Properties of Ethers Questions in English

(C) The reaction involves the acid-catalyzed ring opening of an epoxide.
In the presence of $H^+$,the oxygen atom of the epoxide is protonated,making it a better leaving group.
$HBr$ provides the nucleophile $Br^-$.
For an unsymmetrical epoxide,the nucleophile attacks the more substituted carbon atom because it can better stabilize the partial positive charge in the transition state.
In this case,the tertiary carbon (the one with the $CH_3$ group) is more substituted than the secondary carbon.
Therefore,$Br^-$ attacks the tertiary carbon,and the $C-O$ bond breaks to form a hydroxyl group at the secondary carbon.
The final product is $1-bromo-1-methylcyclohexan-2-ol$.

(B) In alkyl aryl ethers,the bond between the oxygen atom and the aromatic ring has partial double bond character due to resonance,making it stronger than the alkyl-oxygen bond.
Therefore,when reacted with $HI$,the alkyl-oxygen bond breaks,yielding phenol and an alkyl iodide.
Reaction: $C_6H_5-O-CH(CH_3)_2 + HI \xrightarrow{\Delta} C_6H_5-OH + CH_3-CH(I)-CH_3$.

(A) The reaction sequence is as follows:
$1$. Phenol $(C_6H_5OH)$ reacts with sodium $(Na)$ to form sodium phenoxide $(X = C_6H_5ONa)$ and hydrogen gas.
$C_6H_5OH + Na \rightarrow C_6H_5ONa + \frac{1}{2}H_2$
$2$. Sodium phenoxide $(C_6H_5ONa)$ then reacts with ethyl chloride $(CH_3-CH_2-Cl)$ via a Williamson ether synthesis mechanism to form phenetole (ethoxybenzene) $(Y = C_6H_5-O-CH_2-CH_3)$.
$C_6H_5ONa + CH_3-CH_2-Cl \rightarrow C_6H_5-O-CH_2-CH_3 + NaCl$

(C) The reaction of tert-butyl methyl ether with cold $HI$ proceeds via the $S_N1$ mechanism.
Since the tert-butyl group can form a stable tertiary carbocation $(CH_3)_3C^+$,the cleavage occurs such that the iodide ion attacks the tertiary carbon.
The reaction is: $CH_3-C(CH_3)_2-O-CH_3 + HI \xrightarrow{\text{Cold}} (CH_3)_3C-I + CH_3-OH$.
Thus,the products are tert-butyl iodide and methanol.

(D) Williamson's synthesis is a laboratory method for the preparation of symmetrical and unsymmetrical ethers.
It involves the reaction of an alkyl halide with a sodium alkoxide.
The reaction proceeds via an $S_{N}2$ mechanism.
Because it follows the $S_{N}2$ mechanism,it is sensitive to steric hindrance.
Therefore,primary alkyl halides are preferred,while tertiary halides are not used because they undergo elimination reactions to form alkenes instead of ethers.
Thus,all the given statements are correct.

(B) For the preparation of Methoxybenzene $(C_6H_5OCH_3)$,the Williamson ether synthesis is used.
In reaction $(x)$,$C_6H_5Br + CH_3ONa$,the $C-Br$ bond in bromobenzene has partial double bond character due to resonance,making it resistant to nucleophilic substitution. Thus,it does not undergo the reaction.
In reaction $(y)$,$C_6H_5ONa + CH_3Br$,sodium phenoxide acts as a nucleophile and attacks the methyl bromide $(CH_3Br)$ via an $S_N2$ mechanism to form Methoxybenzene. Therefore,only reaction $(y)$ is suitable.

(B) The reaction of an ether with $HI$ involves the protonation of the ether oxygen followed by nucleophilic attack by the iodide ion $(I^-)$.
In the given ether,$C_6H_5-O-CH_2-C_6H_5$ (benzyl phenyl ether),the bond between the oxygen and the benzyl carbon $(CH_2-C_6H_5)$ is weaker than the bond between the oxygen and the phenyl carbon $(C_6H_5-O)$ because the benzyl carbocation is resonance-stabilized.
Upon protonation,the $I^-$ ion attacks the more electrophilic benzyl carbon via an $S_N1$ or $S_N2$ mechanism (favored by the stability of the benzyl carbocation intermediate),leading to the formation of phenol $(C_6H_5OH)$ and benzyl iodide $(C_6H_5CH_2I)$.
Thus,the major products are phenol and benzyl iodide,which corresponds to option $(II)$.

(C) The reaction of chloroethane $(C_2H_5Cl)$ with $Y$ produces $NaCl$ and $Z$. This indicates a nucleophilic substitution reaction where $Y$ is a sodium alkoxide,specifically sodium ethoxide $(C_2H_5ONa)$.
$C_2H_5Cl + C_2H_5ONa \rightarrow C_2H_5OC_2H_5 + NaCl$
Here,$Z$ is diethyl ether $(C_2H_5OC_2H_5)$.
When one mole of diethyl ether reacts with two moles of $HI$,it undergoes cleavage to form two moles of iodoethane and water:
$C_2H_5OC_2H_5 + 2HI \rightarrow 2C_2H_5I + H_2O$
Thus,$Y$ is $C_2H_5ONa$.

(B) The reaction between methyl chloride $(CH_3Cl)$ and silver acetate $(CH_3COOAg)$ is a nucleophilic substitution reaction.
The reaction proceeds as follows:
$CH_3Cl + CH_3COOAg \rightarrow CH_3COOCH_3 + AgCl$
The product formed is methyl acetate $(CH_3COOCH_3)$.

(B) The reaction proceeds as follows:
$1$. Ethanol $(C_2H_5OH)$ reacts with sodium $(Na)$ to form sodium ethoxide $(P)$: $C_2H_5OH + Na \rightarrow C_2H_5ONa + \frac{1}{2}H_2$.
$2$. Sodium ethoxide $(C_2H_5ONa)$ then reacts with methyl chloride $(CH_3-Cl)$ via the Williamson ether synthesis to form ethyl methyl ether $(Q)$: $C_2H_5ONa + CH_3-Cl \rightarrow C_2H_5-O-CH_3 + NaCl$.
Thus,$Q$ is $CH_3-CH_2-O-CH_3$.

(A) The reaction between sodium phenoxide $(C_6H_5ONa)$ and ethyl iodide $(C_2H_5I)$ is a Williamson ether synthesis.
The nucleophilic phenoxide ion attacks the ethyl iodide,displacing the iodide ion to form an ether.
The reaction is: $C_6H_5ONa + C_2H_5I \rightarrow C_6H_5OC_2H_5 + NaI$.
The product $C_6H_5OC_2H_5$ is known as ethyl phenyl ether or phenetole.

(B) $1$. The molecular formula $C_3H_6O$ corresponds to a degree of unsaturation of $1$ $(3 - 6/2 + 1 = 1)$.
$2$. The compound does not react with $2,4-DNP$,which indicates the absence of a carbonyl group (aldehyde or ketone).
$3$. The compound does not react with sodium metal,which indicates the absence of an active hydrogen atom (like in alcohols or phenols).
$4$. Given these conditions,the compound must be an ether with a double bond (unsaturated ether).
$5$. Among the options,$CH_2=CH-OCH_3$ (methoxyethene) fits the molecular formula $C_3H_6O$,lacks a carbonyl group,and lacks an active hydrogen atom.

(C) The reaction of acetic anhydride $(CH_3CO)_2O$ with diethyl ether $(C_2H_5OC_2H_5)$ in the presence of anhydrous $AlCl_3$ is a cleavage reaction of the ether.
$AlCl_3$ acts as a Lewis acid and coordinates with the oxygen atom of the ether.
This facilitates the cleavage of the $C-O$ bond,leading to the formation of an ester.
The reaction proceeds as follows:
$(CH_3CO)_2O + C_2H_5OC_2H_5 \xrightarrow{AlCl_3} 2CH_3COOC_2H_5$
Thus,the product formed is ethyl acetate,which is $CH_3COOCH_2CH_3$.

(A) Anisole $(C_6H_5OCH_3)$ contains a methoxy group $(-OCH_3)$,which is an activating group and ortho/para directing.
When anisole reacts with a nitrating mixture (concentrated $HNO_3$ and concentrated $H_2SO_4$),the electrophile $NO_2^+$ attacks the ortho and para positions of the benzene ring.
This results in the formation of a mixture of $o-$nitroanisole and $p-$nitroanisole.

(C) 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 must be an ether.
Among the given options,$1-$Methoxypropane $(CH_3-O-CH_2-CH_2-CH_3)$ is an ether.
When $1-$Methoxypropane reacts with excess $HI$,it undergoes cleavage:
$CH_3-O-CH_2-CH_2-CH_3 + 2HI \rightarrow CH_3I + CH_3CH_2CH_2I + H_2O$.
This reaction produces two types of alkyl halides: methyl iodide $(CH_3I)$ and $1-$iodopropane $(CH_3CH_2CH_2I)$.

(A) $1$. Diethyl ether reacts with $Cl_2$ in the dark to form $\alpha$-chloro diethyl ether,which is a substituted ether,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.
$4$. Divinyl ether on reduction gives diethyl ether,which upon further reaction with $SOCl_2$ (or other halogenating agents) can yield alkyl halides. Thus,the reaction in option $A$ does not produce an alkyl halide.

(B) Aromatic ethers like $C_6H_5-O-C_6H_5$ (diphenyl ether) do not undergo cleavage by $HI$ even at high temperatures $(525 \ K)$.
This is because the $C-O$ bond in aromatic ethers has partial double bond character due to resonance,making it very strong and difficult to break.
In contrast,alkyl aryl ethers like $C_6H_5OCH_3$ or $C_6H_5OC_3H_7$ undergo cleavage to form phenol and alkyl iodide.
Tetrahydrofuran is an aliphatic cyclic ether and undergoes cleavage easily.

(A) The reaction of diethyl ether with excess of concentrated $HI$ in the presence of red phosphorus at $423 \ K$ is a reduction reaction.
$C_2H_5OC_2H_5 + 2HI \xrightarrow{Red \ P, 423 \ K} 2C_2H_5I + H_2O$
Further,the ethyl iodide formed is reduced by $HI$ in the presence of red phosphorus:
$2C_2H_5I + 2HI \xrightarrow{Red \ P} 2C_2H_6 + 2I_2$
Combining these steps,the overall reaction is:
$C_2H_5OC_2H_5 + 4[H] \xrightarrow{Red \ P + HI} 2C_2H_6 + H_2O$
Comparing this with the given equation,$2X = 2C_2H_6$,so $X = C_2H_6$ (Ethane).

(A) The reaction of phenyl ethyl ether $(C_6H_5-O-C_2H_5)$ with concentrated $HBr$ 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 cannot be easily cleaved.
Therefore,the $O-C_2H_5$ bond breaks,resulting in the formation of phenol $(C_6H_5OH)$ and ethyl bromide $(C_2H_5Br)$.

(B) The reaction of $tert$-butyl methyl ether with $HI$ follows an $S_N1$ mechanism because the $tert$-butyl group can form a stable $tert$-butyl carbocation.
$1$. The oxygen atom of the ether is protonated by $HI$ to form an oxonium ion: $(CH_3)_3C-O(CH_3) + H^+ \rightarrow (CH_3)_3C-O^+(H)(CH_3)$.
$2$. The $C-O$ bond between the $tert$-butyl group and oxygen breaks to form a stable $tert$-butyl carbocation $(CH_3)_3C^+$ and methanol $(CH_3OH)$.
$3$. The iodide ion $(I^-)$ then attacks the $tert$-butyl carbocation to form $tert$-butyl iodide $(CH_3)_3C-I$.
Thus,the products are $tert$-butyl iodide and methanol.

(C) When ethyl hydrogen sulphate $(C_2H_5HSO_4)$ is heated with an excess of ethanol $(C_2H_5OH)$ at $410 \ K$,the reaction proceeds via a nucleophilic substitution mechanism to form diethyl ether $(C_2H_5-O-C_2H_5)$.
The reaction is as follows:
$C_2H_5HSO_4 + C_2H_5OH \xrightarrow{410 \ K} C_2H_5-O-C_2H_5 + H_2SO_4$.
If the temperature were higher (around $443 \ K$),the reaction would favor the formation of ethene (dehydration) instead.

(C) $1$. $KMnO_4$ is a strong oxidizing agent. When alkyl benzenes or compounds with a benzylic hydrogen are treated with basic $KMnO_4$ followed by acidification,they are oxidized to benzoic acid.
$2$. Toluene $(C_6H_5CH_3)$ has a benzylic hydrogen and oxidizes to benzoic acid.
$3$. Acetophenone $(C_6H_5COCH_3)$ has a benzylic hydrogen on the methyl group and oxidizes to benzoic acid.
$4$. Benzyl alcohol $(C_6H_5CH_2OH)$ has benzylic hydrogens and oxidizes to benzoic acid.
$5$. Anisole $(C_6H_5OCH_3)$ does not have a benzylic hydrogen (the carbon attached to the ring is part of an ether linkage and lacks the necessary hydrogen for oxidation to the carboxylic acid group). Therefore,it does not yield benzoic acid under these conditions.

(D) The synthesis of ethers via the Williamson ether synthesis involves the reaction of an alkoxide ion with an alkyl halide.
To synthesize methyl tert-butyl ether,$(CH_3)_3C-O-CH_3$,we must choose reactants that avoid the elimination reaction.
Tertiary alkyl halides undergo $E2$ elimination in the presence of strong bases (alkoxides).
Therefore,we must use a tertiary alkoxide and a primary alkyl halide.
Reaction: $(CH_3)_3C-ONa + CH_3Cl \rightarrow (CH_3)_3C-O-CH_3 + NaCl$.
This corresponds to option $D$.

(C) The correct answer is $(c)$.
Diethyl ether $(CH_3CH_2OCH_2CH_3)$ is resistant to nucleophilic attack by $OH^{-}$ ions because it lacks an electrophilic carbon atom that can be attacked by a nucleophile.
In contrast,carboxylic acid derivatives such as methyl acetate $(CH_3COOCH_3)$ and acetamide $(CH_3CONH_2)$,as well as nitriles like acetonitrile $(CH_3CN)$,contain a carbonyl or cyano carbon atom which is electrophilic and susceptible to nucleophilic attack.
Ethers are generally inert towards bases because the alkoxide ion $(RO^{-})$ is a poor leaving group.

(C) The reaction of anisole $(C_6H_5OCH_3)$ with $HI$ involves the protonation of the oxygen atom,followed by the nucleophilic attack of the iodide ion $(I^-)$.
The $I^-$ ion attacks the less sterically hindered methyl group $(CH_3)$ to form methyl iodide $(CH_3I)$ and phenol $(C_6H_5OH)$.
Therefore,the Assertion is correct as the iodide ion attacks the smaller group.
However,the Reason states that it gives iodobenzene and methyl alcohol,which is incorrect; it actually gives phenol and methyl iodide.
Thus,the Assertion is correct but the Reason is incorrect.

(A) The reaction of ethers with $HI$ involves the cleavage of the $C-O$ bond.
For the first ether,$CH_3-O-C_2H_5$,the reaction follows the $S_N2$ mechanism,where the iodide ion attacks the less sterically hindered alkyl group,forming $CH_3I + C_2H_5OH$.
For the second ether,$(CH_3)_3C-O-CH_3$,the reaction follows the $S_N1$ mechanism because the tert-butyl carbocation is highly stable,leading to the formation of $(CH_3)_3CI + CH_3OH$.

(C) The reaction of $CH_3Br$ (methyl bromide) with sodium $t-$butoxide $((CH_3)_3C-ONa)$ is an $S_N2$ reaction which produces $t-$butyl methyl ether.
In reaction $(d)$,the $t-$butyl bromide undergoes elimination due to steric hindrance and the presence of a strong base,forming isobutylene instead of the ether.

(B) The cleavage of ethers by $HI$ involves the protonation of the ether oxygen followed by a nucleophilic attack by the iodide ion $(I^-)$.
In diaryl ethers like $C_6H_5OC_6H_5$ (diphenyl ether),the lone pair of electrons on the oxygen atom is involved in resonance with both phenyl rings.
This imparts a partial double bond character to the $C-O$ bonds,making them very strong and difficult to break.
Additionally,the $sp^2$ hybridized carbon atoms of the phenyl rings are less susceptible to nucleophilic attack compared to $sp^3$ hybridized carbons.
Therefore,$C_6H_5OC_6H_5$ is resistant to cleavage by $HI$ even at high temperatures.

(A) Electrophilic substitution reactions (such as halogenation,nitration,or Friedel-Crafts reactions) occur on the benzene ring.
Only aromatic ethers (alkyl aryl ethers) contain a benzene ring that can be activated by the alkoxy group $(-OR)$ towards electrophilic substitution.
Among the given options,$C_6H_5OCH_3$ (Anisole) is an aromatic ether,while the others are aliphatic ethers.
Therefore,$C_6H_5OCH_3$ undergoes electrophilic substitution reactions.

(A) The reaction of $C_6H_5CH_2OCH_3$ with $HI$ proceeds via an $S_N1$ mechanism because the benzyl carbocation $(C_6H_5CH_2^+)$ is highly stable due to resonance.
The cleavage of the $C-O$ bond occurs such that the more stable carbocation is formed.
Since the benzyl carbocation is more stable than the methyl carbocation $(CH_3^+)$,the iodide ion attacks the benzyl group to form $C_6H_5CH_2I$,while the methoxide group is protonated to form $CH_3OH$.
Therefore,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) Ethers have two lone pairs of electrons on the oxygen atom,which makes them act as Lewis bases.
In the presence of mineral acids (like $HCl$),the oxygen atom donates a lone pair to the $H^+$ ion,forming an oxonium salt.
The reaction is: $R-O-R + HCl \rightarrow [R_2O^+-H]Cl^-$.
Since the basic behavior is directly caused by the lone pairs on the oxygen atom,the Reason correctly explains the Assertion.

(A) Alkyl aryl ethers are cleaved at the alkyl-oxygen bond because the phenyl-oxygen bond has partial double bond character due to resonance,making it stronger and more stable.
Therefore,the reaction of ethyl phenyl ether $(C_6H_5-O-C_2H_5)$ with $HBr$ proceeds as follows:
$C_6H_5-O-C_2H_5 + HBr \rightarrow C_6H_5OH + C_2H_5Br$
Both the Assertion and the Reason are correct,and the Reason correctly explains why the cleavage occurs at the alkyl-oxygen bond.

(A) The reaction of an ether with excess $HI$ involves the protonation of the ether oxygen followed by nucleophilic attack by $I^-$.
$CH_{3}CH_{2}CH_{2}OC(CH_{3})_{3} + HI \rightarrow CH_{3}CH_{2}CH_{2}OH + (CH_{3})_{3}CI$.
Since $HI$ is in excess,the alcohol formed $(CH_{3}CH_{2}CH_{2}OH)$ further reacts with $HI$ to form an alkyl iodide.
$CH_{3}CH_{2}CH_{2}OH + HI \rightarrow CH_{3}CH_{2}CH_{2}I + H_{2}O$.
The tertiary carbocation formed from the cleavage of the $C-O$ bond in the tert-butyl group is highly stable,leading to the formation of $(CH_{3})_{3}CI$.
Thus,the final products are $CH_{3}CH_{2}CH_{2}I$ and $(CH_{3})_{3}CI$.

(B) The reaction of the given cyclic ether with $HBr$ under heating causes the cleavage of the $C-O$ bond. The $H^+$ protonates the oxygen,and $Br^-$ attacks the less hindered carbon,resulting in the opening of the ring to form $A$ ($2$-($3$-bromopropyl)phenol).
When $A$ is treated with $Na$ in ether,the phenolic $-OH$ group is deprotonated to form a phenoxide ion. This nucleophilic phenoxide ion then undergoes an intramolecular $S_N2$ reaction with the $-CH_2Br$ group (if present) or,in this specific case,the intramolecular cyclization leads to the formation of $1$-tetralol ($5$,$6$,$7$,$8$-tetrahydronaphthalen$-1-$ol) via the reaction of the phenoxide with the alkyl bromide side chain.

(B) $(i)$ The major product of the given reaction is $2$-methylpropene $(CH_2=C(CH_3)_2)$. Sodium ethoxide $(C_2H_5ONa)$ is a strong base,and $t$-butyl chloride is a tertiary $(3^\circ)$ alkyl halide. In the case of tertiary halides,elimination $(E2)$ predominates over substitution $(S_N2)$.
$(ii)$ For the preparation of $t$-butyl ethyl ether,a primary alkyl halide should be reacted with a tertiary alkoxide (Williamson ether synthesis).
$(CH_3)_3C-ONa + C_2H_5-Br \to (CH_3)_3C-O-C_2H_5 + NaBr$.

(N/A) $(i)$ The reaction follows $S_N2$ mechanism because the alkyl groups are primary. The smaller alkyl group forms the alkyl iodide. Products: $CH_3-CH_2-CH(CH_3)-CH_2OH + CH_3CH_2I$.
$(ii)$ The reaction follows $S_N1$ mechanism because the ether contains a tertiary alkyl group. The tertiary carbocation is more stable. Products: $CH_3CH_2CH_2OH + CH_3CH_2-C(CH_3)_2I$.
$(iii)$ The cleavage occurs at the $C-O$ bond where the benzyl carbocation is formed,which is resonance stabilized. Products: $C_6H_5CH_2I + C_6H_5OH$.

(B) Set $(ii)$ is an appropriate set of reactants for the preparation of $1-methoxy-4-nitrobenzene$. The reaction is: $p-nitrophenoxide + CH_3Br \rightarrow 1-methoxy-4-nitrobenzene + NaBr$.
In set $(i)$,the reactant is $1-bromo-4-nitrobenzene$. Aryl halides are generally unreactive towards nucleophilic substitution reactions under ordinary conditions because the $C-X$ bond has partial double bond character due to resonance and the $sp^2$ hybridized carbon atom is more electronegative. Furthermore,$CH_3ONa$ is a strong nucleophile and base,but it cannot easily displace the halide from the aryl ring in this manner compared to the Williamson ether synthesis shown in set $(ii)$.

(N/A) $(i)$ $CH_3-CH_2-CH_2-O-CH_3 + HBr \rightarrow CH_3-CH_2-CH_2-OH + CH_3-Br$
$(ii)$ $\text{Ethoxybenzene} + HBr \rightarrow \text{Phenol} + C_2H_5Br$
$(iii)$ $\text{Ethoxybenzene}$ $\xrightarrow{\text{Conc. } H_2SO_4 / \text{Conc. } HNO_3} \text{4-Ethoxynitrobenzene (Major)} + \text{2-Ethoxynitrobenzene (Minor)}$
$(iv)$ $(CH_3)_3C-O-C_2H_5 \xrightarrow{HI} (CH_3)_3C-I + C_2H_5OH$

(A) $(i)$ $1$-Ethoxy-$2$-methylpropane
$(ii)$ $1$-Chloro-$2$-methoxyethane
$(iii)$ $1$-Methoxy-$4$-nitrobenzene (or $4$-Nitroanisole)
$(iv)$ $1$-Methoxypropane
$(v)$ $1$-Ethoxy-$4,4$-dimethylcyclohexane
$(vi)$ Ethoxybenzene

(N/A) The Williamson synthesis involves the $S_{N}2$ attack of an alkoxide ion on a primary alkyl halide.
For example: $(CH_3)_3CONa + CH_3Cl \rightarrow (CH_3)_3COCH_3 + NaCl$
However,if secondary or tertiary alkyl halides are used instead of primary alkyl halides,elimination competes with substitution.
As a result,alkenes are produced instead of ethers because alkoxides act as both nucleophiles and strong bases.
For example: $(CH_3)_3CCl + NaOCH_3 \rightarrow CH_3-C(CH_3)=CH_2 + CH_3OH + NaCl$

(N/A) $1-$propoxypropane can be synthesized from propan$-1-$ol by acid-catalyzed dehydration. Propan$-1-$ol undergoes dehydration in the presence of protic acids (such as $H_{2}SO_{4}$ or $H_{3}PO_{4}$) at $413 \ K$ to give $1-$propoxypropane.
$2CH_{3}CH_{2}CH_{2}OH \xrightarrow{H^{+}, 413K} CH_{3}CH_{2}CH_{2}-O-CH_{2}CH_{2}CH_{3} + H_{2}O$
The mechanism of this reaction involves the following three steps:
Step $1$: Protonation
$CH_{3}CH_{2}CH_{2}-\ddot{O}H + H^{+} \rightleftharpoons CH_{3}CH_{2}CH_{2}-\overset{+}{O}H_{2}$
Step $2$: Nucleophilic attack
$CH_{3}CH_{2}CH_{2}-\ddot{O}H + CH_{3}CH_{2}CH_{2}-\overset{+}{O}H_{2} \to CH_{3}CH_{2}CH_{2}-\overset{+}{O}(H)-CH_{2}CH_{2}CH_{3} + H_{2}O$
Step $3$: Deprotonation
$CH_{3}CH_{2}CH_{2}-\overset{+}{O}(H)-CH_{2}CH_{2}CH_{3} \to CH_{3}CH_{2}CH_{2}-O-CH_{2}CH_{2}CH_{3} + H^{+}$

(N/A) $(i)$ $CH_3CH_2CH_2-O-CH_2CH_2CH_3 + HI \xrightarrow{373 \ K} CH_3CH_2CH_2-OH + CH_3CH_2CH_2-I$
$(ii)$ $C_6H_5-OCH_3 + HI \rightarrow C_6H_5-OH + CH_3I$
$(iii)$ $C_6H_5CH_2-O-C_2H_5 + HI \rightarrow C_6H_5CH_2-I + C_2H_5OH$

(N/A) $(i)$ In aryl alkyl ethers,due to the $+R$ effect of the alkoxy group,the electron density in the benzene ring increases as shown in the following resonance structures.
Thus,the benzene ring is activated towards electrophilic substitution by the alkoxy group.
$(ii)$ It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result,the incoming electrophilic substituents are directed to the ortho and para positions in the benzene ring.

(N/A) The mechanism of the reaction of $HI$ with methoxymethane involves the following steps:
Step $1$: Protonation of methoxymethane:
$CH_3-O-CH_3 + HI \rightleftharpoons CH_3-O^{+}(H)-CH_3 + I^{-}$
Step $2$: Nucleophilic attack of $I^{-}$ on the protonated ether:
$I^{-} + CH_3-O^{+}(H)-CH_3 \to CH_3-I + CH_3-OH$
Step $3$: When $HI$ is in excess and the reaction is carried out at a high temperature,the methanol formed in the second step reacts with another $HI$ molecule:
$CH_3-OH + HI \rightleftharpoons CH_3-OH_2^{+} + I^{-}$
$I^{-} + CH_3-OH_2^{+} \to CH_3-I + H_2O$

(N/A) $(i)$ Friedel-Crafts alkylation: $C_6H_5OCH_3 + CH_3Cl \xrightarrow{Anhyd. AlCl_3, CS_2} C_6H_4(OCH_3)(CH_3) \text{ (ortho and para isomers)}$.
$(ii)$ Nitration: $C_6H_5OCH_3 + HNO_3 \xrightarrow{H_2SO_4} C_6H_4(OCH_3)(NO_2) \text{ (ortho and para isomers)}$.
$(iii)$ Bromination: $C_6H_5OCH_3 + Br_2 \xrightarrow{CH_3COOH} C_6H_4(OCH_3)(Br) \text{ (ortho and para isomers)}$.
$(iv)$ Friedel-Crafts acetylation: $C_6H_5OCH_3 + CH_3COCl \xrightarrow{Anhyd. AlCl_3} C_6H_4(OCH_3)(COCH_3) \text{ (ortho and para isomers)}$.

(N/A) The common and $\rm {IUPAC}$ names are as follows:

Compound	Common Name	$\rm {IUPAC}$ Name
$(i) \ CH_3OCH_3$	Dimethyl ether	Methoxymethane
$(ii) \ C_2H_5OC_2H_5$	Diethyl ether	Ethoxyethane
$(iii) \ C_6H_5OC_6H_5$	Diphenyl ether	Phenoxybenzene
$(iv) \ CH_3OCH_2CH_2CH_3$	Methyl n-propyl ether	$1$-Methoxypropane
$(v) \ CH_3OCH(CH_3)_2$	Methyl isopropyl ether	$2$-Methoxypropane
$(vi) \ C_6H_5OCH_3$	Methyl phenyl ether (Anisole)	Methoxybenzene
$(vii) \ C_6H_5OCH_2CH_3$	Ethyl phenyl ether (Phenetole)	Ethoxybenzene
$(viii) \ C_6H_5O(CH_2)_6CH_3$	Heptyl phenyl ether	$1$-Phenoxyheptane
$(ix) \ C_6H_5OCH_2CH_2CH(CH_3)_2$	Isobutyl phenyl ether	$1$-Phenoxy$-3-$methylbutane
$(x) \ CH_3OCH_2CH_2OCH_3$	$1$,$2$-Dimethoxyethane	$1$,$2$-Dimethoxyethane
$(xi) \ C_2H_5OCH_2CH_2OC_2H_5$	$1$,$2$-Diethoxyethane	$1$,$2$-Diethoxyethane
$(xii) \ CH_3OCH_2CH_2CH_2OCH_3$	$1$,$3$-Dimethoxypropane	$1$,$3$-Dimethoxypropane
$(xiii) \ CH_3OCH_2CH_2OC_2H_5$	$1$-Ethoxy$-2-$methoxyethane	$1$-Ethoxy$-2-$methoxyethane
$(xiv) \ \text{Cyclohexyl methyl ether}$	Cyclohexyl methyl ether	Methoxycyclohexane

(N/A) The given structure is a substituted cyclohexane ring.
$1$. Identify the parent chain: The ring is a cyclohexane.
$2$. Identify substituents: There are two methyl groups at position $1$ and an ethoxy group $(-OC_2H_5)$ at position $2$.
$3$. $\text{IUPAC}$ naming: Numbering starts from the carbon with the substituents to give the lowest possible locants. Thus,the name is $2-\text{ethoxy}-1,1-\text{dimethylcyclohexane}$.
$4$. Common name: This is an ether where the alkyl groups are an ethyl group and a $2,2-\text{dimethylcyclohexyl}$ group. Thus,the common name is $\text{ethyl } 2,2-\text{dimethylcyclohexyl ether}$.

(N/A) Williamson's ether synthesis is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers.
In this reaction,an alkyl halide is reacted with a sodium alkoxide to form an ether.
The general reaction is: $R-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX$.
This reaction involves an $S_N2$ mechanism.
For the synthesis of unsymmetrical ethers,it is important to choose the reactants such that the alkyl halide is primary $(1^\circ)$ to minimize steric hindrance and favor the $S_N2$ pathway.
If a tertiary alkyl halide is used,elimination (forming an alkene) becomes the major reaction instead of substitution.
Example: The reaction of sodium tert-butoxide with methyl bromide yields tert-butyl methyl ether: $(CH_3)_3C-O^-Na^+ + CH_3-Br \rightarrow (CH_3)_3C-O-CH_3 + NaBr$.

(N/A) The $C-O$ bond in ethers is polar,and due to the angular shape of the ether molecule,it possesses a net dipole moment. Consequently,ethers have slightly higher boiling points than alkanes of comparable molecular mass.
However,ethers have significantly lower boiling points than alcohols of comparable molecular mass. This is because alcohols can form intermolecular hydrogen bonds,whereas ethers cannot.
The order of boiling points for compounds of comparable molecular mass is: $Alkane \approx Ether < Alcohol$.