Properties of alcohol Questions in English

(A) Ethanol is a colorless,volatile liquid with a characteristic smell. It is represented by the chemical formula $CH_3CH_2OH$.
Absolute alcohol is defined as ethanol that contains almost no water,typically having a purity of more than $99 \%$ by weight.
Rectified spirit or rectified alcohol is a mixture of $95 \%$ ethanol and $5 \%$ water.
Denatured alcohol is ethanol that has been made unfit for human consumption by adding additives like methanol ($95 \%$ ethanol $+ 5 \%$ methanol).
Therefore,absolute alcohol refers to $100 \%$ pure ethanol.

(A) Nitroglycerine is chemically known as glyceryl trinitrate. It is formed by the reaction of glycerol with concentrated nitric acid in the presence of concentrated sulfuric acid. Since it is a product of the reaction between an alcohol (glycerol) and an acid (nitric acid),it is classified as a nitrate ester.

(D) The reaction of an organic compound with sodium $(Na)$ to evolve hydrogen gas $(H_2)$ is a characteristic test for the presence of an acidic hydrogen atom,such as that found in the hydroxyl $(-OH)$ group of alcohols.
The chemical equation for the reaction is:
$2R-OH + 2Na \to 2R-ONa + H_2 \uparrow$
Ketones,aldehydes,and tertiary amines do not contain an acidic hydrogen atom capable of reacting with sodium to release hydrogen gas under these conditions.
Therefore,the compound is an alcohol.

(B) The final product is $tert$-butyl alcohol,which is a tertiary alcohol formed by the reaction of a ketone with a Grignard reagent.
Step $1$: The reaction $B \xrightarrow{CH_3MgI / H_2O} CH_3-C(CH_3)(OH)-CH_3$ indicates that $B$ must be acetone $(CH_3-CO-CH_3)$.
Step $2$: The reaction $A \xrightarrow{K_2Cr_2O_7 / \text{dil. } H_2SO_4} B$ shows that $A$ is an alcohol that oxidizes to acetone. Secondary alcohols oxidize to ketones.
Therefore,$A$ is isopropyl alcohol $(CH_3-CH(OH)-CH_3)$.

(C) Rectified spirit contains approximately $95.6 \%$ ethanol and $4.4 \%$ water by mass,which forms a constant boiling azeotropic mixture. To obtain absolute alcohol ($100 \%$ pure ethanol),the water must be removed. This is commonly achieved by the azeotropic distillation method,where benzene is added to the mixture. Benzene forms a ternary azeotrope with water and ethanol,allowing the water to be removed as a distillate,leaving behind absolute alcohol.

(D) Grignard reagents $(RMgX)$ are strong nucleophiles that react with carbonyl compounds to form alcohols.
$1$. Reaction with formaldehyde $(HCHO)$ yields a primary alcohol.
$2$. Reaction with other aldehydes $(RCHO)$ yields a secondary alcohol.
$3$. Reaction with ketones $(R_2CO)$ yields a tertiary alcohol.
$4$. Reaction with nitriles or acid chlorides can yield ketones.
However,Grignard reagents react with esters $(RCOOR')$ to form tertiary alcohols (after two additions) and cannot be used to synthesize the ester itself,as the Grignard reagent would attack the ester carbonyl group.

(A) The formation of diethyl ether from ethanol occurs via the intermolecular dehydration of ethanol in the presence of concentrated $H_2SO_4$ at $413 \ K$. The reaction is: $2C_2H_5OH \xrightarrow{H_2SO_4, 413 \ K} C_2H_5-O-C_2H_5 + H_2O$. Thus,it is a dehydration reaction.

(A) The molecular formula $C_3H_8O$ corresponds to either a primary alcohol $(CH_3CH_2CH_2OH)$,a secondary alcohol $(CH_3CH(OH)CH_3)$,or an ether $(CH_3OCH_2CH_3)$.
Upon oxidation,a primary alcohol yields an aldehyde $(C_3H_6O)$,and a secondary alcohol yields a ketone $(C_3H_6O)$.
Since the product $C_3H_6O$ is formed,$X$ must be an alcohol.
Among the given options,$X$ can be a secondary alcohol (specifically propan$-2-$ol),which oxidizes to propanone $(C_3H_6O)$.

(A) In alcohols,strong intermolecular hydrogen bonding is present due to the high electronegativity of the oxygen atom.
Sulfur is significantly less electronegative than oxygen,and therefore,thiols do not exhibit significant hydrogen bonding.
Stronger intermolecular forces lead to a higher boiling point.
Thus,the boiling point of alcohols is higher than that of the corresponding thiols.

(C) The correct answer is $C$.
Ethyl alcohol $(C_2H_5OH)$ contains a $CH_3CH(OH)-$ group,which gives a positive iodoform test with $NaOH$ and $I_2$,forming a yellow precipitate of iodoform $(CHI_3)$.
$C_2H_5OH + 4I_2 + 6NaOH \rightarrow CHI_3 + HCOONa + 5NaI + 5H_2O$.
Methyl alcohol $(CH_3OH)$ does not contain this group and does not form a yellow precipitate with $NaOH$ and $I_2$.

(A) The molecular formula $C_3H_8O$ corresponds to propanol ($CH_3CH_2CH_2OH$ or $CH_3CH(OH)CH_3$).
Oxidation of a primary alcohol $(R-CH_2OH)$ yields a carboxylic acid $(R-COOH)$ with the same number of carbon atoms.
Here,$C_3H_8O$ (propanol) oxidizes to $C_3H_6O_2$ (propanoic acid,$CH_3CH_2COOH$).
Since the number of carbon atoms remains the same ($3$ carbons),the starting compound $X$ must be a primary alcohol.

(D) The oxidation of a secondary alcohol under vigorous conditions leads to the cleavage of the $C-C$ bond adjacent to the carbonyl group formed after initial oxidation to a ketone.The given products are acetic acid ($CH_3COOH$,$2$ carbons) and propanoic acid ($CH_3CH_2COOH$,$3$ carbons).This implies the original alcohol must have a total of $5$ carbon atoms with the hydroxyl group at the $C-2$ position.The structure $CH_3CH(OH)CH_2CH_2CH_3$ (pentan-$2$-ol) undergoes oxidation to break the bond between $C-2$ and $C-3$,resulting in $CH_3COOH$ and $CH_3CH_2COOH$.

(D) The organic liquid $A$ reacts with conc. $H_2SO_4$ to form a colourless gas that decolourises bromine water and alkaline $KMnO_4$,indicating the gas is an alkene.
Since the gas reacts with one mole of $H_2$,it is ethene $(CH_2=CH_2)$.
The dehydration of ethanol $(C_2H_5OH)$ with conc. $H_2SO_4$ at $443 \ K$ yields ethene.
Ethanol has a pleasant odour and a boiling point of $78\ ^\circ C$.
The reaction is: $CH_3CH_2OH \xrightarrow{Conc. H_2SO_4} CH_2=CH_2 + H_2O$.

(B) Glycerol reacts with $P_4 + I_2$ (which generates $PI_3$ in situ) to initially form $1,2,3-triiodopropane$.
This intermediate is unstable due to the presence of iodine atoms on adjacent carbons and undergoes elimination of $I_2$ to form allyl iodide $(CH_2=CH-CH_2I)$.
$CH_2OH-CHOH-CH_2OH$ $\xrightarrow{P_4/I_2} [CH_2I-CHI-CH_2I]$ $\xrightarrow{-I_2} CH_2=CH-CH_2I$

(A) Isopropyl alcohol $(CH_3-CH(OH)-CH_3)$ is a secondary alcohol.
Upon oxidation,secondary alcohols yield ketones.
Therefore,isopropyl alcohol forms acetone $(CH_3-CO-CH_3)$.

(C) The reaction of alcohols with $ZnCl_2 +$ conc. $HCl$ is known as the $Lucas$ test.
$3^o$ alcohols react immediately to form turbidity.
$2^o$ alcohols react after about $5$ minutes.
$1^o$ alcohols do not react at room temperature.
Among the given options,$(CH_3)_3C-OH$ ($2-$methylpropan$-2-$ol) is a $3^o$ alcohol,which reacts immediately to produce turbidity.

(B) Grignard reagents $(R-Mg-X)$ are highly reactive towards compounds containing active hydrogen atoms,such as alcohols.
Both ethanol $(C_2H_5OH)$ and propanol $(C_3H_7OH)$ contain an acidic hydroxyl group $(-OH)$ that reacts with the Grignard reagent to form an alkane and an alkoxide.
Reaction with ethanol: $C_2H_5OH + R-Mg-X \rightarrow RH + C_2H_5OMgX$
Reaction with propanol: $C_3H_7OH + R-Mg-X \rightarrow RH + C_3H_7OMgX$
Therefore,the correct option is $(b)$.

(B) Dehydrogenation of alcohols involves the removal of hydrogen to form alkenes or carbonyl compounds.
When $2-$propanol $(CH_3-CH(OH)-CH_3)$ is passed over heated copper at $573 \ K$,it undergoes dehydrogenation to form propanone (a ketone).
However,if we consider the dehydration of $1-$propanol or $2-$propanol,propene is formed.
Specifically,the question asks for dehydrogenation,but in the context of standard chemistry problems,the dehydration of $1-$propanol $(CH_3-CH_2-CH_2-OH)$ using catalysts like $Al_2O_3$ at high temperatures yields propene $(CH_3-CH=CH_2)$.
Given the options,$1-$propanol is the precursor that leads to propene via elimination reactions.

(A) Alcohols contain a polar $-OH$ group,which allows for the formation of intermolecular hydrogen bonds between alcohol molecules.
This additional intermolecular force requires more energy to overcome during the phase transition from liquid to gas.
In contrast,alkanes are non-polar and only exhibit weak London dispersion forces.
Therefore,alcohols have significantly higher boiling points than alkanes of comparable molecular mass.

(C) Ethylene glycol $(HO-CH_2-CH_2-OH)$ reacts with periodic acid $(HIO_4)$ to undergo oxidative cleavage of the $C-C$ bond.
This reaction results in the formation of two molecules of formaldehyde $(HCHO)$.
The chemical equation is: $HO-CH_2-CH_2-OH + HIO_4 \rightarrow 2HCHO + HIO_3 + H_2O$.

(B) Primary alcohols are oxidized to aldehydes and subsequently to carboxylic acids using strong oxidizing agents like $K_2Cr_2O_7$ and $H_2SO_4$.
The reaction sequence is: $CH_3CH_2OH$ $\xrightarrow{[O]} CH_3CHO$ $\xrightarrow{[O]} CH_3COOH$.
Since the final product is acetic acid $(CH_3COOH)$,the starting compound $D$ must be ethanol $(CH_3CH_2OH)$.

(B) Ethylene glycol $(HOCH_2-CH_2OH)$ reacts with $PI_3$ to form $1,2$-diiodoethane $(ICH_2-CH_2I)$.
$1,2$-diiodoethane is unstable and undergoes elimination of $I_2$ to form ethylene $(CH_2=CH_2)$.
The overall reaction is: $HOCH_2-CH_2OH + PI_3$ $\rightarrow ICH_2-CH_2I$ $\rightarrow CH_2=CH_2 + I_2$.

(A) Ceric ammonium nitrate test is a characteristic test for alcohols.
When an alcohol $(R-OH)$ reacts with ceric ammonium nitrate,it forms a red or yellow colored complex (ceric alkoxy complex).
Reaction: $R-OH + (NH_4)_2Ce(NO_3)_6 \to Ce(NO_3)_6(ROH)_9 + 2NH_4NO_3 + 4HNO_3$
Therefore,the compound $A$ is an alcohol.

(A) The dehydration of alcohols to alkenes in the presence of concentrated $H_2SO_4$ follows an $E1$ mechanism.
Step $1$: The initiation step involves the protonation of the alcohol molecule by the acid catalyst $(H_2SO_4)$ to form a protonated alcohol (alkyloxonium ion).
$CH_3CH_2OH + H^+ \rightarrow CH_3CH_2OH_2^+$
This step makes the $-OH$ group a better leaving group (as $H_2O$).
Therefore,the correct initiation step is the protonation of the alcohol molecule.

(B) Dehydration of alcohols proceeds via the formation of a carbocation intermediate.
The stability of carbocations follows the order: $3^\circ > 2^\circ > 1^\circ$.
$a.$ $CH_3-CH(CH_3)-CH_2OH$ forms a $1^\circ$ carbocation: $CH_3-CH(CH_3)-CH_2^+$.
$b.$ $CH_3-C(CH_3)(OH)-CH_3$ forms a $3^\circ$ carbocation: $(CH_3)_3C^+$.
$c.$ $CH_3-CH_2-CH_2-CH_2OH$ forms a $1^\circ$ carbocation: $CH_3-CH_2-CH_2-CH_2^+$.
$d.$ $CH_3-CH(OH)-CH_2-CH_3$ forms a $2^\circ$ carbocation: $CH_3-CH^+-CH_2-CH_3$.
Since the tertiary $(3^\circ)$ carbocation is the most stable,tert-butyl alcohol (option $b$) is the correct answer.

(D) The reaction represents the dehydration of ethanol to ethene.
$CH_3CH_2OH \xrightarrow{Al_2O_3, 350 \ ^oC} CH_2 = CH_2 + H_2O$.
$Al_2O_3$ (Alumina) acts as a catalyst for the dehydration of primary alcohols at high temperatures.
Therefore,$X$ is $Al_2O_3$.

(C) The reaction of ethanol with concentrated $H_2SO_4$ depends on the temperature and the ratio of reactants:
$1$. At $110 ^\circ C$,it forms $C_2H_5HSO_4$ (Ethyl hydrogen sulphate).
$2$. At $140 ^\circ C$ $(413 \ K)$,it forms $C_2H_5-O-C_2H_5$ (Diethyl ether).
$3$. At $170 ^\circ C$ $(443 \ K)$,it forms $C_2H_4$ (Ethene or Ethylene).
Acetylene $(C_2H_2)$ is not formed in this reaction.

(C) Alcohols with a smaller number of carbon atoms are more soluble in water due to the formation of hydrogen bonds with water molecules. As the length of the hydrocarbon chain increases,the hydrophobic part dominates,decreasing solubility. Therefore,the statement that alcohols with fewer carbon atoms are less soluble is incorrect.

(D) The reaction sequence is as follows:
$1$. $C_2H_5OH$ $(A) + PCl_5 \to C_2H_5Cl$ $(B) + POCl_3 + HCl$
$2$. $C_2H_5Cl$ $(B) + KCN \to C_2H_5CN + KCl$
$3$. $C_2H_5CN + 2H_2O \xrightarrow{H^+} C_2H_5COOH$ (Propionic acid)
Since the final product is propionic acid $(C_2H_5COOH)$,which contains $3$ carbon atoms,the starting material $A$ must be ethanol $(C_2H_5OH)$ and $B$ must be ethyl chloride $(C_2H_5Cl)$.

(B) The correct answer is $(B)$.
Lower alcohols are typically volatile liquids with a characteristic pungent smell and a burning taste.
As the molecular mass increases,the boiling point of alcohols increases due to increased van der Waals forces.
Lower alcohols are soluble in water due to hydrogen bonding.
Therefore,the statement that higher alcohols are stronger and have a bitter taste is not a general characteristic of alcohols.

(C) The reactivity of alcohols with alkali metals (like $Na$) depends on the acidity of the $O-H$ bond.
The order of reactivity is $1^o > 2^o > 3^o$.
This is because the electron-donating inductive effect ($+I$ effect) of alkyl groups increases from primary to tertiary alcohols,which decreases the polarity of the $O-H$ bond and reduces the acidity.
Therefore,primary alcohols react fastest.

(A) The reactivity of alcohols towards $Na$ metal depends on the acidity of the alcohol,as the reaction involves the release of $H^+$ ions to form alkoxides $(R-O^-Na^+)$.
The acidity of alcohols decreases as the number of alkyl groups attached to the carbon bearing the $-OH$ group increases due to the $+I$ effect of alkyl groups,which destabilizes the alkoxide ion.
Therefore,the order of acidity is: $\text{Primary} > \text{Secondary} > \text{Tertiary}$.
Consequently,the order of reactivity towards $Na$ metal is: $\text{Primary} > \text{Secondary} > \text{Tertiary}$.

(A) is the correct answer.
$CH_3OH + Cl_2 \to \text{No reaction}$
$CH_3OH + HCl \xrightarrow{ZnCl_2} CH_3Cl + H_2O$
$3CH_3OH + PCl_3 \to 3CH_3Cl + H_3PO_3$
$CH_3OH + PCl_5 \to CH_3Cl + POCl_3 + HCl$

(D) Secondary alcohols,such as $CH_3CHOHCH_3$ (propan$-2-$ol),undergo oxidation to form ketones.
$CH_3-CH(OH)-CH_3 \xrightarrow{[O]} CH_3-CO-CH_3$ (propanone).
Primary alcohols like $CH_3CH_2CH_2OH$ form aldehydes,while tertiary alcohols like $(CH_3)_3COH$ are resistant to oxidation under normal conditions.

(D) The boiling point of a compound depends on the strength of intermolecular forces.
$Ethanol$ $(CH_3CH_2OH)$ has a higher molecular mass compared to $Methanol$ $(CH_3OH)$ and exhibits strong intermolecular hydrogen bonding.
$Acetone$ and $Diethyl$ $ether$ have weaker intermolecular forces (dipole-dipole and London dispersion forces,respectively) compared to the hydrogen bonding present in alcohols.
Among the given options,$Ethanol$ has the highest boiling point due to its larger molecular size and strong hydrogen bonding.

(A) The oxidation of ethanol $(C_2H_5OH)$ over platinised asbestos in the presence of excess air leads to the formation of acetaldehyde $(CH_3CHO)$.
The reaction is: $C_2H_5OH + [O] \xrightarrow{Pt, \text{air}} CH_3CHO + H_2O$.

(B) The iodoform test is positive for compounds containing a methyl ketone group $(CH_3-CO-)$ or alcohols that can be oxidized to methyl ketones (containing the $CH_3-CH(OH)-$ group).
Propan$-2-$ol $(CH_3-CH(OH)-CH_3)$ contains the $CH_3-CH(OH)-$ group and thus gives a positive iodoform test.

(B) The reaction of alcohols with conc. $HCl$ (Lucas reagent) proceeds via the formation of a carbocation intermediate.
The reactivity order is $3^\circ > 2^\circ > 1^\circ$ because the stability of the carbocation formed follows the same order.
$(CH_3)_3C-OH$ is a tertiary $(3^\circ)$ alcohol,which forms a stable tertiary carbocation,making it the most reactive among the given options.

(B) Low molecular weight alcohols (such as methanol,ethanol,and propanol) are capable of forming hydrogen bonds with water molecules,making them highly soluble in water. They are also generally miscible with many organic solvents due to the presence of the non-polar alkyl group and the polar hydroxyl group. Therefore,they are soluble in both water and many organic solvents.

(C) Methyl ethyl ketone (butanone) is a ketone with the structure $CH_3-CO-CH_2-CH_3$.
Secondary alcohols are oxidized to ketones.
$2-$butanol $(CH_3-CH(OH)-CH_2-CH_3)$ is a secondary alcohol.
Upon oxidation,$2-$butanol undergoes the following reaction:
$CH_3-CH(OH)-CH_2-CH_3 \xrightarrow{[O]} CH_3-CO-CH_2-CH_3 + H_2O$
Therefore,the correct option is $(C)$.

(A) The dehydration of primary alcohols in the presence of a strong acid like concentrated $H_2SO_4$ at $170 \ ^{o}C$ results in the formation of alkenes.
$R-CH_2-CH_2-OH \xrightarrow{Conc. \ H_2SO_4, \ 170 \ ^{o}C} R-CH=CH_2 + H_2O$
Thus,the correct option is $(A)$.

(A) Primary alcohols $(R-CH_2OH)$ on dehydrogenation with reduced copper $(Cu)$ at $573 \ K$ give aldehydes $(R-CHO)$.
Secondary alcohols $(R-CH(OH)-R')$ on dehydrogenation with reduced copper $(Cu)$ at $573 \ K$ give ketones $(R-CO-R')$.
Therefore,primary and secondary alcohols give aldehydes and ketones respectively.

(C) Primary alcohols on oxidation with strong oxidizing agents like acidified $K_2Cr_2O_7$ are oxidized to carboxylic acids.
Methyl alcohol $(CH_3OH)$ is a primary alcohol.
It undergoes oxidation to form formaldehyde $(HCHO)$,which is further oxidized to formic acid $(HCOOH)$.
The reaction is: $CH_3OH$ $\xrightarrow{[O]} HCHO$ $\xrightarrow{[O]} HCOOH$.

(A) Ethyl alcohol $(CH_3CH_2OH)$ is a primary alcohol.
When it undergoes oxidation with a strong oxidizing agent like $K_2Cr_2O_7$ in an acidic medium,it is first oxidized to acetaldehyde $(CH_3CHO)$ and then further oxidized to acetic acid $(CH_3COOH)$.
The overall reaction is: $CH_3CH_2OH \xrightarrow{[O]} CH_3COOH$.

(A) Lucas test is used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction with Lucas reagent (a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$).

(A) In the esterification reaction (Fischer esterification),the alcohol acts as a nucleophile and attacks the carbonyl carbon of the carboxylic acid.
Steric hindrance plays a crucial role in this reaction.
As the degree of the alcohol increases $(1^o < 2^o < 3^o)$,the steric hindrance around the hydroxyl group increases,which makes the nucleophilic attack more difficult.
Therefore,the reactivity order of alcohols towards esterification is $1^o > 2^o > 3^o$.

(B) . $C_2H_5OH$ gives the iodoform test because it contains a $CH_3CH(OH)-$ group (having an $\alpha$-hydrogen atom attached to the carbon bearing the hydroxyl group).
$CH_3OH$ does not give the iodoform test due to the absence of the required $CH_3CH(OH)-$ structural unit.