Properties of Ethers Questions in English

(C) The chemical formulas for the given ethers are:
Diethyl ether: $C_2H_5-O-C_2H_5$
Methyl $n$-propyl ether: $CH_3-O-C_3H_7$
Metamerism arises due to the difference in the distribution of carbon atoms on either side of the functional group $(-O-)$.
Since the alkyl groups attached to the oxygen atom are different in the two compounds,they are metamers.

(C) When diethyl ether $(C_2H_5-O-C_2H_5)$ is heated with excess concentrated $HI$,it undergoes cleavage to form $2$ moles of ethyl iodide $(C_2H_5I)$ and $1$ mole of water $(H_2O)$.
The reaction is: $C_2H_5-O-C_2H_5 + 2HI \to 2C_2H_5I + H_2O$.

(B) The reaction involves the nucleophilic substitution of ethyl iodide $(C_2H_5I)$ by the phenoxide ion $(C_6H_5O^-)$.
$1$. Phenol $(C_6H_5OH)$ reacts with the ethoxide ion $(C_2H_5O^-)$ in ethanol to form the phenoxide ion $(C_6H_5O^-)$.
$2$. The phenoxide ion acts as a nucleophile and attacks the ethyl iodide $(C_2H_5I)$ via an $S_N2$ mechanism.
$3$. The reaction is: $C_6H_5O^- + C_2H_5I \rightarrow C_6H_5OC_2H_5 + I^-$.
$4$. The product formed is ethyl phenyl ether $(C_6H_5OC_2H_5)$,also known as phenetole.

(B) Alkyl halides react with dry silver oxide $(Ag_2O)$ to form ethers.
The general reaction is: $2R-X + Ag_2O \xrightarrow{\Delta} R-O-R + 2AgX$.
For example,when methyl chloride reacts with dry $Ag_2O$: $2CH_3Cl + Ag_2O \xrightarrow{\Delta} CH_3OCH_3 + 2AgCl$ (methoxy methane).

(C) Ethers are represented by the general formula $R-O-R'$. The oxygen atom in ethers is bonded to two alkyl or aryl groups. Due to the stability of the $C-O$ bond and the lack of a reactive functional group like the hydroxyl group in alcohols,the oxygen atom in ethers is comparatively inert towards most reagents under normal conditions.

(D) An acyl group has the general formula $RCO-$.
$1$. Acid chloride: $RCOCl$ (contains $RCO-$ group).
$2$. Amide: $RCONH_2$ (contains $RCO-$ group).
$3$. Ester: $RCOOR'$ (contains $RCO-$ group).
$4$. Ether: $ROR'$ (does not contain $RCO-$ group).
Therefore,ether does not contain an acyl group.

(C) The oxygen atom in ether is $sp^3$ hybridized,which typically results in a tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.
In ethers,the two lone pairs on the oxygen atom exert repulsion on the bond pairs,which would normally decrease the bond angle.
However,the presence of bulky alkyl groups attached to the oxygen atom causes steric repulsion between them.
This steric repulsion between the alkyl groups counterbalances the lone pair-bond pair repulsion,resulting in a $C-O-C$ bond angle that is slightly larger than the tetrahedral angle,approximately $110^{\circ}$.

(B) According to the Lewis concept,a base is an electron pair donor.
In ethers $(R-O-R)$,the oxygen atom possesses two lone pairs of electrons.
Since the oxygen atom can donate these lone pairs,ether acts as a Lewis base.
Thus,the correct option is $(B)$.

(D) Kolbe's synthesis is a method of synthesizing aromatic $o-$hydroxycarboxylic acids (Salicylic acid) by the action of $CO_2$ on the alkaline salt of phenol.
The preparation of dialkyl zinc from zinc and alkyl iodide is called the Frankland reaction.
In Wurtz's reaction,two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane.
Williamson's synthesis involves the reaction of an alkyl halide with a sodium alkoxide to form symmetrical or unsymmetrical ethers. The reaction is represented as:
$CH_3CH_2-X + CH_3CH_2-O^-Na^+ \rightarrow CH_3CH_2-O-CH_2CH_3 + NaX$
(where $X$ = halide group).
Thus,the structure of diethyl ether is confirmed by Williamson's synthesis.

(C) The given reaction is: $C_2H_5ONa + IC_2H_5 \to C_2H_5OC_2H_5 + NaI$.
This is a reaction between a sodium alkoxide $(C_2H_5ONa)$ and an alkyl halide $(IC_2H_5)$ to form an ether $(C_2H_5OC_2H_5)$.
This method of preparing symmetrical or unsymmetrical ethers is known as $Williamson's$ synthesis.

(C) The reaction between a phenol $(Ar-OH)$ and an alkyl halide $(RX)$ in the presence of an alkali (like $NaOH$) is known as the Williamson ether synthesis.
In this reaction,the alkali converts the phenol into a phenoxide ion $(Ar-O^-)$,which then acts as a nucleophile and attacks the alkyl halide $(RX)$ via an $S_N2$ mechanism to form an alkyl aryl ether $(Ar-O-R)$.
Therefore,the product $A$ is an ether.

(B) Williamson's synthesis is a laboratory method for the preparation of symmetrical and unsymmetrical ethers.
The reaction involves the nucleophilic substitution $(S_N2)$ of an alkyl halide by an alkoxide ion.
The reaction is: $C_2H_5Br + C_2H_5ONa \rightarrow C_2H_5-O-C_2H_5 + NaBr$.
Thus,it is used to prepare $C_2H_5-O-C_2H_5$ (Diethyl ether).

(C) $RX + RONa \to R-O-R + NaX$
This reaction is known as Williamson's synthesis.
In this reaction,an alkyl halide $(RX)$ reacts with a sodium alkoxide $(RONa)$ to form an ether $(R-O-R)$ and a sodium halide $(NaX)$.

(B) Williamson synthesis involves the reaction of an alkoxide with an alkyl halide to form an ether.
For the preparation of ethoxyethane $(CH_3CH_2OCH_2CH_3)$,sodium ethoxide $(CH_3CH_2ONa)$ reacts with ethyl chloride $(CH_3CH_2Cl)$:
$CH_3CH_2ONa + CH_3CH_2Cl \to CH_3CH_2OCH_2CH_3 + NaCl$
Therefore,the correct option is $(B)$.

(B) The reaction of ethyl bromide with dry silver oxide $(Ag_2O)$ is a Williamson ether synthesis type reaction.
$2 C_2H_5Br + Ag_2O (\text{dry}) \to C_2H_5 - O - C_2H_5 + 2 AgBr$
Thus,the product formed is diethyl ether.
If moist $Ag_2O$ is used,it acts as $AgOH$ and produces ethyl alcohol:
$C_2H_5Br + AgOH \to C_2H_5OH + AgBr$

(D) The reaction of a halogenated ether with a Grignard reagent is a standard method to increase the carbon chain length of the ether.
$CH_3OCH_2Cl + CH_3MgBr \rightarrow CH_3OCH_2CH_3 + MgBrCl$
Here,$CH_3OCH_2Cl$ is a halogenated ether and $CH_3MgBr$ is a Grignard reagent,which acts as a nucleophile to replace the halogen atom,resulting in a higher ether.

(B) The industrial preparation of ethylene oxide involves the reaction of ethylene with hypochlorous acid to form ethylene chlorohydrin,which is then treated with a base to yield ethylene oxide.
$H_2C=CH_2 + HOCl \to Cl-CH_2-CH_2-OH$ (Ethylene chlorohydrin)
$Cl-CH_2-CH_2-OH + NaOH \to C_2H_4O$ (Ethylene oxide) $+ NaCl + H_2O$
Note: While ethylene glycol can be produced from ethylene oxide,the direct product of passing ethylene into hypochlorous acid followed by base treatment is ethylene oxide.

(B) The formation of methyl $t$-butyl ether occurs via the Williamson ether synthesis.
This reaction involves the nucleophilic substitution of an alkoxide ion with an alkyl halide.
To obtain methyl $t$-butyl ether,we require sodium $t$-butoxide $(CH_3)_3CONa$ and methyl chloride $(CH_3Cl)$.
The reaction is: $(CH_3)_3CONa + CH_3Cl \rightarrow (CH_3)_3COCH_3 + NaCl$.

(A) The reaction is a $Williamson$ synthesis.
$Methylphenyl$ ether $(Anisole)$ is prepared by the reaction of phenolate ions $(C_6H_5O^-)$ with methyl iodide $(CH_3I)$ via an $S_N2$ mechanism.
$C_6H_5O^- + CH_3I \to C_6H_5OCH_3 + I^-$

(D) When ethers are exposed to air and light for a long period,they undergo auto-oxidation to form explosive peroxides.
For diethyl ether,the reaction is:
$C_2H_5-O-C_2H_5 + O_2 \xrightarrow{hv} CH_3-CH(OOH)-O-C_2H_5$
Thus,the product formed is a peroxide of diethyl ether.

(A) $1$. The reaction of diethyl ether with $Cl_2$ in the dark leads to substitution of $\alpha$-hydrogen atoms,forming $\alpha,\alpha'$-dichlorodiethyl ether,which is not an alkyl halide.
$2$. Diethyl ether reacts with $HI$ to form ethyl iodide $(C_2H_5I)$,which is an alkyl halide.
$3$. Diethyl ether reacts with $PCl_5$ to form ethyl chloride $(C_2H_5Cl)$,which is an alkyl halide.
$4$. Therefore,the reaction of diethyl ether with $Cl_2$ in the dark does not yield an alkyl halide.

(A) When ethers are exposed to air and light for some time,they undergo autoxidation to form explosive peroxides and hydroperoxides. The reaction involves the formation of an ether peroxide by the radical mechanism.
$R-CH_2-O-R' + O_2 \xrightarrow{h\nu} R-CH(OOH)-O-R'$

(C) $CH_3OCH_3$ (dimethyl ether) has a boiling point of $248 \ K$ and $C_2H_5OCH_3$ (ethyl methyl ether) has a boiling point of $281 \ K$,both of which are gases at room temperature $(298 \ K)$.
$C_2H_5OC_2H_5$ (diethyl ether) has a boiling point of $308 \ K$,which makes it a low-boiling liquid at room temperature.

(A) The reaction of diethyl ether $(C_2H_5OC_2H_5)$ with excess $HI$ in the presence of red phosphorus $(P)$ is a reduction reaction.
Step $1$: $C_2H_5OC_2H_5 + 2HI \rightarrow 2C_2H_5I + H_2O$
Step $2$: $2C_2H_5I + 4[H] \xrightarrow{\text{Red } P} 2C_2H_6 + 2I_2$
Thus,$X$ is $C_2H_6$,which is Ethane.

(D) Diethyl ether reacts with atmospheric oxygen in the presence of light to form ether peroxide. This is an auto-oxidation process.
$CH_3-CH_2-O-CH_2-CH_3 + O_2 \xrightarrow{h\nu} CH_3-CH(OOH)-O-CH_2-CH_3$

(A) Diethyl ether $(C_2H_5OC_2H_5)$ reacts with concentrated $HI$ upon heating to undergo cleavage of the $C-O$ bond.
The reaction is: $C_2H_5OC_2H_5 + HI \to C_2H_5OH + C_2H_5I$.
Thus,$HI$ is the reagent used for the decomposition of ethers.

(A) The reaction of phenyl ethyl ether $(C_6H_5-O-C_2H_5)$ with concentrated hydrobromic acid $(HBr)$ involves the cleavage of the $C-O$ bond.
Since the $C-O$ bond between the phenyl group and oxygen has partial double bond character due to resonance,it is stronger and does not break easily.
Therefore,the cleavage occurs at the $O-C_2H_5$ bond,resulting in the formation of phenol $(C_6H_5OH)$ and ethyl bromide $(C_2H_5Br)$.
The reaction is: $C_6H_5OC_2H_5 + HBr \rightarrow C_6H_5OH + C_2H_5Br$.

(B) When excess ethanol $(C_2H_5OH)$ is heated with concentrated sulphuric acid $(H_2SO_4)$ at $413 \ K$ $(140^{\circ}C)$,intermolecular dehydration occurs to form Diethyl ether $(C_2H_5-O-C_2H_5)$.
$2C_2H_5OH \xrightarrow{conc. H_2SO_4, 413 K} C_2H_5-O-C_2H_5 + H_2O$
Conversely,when ethanol is heated with excess concentrated sulphuric acid at a higher temperature of $443 \ K$ $(170^{\circ}C)$,intramolecular dehydration occurs to form Ethene $(C_2H_4)$.
$C_2H_5OH \xrightarrow{conc. H_2SO_4, 443 K} C_2H_4 + H_2O$

(A) The reaction of an alkyl aryl ether with $HI$ involves the cleavage of the $C-O$ bond.
In the given ether,$C_6H_5-O-CH_2-C_6H_5$,the oxygen is attached to a phenyl group $(C_6H_5-)$ and a benzyl group $(-CH_2-C_6H_5)$.
The $C-O$ bond between the oxygen and the phenyl group is stronger due to partial double bond character (resonance),while the $C-O$ bond between the oxygen and the benzyl group is weaker because the resulting carbocation $(C_6H_5CH_2^+)$ is resonance-stabilized.
Therefore,the $HI$ attacks the benzyl carbon,leading to the formation of benzyl iodide $(C_6H_5CH_2I)$ and phenol $(C_6H_5OH)$.

(B) When dimethyl ether $(CH_3-O-CH_3)$ is heated with excess $HI$,the ether linkage is cleaved.
In the first step,one $C-O$ bond breaks to form $CH_3I$ and $CH_3OH$.
In the second step,the excess $HI$ reacts with the formed $CH_3OH$ to produce another molecule of $CH_3I$ and water $(H_2O)$.
The overall reaction is: $CH_3-O-CH_3 + 2HI \to 2CH_3I + H_2O$.

(A) Acetyl chloride $(CH_3COCl)$ is an acylating agent.
It reacts with nucleophiles like alcohols $(R-OH)$,phenols $(Ar-OH)$,and amines $(R-NH_2)$ to form esters and amides,respectively.
Diethyl ether $(C_2H_5-O-C_2H_5)$ is an inert solvent and does not possess an active hydrogen atom (nucleophilic site) to react with acetyl chloride under normal conditions.
Therefore,the correct answer is $A$.

(B) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.
In alkyl aryl ethers,the $O-CH_3$ bond is weaker than the $O-C_6H_5$ bond because the $O-C_6H_5$ bond has partial double bond character due to resonance.
Therefore,the $HI$ attacks the methyl group,leading to the formation of phenol $(C_6H_5OH)$ and methyl iodide $(CH_3I)$.
The reaction is: $C_6H_5OCH_3 + HI \xrightarrow{\text{heat}} C_6H_5OH + CH_3I$.

(C) Etherates are coordination complexes formed when ethers (acting as Lewis bases due to the lone pairs on oxygen) donate a pair of electrons to a Lewis acid (like $BF_3$).
For example,$BF_3 + (C_2H_5)_2O \rightarrow BF_3 \cdot O(C_2H_5)_2$.

(D) Due to inter-molecular hydrogen bonding in alcohols,the boiling point of alcohols is much higher than that of ethers.
Because ethers do not form inter-molecular hydrogen bonds,they have weaker intermolecular forces,making them more volatile.

(A) When ethers are stored in the presence of atmospheric oxygen,they slowly oxidize to produce hydroperoxides and dialkyl peroxides,both of which are highly unstable and explosive.
This process of oxidation by atmospheric oxygen is known as autoxidation.
The reaction is represented as:
$R-O-CH_2-R + O_2 \rightarrow R-O-CH(OOH)-R + R-O-O-CH_2-R$
Thus,the explosion is caused by the formation of peroxides.

(B) Sodium $(Na)$ reacts with compounds containing acidic hydrogen atoms (like $-OH$ or $-COOH$ groups) to release hydrogen gas $(H_2)$.
$C_2H_5OH$ (ethanol),$CH_3COOH$ (acetic acid),and $CH_3-CH(OH)-CH_3$ (propan$-2-$ol) all contain an acidic hydrogen atom attached to an oxygen atom.
$CH_3-O-CH_3$ (dimethyl ether) is an ether and does not contain any replaceable hydrogen atom attached to an oxygen atom.
Therefore,it does not react with sodium metal.

(B) The reaction of ethers with $HI$ depends on the stability of the carbocation formed.
In the case of methyl-tert-butyl ether,the protonation of the oxygen atom occurs first.
$(CH_3)_3C-O-CH_3 + HI \to (CH_3)_3C-O^+(H)-CH_3 + I^-$.
Since the tert-butyl group can form a stable $3^{\circ}$ carbocation,the $I^-$ ion attacks the tert-butyl group via an $S_N1$ mechanism,leading to the formation of tert-butyl iodide and methanol.
$(CH_3)_3C-O^+(H)-CH_3 \to (CH_3)_3C^+ + CH_3OH$.
$(CH_3)_3C^+ + I^- \to (CH_3)_3CI$.
Thus,the products are $(CH_3)_3CI$ and $CH_3OH$.

(D) $2,4-$dinitrophenylhydrazine $(2,4-DNP)$ is a reagent used to detect carbonyl groups (aldehydes and ketones). Since the compound does not react with $2,4-DNP$,it is not an aldehyde or a ketone.
$Na$ metal reacts with compounds containing active hydrogen atoms,such as alcohols $(R-OH)$. Since the compound does not react with $Na$,it does not contain an active hydrogen atom (i.e.,it is not an alcohol).
Among the options:
$A$) Acetone is a ketone.
$B$) Acetaldehyde is an aldehyde.
$C$) $CH_3OH$ is an alcohol.
$D$) $CH_3OCH_3$ (dimethyl ether) is an ether,which does not contain a carbonyl group nor an active hydrogen atom. Therefore,it does not react with $2,4-DNP$ or $Na$.

(A) The reaction of an alkyl aryl ether with $HBr$ involves the protonation of the ether oxygen atom to form a protonated ether (oxonium salt).
The halide ion $(Br^-)$ then performs a nucleophilic attack on the less sterically hindered alkyl group (in this case,the methyl group attached to the oxygen) via an $S_N2$ mechanism.
This results in the cleavage of the $C-O$ bond between the oxygen and the methyl group,producing a phenol ($m$-methylphenol) and an alkyl halide $(CH_3Br)$.

(D) Diethyl ether $(C_2H_5-O-C_2H_5)$ was historically used as a general anaesthetic in medicine.
Therefore,the correct option is $(D)$.

(D) Ether $(Diethyl \ ether)$ is an industrial solvent used for oils,fats,gum,resin,perfume,etc.
$Diethyl \ ether$ produces cooling on evaporation.
Hence,it is used as a refrigerant.
$Diethyl \ ether$ was also widely used as an anesthetic agent.
Therefore,all the given options are correct.

(C) The reaction of methoxybenzene $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.
Since the $C_6H_5-O$ bond has partial double bond character due to resonance,it is stronger and harder to break than the $CH_3-O$ bond.
Therefore,the $HI$ attacks the $CH_3$ group,resulting in the formation of methyl iodide $(CH_3I)$ and phenol $(C_6H_5OH)$.
The reaction is: $C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$.

(A) When ethers are kept in contact with air for a long time,they undergo autoxidation to form explosive peroxides.
For diethyl ether $(C_2H_5 - O - C_2H_5)$,the reaction with atmospheric oxygen leads to the formation of $1$-ethoxyethyl hydroperoxide.
The reaction is: $C_2H_5 - O - C_2H_5 \xrightarrow{O_2, \text{light}} C_2H_5 - O - CH(CH_3) - O - OH$.

(C) The oxidation of alkyl benzenes or their derivatives with side chains having at least one benzylic hydrogen atom using alkaline $KMnO_4$ followed by acidification yields benzoic acid.
$1$. Benzyl alcohol $(C_6H_5CH_2OH)$ oxidizes to benzoic acid.
$2$. Acetophenone $(C_6H_5COCH_3)$ oxidizes to benzoic acid.
$3$. Toluene $(C_6H_5CH_3)$ oxidizes to benzoic acid.
$4$. Anisole $(C_6H_5OCH_3)$ is an ether. The methoxy group attached to the benzene ring is stable under these conditions and does not undergo oxidation to a carboxylic acid group. Therefore,it does not yield benzoic acid.

(C) The correct answer is $C$.
Dimethyl ether $(CH_3-O-CH_3)$ is an ether.
Ethers are generally resistant to nucleophilic attack by hydroxyl ions $(OH^-)$ because the lone pairs on the oxygen atom and the electron-donating effect of the alkyl groups create high electron density around the oxygen,which repels the incoming nucleophile.
In contrast,esters $(CH_3COOCH_3)$,nitriles $(CH_3CN)$,and amides $(CH_3CONH_2)$ contain a carbonyl or cyano group that is highly susceptible to nucleophilic attack.

(D) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ involves the cleavage of the $C-O$ bond.
Since the $C_6H_5-O$ bond has partial double bond character due to resonance,it is stronger and harder to break than the $CH_3-O$ bond.
Therefore,the $HI$ attacks the methyl group,resulting in the formation of methyl iodide $(CH_3I)$ and phenol $(C_6H_5OH)$.
The reaction is: $C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I$.

(A) The reaction of acetic anhydride with diethyl ether in the presence of anhydrous $AlCl_3$ is a cleavage reaction of the ether.
The reaction proceeds as follows:
$(CH_3CO)_2O + (C_2H_5)_2O \xrightarrow{AlCl_3} 2 CH_3COOC_2H_5$
The product formed is ethyl acetate.

(A) The molecular formula $C_4H_{10}O$ corresponds to either an alcohol or an ether.
Since the compound does not react with sodium metal,it cannot be an alcohol; therefore,it must be an ether.
When an ether reacts with excess $HI$,it undergoes cleavage to form alkyl halides.
The problem states that only one type of alkyl halide is formed,which implies that the ether must be symmetrical.
Ethoxyethane $(C_2H_5-O-C_2H_5)$ is a symmetrical ether.
The reaction is: $C_2H_5OC_2H_5 + 2HI \to 2C_2H_5I + H_2O$.

(C) The reaction of diethyl ether $(C_2H_5-O-C_2H_5)$ with concentrated $HI$ at $0^o C$ proceeds via the protonation of the ether oxygen atom.
Step $1$: Protonation of ether: $C_2H_5-O-C_2H_5 + HI \rightarrow C_2H_5-O^+(H)-C_2H_5 + I^-$.
Step $2$: Nucleophilic attack by $I^-$ on one of the ethyl groups: $I^- + C_2H_5-O^+(H)-C_2H_5 \rightarrow C_2H_5I + C_2H_5OH$.
Since both alkyl groups are identical,the reaction produces one molecule of ethyl iodide $(C_2H_5I)$ and one molecule of ethanol $(C_2H_5OH)$.
Therefore,both products are formed.

(B) Phosphorus pentachloride $(PCl_5)$ is commonly used to convert alcohols or ethers into alkyl chlorides.
Specifically,the reaction with dimethyl ether is: $CH_3-O-CH_3 + PCl_5 \rightarrow 2CH_3Cl + POCl_3$.
Therefore,the correct transformation is $H_3C-O-CH_3 \rightarrow 2CH_3Cl$.