Properties of alcohol Questions in English

(C) The reaction between a $Grignard \ reagent$ $(C_2H_5MgBr)$ and an alcohol $(CH_3OH)$ is an acid-base reaction.
$C_2H_5MgBr + CH_3OH \rightarrow C_2H_6 + Mg(OCH_3)Br$.
Here,the $Ethyl$ group of the $Grignard \ reagent$ abstracts the acidic proton from the $Methanol$ to form $Ethane$ $(C_2H_6)$.

(B) $CH_3-CH(OH)-CH_2-CH_3 \xrightarrow{H^+} CH_3-CH=CH-CH_3 + H_2O$
According to Saytzeff's rule,in elimination reactions,the more substituted alkene (the one with fewer hydrogen atoms on the double-bonded carbons) is the major product.

(C) $CH_2 = CH_2 \xrightarrow{HBr} CH_3-CH_2Br$ ($X$ is ethyl bromide).
$CH_3-CH_2Br \xrightarrow{\text{Hydrolysis}} CH_3-CH_2OH$ ($Y$ is ethanol).
$CH_3-CH_2OH \xrightarrow[I_2 \text{ excess}]{Na_2CO_3} CHI_3$ ($Z$ is iodoform).
This is the iodoform test,which gives a yellow precipitate of $CHI_3$.

(B) The dehydration of butan$-2-$ol with $H^+/H_2O$ (acid-catalyzed dehydration) yields a mixture of alkenes $(X)$.
Butan$-2-$ol $\xrightarrow{H^+}$ but$-1-$ene + but$-2-$ene (cis and trans).
Thus,the mixture $X$ consists of $3$ alkenes: but$-1-$ene,cis-but$-2-$ene,and trans-but$-2-$ene.
When these react with $Br_2$:
$1$. But$-1-$ene gives $1,2$-dibromobutane (one chiral center,$2$ enantiomers).
$2$. Cis-but$-2-$ene gives meso-$2,3$-dibromobutane ($1$ isomer).
$3$. Trans-but$-2-$ene gives a racemic mixture of $(2R,3R)$- and $(2S,3S)$-$2,3$-dibromobutane ($2$ enantiomers).
Total isomers = $2 + 1 + 2 = 5$ isomers of $C_4H_8Br_2$.
Therefore,the number of alkenes $X$ is $3$.

(B) Phenylmagnesium bromide $(C_6H_5MgBr)$ is a Grignard reagent,which acts as a strong base.
When it reacts with methanol $(CH_3OH)$,which contains an acidic hydrogen atom,an acid-base reaction occurs.
The reaction is: $C_6H_5MgBr + CH_3OH \rightarrow C_6H_6 + Mg(OCH_3)Br$.
Here,$C_6H_6$ is benzene and $Mg(OCH_3)Br$ is methoxymagnesium bromide.

(D) $1$. Hydration of ethyne $(CH \equiv CH)$ in the presence of $Hg^{2+}/H_2SO_4$ (Kucherov reaction) yields acetaldehyde $(CH_3CHO)$ as product $B$.
$2$. Reaction of acetaldehyde $(CH_3CHO)$ with Grignard reagent $(CH_3MgX)$ followed by hydrolysis $(H_2O)$ yields isopropyl alcohol $(CH_3-CH(OH)-CH_3)$ as product $C$.
$3$. Oxidation $([O])$ of isopropyl alcohol (a secondary alcohol) yields acetone $(CH_3-CO-CH_3)$ as the final product $D$.

(B) The dehydration of alcohols involves the removal of a water molecule $(H_2O)$ from the alcohol to form an alkene.
This process is a type of $\beta$-elimination reaction where a hydrogen atom is removed from the $\beta$-carbon and a hydroxyl group $(-OH)$ is removed from the $\alpha$-carbon,resulting in the formation of a double bond.

(A) When ethyl alcohol $(CH_3CH_2OH)$ is heated with red phosphorus and $HI$,it undergoes reduction to form the corresponding alkane.
The chemical reaction is:
$CH_3CH_2OH + 2HI \xrightarrow{\text{Red } P} CH_3CH_3 + H_2O + I_2$
Thus,the product obtained is ethane $(C_2H_6)$.

(D) The acidity of alcohols is determined by the stability of the conjugate base (alkoxide ion).
Electron-withdrawing groups ($-I$ effect) stabilize the negative charge on the oxygen atom,thereby increasing acidity.
The strength of the $-I$ effect depends on the electronegativity of the substituent and its distance from the $-OH$ group.
Fluorine $(F)$ has a stronger $-I$ effect than Iodine $(I)$.
In option $(D)$,the fluorine atom is on the carbon directly attached to the $-OH$ group (alpha-carbon),which provides the strongest $-I$ effect due to the shortest distance.
Therefore,$CH_3-CH_2-CH_2-CHF-OH$ is the most acidic.

(A) primary alcohol contains the functional group $-CH_2OH$.
For the formula $C_4H_{10}O$,we remove one $-CH_2OH$ group,leaving a $C_3H_7$ alkyl group.
The possible isomers for the $C_3H_7$ group are:
$1$. $CH_3-CH_2-CH_2-$ (n-propyl)
$2$. $(CH_3)_2CH-$ (isopropyl)
Attaching the $-CH_2OH$ group to these gives:
$1$. $CH_3-CH_2-CH_2-CH_2OH$ (butan$-1-$ol)
$2$. $(CH_3)_2CH-CH_2OH$ ($2$-methylpropan$-1-$ol)
Thus,there are $2$ possible primary alcohols.

(A) The 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 due to inductive effect and hyperconjugation.
In option $A$,the structure is $CH_3-CH_2-C(OH)(CH_3)-CH_2-CH_3$,which is a $3^\circ$ alcohol.
Therefore,it undergoes dehydration most easily.

(D) The acidity of a compound depends on the stability of its conjugate base.
$1$. For $HC \equiv CH$,the conjugate base is $HC \equiv C^-$,where the negative charge is on an $sp$ hybridized carbon atom.
$2$. For $C_6H_6$,the conjugate base is $C_6H_5^-$,where the negative charge is on an $sp^2$ hybridized carbon atom.
$3$. For $C_2H_6$,the conjugate base is $C_2H_5^-$,where the negative charge is on an $sp^3$ hybridized carbon atom.
$4$. For $CH_3OH$,the conjugate base is $CH_3O^-$,where the negative charge is on an oxygen atom.
Since oxygen is more electronegative than carbon,the $CH_3O^-$ ion is the most stable conjugate base among the given options.
Therefore,$CH_3OH$ is the strongest acid.

(D) The molecular formula $C_5H_{12}O$ corresponds to a saturated alcohol with the general formula $C_nH_{2n+1}OH$.
To find the number of structural isomers,we determine the possible carbon skeletons for a pentyl group $(-C_5H_{11})$:
$1$. $CH_3-CH_2-CH_2-CH_2-CH_2-OH$ (pentan$-1-$ol)
$2$. $CH_3-CH_2-CH_2-CH(OH)-CH_3$ (pentan$-2-$ol)
$3$. $CH_3-CH_2-CH(OH)-CH_2-CH_3$ (pentan$-3-$ol)
$4$. $(CH_3)_2CH-CH_2-CH_2-OH$ ($3$-methylbutan$-1-$ol)
$5$. $(CH_3)_2CH-CH(OH)-CH_3$ ($3$-methylbutan$-2-$ol)
$6$. $CH_3-CH_2-CH(CH_3)-CH_2-OH$ ($2$-methylbutan$-1-$ol)
$7$. $CH_3-C(OH)(CH_3)-CH_2-CH_3$ ($2$-methylbutan$-2-$ol)
$8$. $(CH_3)_3C-CH_2-OH$ ($2$,$2$-dimethylpropan$-1-$ol)
Thus,there are $8$ possible structural isomers.

(D) The acidity of organic compounds depends on the stability of the conjugate base formed after the removal of a proton.
In $HC \equiv CH$,the carbon atom is $sp$ hybridized,which has $50\%$ $s$-character,making it the most electronegative among the given hydrocarbons.
$CH_3OH$ is an alcohol where the proton is attached to an oxygen atom,which is more electronegative than carbon.
Comparing the acidity: $CH_3OH$ $(pK_a \approx 15.5)$ is significantly more acidic than $HC \equiv CH$ $(pK_a \approx 25)$,$C_6H_6$ $(pK_a \approx 43)$,and $C_2H_6$ $(pK_a \approx 50)$.

(C) The primary alcohol isomers for the molecular formula $C_5H_{11}OH$ are as follows:
$(i) \ CH_3-CH_2-CH_2-CH_2-CH_2OH$ (Pentan$-1-$ol)
$(ii) \ CH_3-CH_2-CH(CH_3)-CH_2OH$ ($2$-Methylbutan$-1-$ol)
$(iii) \ CH_3-CH(CH_3)-CH_2-CH_2OH$ ($3$-Methylbutan$-1-$ol)
$(iv) \ (CH_3)_3C-CH_2OH$ ($2$,$2$-Dimethylpropan$-1-$ol)
There are a total of $4$ primary alcohol isomers.

(D) All the given reagents,$H_2SO_4$,$Al_2O_3$,and $P_2O_5$,can act as dehydrating agents for alcohols to facilitate the elimination of water.

(A) The oxidation of primary alcohols like ethyl alcohol $(CH_3CH_2OH)$ with a strong oxidizing agent like acidified $K_2Cr_2O_7$ proceeds through an aldehyde intermediate to form a carboxylic acid.
$CH_3CH_2OH + 2[O] \xrightarrow{K_2Cr_2O_7/H^+} CH_3COOH + H_2O$
Thus,the final product is acetic acid.

(A) The reaction of ethyl alcohol $(CH_3CH_2OH)$ with thionyl chloride $(SOCl_2)$ is the best method for preparing alkyl chlorides because the by-products ($SO_2$ and $HCl$) are gases and escape,leaving pure ethyl chloride $(CH_3CH_2Cl)$.
$CH_3CH_2OH + SOCl_2 \to CH_3CH_2Cl + SO_2 + HCl$

(A) The reactivity of alcohols in reactions involving the cleavage of the $C-OH$ bond follows the order:
Tertiary > Secondary > Primary
This order is based on the stability of the carbocation intermediate formed during the reaction.

(A) The reaction of ethanol $(A)$ with sodium metal produces sodium ethoxide $(B)$ and hydrogen gas:
$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2$
When ethanol $(A)$ is heated with concentrated $H_2SO_4$ at $413 \ K$,it undergoes intermolecular dehydration to form diethyl ether:
$2C_2H_5OH \xrightarrow{conc. H_2SO_4, 413 \ K} C_2H_5-O-C_2H_5 + H_2O$
Thus,$(A)$ is $C_2H_5OH$ and $(B)$ is $C_2H_5ONa$.

(D) The reaction of ethyl alcohol $(C_2H_5OH)$ with chlorine $(Cl_2)$ in the presence of $NaOH$ is a haloform reaction.
First,$C_2H_5OH$ is oxidized to acetaldehyde $(CH_3CHO)$ by $Cl_2$.
Then,acetaldehyde undergoes chlorination to form chloral $(CCl_3CHO)$.
Finally,chloral reacts with $NaOH$ to produce chloroform $(CHCl_3)$ and sodium formate $(HCOONa)$.
The overall reaction is: $C_2H_5OH + 4Cl_2 + 6NaOH \to CHCl_3 + HCOONa + 5NaCl + 5H_2O$.

(D) Ethanol $(C_2H_5OH)$ is the type of alcohol used in alcoholic beverages.

(D) The reaction between ethanol and phosphorus tribromide is given by the balanced chemical equation:
$3C_2H_5OH + PBr_3 \rightarrow 3C_2H_5Br + H_3PO_3$
Here,$3 \ mol$ of ethanol reacts with $1 \ mol$ of $PBr_3$ to produce $3 \ mol$ of bromoethane and $1 \ mol$ of phosphorous acid $(H_3PO_3)$.
Therefore,$X$ is $H_3PO_3$.

(D) Since the compound reacts with sodium,it must be an alcohol.
Upon oxidation,it yields a carbonyl compound that does not reduce Tollen's reagent,which indicates that the product is a ketone.
Primary alcohols oxidize to aldehydes (which reduce Tollen's reagent),whereas secondary alcohols oxidize to ketones (which do not reduce Tollen's reagent).
Therefore,the original substance must be a secondary alcohol,which is $sec$-butyl alcohol $(butan-2-ol)$.
Reaction: $CH_3CH_2CH(OH)CH_3 \xrightarrow{[O]} CH_3CH_2COCH_3$ $(butan-2-one)$.

(A) The reaction of alcohol with concentrated $H_2SO_4$ involves the protonation of the alcohol group.
$R-OH + H^+ \to R-OH_2^+$.
This protonated alcohol then loses a water molecule to form a carbocation (carbonium ion) intermediate:
$R-OH_2^+ \to R^+ + H_2O$.
Thus,the carbonium ion is the key intermediate formed during the dehydration process.

(A) Lower molecular weight alcohols are soluble in water due to the formation of intermolecular hydrogen bonds with water molecules.

(A) When primary $(1^o)$ alcohols are passed over red-hot copper at $573 \ K$ $(300 \ ^oC)$,they undergo dehydrogenation to form aldehydes.
$R-CH_2OH \xrightarrow{Cu/573 \ K} R-CHO + H_2$
When secondary $(2^o)$ alcohols are passed over red-hot copper at $573 \ K$,they undergo dehydrogenation to form ketones.
$R_2CHOH \xrightarrow{Cu/573 \ K} R_2C=O + H_2$
Thus,primary and secondary alcohols yield aldehydes and ketones respectively.

(D) The most suitable method to remove trace amounts of water from ethanol is by reacting it with $Mg$ metal.
$Mg$ reacts with water to form $MgO$ and $H_2$ gas,while it does not react with pure ethanol under these conditions.
The reaction is: $Mg + H_2O \rightarrow MgO + H_2$.
This process is commonly used to prepare absolute alcohol.

(D) The substance is soluble in concentrated $H_2SO_4$,which indicates it could be an alcohol,alkene,or ether.
Since it does not decolorize bromine in $CCl_4$,it is not an alkene.
Ethers are generally resistant to oxidation by chromic anhydride in aqueous $H_2SO_4$ (Jones reagent).
Primary alcohols are oxidized very rapidly by chromic anhydride in aqueous $H_2SO_4$ (Jones oxidation),which matches the observed color change (orange $Cr^{6+}$ to green/blue $Cr^{3+}$).
Therefore,the substance is a primary alcohol.

(D) $PCC$ $(CrO_3 \cdot C_5H_5N \cdot HCl)$ is the most suitable reagent for the oxidation of primary alcohols to aldehydes because it stops the oxidation at the aldehyde stage and does not further oxidize it to a carboxylic acid.

(B) $1$. Oxidation of isopropyl alcohol $((CH_3)_2CHOH)$ with $PCC$ gives acetone $(CH_3COCH_3)$ as $X$.
$2$. Nucleophilic addition of methylmagnesium bromide $(CH_3MgBr)$ to acetone followed by acid hydrolysis $(H_2O)$ yields tert-butyl alcohol $((CH_3)_3COH)$ as $Y$.
$3$. The reaction sequence is: $(CH_3)_2CHOH$ $\xrightarrow{PCC} CH_3COCH_3$ $\xrightarrow[(ii) H_2O]{(i) CH_3MgBr} (CH_3)_3COH$.

(D) The reaction of a Grignard reagent $(RMgX)$ with ethylene oxide (epoxide) involves the nucleophilic attack of the $R^-$ group on one of the carbon atoms of the epoxide ring,leading to ring opening.
$RMgX + CH_2-CH_2(O) \to R-CH_2-CH_2-OMgX$
Subsequent hydrolysis of the intermediate alkoxide yields a primary alcohol:
$R-CH_2-CH_2-OMgX + H_2O \to R-CH_2-CH_2-OH + Mg(OH)X$
Thus,the final product is $RCH_2CH_2OH$.

(B) Step $1$: $C_2H_5OH + PBr_3 \rightarrow C_2H_5Br (X) + H_3PO_3$.
Step $2$: $C_2H_5Br + KOH (alc.) \rightarrow CH_2=CH_2 (Y) + KBr + H_2O$.
Step $3$: $CH_2=CH_2 + H_2SO_4 \rightarrow CH_3CH_2OSO_3H$,followed by hydrolysis with $H_2O/\Delta$ gives $CH_3CH_2OH (Z) + H_2SO_4$.
Thus,$Z$ is $CH_3-CH_2OH$.

(C) The statement in option $C$ is incorrect. Lower molecular mass alcohols are highly soluble in water due to hydrogen bonding,and their solubility decreases as the hydrocarbon chain length (molecular mass) increases. They are generally soluble in organic solvents.

(C) Tertiary alcohols do not undergo dehydrogenation easily. When passed over heated copper at $573 \ K$,they undergo dehydration to form alkenes.

(A) The density of ethyl alcohol $(C_2H_5OH)$ is approximately $0.789 \ g/mL$,whereas the density of water is $1.00 \ g/mL$. Therefore,ethyl alcohol is lighter than water,making statement $A$ incorrect.

(A) The boiling points of alcohols are higher than those of their corresponding alkanes due to the presence of intermolecular hydrogen bonding.
Butan-$2$-ol is an alcohol,whereas Ethane,Butane,and Pentane are alkanes.
Therefore,Butan-$2$-ol has the highest boiling point.

(D) To give the iodoform test,a compound must contain either the $CH_3-CH(OH)-$ group or the $CH_3-C(=O)-$ group.
$1$. Ethanol $(CH_3CH_2OH)$ contains the $CH_3-CH(OH)-$ group.
$2$. Ethanal $(CH_3CHO)$ contains the $CH_3-C(=O)-$ group.
$3$. Iso-propyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3-CH(OH)-$ group.
$4$. Benzyl alcohol $(C_6H_5CH_2OH)$ does not contain either of these groups.
Therefore,benzyl alcohol will not give the iodoform test.

(A) The boiling points of isomeric alcohols follow the order: $1^o > 2^o > 3^o$.
This is because as the branching increases,the surface area of the molecule decreases,which leads to weaker van der Waals forces of attraction,resulting in a lower boiling point.

(C) The reaction of ethanol (an alcohol) with iodine $(I_2)$ in the presence of sodium hydroxide $(NaOH)$ is known as the iodoform test.
The chemical reaction is:
$CH_3CH_2OH + 4I_2 + 6NaOH \to CHI_3 + HCOONa + 5NaI + 5H_2O$.
This reaction produces iodoform $(CHI_3)$,which is a yellow precipitate.
Therefore,option $C$ is correct.

(C) The order of reactivity of alcohols towards esterification is $1^o > 2^o > 3^o$.
This is due to the increasing steric hindrance in the alcohols as we move from primary to tertiary alcohols.

(A) Nucleophilic reactivity is influenced by both basicity and steric hindrance.
$CH_3O^{-}$ is the most reactive nucleophile among the given options because it has the least steric hindrance,allowing it to easily approach the electrophilic center.
While $(CH_3)_3CO^{-}$ is a stronger base,its high steric bulk significantly reduces its nucleophilicity compared to $CH_3O^{-}$.

(D) Glycerol contains three $-OH$ groups.
The structure is $CH_2(OH)-CH(OH)-CH_2(OH)$.
$\therefore$ It is a trihydric alcohol.

(D) Ethanol mixed with a small amount of methanol is known as $Methylated \ spirit$.

(A) Alcoholic beverages are primarily composed of $C_2H_5OH$ (ethanol),which is produced by the fermentation of sugars.

(A) Ethylene glycol is commonly used as an antifreeze in automobile radiators to lower the freezing point of the coolant.

(D) The structure of glycerol is $CH_2(OH)-CH(OH)-CH_2(OH)$.
It contains three $-OH$ groups.
The two terminal $-OH$ groups are attached to primary carbon atoms,making them primary $-OH$ groups.
The central $-OH$ group is attached to a secondary carbon atom,making it a secondary $-OH$ group.
Thus,glycerol contains two primary $-OH$ groups and one secondary $-OH$ group.

(B) Glycerol contains three $-OH$ groups,which allow for extensive intermolecular $H$-bonding,leading to a higher boiling point compared to propanal.

(D) The structure of glycerol is $CH_2(OH)-CH(OH)-CH_2(OH)$.
It contains two primary hydroxyl groups and one secondary hydroxyl group.
Therefore,it is not a tertiary alcohol; it is a trihydric alcohol.

(D) To obtain absolute alcohol from rectified spirit,the traces of water present in it are removed by distillation over $Mg(OC_2H_5)_2$.
The reaction is: $Mg(OC_2H_5)_2 + 2H_2O \to Mg(OH)_2 + 2C_2H_5OH$.