reacting with $Mg$ metal. The $Mg$ reacts with water to form $Mg(OH)_2$ and $H_2$ gas,which can then be separated,yielding absolute ethanol.

(C) Dehydration of alcohols can be carried out using various catalysts depending on the temperature and conditions. Aluminium oxide $(Al_2O_3)$ acts as a dehydrating agent at high temperatures.
$C_2H_5OH \xrightarrow[250 ^\circ C]{Al_2O_3} C_2H_5OC_2H_5 \text{ (Diethyl ether)} + H_2O$
$C_2H_5OH \xrightarrow[350 ^\circ C]{Al_2O_3} C_2H_4 \text{ (Ethene)} + H_2O$

(B) The iodoform test is positive for compounds containing a methyl ketone group $(CH_3-CO-R)$ or a methyl carbinol group $(CH_3-CH(OH)-R)$.
$1-phenyl\ ethanol$ $(Ph-CH(OH)-CH_3)$ contains the methyl carbinol group,so it gives a positive iodoform test,forming a yellow precipitate of $CHI_3$ when treated with $I_2$ and $NaOH$.

(B) The acidic nature of alcohols is demonstrated by their reaction with active metals like sodium $(Na)$.
Ethyl alcohol reacts with sodium metal to form sodium ethoxide and releases hydrogen gas:
$2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2$

(C) The correct answer is $(C)$. Lower members of alcohols are soluble in water due to the formation of hydrogen bonds with water molecules. As the molecular mass increases,the hydrophobic alkyl group increases in size,which causes the solubility to decrease,not increase.

(A) Lucas reagent is a solution of anhydrous $ZnCl_2$ in concentrated $HCl$. The Lucas test is used to differentiate between primary,secondary,and tertiary alcohols.
$1$. Tertiary alcohols react immediately with Lucas reagent at room temperature to form turbidity.
$2$. Secondary alcohols react within $5$ to $10$ minutes to form turbidity.
$3$. Primary alcohols do not react appreciably with Lucas reagent at room temperature.

(D) The correct answer is $(d)$.
Anhydrous $CaCl_2$ cannot be used to dry alcohols like methanol or ethanol because it reacts with them to form addition compounds (solvates).
For example,$CaCl_2$ forms $CaCl_2 \cdot 4CH_3OH$ with methanol and $CaCl_2 \cdot 4C_2H_5OH$ with ethanol.

(A) Lucas test is used to distinguish between $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ alcohols.
Lucas reagent is a solution of anhydrous $ZnCl_2$ in concentrated $HCl$.
The Lucas test in alcohols is a test to differentiate between primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ alcohols.
Tertiary alcohols react immediately with Lucas reagent.
Secondary alcohols react within $10-15 \text{ minutes}$ (depending on their solubility).
Primary alcohols do not react appreciably with Lucas reagent at room temperature.

(B) Alcohols undergo dehydration (removal of water) to form an alkene.
To form an alkene,we need at least two carbon atoms.
Methanol $(CH_3OH)$ has only one carbon atom.
Therefore,it cannot form an alkene upon dehydration and does not yield a stable compound in this context.

(B) When methanol vapours are passed over heated copper at $573 \ K$,it undergoes catalytic dehydrogenation/oxidation to form formaldehyde and water vapour.
The chemical reaction is: $CH_3OH + \frac{1}{2} O_2 \xrightarrow{Cu, 573 \ K} HCHO + H_2O$.

(A) In the esterification reaction,an alcohol reacts with a carboxylic acid in the presence of an acid catalyst to form an ester and water.
Specifically,the $OH$ group of the alcohol is replaced by the acyloxy group $(RCOO^-)$ from the acid.
For the reaction between acetic acid and ethanol: $CH_3COOH + C_2H_5OH \to CH_3COOC_2H_5 + H_2O$.
Here,the $OH$ group of the alcohol is replaced by the $CH_3COO$ group.

(A) The compound $A$ is a primary alcohol $(CH_3CH_2OH)$.
Step $1$: Oxidation of primary alcohol $CH_3CH_2OH$ gives acetaldehyde $(CH_3CHO)$.
Step $2$: Further oxidation of $CH_3CHO$ gives acetic acid $(CH_3COOH)$.
Step $3$: Acetaldehyde $(CH_3CHO)$ reacts with ammoniacal $AgNO_3$ (Tollens' reagent) to produce a silver mirror,which is a characteristic test for aldehydes.
Therefore,$A$ is a primary alcohol.

(A) In the Victor Meyer test,primary alcohols $(1^\circ)$ produce a red color.
$C_2H_5OH$ (ethanol) is a primary alcohol.
The reaction sequence involves the formation of nitrolic acid,which turns red upon the addition of $NaOH$.
Secondary alcohols $(2^\circ)$ give a blue color,and tertiary alcohols $(3^\circ)$ remain colorless.

(B) When ethanol $(C_2H_5OH)$ is heated with concentrated $H_2SO_4$ at $140\,^{\circ}C$ $(413 \ K)$ in excess,intermolecular dehydration occurs to form diethyl ether.
The reaction is: $2C_2H_5OH \xrightarrow{\text{Conc. } H_2SO_4, 140\,^{\circ}C} C_2H_5-O-C_2H_5 + H_2O$.
Thus,the product is diethyl ether $(CH_3CH_2-O-CH_2CH_3)$.

(B) When secondary alcohols are passed over heated copper at $573 \ K$,dehydrogenation occurs to form ketones. Propan$-2-$ol $(CH_3-CH(OH)-CH_3)$ undergoes dehydrogenation to form acetone $(CH_3-CO-CH_3)$.
$CH_3-CH(OH)-CH_3 \xrightarrow{Cu, 573 \ K} CH_3-CO-CH_3 + H_2$

(C) Glycerol reacts with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ (nitrating mixture) to undergo nitration,forming glycerol trinitrate,which is also known as nitroglycerin.
The reaction is:
$CH_2OH-CHOH-CH_2OH + 3HNO_3 \xrightarrow{\text{conc. } H_2SO_4} CH_2(ONO_2)-CH(ONO_2)-CH_2(ONO_2) + 3H_2O$

(B) . The solubility of lower alcohols like methanol $(CH_3OH)$ and ethanol $(C_2H_5OH)$ in water is due to their ability to form intermolecular hydrogen bonds with water molecules.
The oxygen atom of the alcohol group carries a partial negative charge $(\delta-)$ and the hydrogen atom carries a partial positive charge $(\delta+)$,allowing them to interact with the water molecules through hydrogen bonding: $R-O^{\delta-}-H^{\delta+} \dots O^{\delta-}(H)-H^{\delta+}$.
Thus,option $B$ is correct.

(C) When ethylene glycol is distilled with fuming $H_2SO_4$,it undergoes intermolecular dehydration to form $1,4$-dioxan.
The reaction is as follows:
$2HO-CH_2-CH_2-OH \xrightarrow{\text{fuming } H_2SO_4} C_4H_8O_2 + 2H_2O$
The product obtained is $1,4$-dioxan,which corresponds to option $(C)$.

(B) Dehydration of alcohols proceeds via the formation of a carbocation (carbonium ion). The stability of carbocations follows the order: $3^\circ > 2^\circ > 1^\circ$.
$1.$ Option $(a)$ is a primary alcohol and forms a $1^\circ$ carbocation: $CH_3-CH(CH_3)-CH_2^+$.
$2.$ Option $(b)$ is a tertiary alcohol and forms a $3^\circ$ carbocation: $(CH_3)_3C^+$.
$3.$ Option $(c)$ is a primary alcohol and forms a $1^\circ$ carbocation: $CH_3-CH_2-CH_2-CH_2^+$.
$4.$ Option $(d)$ is an alkane and does not undergo dehydration in this context.
Since the $3^\circ$ carbocation is the most stable,tert-butyl alcohol (Option $B$) is the correct answer.

(D) . In ethanol $(CH_3CH_2OH)$,the $O-H$ bond is the most polar due to the high electronegativity of the oxygen atom.
Therefore,it undergoes heterolytic fission to release a proton $(H^+)$ and an ethoxide ion $(CH_3CH_2O^-)$.
$CH_3CH_2OH \rightarrow CH_3CH_2O^- + H^+$

(B) The correct answer is $(B)$.
$C_2H_5OH$ (ethanol) is soluble in water because it can form hydrogen bonds with water molecules.
$CS_2$,$CCl_4$,and $CHCl_3$ are non-polar or weakly polar organic solvents that do not form hydrogen bonds with water,making them insoluble.

(C) The solubility of alcohols in water depends on the extent of hydrogen bonding and the hydrophobic nature of the alkyl chain.
As the branching of the alkyl chain increases,the surface area of the non-polar hydrocarbon part decreases,which reduces the hydrophobic effect and increases solubility in water.
Among the given isomers of butyl alcohol $(C_4H_9OH)$,tertiary butyl alcohol ($CH_3)_3COH$ has the most branching,making it the most compact and thus the most soluble in water.

(B) The iodoform test is positive for compounds containing the $CH_3-CO-$ group or the $CH_3-CH(OH)-$ group.
$1.$ $CH_3-CH_2-OH$ (Ethanol) contains the $CH_3-CH(OH)-$ group,so it gives a positive test.
$2.$ $CH_3-CH_2-CH_2-OH$ ($1$-Propanol) does not contain the $CH_3-CH(OH)-$ group,so it gives a negative test.
$3.$ $C_6H_5-CH(OH)-CH_3$ ($1$-Phenylethanol) contains the $CH_3-CH(OH)-$ group,so it gives a positive test.
$4.$ $CH_3-CH(OH)-CH_3$ ($2$-Propanol) contains the $CH_3-CH(OH)-$ group,so it gives a positive test.

(A) Alcohols $(R-OH)$ contain an acidic hydrogen atom attached to an oxygen atom.
When an alcohol reacts with a Grignard reagent $(R'-MgX)$,the Grignard reagent acts as a strong base and abstracts the acidic proton from the alcohol.
The general reaction is:
$R-OH + R'-MgX \rightarrow R'-H + Mg(OR)X$
Here,$R'-H$ is an alkane.
For example,as shown in the image:
$CH_3OH + C_2H_5MgBr \rightarrow C_2H_6 + Mg(OCH_3)Br$
Thus,the product formed is an alkane.
Therefore,the correct option is $(A)$.

(C) The solubility of lower alcohols in water is primarily due to their ability to form intermolecular $H$-bonds with water molecules.
When alcohol is added to water,the oxygen atom of the hydroxyl group $(-OH)$ in alcohol forms a hydrogen bond with the hydrogen atom of water,and the hydrogen atom of the alcohol's hydroxyl group forms a hydrogen bond with the oxygen atom of water.
This interaction is represented as: $R-O^{\delta -}-H^{\delta +} \cdots O^{\delta -}(H)-H^{\delta +}$.

(C) The reaction of ethanol $(CH_3CH_2OH)$ with chlorine $(Cl_2)$ involves both oxidation and chlorination.
Initially,ethanol is oxidized to acetaldehyde $(CH_3CHO)$,which then undergoes chlorination to form chloral $(CCl_3CHO)$:
$CH_3CH_2OH + 2Cl_2 \to CCl_3CHO + 3HCl$
Thus,the final product is chloral $(CCl_3CHO)$.

(C) The reaction of alcohols with acetylene $(HC \equiv CH)$ in the presence of mercury salts (like $Hg^{2+}$) as a catalyst leads to the formation of vinyl ethers.
The general reaction is: $R-OH + HC \equiv CH \xrightarrow{Hg^{2+}} R-O-CH=CH_2$.

(D) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. $CH_3CHO$ (Acetaldehyde) contains the $CH_3CO-$ group and gives a positive iodoform test.
$2$. $CH_3CH_2OH$ (Ethanol) contains the $CH_3CH(OH)-$ group and gives a positive iodoform test.
$3$. Isopropyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group and gives a positive iodoform test.
$4$. Benzyl alcohol $(C_6H_5CH_2OH)$ does not contain either of these groups,so it gives a negative iodoform test.

(B) The substance $C_4H_{10}O$ is an alcohol.
On oxidation,it gives $C_4H_8O$ (a ketone or aldehyde).
Since the product gives a positive iodoform test,it must be a methyl ketone $(CH_3CO-R)$.
$2-$butanol $(CH_3CH(OH)CH_2CH_3)$ on oxidation gives butanone $(CH_3COCH_2CH_3)$,which is a methyl ketone and gives a positive iodoform test and forms an oxime.
Also,$2-$butanol on dehydration with conc. $H_2SO_4$ gives $C_4H_8$ (but$-2-$ene).
Thus,the compound is $2-$butanol.

(B) Ethylene glycol $(HO-CH_2-CH_2-OH)$ reacts with excess phosphorus pentachloride $(PCl_5)$ to replace both hydroxyl groups with chlorine atoms,forming $1, 2-$ dichloroethane $(Cl-CH_2-CH_2-Cl)$.
$HO-CH_2-CH_2-OH + 2PCl_5 \rightarrow Cl-CH_2-CH_2-Cl + 2POCl_3 + 2HCl$.

(B) $($ $B$ $)$ $C_2H_5OH$ (ethanol) is a very weak acid,hence it does not react with $NaOH$.
However,it reacts with metallic sodium.
Picric acid is a strong acid due to the electron-withdrawing effect of three $-NO_2$ groups and reacts with $NaOH$.
$CH_3CONH_2$ (acetamide) is amphoteric and reacts with strong bases.
$CH(CN)_3$ (tricyanomethane) is a strong acid due to the electron-withdrawing effect of three $-CN$ groups and reacts with $NaOH$.

(A) When glycerol is heated with oxalic acid at $110^\circ C$,glycerol monooxalate is formed first,which then undergoes decarboxylation to give glycerol monoformate as the final product.
$CH_2OH-CHOH-CH_2OH + COOH-COOH \xrightarrow{110^\circ C} CH_2(OOCH)-CHOH-CH_2OH + CO_2$

(A) The formation of a yellow precipitate on heating a compound with an alkaline solution of iodine is known as the iodoform reaction.
This test is positive for compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$CH_3OH$ (methanol) does not contain these groups and therefore does not respond to the iodoform test.
$CH_3CH_2OH$ (ethanol),$CH_3CH(OH)CH_3$ (propan$-2-$ol),and $CH_3CH_2CH(OH)CH_3$ (butan$-2-$ol) all contain the $CH_3CH(OH)-$ group and will form a yellow precipitate of iodoform $(CHI_3)$.

(D) The reaction with $I_2$ and $NaOH$ is known as the iodoform test.
Compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group attached to a carbon or hydrogen atom give a positive iodoform test,resulting in a yellow precipitate of $CHI_3$.
Among the given options,$CH_3CH_2OH$ (ethanol) contains the $CH_3CH(OH)-$ group.
The reaction is:
$CH_3CH_2OH + 4I_2 + 6NaOH \to CHI_3 (\text{yellow ppt.}) + 5NaI + CH_3COONa + 5H_2O$
Therefore,the correct option is $D$.

(D) The reaction of alcohols with hydrogen halides $(HX)$ follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$ alcohols.
This is because the reaction proceeds via the formation of a carbocation intermediate,and the stability of the carbocation follows the order: $3^{\circ} > 2^{\circ} > 1^{\circ}$.
$2$-methyl propane-$2$-ol is a tertiary $(3^{\circ})$ alcohol,which forms a stable tertiary carbocation,making it the most reactive towards $HBr$.

(C) Ethyl alcohol $(CH_3CH_2OH)$ undergoes dehydration when heated with concentrated $H_2SO_4$ at $170 \, ^\circ C$ to form ethylene $(C_2H_4)$.
The chemical reaction is:
$CH_3CH_2OH \xrightarrow[170 \, ^\circ C]{\text{conc. } H_2SO_4} CH_2=CH_2 + H_2O$

(D) The dehydration of $2-$butanol $(CH_3-CH_2-CH(OH)-CH_3)$ in the presence of an acid catalyst follows $E1$ mechanism.
According to $Saytzeff$ rule, the more substituted alkene is the major product.
$CH_3-CH_2-CH(OH)-CH_3$ $\xrightarrow{H^+, \Delta} CH_3-CH_2-CH=CH_2 \text{ (} 1-\text{butene, minor)} + CH_3-CH=CH-CH_3 \text{ (} 2-\text{butene, major)}$.
Thus, both $1-$butene and $2-$butene are formed.

(C) When vapours of a tertiary alcohol are passed over hot reduced copper at $300^{\circ} C$,dehydration occurs instead of dehydrogenation,leading to the formation of an alkene.
For example,tert-butyl alcohol undergoes dehydration:
$CH_3-C(OH)(CH_3)-CH_3 \xrightarrow{Cu, 300^{\circ} C} CH_3-C(CH_3)=CH_2 + H_2O$
Primary alcohols undergo dehydrogenation to form aldehydes,and secondary alcohols undergo dehydrogenation to form ketones.

(A) Grignard reagents $(R-MgX)$ react with esters $(R-COOR')$ in excess to form tertiary alcohols after hydrolysis.
Specifically,the reaction of an ester with two equivalents of a Grignard reagent involves the formation of a ketone intermediate,which then reacts with the second equivalent of the Grignard reagent to form an alkoxide,yielding a tertiary alcohol upon hydrolysis.
Aldehydes (except formaldehyde) yield secondary alcohols,and primary alcohols are not typically formed from Grignard reactions with carbonyl compounds in this manner.