Properties of Ethers Questions in English

(A) The reaction between sodium ethoxide and ethyl iodide is a $Williamson$ synthesis reaction.
The chemical equation is:
$C_2H_5ONa + C_2H_5I \rightarrow C_2H_5OC_2H_5 + NaI$
Here,$C_2H_5OC_2H_5$ is diethyl ether,which belongs to the class of compounds known as ethers.

(A) When ethanol vapor is passed over $Al_2O_3$ at $240 - 250 \, ^\circ C$,it undergoes intermolecular dehydration to form diethyl ether.
The chemical reaction is:
$2C_2H_5OH \xrightarrow{Al_2O_3, 240 - 250 \, ^\circ C} C_2H_5-O-C_2H_5 + H_2O$
Thus,the product is diethyl ether (ethyl ether).

(D) The reaction between sodium phenoxide $(C_6H_5ONa)$ and ethyl iodide $(C_2H_5I)$ is a Williamson ether synthesis reaction.
In this reaction,the nucleophilic phenoxide ion attacks the ethyl iodide,displacing the iodide ion to form ethyl phenyl ether,which is commonly known as phenetole $(C_6H_5OC_2H_5)$.
The chemical equation is: $C_6H_5ONa + C_2H_5I \rightarrow C_6H_5OC_2H_5 + NaI$.

(C) The reaction of diethyl ether with acetic anhydride in the presence of anhydrous $AlCl_3$ is a cleavage reaction of the ether linkage.
The reaction proceeds as follows:
$(C_2H_5)_2O + (CH_3CO)_2O \xrightarrow{AlCl_3 (anhyd.)} 2CH_3COOC_2H_5$
The product formed is ethyl acetate $(CH_3COOCH_2CH_3)$.

(C) When diethyl ether $(CH_3CH_2-O-CH_2CH_3)$ reacts with an excess of chlorine $(Cl_2)$ in the presence of sunlight $(hv)$,all the hydrogen atoms are replaced by chlorine atoms.
The chemical reaction is:
$CH_3CH_2-O-CH_2CH_3 + 10Cl_2 \xrightarrow{hv} CCl_3CCl_2-O-CCl_2CCl_3 + 10HCl$
Thus,the final product is perchlorodiethyl ether,which is $CCl_3CCl_2-O-CCl_2CCl_3$.

(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.
$1-$Methoxypropane $(CH_3-O-CH_2CH_2CH_3)$ reacts with excess $HI$ as follows:
$CH_3-O-CH_2CH_2CH_3 + 2HI \rightarrow CH_3I + CH_3CH_2CH_2I + H_2O$.
This reaction produces two distinct alkyl halides: methyl iodide and propyl iodide.

(B) The cleavage of ethers with $HI$ involves the formation of an alkyl halide and an alcohol. In the case of alkyl aryl ethers,the bond between the oxygen and the alkyl group is cleaved because the bond between the oxygen and the aryl group has partial double bond character due to resonance,making it stronger and harder to break. However,diaryl ethers,where both groups attached to the oxygen are aryl groups,do not undergo cleavage by $HI$ even at high temperatures like $525 \ K$ because both $C-O$ bonds have partial double bond character. Option $B$ represents a diaryl ether,specifically $4-methylphenyl$ phenyl ether.

(D) Ethers are generally inert towards nucleophilic attack because they lack a reactive site or a good leaving group that can be displaced by a nucleophile like $OH^{-}$ under normal conditions.

(A) The reaction of diethyl ether with excess $HI$ is a cleavage reaction.
${C_2H_5}O{C_2H_5} + 2HI \rightarrow 2{C_2H_5}I + {H_2}O$
Further reduction of ethyl iodide with $HI$ (acting as a reducing agent) gives ethane:
${C_2H_5}I + HI \rightarrow {C_2H_6} + I_2$
Combining these steps,the overall reaction is:
${C_2H_5}O{C_2H_5} + 4[H] \xrightarrow{HI} 2{C_2H_6} + {H_2}O$
Therefore,$X$ is ${C_2H_6}$ (Ethane).

(C) Williamson synthesis involves the reaction of an alkoxide ion with an alkyl halide.
For the preparation of ethoxyethane $(C_2H_5-O-C_2H_5)$,sodium ethoxide $(C_2H_5ONa)$ reacts with ethyl bromide $(C_2H_5Br)$:
$C_2H_5ONa + C_2H_5Br \xrightarrow{\Delta} C_2H_5-O-C_2H_5 + NaBr$
Thus,heating sodium ethoxide with ethyl bromide is the correct method.

(B) Electrophilic substitution reactions in ethers occur in the aromatic ring.
$C_6H_5OCH_3$ (Anisole) contains an aromatic ring,which activates the ring towards electrophilic substitution due to the electron-donating effect of the $-OCH_3$ group.
Aliphatic ethers like $CH_3OC_2H_5$,$CH_3OCH_3$,and $C_2H_5-O-C_2H_5$ do not contain an aromatic ring and therefore do not undergo typical electrophilic aromatic substitution.

(D) Ethers are generally inert towards bases,reducing agents,and many oxidizing agents under normal conditions. Among the given options,they are most notably inert towards bases.

(B) The reaction of tert-butyl methyl ether with $HI$ proceeds via an $S_N1$ mechanism because the tert-butyl carbocation is highly stable.
$CH_3-C(CH_3)_2-OCH_3 + HI \to CH_3-C(CH_3)_2-I + CH_3OH$.
The protonation of the ether oxygen is followed by the cleavage of the $C-O$ bond to form the stable tert-butyl carbocation,which then reacts with the iodide ion to form tert-butyl iodide and methanol.

(C) Anisole undergoes electrophilic aromatic substitution (nitration) when treated with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$.
The methoxy group $(-OCH_3)$ is an ortho- and para-directing group due to its $+R$ effect.
Therefore,the nitration of anisole yields a mixture of $o-$nitroanisole and $p-$nitroanisole as the major products.

(C) Alkaline $KMnO_4$ is a strong oxidizing agent that oxidizes alkyl side chains attached to a benzene ring to a carboxylic acid group $(-COOH)$.
$1$. Toluene $(C_6H_5CH_3)$ oxidizes to benzoic acid $(C_6H_5COOH)$.
$2$. Acetophenone $(C_6H_5COCH_3)$ oxidizes to benzoic acid.
$3$. Benzyl alcohol $(C_6H_5CH_2OH)$ oxidizes to benzoic acid.
$4$. Anisole $(C_6H_5OCH_3)$ is an ether. Ethers are generally inert towards oxidation by alkaline $KMnO_4$ under these conditions,as the methoxy group does not possess the required benzylic hydrogen atoms or the reactivity to be oxidized to a carboxyl group.

(C) In ethers,the oxygen atom is $sp^3$ hybridized. Due to the repulsion between the two bulky alkyl groups attached to the oxygen atom,the $C - O - C$ bond angle is slightly greater than the tetrahedral angle of $109^o28'$,which is approximately $110^o$.

(D) The synthesis of methyl tert-butyl ether follows the Williamson ether synthesis,which involves the reaction of a sodium alkoxide with a primary alkyl halide.
$(CH_3)_3CONa + CH_3Cl \to (CH_3)_3COCH_3 + NaCl$.
If a tertiary alkyl halide is used with a sodium alkoxide,the elimination reaction dominates over substitution,leading to the formation of an alkene instead of an ether.
Therefore,the correct reaction is $(CH_3)_3CONa + CH_3Cl$.

(D) Williamson synthesis involves the reaction of an alkoxide ion with an alkyl halide. $S_N2$ mechanism is preferred for this reaction. Tertiary alkyl halides undergo elimination reactions rather than substitution when treated with strong bases like alkoxides. Therefore,di-tert-butyl ether cannot be prepared by this method because it would require the reaction of sodium tert-butoxide with tert-butyl halide,which leads to the formation of isobutylene via an elimination reaction.

(C) Alkaline $KMnO_4$ is a strong oxidizing agent that oxidizes alkyl side chains attached to a benzene ring to a carboxylic acid group $(-COOH)$.
Benzyl alcohol $(C_6H_5CH_2OH)$,Toluene $(C_6H_5CH_3)$,and Acetophenone $(C_6H_5COCH_3)$ all contain a side chain attached to the benzene ring that can be oxidized to benzoic acid.
Anisole $(C_6H_5OCH_3)$ contains a methoxy group $(-OCH_3)$ attached to the benzene ring. The $C-O$ bond in the methoxy group is stable towards oxidation by alkaline $KMnO_4$ under these conditions,so it does not yield benzoic acid.

(D) Acid derivatives like $CH_3COOCH_3$ (methyl acetate),$CH_3CN$ (acetonitrile),and $CH_3CONH_2$ (acetamide) contain an electrophilic carbonyl carbon atom,which is susceptible to nucleophilic attack by the $OH^-$ ion.
Dimethyl ether $(CH_3OCH_3)$ does not contain an electrophilic center or a leaving group,making it resistant to nucleophilic attack by the hydroxide ion.

(C) $C_2H_5I$ and $Ag_2O$ (moist) react to produce diethyl ether as follows:
$2C_2H_5I + Ag_2O \to C_2H_5-O-C_2H_5 + 2AgI$
Thus,the product formed is $C_2H_5-O-C_2H_5$ (diethyl ether).

(A) The reaction of $CH_3CH_2CH=CH_2$ with $HBr$ in the presence of $H_2O_2$ (peroxide effect or Kharasch effect) follows anti-Markovnikov addition to give $CH_3CH_2CH_2CH_2Br$ $(Y)$.
Then,$CH_3CH_2CH_2CH_2Br$ reacts with sodium ethoxide $(C_2H_5ONa)$ via the Williamson ether synthesis mechanism to form $CH_3(CH_2)_3 - O - CH_2CH_3$ $(Z)$.

(B) When phenyl methyl ether $(Ph-O-CH_3)$ is heated with $HI$,the reaction proceeds via the protonation of the ether oxygen followed by a nucleophilic attack by the iodide ion $(I^-)$ on the less sterically hindered methyl group via an $S_N2$ mechanism.
The reaction is as follows:
$Ph-O-CH_3 + HI \rightarrow Ph-OH + CH_3I$
The products formed are phenol $(Ph-OH)$ and methyl iodide $(CH_3I)$. Among the given options,phenol is the correct product.

(C) The given reaction involves the formation of an ether from an alkoxide ion and an alkyl halide.
First,cyclopentanol reacts with $NaH$ to form sodium cyclopentoxide.
Then,sodium cyclopentoxide reacts with methyl iodide $(Me-I)$ via an $S_N2$ mechanism to form methyl cyclopentyl ether.
This is a classic example of the Williamson ether synthesis reaction,represented by the general equation: $R-X + R'-ONa \longrightarrow R-O-R' + NaX$.

(C) The reaction of a sodium alkoxide with an alkyl halide to form an ether is known as $Williamson$ synthesis.
This reaction proceeds via an $S_N2$ mechanism where the alkoxide ion acts as a nucleophile and attacks the alkyl halide.
The given reaction is: $CH_3-C(CH_3)_2-ONa + CH_3-CH_2-Cl \rightarrow CH_3-C(CH_3)_2-O-CH_2-CH_3 + NaCl$.

(B) The reaction of phenol $(C_6H_5OH)$ with sodium hydroxide $(NaOH)$ forms sodium phenoxide $(C_6H_5ONa)$.
Sodium phenoxide then undergoes a nucleophilic substitution reaction $(S_N2)$ with methyl iodide $(CH_3I)$ to produce anisole $(C_6H_5OCH_3)$.
This process is an example of Williamson's ether synthesis.

(A) When an ether contains a tertiary alkyl group,the reaction with hot concentrated $HI$ proceeds via the $S_N1$ mechanism.
In the case of tert-butyl methyl ether $(CH_3-C(CH_3)_2-O-CH_3)$,the $C-O$ bond cleavage occurs such that the stable tertiary carbocation $(CH_3-C^+(CH_3)_2)$ is formed.
This carbocation then reacts with $I^-$ to form tert-butyl iodide,while the methoxy group is protonated to form methyl alcohol $(CH_3OH)$.
In contrast,for primary and secondary alkyl ethers,the reaction follows the $S_N2$ mechanism,where the nucleophile $I^-$ attacks the less sterically hindered carbon (the methyl group),resulting in the formation of methyl iodide $(CH_3I)$ and the corresponding alcohol.

(C) The reaction of an ether with $HI$ follows the $S_N2$ mechanism when the alkyl groups are primary or secondary.
The iodide ion $(I^-)$ attacks the less sterically hindered alkyl group.
In isobutyl ethyl ether $(CH_3-CH(CH_3)-CH_2-O-CH_2-CH_3)$,the ethyl group is less sterically hindered than the isobutyl group.
Therefore,the $I^-$ ion attacks the ethyl group,resulting in the formation of isobutyl alcohol $(CH_3-CH(CH_3)-CH_2OH)$ and ethyl iodide $(CH_3CH_2I)$.

(A) In the reaction of an unsymmetrical ether with $HI$,if the alkyl groups are primary or secondary,the reaction follows the $S_N2$ mechanism.
The iodide ion $(I^{-})$ attacks the smaller (less sterically hindered) alkyl group.
In $CH_3-O-CH(CH_3)_2$,the methyl group $(-CH_3)$ is smaller than the isopropyl group $(-CH(CH_3)_2)$.
Therefore,the $I^{-}$ ion attacks the methyl group,leading to the formation of $CH_3I$ and $(CH_3)_2CHOH$ as the major products.

(D) The reaction between a Grignard reagent $(PhMgBr)$ and an ether $(EtOEt)$ does not proceed under standard conditions. Grignard reagents are nucleophilic and basic,but ethers are generally inert towards them because they lack an acidic proton or a suitable leaving group that can be displaced by the phenyl nucleophile. Therefore,no reaction occurs.

(A) The reaction of an ether with $H_3O^+$ (acid-catalyzed cleavage) typically follows an $S_N2$ mechanism for primary and secondary alkyl groups.
In the given ether,both alkyl groups attached to the oxygen are secondary.
Cleavage occurs at the $C-O$ bond,resulting in the formation of two alcohols.
Since the reaction proceeds via an $S_N2$ mechanism,the configuration at the chiral center where the bond is broken will be inverted.
For the left side $(CH(Me)(D)-O-)$,the $C-O$ bond cleavage leads to $CH(Me)(D)OH$ with inversion.
For the right side $(-O-CH(Me)(Et))$,the $C-O$ bond cleavage leads to $CH(Me)(Et)OH$ with inversion.
Option $A$ represents the alcohol from the left side without inversion.
Option $B$ represents the alcohol from the right side without inversion.
Option $C$ represents the alcohol from the left side with inversion.
Option $D$ represents the alcohol from the right side with inversion.
Since the reaction involves $S_N2$ inversion,the products formed are the inverted alcohols. Therefore,the alcohols with the original configuration (non-inverted) are not present in the major product mixture.

(A) The reaction of cyclic ethers with $HI$ involves the protonation of the oxygen atom followed by the nucleophilic attack of the iodide ion $(I^-)$.
In the case of the given cyclic ether (tetrahydrofuran derivative),the protonation occurs at the oxygen atom.
The iodide ion $(I^-)$ then attacks the less hindered carbon atom to open the ring.
Since both $HI$ (anhydrous) and $conc. HI$ provide the same reactive species ($H^+$ and $I^-$) for this ring-opening reaction,the product formed in both cases is the same,which is $2$-iodocyclopentanol.

(A) The reaction involves the nucleophilic ring opening of an epoxide ($2$-tert-butyl-oxirane) by the methanethiolate ion $(CH_3S^-)$.
In a protic solvent like $C_2H_5OH$,the $S_N2$ mechanism dominates.
The nucleophile $CH_3S^-$ attacks the less sterically hindered carbon atom of the epoxide ring,which is the $CH_2$ group.
The epoxide ring opens,and the resulting alkoxide ion is protonated by the solvent to yield the final product,$CH_3S-CH_2-CH(OH)-C(CH_3)_3$.

(C) When ethyl bromide $(CH_3CH_2Br)$ is treated with dry silver oxide $(Ag_2O)$,it undergoes a Williamson-like ether synthesis reaction to form diethyl ether,also known as ethoxy ethane.
$2CH_3CH_2Br + Ag_2O \rightarrow CH_3CH_2OCH_2CH_3 + 2AgBr$

(C) The reaction involves the nucleophilic attack of the ethoxide ion $(C_2H_5O^-)$ on the less sterically hindered terminal carbon $(^{14}CH_2)$ of the epoxide ring in epichlorohydrin.
This opens the epoxide ring to form an alkoxide intermediate: $Cl-CH_2-CH(O^-)-^{14}CH_2-OC_2H_5$.
Subsequently,the alkoxide oxygen performs an intramolecular $S_N2$ attack on the $CH_2$ group attached to the chlorine atom,displacing the chloride ion $(Cl^-)$.
This cyclization reforms an epoxide ring,resulting in the final product: $CH_2(O)CH-^{14}CH_2-OC_2H_5$.

(C) Compound $I$ (tetrahydropyran) is a cyclic ether. Upon reaction with excess $HI$,the ether ring undergoes cleavage.
In the first step,the protonated ether is attacked by $I^-$ to form $HO-(CH_2)_5-I$.
In the second step,the hydroxyl group is converted to an iodide by $HI$,resulting in the formation of $I-(CH_2)_5-I$ ($1$,$5$-diiodopentane).
Compound $II$ contains a phenolic ether linkage (where the oxygen is attached to an aromatic ring). Cleavage of the $C(aryl)-O$ bond is extremely difficult due to the partial double bond character of the $C-O$ bond and the stability of the aromatic ring,so it does not form a di-iodide under standard conditions.

(C) In the reaction of an unsymmetrical ether with $HI$,the iodide ion $(I^-)$ attacks the smaller alkyl group via $S_N2$ mechanism due to less steric hindrance.
$CH_3-O-C_2H_5 + HI \rightarrow CH_3I + C_2H_5OH$

(C) The formation of $diethyl$ ether hydroperoxide occurs via the autoxidation of $diethyl$ ether in the presence of atmospheric oxygen.
This process involves the abstraction of a hydrogen atom from the $\alpha$-carbon of the ether by a radical initiator,followed by the addition of oxygen to form a peroxy radical.
This is a $free \ radical \ chain \ mechanism$ that proceeds through a $free \ radical \ substitution$ pathway.

(C) The carbonylation of ethers involves the reaction with carbon monoxide $(CO)$ in the presence of a catalyst (like $BF_3$ or metal carbonyls) to form esters.
The general reaction is:
$R - O - R + CO \xrightarrow{\text{catalyst}} R - CO - O - R$
Thus,the product formed is an alkyl alkanoate.

(D) The mechanism of ether cleavage by $HI$ depends on the nature of the alkyl groups ($R$ and $R'$).
If the alkyl groups are primary or secondary,the reaction follows the $S_N2$ mechanism.
If one of the alkyl groups is tertiary,it follows the $S_N1$ mechanism.
Since the nature of $R$ and $R'$ is not specified,and the reaction conditions (such as solvent polarity and temperature) significantly influence the pathway,neither statement $A$ nor $B$ is universally true for the given general reaction.

(B) When an ether $(R-O-R)$ is heated with phosphorus pentasulphide $(P_2S_5)$,the oxygen atom of the ether is replaced by a sulphur atom to form a dialkyl sulphide $(R-S-R)$.
The reaction is represented as:
$5 R-O-R + P_2S_5 \rightarrow 5 R-S-R + P_2O_5$

(D) The reaction of an ether with $HI$ involves the protonation of the ether oxygen atom to form an oxonium ion.
In the case of $Ph-O-CH_2-Ph$,the cleavage occurs such that the more stable carbocation is formed.
The $Ph-CH_2^+$ (benzyl carbocation) is stabilized by resonance,whereas the $Ph^+$ (phenyl carbocation) is highly unstable.
Therefore,the $I^-$ ion attacks the $CH_2$ group of the benzyl moiety,leading to the formation of $Ph-OH$ and $Ph-CH_2-I$.

(B) Williamson ether synthesis involves the reaction of an alkoxide ion with an alkyl halide via an $S_N2$ mechanism.
For effective synthesis,the alkyl halide must be primary $(1^\circ)$ to avoid competing elimination reactions.
In option $B$,the reaction between sodium phenoxide $(C_6H_5ONa)$ and methyl iodide $(CH_3I)$ is highly effective because $CH_3I$ is a primary alkyl halide and does not undergo elimination.
Option $A$ is not possible because aryl halides do not undergo $S_N2$ reactions.
Option $C$ involves a tertiary alkyl group,which would lead to elimination (forming an alkene) rather than substitution.
Option $D$ involves a vinyl halide,which is unreactive towards $S_N2$ substitution.

(B) The reaction involves a nucleophilic substitution $(S_N2)$ on benzyl chloride $(C_6H_5CH_2Cl)$.
There are two nucleophiles present: phenoxide ion $(C_6H_5O^-)$ and ethoxide ion $(C_2H_5O^-)$.
Ethoxide ion $(C_2H_5O^-)$ is a stronger nucleophile than the phenoxide ion $(C_6H_5O^-)$ because the negative charge on the phenoxide ion is delocalized over the benzene ring,making it less available for attack.
Therefore,the ethoxide ion will preferentially attack the electrophilic carbon of benzyl chloride to form benzyl ethyl ether $(C_6H_5CH_2OC_2H_5)$ as the major product.

(A) The reaction involves the cleavage of an ether with $HI$. The ether is $3-(1-ethoxyethyl)phenol$.
When $HI$ reacts with this ether,the oxygen atom of the ether is protonated by $H^+$.
This forms a stable carbocation at the benzylic position (specifically,a secondary benzylic carbocation stabilized by the benzene ring).
Since the carbocation is stable,the reaction proceeds via an $S_N1$ mechanism.
The iodide ion $(I^-)$ attacks the carbocation to form $3-(1-iodoethyl)phenol$,and the ethyl group is released as ethanol $(C_2H_5OH)$.
Thus,the major products are $3-(1-iodoethyl)phenol$ and ethanol.

(C) The compound is methyl tert-butyl ether,$(CH_3)_3C-O-CH_3$.
To prepare ethers,the Williamson ether synthesis is the most effective method.
It involves the reaction of an alkoxide ion with a primary alkyl halide.
If a tertiary alkyl halide is used (as in option $A$),elimination (dehydrohalogenation) occurs instead of substitution,leading to an alkene.
Therefore,the correct approach is to use a tertiary alkoxide $(CH_3)_3C-OK$ and a primary alkyl halide $(CH_3-Br)$ to undergo $S_N2$ reaction.
Reaction: $(CH_3)_3C-OK + CH_3-Br \to (CH_3)_3C-O-CH_3 + KBr$.

(C) When ethanol $(CH_3CH_2OH)$ is heated with concentrated $H_2SO_4$ at $140^\circ C$ $(413 \ K)$,it undergoes intermolecular dehydration to form diethyl ether (ethoxy ethane).
The reaction is as follows:
$2CH_3CH_2OH \xrightarrow{conc. H_2SO_4, 140^\circ C} CH_3CH_2-O-CH_2CH_3 + H_2O$
Mechanism:
$1$. Protonation of ethanol: $CH_3CH_2OH + H^+ \rightarrow CH_3CH_2OH_2^+$
$2$. Nucleophilic attack: $CH_3CH_2OH_2^+ + CH_3CH_2OH \rightarrow CH_3CH_2-O^+(H)-CH_2CH_3 + H_2O$
$3$. Deprotonation: $CH_3CH_2-O^+(H)-CH_2CH_3 \rightarrow CH_3CH_2-O-CH_2CH_3 + H^+$

(B) The reaction of an ether with $HBr$ involves the protonation of the ether oxygen atom by $H^+$,followed by the nucleophilic attack of $Br^-$ on the carbon atom.
In the given ether,$Ph-CH_2-O-CH(CH_3)_2$,the benzyl group $(Ph-CH_2-)$ is attached to the oxygen. The carbocation formed at the benzylic position $(Ph-CH_2^+)$ is stabilized by resonance.
Therefore,the $Br^-$ nucleophile attacks the benzylic carbon to form benzyl bromide $(Ph-CH_2Br)$,and the isopropyl group forms isopropyl alcohol $(CH_3-CH(OH)-CH_3)$.
The reaction is: $Ph-CH_2-O-CH(CH_3)_2 + HBr \rightarrow Ph-CH_2Br + CH_3-CH(OH)-CH_3$.

(D) The Williamson ether synthesis involves the reaction of an alkoxide ion with a primary alkyl halide via an $S_N2$ mechanism to form an ether.
$(A)$ $CH_3ONa + CH_3-C(CH_3)_2-Cl \rightarrow$ This is a tertiary alkyl halide. The strong base $CH_3O^-$ will cause elimination $(E2)$ to form an alkene (isobutylene) as the major product,not an ether.
$(B)$ $CH_3-CH_2-ONa + C_6H_5-Cl \rightarrow$ Aryl halides do not undergo $S_N2$ reactions under normal conditions due to the partial double bond character of the $C-Cl$ bond and steric hindrance. No reaction occurs.
$(C)$ $CH_3-CH_2-ONa + CH_3-C(Cl)(CH_3)-CH_2-CH_3 \rightarrow$ This is a tertiary alkyl halide. Similar to $(A)$,elimination $(E2)$ will be the major pathway,yielding an alkene.
$(D)$ $CH_3ONa + C_6H_5-CH_2-Br \rightarrow$ This is a primary benzylic halide. Benzylic halides are highly reactive towards $S_N2$ reactions. The $CH_3O^-$ nucleophile will attack the $CH_2$ group,displacing $Br^-$ to form benzyl methyl ether $(C_6H_5-CH_2-OCH_3)$ as the major product.
Therefore,option $(D)$ is the correct reaction.