Properties of alcohol Questions in English

(B) When wine is exposed to air,the ethanol $(C_2H_5OH)$ present in it undergoes oxidation to form acetic acid $(CH_3COOH)$.
This process is catalyzed by certain bacteria like Acetobacter aceti,which leads to the sour taste.

(C) $C_2H_5OH$ gives the iodoform test because it contains the $CH_3CH(OH)-$ group,whereas $CH_3OH$ does not give this test.
Therefore,the iodoform test is used to distinguish or detect the presence of $C_2H_5OH$ in $CH_3OH$.

(C) The reaction of ethanol with concentrated $H_2SO_4$ depends on temperature and concentration.
At $413 \ K$,it produces diethyl ether.
At $443 \ K$,it produces ethylene.
Ethyl hydrogen sulphate is an intermediate formed at lower temperatures.
Acetylene $(C_2H_2)$ is not produced in this reaction.

(A) The dehydration of ethanol with concentrated $H_2SO_4$ at $160 - 170 \, ^\circ C$ yields ethylene.
The chemical equation is:
$C_2H_5OH \xrightarrow[160 - 170 \, ^\circ C]{H_2SO_4} CH_2 = CH_2 + H_2O$
Thus,$X$ is ethanol $(C_2H_5OH)$.

(D) Ethyl alcohol $(C_2H_5OH)$ can be converted into:
$1$. Ethyl acetate by reaction with acetic acid (esterification).
$2$. Ethylene by dehydration using concentrated $H_2SO_4$.
$3$. Acetic acid by oxidation using oxidizing agents like $KMnO_4$ or $K_2Cr_2O_7$.
Therefore,all of these can be prepared from ethyl alcohol.

(D) The chemical reaction is: $C_2H_5OH \to CH_3CHO$.
In this reaction,the primary alcohol (ethyl alcohol) loses hydrogen atoms to form an aldehyde (acetaldehyde).
Since the removal of hydrogen is defined as oxidation,this process is an example of oxidation.

(C) The dehydration of ethanol to diethyl ether occurs via an intermolecular dehydration process.
When ethanol is heated with concentrated $H_2SO_4$ at $410 \, K$,it undergoes intermolecular dehydration to form diethyl ether.
The reaction is:
$2C_2H_5OH \xrightarrow[410 \, K]{Conc. H_2SO_4} C_2H_5-O-C_2H_5 + H_2O$
Thus,the correct reagent is concentrated $H_2SO_4$.

(A) Dehydrogenation of secondary alcohols yields ketones.
$2-$propanol $(CH_3-CH(OH)-CH_3)$ undergoes dehydrogenation in the presence of copper $(Cu)$ catalyst at $573 \ K$ to form propanone $(CH_3-CO-CH_3)$ and hydrogen gas $(H_2)$.
Reaction: $CH_3-CH(OH)-CH_3 \xrightarrow{Cu, 573 \ K} CH_3-CO-CH_3 + H_2$.

(B) When glycerol reacts with excess $HI$,it undergoes substitution and elimination reactions to form allyl iodide,which then reacts further to form isopropyl iodide. The intermediate $CH_2OH-CHI-CH_2OH$ is not a product of this reaction sequence.
The reaction sequence is:
$CH_2OH-CHOH-CH_2OH + 3HI \rightarrow CH_2I-CHI-CH_2I + 3H_2O$
$CH_2I-CHI-CH_2I \rightarrow CH_2=CH-CH_2I + I_2$ (Allyl iodide)
$CH_2=CH-CH_2I + HI \rightarrow CH_3-CH(I)-CH_3$ (Isopropyl iodide)
Thus,$CH_2OH-CHI-CH_2OH$ is not formed.

(B) The dehydration of alcohols to alkenes proceeds via the following mechanism:
Step $1$: Protonation of the alcohol molecule.
$R-OH + H^+ \rightarrow R-OH_2^+$
This is the first step where the oxygen atom of the alcohol gets protonated by the acid $(H_2SO_4)$ to form an alkyloxonium ion.

(A) The ease of acid-catalyzed esterification of alcohols depends on the steric hindrance around the carbon atom attached to the hydroxyl group.
The order of reactivity is: $1^o \text{ alcohol} > 2^o \text{ alcohol} > 3^o \text{ alcohol}$.
In the given options,$CH_3(CH_2)_2OH$ is a primary $(1^o)$ alcohol,$CH_3CH_2CH(OH)CH_3$ is a secondary $(2^o)$ alcohol,and $(CH_3)_3COH$ is a tertiary $(3^o)$ alcohol.
Therefore,the correct decreasing order is: $CH_3(CH_2)_2OH > CH_3CH_2CH(OH)CH_3 > (CH_3)_3COH$.

(D) The reaction of alcohol with a strong base like sodalime (or sodium metal/amide) involves the cleavage of the $O - H$ bond.
$R - OH + NaNH_2 \to R - ONa + NH_3$
Since the electronegativity difference between oxygen and hydrogen is maximum,the $O - H$ bond undergoes heterolytic cleavage most readily.

(B) Dehydration of alcohols proceeds via the formation of a carbocation intermediate. The stability of carbocations follows the order: $3^o > 2^o > 1^o$.
Among the given options,$(CH_3)_3COH$ is a tertiary alcohol,which upon dehydration forms a tertiary carbocation $(CH_3)_3C^+$.
This tertiary carbocation is the most stable due to the inductive effect and hyperconjugation of nine $\alpha$-hydrogen atoms.

(D) The acidity of a compound depends on the stability of its conjugate base.
$1$. $CH_3OH$ (methanol) has an acidic proton attached to an electronegative oxygen atom,forming a stable methoxide ion $(CH_3O^-)$.
$2$. $HC \equiv CH$ (ethyne) has an $sp$ hybridized carbon,but it is less acidic than an alcohol.
$3$. $C_6H_6$ (benzene) and $C_2H_6$ (ethane) are extremely weak acids.
Therefore,$CH_3OH$ is the strongest acid among the given options.

(C) $1-propanol$ $(CH_3CH_2CH_2OH)$ on heating with $Cu$ at $573 \ K$ undergoes dehydrogenation to form propanal $(CH_3CH_2CHO)$,which is an aldehyde.
Aldehydes give a positive test with Fehling's solution.
$2-propanol$ $(CH_3CH(OH)CH_3)$ on heating with $Cu$ at $573 \ K$ undergoes dehydrogenation to form propanone $(CH_3COCH_3)$,which is a ketone.
Ketones do not give a positive test with Fehling's solution.
Other reagents like $KMnO_4$ or acidic dichromate oxidize $1-propanol$ to propanoic acid,which does not react with Fehling's solution,and concentrated $H_2SO_4$ causes dehydration to propene.

(C) The dehydration of alcohols with concentrated $H_2SO_4$ proceeds via the formation of a carbocation intermediate.
Substituents that donate electron density to the carbocation center stabilize it,thereby increasing the rate of dehydration.
The $-OCH_3$ group is a strong electron-donating group due to its $+M$ (mesomeric) effect,which significantly stabilizes the resulting carbocation.
In contrast,$-NO_2$ and $-Cl$ are electron-withdrawing groups that destabilize the carbocation.
Therefore,$p-CH_3OC_6H_4CH(OH)CH_3$ undergoes dehydration most easily.

(D) The reaction of $3-$phenylpropane $-1,2-$diol with $H_2SO_4$ involves the dehydration of the vicinal diol.
Under acidic conditions and heating,the diol undergoes intramolecular dehydration to form an epoxide (oxirane derivative).
The hydroxyl group at the $C-1$ position is protonated,followed by the nucleophilic attack of the $C-2$ hydroxyl group on the $C-1$ carbon,leading to the elimination of water and the formation of the epoxide ring.
The product is $2-$benzyl oxirane.

(A) Heated copper $(Cu)$ at $300\,^oC$ acts as a dehydrogenating agent for primary and secondary alcohols,and a dehydrating agent for tertiary alcohols.
$1.$ Primary alcohols $(R-CH_2OH)$ are dehydrogenated to aldehydes.
$2.$ Secondary alcohols $(R_2CHOH)$ are dehydrogenated to ketones.
$3.$ Tertiary alcohols $(R_3COH)$ undergo dehydration to form alkenes (olefins).
$4.$ Phenol does not undergo these specific reactions with heated copper to form benzyl alcohol; therefore,the reaction $Phenol \rightarrow Benzyl \ alcohol$ is incorrect.

(C) The stability order of carbocations is as follows:
$3^o > 2^o > 1^o$.
During the dehydration of alcohols,the rate-determining step involves the formation of a carbocation intermediate.
$3^o$-Butyl alcohol (tert-butyl alcohol) forms a tertiary carbocation,which is the most stable among the given options due to the inductive effect and hyperconjugation of the surrounding alkyl groups.

(A) The reaction of alcohols with hydrogen halides $(HX)$ follows the order: $3^o > 2^o > 1^o$ due to the stability of the carbocation intermediate formed during the reaction.
$2-$Methylpropan-$2-$ol is a $3^o$ alcohol:
$(CH_3)_3C-OH$
Since it forms a stable tertiary carbocation,it reacts most rapidly with hydrogen halides.

(A) When ethyl alcohol $(C_2H_5OH)$ is heated with concentrated $H_2SO_4$ at a temperature of $160-170 \ ^\circ C$,dehydration occurs to form ethene $(C_2H_4)$:
$C_2H_5OH \xrightarrow[160-170 \ ^\circ C]{\text{conc. } H_2SO_4} CH_2=CH_2 + H_2O$

(A) The Lucas reagent (a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$) is used to distinguish between $1^o, 2^o$,and $3^o$ alcohols.
The reaction is: $R-OH + HCl \xrightarrow{ZnCl_2} RCl + H_2O$
The three types of alcohols react with this reagent at different rates:
$3^o$ Alcohol (immediate turbidity) > $2^o$ Alcohol (turbidity in $5-10$ minutes) > $1^o$ Alcohol (no turbidity at room temperature).

(A) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
$1$. $CH_3CH_2OH$ (Ethanol) contains the $CH_3CH(OH)-$ group,so it gives a positive iodoform test.
$2$. $CH_3OH$ (Methanol) does not contain this group.
$3$. $(CH_2OH)_2$ (Ethane$-1,2-$diol) does not contain this group.
The reaction for ethanol is:
$CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3 + HCOONa + 5NaI + 5H_2O$

(B) Conc. $HCl$,conc. $HBr$,and conc. $HCl + ZnCl_2$ are all nucleophilic reagents that convert alcohols into alkyl halides. Conc. $H_3PO_4$ is a good dehydrating agent that converts alcohols into alkenes via the elimination reaction.

(B) Phenylmagnesium bromide $(C_6H_5MgBr)$ is a Grignard reagent,which acts as a strong base. Methanol $(CH_3OH)$ contains an acidic proton attached to the oxygen atom.
When they react,the Grignard reagent abstracts the acidic proton from methanol to form benzene $(C_6H_6)$.
The remaining part forms the magnesium salt,methoxymagnesium bromide $(Mg(OCH_3)Br)$.
The reaction is: $C_6H_5MgBr + CH_3OH \to C_6H_6 + Mg(OCH_3)Br$.

(D) $CH_3CH_2OH \xrightarrow{P + I_2} CH_3CH_2I$ $(A)$
$CH_3CH_2I \xrightarrow{Mg} CH_3CH_2MgI$ $(B)$
$CH_3CH_2MgI + HCHO \rightarrow CH_3CH_2CH_2OMgI$ $(C)$
$CH_3CH_2CH_2OMgI \xrightarrow{H_2O} CH_3CH_2CH_2OH$ $(D)$
The final product $D$ is $n$-propyl alcohol (propan$-1-$ol).

(A) The dehydration of ethanol to form diethyl ether occurs in the presence of concentrated $H_2SO_4$ at $413 \ K$.
$2C_2H_5OH \xrightarrow{H_2SO_4, 413 \ K} C_2H_5-O-C_2H_5 + H_2O$

(A) Isopropyl alcohol is a secondary alcohol. Oxidation of secondary alcohols with oxidizing agents like $K_2Cr_2O_7 / H_2SO_4$ yields ketones.
The reaction is:
$CH_3-CH(OH)-CH_3 + [O] \xrightarrow{K_2Cr_2O_7 / H_2SO_4} CH_3-CO-CH_3 + H_2O$
The product formed is $CH_3-CO-CH_3$,which is known as acetone.

(B) The compound $C_4H_{10}O$ is $2-butanol$ $(CH_3CH(OH)CH_2CH_3)$.
On oxidation,$2-butanol$ gives $2-butanone$ $(CH_3COCH_2CH_3)$,which is a ketone.
$2-butanone$ contains a $CH_3CO-$ group,so it gives a positive iodoform test and forms an oxime with $NH_2OH$.
On dehydration with concentrated $H_2SO_4$,$2-butanol$ gives $2-butene$ $(CH_3CH=CHCH_3)$ as the major product.
Thus,the correct structure is $CH_3CH(OH)CH_2CH_3$.

(A) When secondary alcohols like $2-propanol$ are passed over heated copper at $300\,^oC$,they undergo dehydrogenation to form ketones.
The reaction is: $CH_3-CH(OH)-CH_3 \xrightarrow{Cu/300\,^oC} CH_3-CO-CH_3 + H_2$.
Thus,$2-propanol$ gives $Acetone$.

(D) Methyl formate $(HCOOCH_3)$ reacts with $CH_3MgI$ in two steps.
Step $1$: $HCOOCH_3 + CH_3MgI \to CH_3COCH_3 + Mg(OCH_3)I$ (Acetone is formed).
Step $2$: The excess $CH_3MgI$ reacts with the formed acetone: $CH_3COCH_3 + CH_3MgI \to (CH_3)_3COMgI$.
Step $3$: Hydrolysis: $(CH_3)_3COMgI + H_2O/H^+ \to (CH_3)_3COH + Mg(OH)I$ (tert-butyl alcohol).
Note: If the question implies only one equivalent of $CH_3MgI$,the product would be acetaldehyde,but with excess $CH_3MgI$,the final product is tert-butyl alcohol. Given the options,if the reaction is meant to be with a different ester or limited reagent,isopropyl alcohol ($2^o$ alcohol) is formed from $HCOOCH_3$ only if the stoichiometry is specific. However,based on standard textbook reactions for methyl formate with excess Grignard reagent,tert-butyl alcohol is the product. Re-evaluating the provided options,$D$ is the intended answer for $2^o$ alcohol formation from specific formates.

(C) Ethyl alcohol $(CH_3CH_2OH)$ undergoes controlled oxidation in the presence of an aqueous oxidizing agent like $KMnO_4$ or $K_2Cr_2O_7$ to form acetaldehyde $(CH_3CHO)$.
$CH_3CH_2OH + [O] \xrightarrow{KMnO_{4(aq)}} CH_3CHO + H_2O$

(D) When secondary alcohols are passed over heated copper,they undergo dehydrogenation to form ketones.
$CH_3-CH(OH)-CH_3 \xrightarrow{Cu, 573 K} CH_3-CO-CH_3 + H_2$
Isopropyl alcohol (a secondary alcohol) gives acetone as the product.

(D) The compound $X$ is $CH_3CH(OH)CH_3$ (isopropyl alcohol).
When $CH_3CH(OH)CH_3$ is oxidized with acidic $K_2Cr_2O_7$,it forms $CH_3COCH_3$ (acetone) as product $Y$.
Acetone contains the $CH_3CO-$ group,which gives a positive iodoform test with $I_2$ and $Na_2CO_3$ (or $NaOH$),producing yellow crystals of triiodomethane $(CHI_3)$.

(C) The oxidation of secondary alcohols yields ketones. $2-$butanol is a secondary alcohol. The reaction is as follows:
$CH_3-CH(OH)-CH_2-CH_3 + [O] \xrightarrow{K_2Cr_2O_7/H_2SO_4} CH_3-CO-CH_2-CH_3 + H_2O$
Here,$2-$butanol is oxidized to ethyl methyl ketone (butanone).

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

(A) The fermentation of ethanol $(C_2H_5OH)$ using bacteria like $Acetobacter \, aceti$ produces acetic acid $(CH_3COOH)$.
$CH_3CH_2OH \xrightarrow{\text{fermentation}} CH_3COOH$
Thus,ethanol is the correct substance.

(C) The $-OH$ group in alcohols and the $-COOH$ group in carboxylic acids can be replaced by a $-Cl$ atom using $PCl_5$ (Phosphorus pentachloride).
For alcohols: $R-OH + PCl_5 \rightarrow R-Cl + POCl_3 + HCl$.
For carboxylic acids: $R-COOH + PCl_5 \rightarrow R-COCl + POCl_3 + HCl$.

(C) The reaction sequence is as follows:
$CH_3-CH(OH)-CH_3 \xrightarrow{PBr_3} CH_3-CH(Br)-CH_3$ $[A]$
$CH_3-CH(Br)-CH_3 \xrightarrow{Mg} CH_3-CH(MgBr)-CH_3$ $[B]$
$CH_3-CH(MgBr)-CH_3 \xrightarrow{\text{Oxirane}} (CH_3)_2CH-CH_2-CH_2-OMgBr$ $[C]$
$(CH_3)_2CH-CH_2-CH_2-OMgBr \xrightarrow{H_2O, -Mg(OH)Br} (CH_3)_2CH-CH_2-CH_2-OH$ $[D]$
Compound $[D]$ is $3-\text{methylbutan}-1-\text{ol}$.

(B) The dehydration of alcohols proceeds via the formation of a carbocation intermediate. The ease of dehydration depends on the stability of the carbocation formed after the loss of the water molecule.
$1$. Compound $III$ forms a conjugated carbocation (allylic and conjugated with the other double bond),which is the most stable.
$2$. Compound $II$ forms an allylic carbocation,which is more stable than the simple secondary carbocation formed from $I$.
$3$. Compound $I$ forms a simple secondary carbocation.
Thus,the stability order of the carbocations is $III > II > I$,which is also the order of ease of dehydration. Therefore,the correct order is $III > II > I$.

(B) $PCl_5$ (phosphorus pentachloride) is a common reagent in organic chemistry used to convert hydroxyl groups $(-OH)$ into chloro groups $(-Cl)$.
For example,$R-OH + PCl_5 \rightarrow R-Cl + POCl_3 + HCl$.
It also reacts with carbonyl compounds to replace the oxygen atom with two chlorine atoms.

(B) The correct answer is $B$.
The solubility of alcohols in water decreases as the molecular mass increases.
As the size of the alkyl group increases,the hydrophobic character of the molecule increases.
Therefore,the solubility of alkanols in water decreases in the order: $Propanol > Butanol > Pentanol$.

(D) The correct order is $D$.
The solubility of alcohols in water decreases as the size of the hydrophobic alkyl group increases.
This is because the larger alkyl group hinders the formation of hydrogen bonds with water molecules.
Therefore,the order of solubility is $Ethanol > n-propanol > n-butyl alcohol$.

(B) Compounds containing the $CH_3CO-$ group or those that can be oxidized to this group (like $CH_3CH(OH)-$ group) give a positive iodoform test.
Ethyl alcohol $(CH_3CH_2OH)$ is oxidized to acetaldehyde $(CH_3CHO)$ by $I_2/NaOH$,which then reacts to form iodoform $(CHI_3)$.
The reaction is:
$CH_3CH_2OH$ $\xrightarrow{I_2/NaOH} CH_3CHO$ $\xrightarrow{I_2/NaOH} CI_3CHO$ $\xrightarrow{NaOH} CHI_3 + HCOONa$

(A) The dehydration of $1$-cyclohexylbutan-$1$-ol involves the formation of a carbocation intermediate. Upon protonation of the $-OH$ group and loss of water,a tertiary carbocation is formed at the carbon attached to the cyclohexane ring. This carbocation can undergo elimination by losing a proton from adjacent carbons to form various alkenes. The product shown in option $A$ (where the double bond is inside the ring) is a possible product. However,the structure shown in the solution image (a double bond between the ring and the side chain) is not a possible product of this specific dehydration reaction because it would require the formation of a less stable alkene or a different carbocation rearrangement that is not favored here.

(B) The iodoform test is positive for alcohols containing the $CH_3CH(OH)-$ group.
For the molecular formula $C_6H_{14}O$,the isomeric alcohols that contain this structural unit are:
$1.$ $CH_3CH_2CH_2CH_2CH(OH)CH_3$ ($2$-Hexanol)
$2.$ $CH_3CH_2CH(CH_3)CH(OH)CH_3$ ($3$-Methyl-$2$-pentanol)
$3.$ $(CH_3)_2CHCH_2CH(OH)CH_3$ ($4$-Methyl-$2$-pentanol)
$4.$ $(CH_3)_3CCH(OH)CH_3$ ($3,3$-Dimethyl-$2$-butanol)
Thus,there are $4$ such isomeric alcohols.

(B) Ethylene glycol (commonly known as $Glycol$) is widely used as an antifreeze in automobile radiators because it lowers the freezing point of water and raises its boiling point.

(B) When glycerol is treated with excess $HI$,it produces $2$-iodopropane. The reaction proceeds through several steps:
$1.$ Glycerol reacts with $3$ moles of $HI$ to form an unstable triiodide: $CH_2(OH)CH(OH)CH_2(OH) + 3HI \rightarrow [CH_2I-CHI-CH_2I] + 3H_2O$.
$2.$ The unstable triiodide loses $I_2$ to form allyl iodide: $[CH_2I-CHI-CH_2I] \rightarrow CH_2=CH-CH_2I + I_2$.
$3.$ Since $HI$ is in excess,allyl iodide reacts further to form $1,2$-diiodopropane,which loses $I_2$ to form propene: $CH_2=CH-CH_2I + HI$ $\rightarrow [CH_3-CHI-CH_2I]$ $\rightarrow CH_3-CH=CH_2 + I_2$.
$4.$ Finally,propene reacts with $HI$ (Markovnikov addition) to give $2$-iodopropane: $CH_3-CH=CH_2 + HI \rightarrow CH_3-CHI-CH_3$.

(C) $1$. $CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \ (X)$
$2$. $CH_3CH_2Br \xrightarrow{alc. KOH} CH_2=CH_2 \ (Y) \text{ (Dehydrohalogenation)}$
$3$. $CH_2=CH_2$ $\xrightarrow{H_2SO_4} CH_3CH_2OSO_3H$ $\xrightarrow{H_2O, \text{heat}} CH_3CH_2OH \ (Z) \text{ (Hydration of ethene)}$
Thus,the product $Z$ is ethanol.

(C) Periodic acid $(HIO_4)$ is a selective oxidizing agent used for the oxidative cleavage of vicinal diols ($1,2$-diols).
Ethylene glycol $(HO-CH_2-CH_2-OH)$ is a vicinal diol that reacts with $HIO_4$ to produce two molecules of formaldehyde $(HCHO)$.
$HO-CH_2-CH_2-OH + HIO_4 \rightarrow 2HCHO + HIO_3 + H_2O$.