601. Which reagent commonly cleaves ethers on heating to give alcohols and alkyl halides?
ⓐ. \(NaBH_4\)
ⓑ. \(HI\)
ⓒ. \(NaOH\)
ⓓ. \(Br_2\) water
Correct Answer: \(HI\)
Explanation: Ethers are comparatively less reactive, but they can be cleaved by strong hydrogen halides such as \(HI\) and \(HBr\) on heating. Protonation of ether oxygen first makes the \(C-O\) bond more susceptible to cleavage. Iodide ion then attacks an alkyl group to give an alkyl iodide. This reaction is an important chemical property of ethers.
602. Which general equation represents cleavage of a simple ether by \(HI\)?
ⓐ. \[R-O-R' + HI \rightarrow R-I + R'-OH\]
ⓑ. \[R-OH + HI \rightarrow R-O-R' + H_2O\]
ⓒ. \[R-CHO + HI \rightarrow R-CH_2OH + I_2\]
ⓓ. \[R-ONa + HI \rightarrow R-O-R' + NaI\]
Correct Answer: \[R-O-R' + HI \rightarrow R-I + R'-OH\]
Explanation: \(\textbf{Reaction type:}\)
Ether cleavage by \(HI\) breaks one \(C-O\) bond of the ether.
\(\textbf{First step:}\)
The ether oxygen is protonated by \(HI\), making the ether linkage easier to cleave.
\(\textbf{Nucleophile:}\)
\(I^-\) attacks an alkyl carbon.
\(\textbf{Product pattern:}\)
One side becomes an alkyl iodide, while the other side becomes an alcohol.
\(\textbf{General equation:}\)
\[
R-O-R' + HI \rightarrow R-I + R'-OH
\]
603. What is formed when methoxyethane, \(CH_3OC_2H_5\), is heated with one mole of \(HI\)?
ⓐ. \(CH_3I\) and \(C_2H_5OH\)
ⓑ. \(C_2H_5I\) and \(CH_3OH\)
ⓒ. \(CH_3OH\) and \(C_2H_5OH\)
ⓓ. \(CH_3I\) and \(C_2H_5I\)
Correct Answer: \(CH_3I\) and \(C_2H_5OH\)
Explanation: In cleavage of an unsymmetrical ether by \(HI\), iodide ion usually attacks the less hindered alkyl group when both groups are primary or methyl-type. In methoxyethane, the methyl group is less hindered than the ethyl group. Therefore \(CH_3I\) and \(C_2H_5OH\) are formed initially. With excess \(HI\), the alcohol may further convert into an alkyl iodide.
604. Which products are obtained when diethyl ether is heated with excess \(HI\)?
ⓐ. \(C_2H_5OH\) and \(CH_3I\)
ⓑ. Only \(C_2H_5I\) as organic halide product
ⓒ. \(CH_3OH\) and \(C_2H_5Br\)
ⓓ. Only \(C_3H_7I\) as organic halide product
Correct Answer: Only \(C_2H_5I\) as organic halide product
Explanation: \(\textbf{Ether given:}\)
Diethyl ether is \(C_2H_5OC_2H_5\).
\(\textbf{First cleavage:}\)
One \(C-O\) bond breaks to form \(C_2H_5I\) and \(C_2H_5OH\).
\(\textbf{Excess HI:}\)
The ethanol formed reacts further with \(HI\).
\(\textbf{Second conversion:}\)
\[
C_2H_5OH + HI \rightarrow C_2H_5I + H_2O
\]
\(\textbf{Overall result:}\)
Excess \(HI\) converts both ethyl groups into ethyl iodide.
605. Which statement correctly describes the first step in acid cleavage of ethers?
ⓐ. Direct oxidation of ether oxygen to carbonyl oxygen
ⓑ. Protonation of ether oxygen by strong acid
ⓒ. Formation of sodium alkoxide by \(NaOH\)
ⓓ. Reduction of ether by \(NaBH_4\)
Correct Answer: Protonation of ether oxygen by strong acid
Explanation: Ether oxygen has lone pairs and can accept a proton from a strong acid such as \(HI\) or \(HBr\). Protonation gives an oxonium ion, which has a more reactive \(C-O\) bond. This activation allows halide ion to attack and cleave the ether. Without protonation, ordinary ethers are much less reactive toward cleavage.
606. Which product pair is expected from cleavage of \(CH_3OCH_3\) with excess \(HI\)?
ⓐ. \(CH_3OH\) and \(CH_3I\)
ⓑ. \(2CH_3I\) and \(H_2O\)
ⓒ. \(CH_3CHO\) and \(I_2\)
ⓓ. \(CH_4\) and \(CH_3OH\)
Correct Answer: \(2CH_3I\) and \(H_2O\)
Explanation: \(\textbf{Ether:}\)
\(CH_3OCH_3\) is dimethyl ether.
\(\textbf{Initial cleavage:}\)
One molecule of \(HI\) gives \(CH_3I\) and \(CH_3OH\).
\(\textbf{Excess HI:}\)
The methanol formed reacts further with \(HI\) to give another molecule of \(CH_3I\).
\(\textbf{Overall equation:}\)
\[
CH_3OCH_3 + 2HI \rightarrow 2CH_3I + H_2O
\]
\(\textbf{Final Answer:}\)
The complete cleavage products are \(2CH_3I\) and \(H_2O\).
607. Which ether is cleaved by \(HI\) to give methyl iodide and phenol?
ⓐ. \(C_6H_5OCH_3\)
ⓑ. \(C_6H_5OC_2H_5\)
ⓒ. \(C_2H_5OC_2H_5\)
ⓓ. \(CH_3OCH_2CH_3\)
Correct Answer: \(C_6H_5OCH_3\)
Explanation: \(C_6H_5OCH_3\) is anisole, an aryl methyl ether. In cleavage by \(HI\), the alkyl-oxygen bond breaks more readily than the aryl-oxygen bond. Iodide attacks the methyl group to form \(CH_3I\). The aryl part remains as phenol, \(C_6H_5OH\).
608. Which equation correctly represents cleavage of anisole by \(HI\)?
ⓐ. \[C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I\]
ⓑ. \[C_6H_5OCH_3 + HI \rightarrow C_6H_5I + CH_3OH\]
ⓒ. \[C_6H_5OH + CH_3I \rightarrow C_6H_5OCH_3 + HI\]
ⓓ. \[C_6H_5OCH_3 + HI \rightarrow C_6H_6 + CH_3OI\]
Correct Answer: \[C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I\]
Explanation: \(\textbf{Ether type:}\)
Anisole is an aryl alkyl ether.
\(\textbf{Bond cleaved:}\)
The \(O-CH_3\) bond breaks, not the aryl \(C-O\) bond.
\(\textbf{Reason:}\)
The aryl \(C-O\) bond has partial double-bond character and is difficult to cleave by ordinary \(S_N2\)-type attack.
\(\textbf{Product formation:}\)
\[
C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I
\]
\(\textbf{Final Answer:}\)
Anisole gives phenol and methyl iodide.
609. Why does anisole give phenol rather than iodobenzene on treatment with \(HI\)?
ⓐ. Phenol has no oxygen atom after the reaction.
ⓑ. The methyl \(C-O\) bond is stronger than the aryl \(C-O\) bond.
ⓒ. Iodide ion attacks the aromatic carbon by normal \(S_N2\) substitution.
ⓓ. Resonance strengthens the aryl \(C-O\) bond against cleavage.
Correct Answer: Resonance strengthens the aryl \(C-O\) bond against cleavage.
Explanation: In anisole, oxygen is directly attached to the aromatic ring as well as to a methyl group. The aryl \(C-O\) bond has partial double-bond character due to resonance with the ring. Direct \(S_N2\) attack at the aromatic carbon is not favoured. Therefore the \(O-CH_3\) bond is cleaved, giving \(CH_3I\) and phenol.
610. Which product is formed from the alkyl part when phenetole, \(C_6H_5OC_2H_5\), is heated with \(HI\)?
ⓐ. \(C_6H_5I\)
ⓑ. \(C_2H_5I\)
ⓒ. \(CH_3I\)
ⓓ. \(C_6H_5CH_2I\)
Correct Answer: \(C_2H_5I\)
Explanation: Phenetole is ethoxybenzene, \(C_6H_5OC_2H_5\). In aryl alkyl ethers, \(HI\) cleaves the alkyl-oxygen bond rather than the aryl-oxygen bond. The ethyl group is converted into ethyl iodide, \(C_2H_5I\). The aromatic oxygen part becomes phenol.
611. Which product pair is obtained from \(C_6H_5OC_2H_5\) on heating with \(HI\)?
ⓐ. \(C_6H_5I\) and \(C_2H_5OH\)
ⓑ. \(C_6H_5OH\) and \(C_2H_5I\)
ⓒ. \(C_6H_6\) and \(C_2H_5OI\)
ⓓ. \(C_6H_5CH_2OH\) and \(HI\)
Correct Answer: \(C_6H_5OH\) and \(C_2H_5I\)
Explanation: \(C_6H_5OC_2H_5\) is an aryl ethyl ether. Hydrogen iodide protonates the ether oxygen and iodide ion attacks the ethyl group. The aryl \(C-O\) bond is not cleaved easily because of resonance and \(sp^2\) aryl carbon character. Thus phenol and ethyl iodide are formed.
612. In cleavage of mixed ether \(CH_3OC(CH_3)_3\) by \(HI\), which product pair is most likely when a tertiary group is present?
ⓐ. \(CH_3I\) and \((CH_3)_3COH\)
ⓑ. \(CH_3OH\) and \((CH_3)_3CI\)
ⓒ. \(CH_3OH\) and \((CH_3)_2C=CH_2\) only
ⓓ. \(CH_3CH_2I\) and \(CH_3OH\)
Correct Answer: \(CH_3OH\) and \((CH_3)_3CI\)
Explanation: In mixed ethers containing a tertiary alkyl group, cleavage often occurs through formation of the more stable tertiary carbocation-like species. The tertiary side then gives the tertiary alkyl iodide. The methoxy side becomes methanol after cleavage. Therefore \(CH_3OC(CH_3)_3\) gives \(CH_3OH\) and \((CH_3)_3CI\) as the expected product pair.
613. Which factor controls cleavage direction in mixed ethers containing primary and tertiary alkyl groups?
ⓐ. Formation of a more stable tertiary intermediate
ⓑ. Formation of a less stable primary carbocation-like intermediate
ⓒ. Complete oxidation of the ether oxygen
ⓓ. Replacement of all hydrogens by iodine
Correct Answer: Formation of a more stable tertiary intermediate
Explanation: Mixed ether cleavage depends on structure. If both groups are primary or methyl, halide ion usually attacks the less hindered side. If one group is tertiary, cleavage tends to occur in a way that forms a stable tertiary carbocation-like intermediate. This is why tertiary alkyl iodide can be formed from tertiary alkyl ethers.
614. Which statement about ether cleavage by \(HI\) is correct?
ⓐ. Aryl \(C-O\) bonds cleave more easily than methyl \(C-O\) bonds.
ⓑ. Ether oxygen is first protonated in acid.
ⓒ. \(HI\) oxidises ethers to carboxylic acids.
ⓓ. Ethers react with \(HI\) only to form alkenes.
Correct Answer: Ether oxygen is first protonated in acid.
Explanation: Ether cleavage by hydrogen halides begins with protonation of the ether oxygen. This creates an oxonium ion, making the adjacent \(C-O\) bonds more reactive. The halide ion then participates in bond cleavage. The exact product depends on whether the ether is symmetrical, unsymmetrical, aryl alkyl, or contains a tertiary alkyl group.
615. Which product is expected when \(CH_3CH_2OCH_3\) is cleaved by \(HBr\) under ordinary cleavage conditions?
ⓐ. \(CH_3Br\) and \(CH_3CH_2OH\)
ⓑ. \(CH_3CH_2Br\) and \(CH_3OH\)
ⓒ. \(CH_3CH_2CH_2Br\) only
ⓓ. \(CH_3CHO\) and \(CH_3Br\)
Correct Answer: \(CH_3Br\) and \(CH_3CH_2OH\)
Explanation: \(CH_3CH_2OCH_3\) is methoxyethane written in the reverse direction. In cleavage by \(HBr\), bromide ion attacks the less hindered methyl group when both possible attack sites are methyl or primary. This gives methyl bromide, \(CH_3Br\), and ethanol, \(CH_3CH_2OH\). With excess hydrogen halide, the alcohol can be further converted to an alkyl halide.
616. Which statement explains why excess \(HI\) can convert both alkyl groups of a simple dialkyl ether into alkyl iodides?
ⓐ. The first alcohol product reacts further with \(HI\).
ⓑ. \(HI\) first oxidises the ether to an aldehyde.
ⓒ. Water converts all alkyl iodides back to ethers.
ⓓ. The ether oxygen becomes a permanent metal oxide.
Correct Answer: The first alcohol product reacts further with \(HI\).
Explanation: Initial ether cleavage by \(HI\) often gives one alkyl iodide and one alcohol. If excess \(HI\) is present, the alcohol product can also react with \(HI\). Its \(OH\) group is replaced by iodine, forming another alkyl iodide and water. This is why a symmetrical ether can give two equivalents of alkyl iodide under excess \(HI\).
617. Which equation represents complete cleavage of diethyl ether by excess \(HI\)?
ⓐ. \[C_2H_5OC_2H_5 + I_2 \rightarrow C_2H_5I + C_2H_5OI\]
ⓑ. \[C_2H_5OC_2H_5 + HI \rightarrow C_2H_5OH + C_2H_5OH\]
ⓒ. \[C_2H_5OC_2H_5 + H_2 \rightarrow 2C_2H_5OH\]
ⓓ. \[C_2H_5OC_2H_5 + 2HI \rightarrow 2C_2H_5I + H_2O\]
Correct Answer: \[C_2H_5OC_2H_5 + 2HI \rightarrow 2C_2H_5I + H_2O\]
Explanation: \(\textbf{First equivalent of HI:}\)
Diethyl ether cleaves to give \(C_2H_5I\) and \(C_2H_5OH\).
\(\textbf{Second equivalent of HI:}\)
The ethanol product reacts further to form another \(C_2H_5I\).
\(\textbf{Water formation:}\)
The oxygen ultimately appears in water.
\(\textbf{Overall equation:}\)
\[
C_2H_5OC_2H_5 + 2HI \rightarrow 2C_2H_5I + H_2O
\]
\(\textbf{Final Answer:}\)
Complete cleavage gives two moles of ethyl iodide per mole of ether.
618. Which aryl alkyl ether gives phenol and methyl bromide on cleavage with \(HBr\)?
ⓐ. \(C_6H_5OCH_3\)
ⓑ. \(C_6H_5OC_2H_5\)
ⓒ. \(C_2H_5OC_2H_5\)
ⓓ. \(CH_3OCH_3\)
Correct Answer: \(C_6H_5OCH_3\)
Explanation: \(C_6H_5OCH_3\) is anisole. On cleavage with \(HBr\), the alkyl-oxygen bond is broken, producing methyl bromide from the methyl group. The aryl-oxygen bond remains intact, so the aromatic part becomes phenol. This product pattern is characteristic of aryl alkyl ether cleavage by hydrogen halides.
619. Which statement best compares cleavage of \(C_2H_5OC_2H_5\) and \(C_6H_5OC_2H_5\) by \(HI\)?
ⓐ. Both give only benzene as the organic product.
ⓑ. Diethyl ether gives ethyl iodide; phenetole gives phenol plus ethyl iodide.
ⓒ. Diethyl ether gives ethyl iodide; phenetole gives benzyl alcohol plus ethyl iodide.
ⓓ. Both resist \(HI\) completely under heating.
Correct Answer: Diethyl ether gives ethyl iodide; phenetole gives phenol plus ethyl iodide.
Explanation: Diethyl ether is a dialkyl ether and can be cleaved by excess \(HI\) to give ethyl iodide. Phenetole is an aryl alkyl ether, \(C_6H_5OC_2H_5\). In phenetole, the alkyl-oxygen bond breaks, giving ethyl iodide, while the aryl side becomes phenol. The aryl \(C-O\) bond is much harder to cleave in this reaction.
620. Which condition is most important for efficient cleavage of ethers by hydrogen halides?
ⓐ. Heating with concentrated \(HCl\) or \(HBr\)
ⓑ. Heating with concentrated \(HI\) or \(HBr\)
ⓒ. Cold dilute \(HNO_3\)
ⓓ. Neutral water only
Correct Answer: Heating with concentrated \(HI\) or \(HBr\)
Explanation: Ethers are not easily cleaved by weak or neutral reagents. Strong hydrogen halides such as \(HI\) and \(HBr\), especially on heating, protonate the ether oxygen and provide a good halide nucleophile. This combination allows \(C-O\) bond cleavage. \(NaOH\), water, or dilute nitrating conditions do not normally cleave simple ethers efficiently.
