CBSE Class 12 Chemistry Ethers Worksheet

Give reasons for the following:-

Question. The c-o-c bond angle in ethers is larger than the tetrahedral value.
Answer: This is due to the repulsive steric interactions between the two bulky alkyl groups attached to the central oxygen atom, which pushes the \( \text{C}-\text{O}-\text{C} \) bond angle outward (to approximately \( 111.7^\circ \)), making it slightly larger than the normal tetrahedral angle of \( 109.5^\circ \).

Question. Ethers are miscible in water just as alcohol’s are
Answer: Ethers have lone pairs of electrons on the electronegative oxygen atom, which allow them to form intermolecular hydrogen bonds with water molecules. Consequently, lower ethers exhibit water solubility comparable to that of alcohols of similar molecular mass.

Question. Boiling pts. Of ethers are comparable to those of ethers of comparable molecular mass.
Answer: Ethers do not have a hydrogen atom directly bonded to oxygen, so they cannot form intermolecular hydrogen bonds among themselves. They are held together only by weak dipole-dipole attractions and London dispersion forces, resulting in boiling points that are much lower than isomeric alcohols and comparable to alkanes of similar molecular mass.

Question. Ethers have a net dipole moment even if they are symmetrical in structure.
Answer: Ethers possess a bent (angular) \( \text{V} \)-shaped geometry around the \( sp^3 \)-hybridized oxygen atom due to the presence of two lone pairs. Because of this non-linear geometry, the individual bond dipoles of the two polar \( \text{C}-\text{O} \) bonds do not cancel each other out, resulting in a net dipole moment.

Question. For the preparation of t-butyl methyl ethers, methyl bromide and sodium t-butoxide is used. Reaction of sodium methoxide and t-butyl bromide can’t be used.
Answer: Williamson ether synthesis involves an \( \text{S}_\text{N}2 \) displacement. Methyl bromide is a primary halide that readily undergoes substitution with the bulky sodium \( t \)-butoxide. Conversely, \( t \)-butyl bromide is a tertiary alkyl halide, and sodium methoxide is both a nucleophile and a strong base; with a tertiary halide, elimination (\( \text{E}2 \)) predominates over substitution, producing an alkene (2-methylpropene) instead of an ether.

Question. Acid catalysed formation of ethers from alcohols is not appropriate for preparation of unsymmetrical ethers.
Answer: When an acid-catalysed dehydration is carried out using an equimolar mixture of two different alcohols, it produces a statistical mixture of three different ethers (two symmetrical ethers and one unsymmetrical ether). Because these ethers have very close boiling points, their separation is extremely difficult.

Question. Anisole is prepared by reacting sodium phenoxide with methyl bromide and not bromobenzene and sodium methoxide.
Answer: In bromobenzene, the lone pair on the bromine atom is in resonance with the benzene ring, imparting partial double-bond character to the \( \text{C}-\text{Br} \) bond and making the \( sp^2 \)-hybridized carbon resistant to nucleophilic substitution. On the other hand, methyl bromide is a primary alkyl halide and readily undergoes an \( \text{S}_\text{N}2 \) reaction with the phenoxide ion to yield anisole.

Question. When alkyl aryl ethers are cleaved with halogen acid, the yield is phenol and a molecule of alkyl halide and not phenyl halide and alcohol.
Answer: Due to resonance in alkyl aryl ethers, the bond between oxygen and the \( sp^2 \)-hybridized aryl carbon acquires partial double-bond character and is significantly stronger than the \( sp^3 \) \( \text{C}-\text{O} \) alkyl bond. Hence, nucleophilic attack by the halide ion occurs exclusively at the weaker alkyl carbon, cleaving the alkyl \( \text{C}-\text{O} \) bond to yield phenol and an alkyl halide.

Question. When ethers with \( 1^\circ \) and \( 2^\circ \) alkyl groups are cleaved with excess of HI under drastic conditions, alkyl iodide is formed with lower alkyl group.
Answer: The cleavage of ethers possessing primary and secondary alkyl groups proceeds predominantly by an \( \text{S}_\text{N}2 \) mechanism. In the initial step, the iodide ion (\( \text{I}^- \)) attacks the less sterically hindered carbon atom (the lower alkyl group), selectively forming the alkyl iodide from the smaller group.

Question. When tert-butyl methyl ether is hydrolysed with HI under drastic conditions, the products are methanol and t-butyl iodide and not butyl alcohol and methyl iodide.
Answer: The reaction proceeds via an \( \text{S}_\text{N}1 \) mechanism because cleavage of the protonated ether produces an exceptionally stable tertiary carbocation (the \( t \)-butyl cation). The iodide ion rapidly attacks this stable carbocation to form \( t \)-butyl iodide, leaving behind methanol.

Question. Anisole undergoes bromination in absence of halogen carrier also.
Answer: The methoxy group (\( -\text{OCH}_3 \)) attached to the benzene ring is strongly activating toward electrophilic aromatic substitution due to the \( +R \) (resonance) effect. It increases electron density significantly at the ortho and para positions, allowing electrophilic bromination to occur rapidly even without a Lewis acid catalyst (halogen carrier).

Question. Reactivity of HX towards ethers follows the Order \( \longrightarrow \text{HI} > \text{HBr} > \text{HCl} \)
Answer: The cleavage of ethers involves both acid protonation of the ether oxygen and nucleophilic attack by the halide ion. Since the bond dissociation energy decreases in the order \( \text{HCl} > \text{HBr} > \text{HI} \), the acid strength follows the order \( \text{HI} > \text{HBr} > \text{HCl} \), and the nucleophilicity of the halide ions follows \( \text{I}^- > \text{Br}^- > \text{Cl}^- \). Therefore, \( \text{HI} \) is the most reactive acid.

Question. Ethers are chemically inert.
Answer: Ethers do not contain active, acidic hydrogen atoms (unlike alcohols) or electrophilic carbonyl groups. The \( \text{C}-\text{O} \) single bonds are strong and stable, and the oxygen atom is sterically shielded by alkyl groups, making ethers resistant to attack by bases, reducing agents, and oxidizing agents under ordinary conditions.

Question. Preparation of ethers by acid dehydration of \( 2^\circ \) and \( 3^\circ \) alcohols is not a suitable method.
Answer: In secondary and tertiary alcohols, steric hindrance around the functional group and the high stability of secondary and tertiary carbocations cause acid-catalysed elimination (\( \text{E}1 \)) to predominate over bimolecular nucleophilic substitution, producing alkenes as the major products instead of ethers.

Question. Di-isopropyl ether and di-tert butyl ether cannot prepared in good yield by Williamson’s synthesis
Answer: The synthesis of these ethers would require the reaction between secondary or tertiary alkoxides and secondary or tertiary alkyl halides. Because alkoxides are strong bases and these halides are sterically hindered, elimination (\( \text{E}2 \)) takes precedence over nucleophilic substitution, yielding predominantly alkenes.

Question. Ethers can only be cleaved by acid not bases. :-
Answer: In nucleophilic cleavage by a base, the departing leaving group would be an alkoxide ion (\( \text{RO}^- \)), which is a strong base and a very poor leaving group. Acid protonates the ether oxygen to form an oxonium ion, transforming the leaving group into a neutral alcohol molecule (\( \text{ROH} \)), which is a good leaving group and easily displaced by nucleophiles.