Properties of Ethers Questions in English

(A) $(i)$ The reaction of an ether with $HI$ produces an alcohol and an alkyl iodide. In the presence of a $3^{\circ}$ carbon atom in one of the alkyl groups of the ether,the iodide ion $(-I)$ attacks the more substituted carbon atom (the $3^{\circ}$ carbon) via the $S_N1$ mechanism.
$(ii)$ This occurs because the $3^{\circ}$ carbocation formed during the reaction is highly stable compared to $1^{\circ}$ or $2^{\circ}$ carbocations,and the $S_N2$ mechanism is hindered by significant steric hindrance at the $3^{\circ}$ carbon.
$(iii)$ In the given reaction,the ether is $2-ethoxy-2-methylbutane$. The $HI$ cleaves the bond between the oxygen and the $3^{\circ}$ carbon,leading to the formation of ethanol $(CH_3CH_2OH)$ and $2-iodo-2-methylbutane$ $((CH_3)_2(CH_3CH_2)C-I)$ via the $S_N1$ mechanism.
Thus,option $A$ is the correct answer.

(A) Anisole $(C_6H_5OCH_3)$ undergoes Friedel-Crafts acylation when treated with acetyl chloride $(CH_3COCl)$ in the presence of anhydrous $AlCl_3$.
The methoxy group $(-OCH_3)$ is an ortho/para-directing group.
Therefore,the reaction yields a mixture of ortho-substituted and para-substituted products.
The products formed are $o-$methoxyacetophenone and $p-$methoxyacetophenone.

(C) The reaction of an ether with $HI$ proceeds via an $S_N1$ mechanism when one of the alkyl groups is tertiary,as it can form a stable carbocation.
In the given ether,$CH_3-CH_2-CH_2-O-C(CH_3)_2-CH_2-CH_3$,the oxygen atom is attached to a primary alkyl group ($n$-propyl) and a tertiary alkyl group ($2$-methylbutan$-2-$yl).
When $HI$ is added,the oxygen gets protonated. The bond between the oxygen and the tertiary carbon breaks to form a stable tertiary carbocation,which then reacts with the iodide ion $(I^-)$ to form an alkyl iodide,and the primary group forms an alcohol.
Therefore,the products are $CH_3-CH_2-CH_2-OH$ (propan$-1-$ol) and $I-C(CH_3)_2-CH_2-CH_3$ ($2$-iodo$-2-$methylbutane).
Comparing this with the given options,option $C$ represents the correct products.

(D) When an alkyl aryl ether is reacted with a halogen acid $(HI)$,the cleavage of the $C-O$ bond occurs such that the alkyl group forms an alkyl halide and the aryl group forms a phenol.
In the reaction of ethyl phenyl ether $(C_6H_5-O-CH_2CH_3)$ with $HI$,the $C-O$ bond between the oxygen and the ethyl group breaks.
$C_6H_5-O-CH_2CH_3 + HI \xrightarrow{\Delta} C_6H_5OH + CH_3CH_2I$
Thus,$Y$ is $C_6H_5OH$ (phenol) and $Z$ is $CH_3CH_2I$ (ethyl iodide).

(A) The $-OCH_3$ group is an electron-donating group and is ortho/para directing. In aromatic electrophilic substitution,the para position is favored due to steric hindrance at the ortho position.
Step-$1$: Friedel-Crafts Acylation of anisole with $CH_3COCl$ and anhydrous $AlCl_3$ yields $4$-methoxyacetophenone as the major product.
Step-$2$: Clemmensen reduction of $4$-methoxyacetophenone using $Zn-Hg$ and concentrated $HCl$ reduces the acetyl group to an ethyl group,forming $4$-ethylanisole.
Step-$3$: Cleavage of the ether linkage in $4$-ethylanisole with $HI$ yields $4$-ethylphenol and methyl iodide $(CH_3I)$.
Thus,the major product is $4$-ethylphenol.

(A) The reaction of $1$-methoxy$-3-$($2$-methoxyethyl)benzene with excess $HI$ at high temperature involves the cleavage of both ether groups.
$1$. The first $S_{N}2$ reaction occurs at the primary alkyl group of the side chain,where the methoxy group is protonated and then attacked by the iodide ion $(I^-)$ to form an alcohol and $CH_3I$.
$2$. The second $S_{N}2$ reaction occurs at the methyl group attached to the phenolic oxygen,where the methoxy group is protonated and then attacked by the iodide ion to form phenol and $CH_3I$.
$3$. The third $S_{N}2$ reaction involves the conversion of the primary alcohol formed in the first step into an alkyl iodide by reaction with $HI$.
Thus,there are a total of $3$ $S_{N}2$ reactions involved in the complete conversion.

(A) In methoxy methane $(CH_3-O-CH_3)$,the oxygen atom is $sp^3$ hybridized.
Due to the presence of two bulky $-CH_3$ groups,there is significant steric repulsion between them.
This repulsion causes the $C-O-C$ bond angle to increase from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $111.7^{\circ}$.

(C) In diethyl ether $(CH_3CH_2-O-CH_2CH_3)$,the oxygen atom is bonded to two carbon atoms and has two lone pairs of electrons.
According to the steric number rule,the steric number is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 2 + 2 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(D) The molecular formula $C_4H_{10}O$ corresponds to ethers with the general formula $R-O-R'$.
For an unsymmetrical ether,the alkyl groups $R$ and $R'$ must be different.
The possible unsymmetrical ethers for $C_4H_{10}O$ are:
$1.$ $1-$methoxypropane $(CH_3-O-CH_2-CH_2-CH_3)$
$2.$ $2-$methoxypropane $(CH_3-O-CH(CH_3)_2)$
Among the given options,$1-$methoxypropane is the correct $IUPAC$ name.

(D) The molecular formula $C_3H_8O$ corresponds to either an alcohol or an ether. Since it reacts with $HI$ to form two products ($X$ and $Y$),it must be an ether. The reaction is: $CH_3-CH_2-O-CH_3 + 2HI \rightarrow CH_3-I (X) + CH_3-CH_2-I (Y) + H_2O$.
When ethyl iodide $(Y)$ is boiled with aqueous alkali $(NaOH)$,it forms ethanol $(CH_3CH_2OH)$.
However,the iodoform test is given by compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group. Ethanol $(CH_3CH_2OH)$ gives a positive iodoform test. Thus,$(A)$ is methoxyethane.

(A) Step $1$: Dimethyl ketone is acetone,$CH_3COCH_3$. Reaction with $CH_3MgCl$ followed by $H_2O$ (Grignard reaction) yields $tert$-butyl alcohol,$X = (CH_3)_3COH$.
Step $2$: Reaction of $X$ with $Na$ gives the alkoxide $(CH_3)_3CONa$. Subsequent reaction with $CH_3Br$ (Williamson ether synthesis) yields $tert$-butyl methyl ether,$Y = (CH_3)_3COCH_3$.
Step $3$: In $Y$,the structure is $(CH_3)_3C-O-CH_3$. The carbons are: three methyl carbons attached to the central carbon,the central quaternary carbon,and the methyl carbon attached to oxygen. All $5$ carbons are $sp^3$ hybridized.

(C) The reaction of benzyl phenyl ether with $HI$ proceeds via the protonation of the ether oxygen atom.
Following protonation,the bond between the oxygen and the benzylic carbon breaks to form a stable benzylic carbocation $(C_6H_5CH_2^+)$ and phenol $(C_6H_5OH)$.
The benzylic carbocation is resonance-stabilized.
Finally,the iodide ion $(I^-)$ attacks the benzylic carbocation to form benzyl iodide $(C_6H_5CH_2I)$.
Thus,the major products are benzyl iodide and phenol.
This corresponds to option $C$.

(A) Step $1$: Reaction of tert-butyl alcohol with $Na$ gives sodium tert-butoxide $(P)$:
$(CH_3)_3 COH Na \rightarrow (CH_3)_3 CONa \frac{1}{2} H_2$
Step $2$: Reaction of $P$ with methyl bromide $(CH_3 Br)$ via $S_N2$ mechanism gives tert-butyl methyl ether $(Q)$:
$(CH_3)_3 CONa CH_3 Br \rightarrow (CH_3)_3 COCH_3 NaBr$
Step $3$: Cleavage of ether $Q$ with $HI$ at high temperature $(\Delta)$:
Since the ether contains a tertiary alkyl group,the reaction proceeds via an $S_N1$ mechanism.
The protonated ether undergoes cleavage to form a stable tertiary carbocation $(CH_3)_3 C $,which then reacts with $I^-$ to form tert-butyl iodide $(R = (CH_3)_3 CI)$ and methanol $(S = CH_3 OH)$.

(B) The reaction of ethyl chloride $(C_2H_5Cl)$ with sodium ethoxide $(C_2H_5ONa)$ is a Williamson ether synthesis,which produces diethyl ether $(C_2H_5-O-C_2H_5)$ as compound $A$.
The reaction is: $C_2H_5Cl + C_2H_5ONa \rightarrow C_2H_5-O-C_2H_5 + NaCl$.
Diethyl ether is also obtained by the intermolecular dehydration of ethyl alcohol $(C_2H_5OH)$ in the presence of concentrated $H_2SO_4$ at $140^{\circ}C$ $(413 \ K)$:
$2C_2H_5OH \xrightarrow{conc. H_2SO_4, 140^{\circ}C} C_2H_5-O-C_2H_5 + H_2O$.

(D) Alkyl halides react with sodium alkoxide to give ether. This is known as Williamson's ether synthesis.
$C_2H_5Cl + C_2H_5ONa \xrightarrow{\Delta} C_2H_5OC_2H_5 + NaCl$
In option $D$,the reaction between ethyl chloride and sodium ethoxide produces diethyl ether,which is an ether.

(C) The reaction of chloroethane $(C_2H_5Cl)$ with sodium ethoxide $(C_2H_5ONa)$ is known as the Williamson ether synthesis.
In this reaction,the ethoxide ion $(C_2H_5O^-)$ acts as a nucleophile and attacks the chloroethane molecule,displacing the chloride ion to form diethyl ether $(C_2H_5OC_2H_5)$.
The chemical equation is:
$C_2H_5Cl + C_2H_5ONa \rightarrow C_2H_5OC_2H_5 + NaCl$
Therefore,$X$ is $C_2H_5ONa$.

(D) For ether $A$ (phenyl $n$-propyl ether): The $C-O$ bond between the phenyl group and oxygen cannot be cleaved because the $C-O$ bond has partial double bond character due to resonance. Therefore,cleavage occurs at the alkyl side,yielding phenol and $n$-propyl bromide.
For ether $B$ ($tert$-butyl $n$-propyl ether): The cleavage occurs to form the more stable carbocation. The $tert$-butyl group forms a $3^{\circ}$ carbocation,which is much more stable than the $n$-propyl carbocation. Thus,the $C-O$ bond between the $tert$-butyl group and oxygen is cleaved,yielding $tert$-butyl bromide and $n$-propanol.
Therefore,the bromides formed are $n$-propyl bromide and $tert$-butyl bromide.

(B) The reaction of an alkyl aryl ether with $HI$ involves the protonation of the ether oxygen atom followed by the nucleophilic attack of the iodide ion $(I^-)$ on the less sterically hindered alkyl group. In this case,the cyclohexyl group is the alkyl group and the phenyl group is the aryl group. The $I^-$ ion attacks the cyclohexyl carbon,leading to the formation of cyclohexyl iodide $(X)$ and phenol $(Y)$.

(D) The reaction of alkyl aryl ethers with $HI$ involves the cleavage of the $O-alkyl$ bond.
This occurs because the $O-aryl$ bond possesses partial double bond character due to resonance,making it stronger and harder to break.
Therefore,the ether $C_6H_5-O-CH_2CH_3$ reacts with $HI$ to form phenol $(C_6H_5OH)$ and ethyl iodide $(CH_3CH_2I)$.
Thus,the correct option is $D$.

(B) The reaction of diethyl ether with carbon monoxide in the presence of a Lewis acid catalyst like $BF_3$ at high pressure and temperature is a carbonylation reaction.
$C_2H_5-O-C_2H_5 + CO \xrightarrow[500 \text{ atm}]{BF_3 / 150^{\circ}C} C_2H_5COOC_2H_5$
The product $(A)$ formed is ethyl propionate.

(D) The reaction of diethyl ether $(C_2H_5OC_2H_5)$ with carbon monoxide $(CO)$ in the presence of a Lewis acid catalyst like $BF_3$ at high temperature $(150^{\circ}C)$ and high pressure $(500 \text{ atm})$ is a carbonylation reaction.
This reaction results in the insertion of $CO$ into the $C-O$ bond of the ether to form an ester.
The product formed is ethyl propionate $(C_2H_5COOC_2H_5)$.

(A) When diethyl ether reacts with cold $HI$,it undergoes cleavage to form ethyl alcohol and ethyl iodide as follows:
$C_2H_5-O-C_2H_5 + HI \rightarrow C_2H_5OH + C_2H_5I$

(C) The reaction of ethers with carbon monoxide $(CO)$ in the presence of a Lewis acid catalyst like $BF_3$ at high temperature $(150^{\circ} C)$ and high pressure $(500 \ atm)$ is a carbonylation reaction.
Diethyl ether $(C_2H_5OC_2H_5)$ reacts with $CO$ to form ethyl propionate $(C_2H_5COOC_2H_5)$ according to the following equation:
$C_2H_5OC_2H_5 + CO \xrightarrow{BF_3, 150^{\circ} C, 500 \ atm} C_2H_5COOC_2H_5$.

(C) The chemical formula of diethyl ether is $C_2H_5-O-C_2H_5$.
In this molecule,the oxygen atom is bonded to two carbon atoms and possesses two lone pairs of electrons.
The steric number of the oxygen atom is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 2 + 2 = 4$.
$A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

(D) Sodium metal reacts with compounds containing acidic hydrogen atoms (like $-OH$ or $-COOH$ groups) to evolve hydrogen gas.
Phenol $(C_6H_5OH)$,ethanol $(C_2H_5OH)$,and benzoic acid $(C_6H_5COOH)$ all contain acidic hydrogen atoms.
Anisole $(C_6H_5OCH_3)$ is an ether and does not contain any acidic hydrogen atom,therefore it does not react with sodium metal.

(B) Step $1$: The first reaction is the Williamson ether synthesis.
$C_2H_5I + NaOC_2H_5 \longrightarrow C_2H_5OC_2H_5 (X) + NaI$
Here,$X$ is diethyl ether $(C_2H_5OC_2H_5)$.
Step $2$: The second reaction is the cleavage of ether by excess hydroiodic acid $(HI)$.
$C_2H_5OC_2H_5 + 2HI \xrightarrow{\Delta} 2C_2H_5I (Y) + H_2O$
Comparing this with the given reaction $X + 2HI \xrightarrow{\Delta} 2Y + H_2O$,we find that $Y$ is $C_2H_5I$.

(B) $Me_{3}CONa + EtCl \rightarrow Me_{3}COEt + NaCl$
Williamson synthesis involves the reaction of an alkoxide with a primary alkyl halide to form an ether.
Since $EtCl$ is a primary $(1^{\circ})$ alkyl halide and $Me_{3}CONa$ is a bulky alkoxide,this reaction proceeds via an $S_{N}2$ mechanism to yield the desired ether.
Option $A$ would result in elimination $(E2)$ because $Me_{3}CCl$ is a tertiary alkyl halide.
Therefore,the correct option is $B$.

(B) Methoxybenzene (anisole) on treatment with $HI$ produces phenol and methyl iodide.
This reaction occurs because the $C_{aryl}-O$ bond is stronger than the $C_{alkyl}-O$ bond.
The carbon of the phenyl group is $sp^{2}$-hybridised,which gives the $C_{aryl}-O$ bond a partial double bond character,making it resistant to cleavage.
Consequently,the nucleophilic iodide ion attacks the less sterically hindered methyl group,resulting in the formation of phenol and methyl iodide $(CH_3I)$.

(D) $1$. The mixed ether $(P)$ reacts with excess hot concentrated $HI$ to form two alkyl iodides.
$2$. These alkyl iodides,upon treatment with aqueous $NaOH$,yield alcohols $(Q)$ and $(R)$.
$3$. Both $(Q)$ and $(R)$ give a yellow precipitate with $NaOI$ (iodoform test),which implies both must contain a $CH_3CH(OH)-$ group or be ethanol $(CH_3CH_2OH)$.
$4$. Ethyl sec-butyl ether $(CH_3CH_2-O-CH(CH_3)CH_2CH_3)$ on cleavage with $HI$ gives ethyl iodide $(CH_3CH_2I)$ and sec-butyl iodide $(CH_3CH(I)CH_2CH_3)$.
$5$. Treatment with $NaOH$ converts these to ethanol $(CH_3CH_2OH)$ and sec-butanol $(CH_3CH(OH)CH_2CH_3)$.
$6$. Both ethanol and sec-butanol give a positive iodoform test with $NaOI$ (forming yellow $CHI_3$ precipitate).
$7$. Thus,the ether $(P)$ is ethyl sec-butyl ether.

(D) The reaction involves the cleavage of an ether,benzyl phenyl ether $(C_6H_5-O-CH_2-C_6H_5)$,with hydrogen iodide $(HI)$.
In the reaction of an alkyl aryl ether with $HI$,the cleavage occurs such that the alkyl group forms an alkyl iodide and the aryl group forms a phenol.
This is because the $O-C_{aryl}$ bond has partial double bond character due to resonance and is stronger,while the $O-C_{alkyl}$ bond is weaker and undergoes nucleophilic substitution.
Here,the $O-CH_2C_6H_5$ bond breaks to form benzyl iodide $(C_6H_5CH_2I)$ and phenol $(C_6H_5OH)$.
Therefore,Statement $I$ is false because the products are benzyl iodide and phenol,not benzyl alcohol and iodobenzene.
Statement $II$ is true because the $O-CH_2$ bond is indeed the one that undergoes cleavage.

(D) The reaction of an ether with $HI$ involves the protonation of the ether oxygen atom,followed by a nucleophilic attack by the iodide ion $(I^-)$.
In the given ether,benzyl phenyl ether $(C_6H_5-O-CH_2-C_6H_5)$,the oxygen atom is attached to a phenyl group and a benzyl group.
The cleavage occurs such that the more stable carbocation is formed. The benzyl carbocation $(C_6H_5CH_2^+)$ is resonance-stabilized and more stable than the phenyl carbocation.
Therefore,the $C-O$ bond on the benzylic side is cleaved,leading to the formation of benzyl iodide $(C_6H_5CH_2I)$ and phenol $(C_6H_5OH)$.
Statement $I$ claims the products are benzyl alcohol and iodobenzene,which is incorrect.
Statement $II$ correctly states that the $-O-CH_2-$ bond is cleaved.
Thus,Statement $I$ is false and Statement $II$ is true.