Ethers
Class 12 Chemistry Notes • NEB Syllabus 2083
Syllabus, 17 solved multiple choice questions, the full textbook exercise with solutions and diagrams, handwritten and typed PDF notes and frequently asked questions on Williamson synthesis, reactions of ethoxyethane and uses of ethers.
1. Syllabus
Unit 12: Ethers
These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Practise the multiple choice questions first, then work through the solved numerical and subjective questions, and finish with the PDF notes for quick revision.
2. Multiple Choice Questions
Question 1. Which of the following could be the product Y? \(\mathrm{CH_3-O-CH_3}\) + excess HI gives Y
- a) Methanol
- b) Methanol and Iodomethane
- c) Methanal and iodomethane
- d) Iodomethane
Correct Answer: d) Iodomethane
Reason: With excess HI both C-O bonds are broken. The first mole gives iodomethane and methanol, \(\mathrm{CH_3OCH_3+HI\longrightarrow CH_3I+CH_3OH}\), and the methanol formed reacts with a second mole of HI, \(\mathrm{CH_3OH+HI\longrightarrow CH_3I+H_2O}\). Overall \(\mathrm{CH_3OCH_3+2HI\longrightarrow2CH_3I+H_2O}\). Option b is the product with only one mole of HI.
Question 2. Which of the following is correct about the information given in the following table?
| Reactant(s) | Condition(s) | Product(s) |
|---|---|---|
| (w) | Dry ether/heat | \(\mathrm{CH_3-CH_2-O-CH_2-CH_3}\) |
| Ethanol | (x) | \(\mathrm{CH_3-CH_2-O-CH_2-CH_3}\) |
| \(\mathrm{CH_3-CH_2-O-CH_2-CH_3}\) | Excess HI | (y) |
| (z) | Sunlight / Chlorine | \(\mathrm{CCl_3-CCl_2-O-CCl_2-CCl_3}\) + HCl |
- a) w – sodium methoxide and chloroethane, x – conc. \(\mathrm{H_2SO_4}\)/heat, y – iodoethane, z – ethoxyethane
- b) w – sodium ethoxide and chloromethane, x – conc. \(\mathrm{H_2SO_4}\)/heat, y – iodoethane, z – ethoxyethane
- c) w – sodium ethoxide and chloroethane, x – conc. \(\mathrm{H_2SO_4}\), y – iodoethane, z – ethoxyethane
- d) w – sodium ethoxide and chloroethane, x – conc. \(\mathrm{H_2SO_4}\)/heat, y – iodoethane, z – ethoxyethane
Correct Answer: d) w – sodium ethoxide and chloroethane, x – conc. \(\mathrm{H_2SO_4}\)/heat, y – iodoethane, z – ethoxyethane
Reason: w: Williamson synthesis of ethoxyethane needs \(\mathrm{C_2H_5ONa+C_2H_5Cl}\) (sodium methoxide or chloromethane would give methoxyethane). x: ethanol gives ethoxyethane with conc. \(\mathrm{H_2SO_4}\) on heating to 413 K. y: ethoxyethane with excess HI gives iodoethane. z: perchlorodiethyl ether is formed from ethoxyethane and chlorine in sunlight. Option c leaves out the heat that the reaction needs.
Question 3. What is Z in the following reaction? \(\mathrm{CH_3CH_2OCH_2CH_3}\) + conc. HCl (cold) gives Z
- a) Ethanol + ethyl chloride
- b) Oxonium chloride salt of ethoxyethane
- c) Ethanol only
- d) Chloride salt of ethoxyethane
Correct Answer: b) Oxonium chloride salt of ethoxyethane
Reason: The oxygen of an ether has two lone pairs, so the ether is a Lewis base. In the cold it takes a proton from conc. HCl to form diethyloxonium chloride, \(\mathrm{[(C_2H_5)_2OH]^+Cl^-}\), and no C-O bond is broken. Cleavage into ethanol and ethyl chloride needs hot, concentrated acid (HI is the most effective).
Question 4. Which of the following is the most suitable requirement for the dehydration of ethanol to form an ether?
- a) Sulphuric acid at 413 K
- b) Sulphuric acid at 443 K
- c) Sodium hydroxide at 413 K
- d) Sodium hydroxide at 443 K
Correct Answer: a) Sulphuric acid at 413 K
Reason: Ethanol (in excess) with conc. \(\mathrm{H_2SO_4}\) at 413 K (140 °C) loses water between two molecules and forms ethoxyethane. At 443 K (170 °C) the acid removes water from a single molecule and ethene is formed. NaOH is a base and does not dehydrate alcohols.
Question 5. Which of the following is not favourable for the proper dehydration of alcohol to form ether?
- a) Good temperature control during reaction
- b) Excess of alcohol
- c) Presence of protic acids
- d) Presence of bulky alkyl groups in the alcohol
Correct Answer: d) Presence of bulky alkyl groups in the alcohol
Reason: Ether formation needs a second alcohol molecule to attack the protonated alcohol (an \(\mathrm{S_N2}\) step). Bulky alkyl groups (secondary and tertiary alcohols) block this attack and form stable carbocations, so the alcohol loses H⁺ and gives an alkene instead. Temperature control, excess alcohol and a protic acid all favour the ether.
Question 6. What is the major product formed when ethanol is dehydrated with concentrated \(\mathrm{H_2SO_4}\) at 413 K?
- a) Ethene
- b) Methoxymethane
- c) Methoxyethane
- d) Ethoxyethane
Correct Answer: d) Ethoxyethane
Reason: \(\mathrm{2CH_3CH_2OH\xrightarrow{conc.\ H_2SO_4,\ 413\ K}CH_3CH_2OCH_2CH_3+H_2O}\). Methoxymethane and methoxyethane would need methanol, and ethene is formed at 443 K.
Question 7. What would be the product when sodium methoxide is heated with bromoethane?
- a) Methoxymethane
- b) Methoxyethane
- c) Ethoxyethane
- d) Diethyl ether
Correct Answer: b) Methoxyethane
Reason: This is Williamson synthesis: the methoxide ion attacks the carbon of bromoethane and displaces bromide (SN2). \(\mathrm{CH_3ONa+CH_3CH_2Br\longrightarrow CH_3OCH_2CH_3+NaBr}\). The ether has a methyl group (from methoxide) and an ethyl group (from bromoethane).
Question 8. What would be the product of the reaction between tert-butyl chloride and sodium ethoxide?
- a) tert-Butyl ethyl ether
- b) tert-Butyl methyl ether
- c) 2-Methylprop-1-ene
- d) Butene
Correct Answer: c) 2-Methylprop-1-ene
Reason: A tertiary alkyl halide cannot be attacked from the back (steric hindrance), and ethoxide is a strong base, so it removes an H⁺ from a neighbouring methyl group: \(\mathrm{(CH_3)_3CCl+C_2H_5ONa\longrightarrow(CH_3)_2C=CH_2+C_2H_5OH+NaCl}\). For tert-butyl ethyl ether the right choice is sodium tert-butoxide with chloroethane.
Question 9. Which of the following is correct about the boiling point of ethers and alcohols of comparable molecular mass?
- a) Ethers have lower boiling points than alcohols
- b) Ethers have similar boiling points to alcohols
- c) Ethers have slightly higher boiling points than alcohols
- d) Ethers have much higher boiling points than alcohols
Correct Answer: a) Ethers have lower boiling points than alcohols
Reason: Alcohol molecules are linked by hydrogen bonds, but ether molecules cannot form them with each other. Examples: methoxymethane (M = 46) boils at −24 °C and ethanol (M = 46) at 78 °C; ethoxyethane (M = 74) boils at 34.6 °C and butan-1-ol (M = 74) at 117.7 °C.
Question 10. Which of the following compounds is a functional isomer of \(\mathrm{CH_3-CH_2-O-CH_2-CH_3}\)?
- a) \(\mathrm{CH_3-CH_2-O-CH_2-CH_3}\)
- b) \(\mathrm{CH_3-CH_2-CH_2-CHO}\)
- c) \(\mathrm{CH_3-CH_2-CH_2-CH_2OH}\)
- d) \(\mathrm{CH_3-CH_2-CO-CH_3}\)
Correct Answer: c) \(\mathrm{CH_3-CH_2-CH_2-CH_2OH}\)
Reason: Ethoxyethane is \(\mathrm{C_4H_{10}O}\). Butan-1-ol has the same formula but the -OH functional group, so it is a functional isomer. Option a is the same compound, and butanal and butanone (b, d) are \(\mathrm{C_4H_8O}\), a different formula.
Question 11. Which of the following is the least reactive functional group?
- a) Alcohols
- b) Ethers
- c) Aldehydes
- d) Ketones
Correct Answer: b) Ethers
Reason: An ether has no O-H bond and no C=O bond, so it is not attacked by bases, reducing agents, oxidising agents or active metals such as sodium. It is cleaved only by hot concentrated HI or HBr, which is why ethers are used as inert solvents.
Question 12. One molecule of dialkyl ether produces how many molecules of alkyl halides with an excess of halogen acid?
- a) 1
- b) 2
- c) 3
- d) 4
Correct Answer: b) 2
Reason: \(\mathrm{R-O-R^\prime+2HX\longrightarrow RX+R^\prime X+H_2O}\). The first HX breaks the ether into an alkyl halide and an alcohol, and the alcohol is converted into a second alkyl halide by the excess acid.
Question 13. Which of the following is the major product of Friedel-Crafts acylation of anisole with ethanoyl chloride?
Figure: Friedel-Crafts acylation of anisole. The –OCH₃ group directs the incoming acyl group to the para position, which gives 4-methoxyacetophenone as the major product
- a) 2-Methoxytoluene
- b) 4-Methoxytoluene
- c) 2-Methoxyacetophenone
- d) 4-Methoxyacetophenone
Correct Answer: d) 4-Methoxyacetophenone
Reason: \(\mathrm{C_6H_5OCH_3+CH_3COCl\xrightarrow{anhyd.\ AlCl_3}CH_3OC_6H_4COCH_3+HCl}\). Acylation puts a ketone group (-COCH₃) on the ring, so the toluene derivatives (a, b) are wrong. The -OCH₃ group is activating and ortho/para directing, and the para isomer is major because the ortho position is crowded.
Question 14. Which of the following compounds is used to detect the peroxy bond in ethers?
- a) \(\mathrm{Na_2SO_4}\)
- b) \(\mathrm{NiSO_4}\)
- c) \(\mathrm{FeCl_3}\)
- d) \(\mathrm{FeSO_4}\)
Correct Answer: d) \(\mathrm{FeSO_4}\)
Reason: A peroxide oxidises iron(II) to iron(III): \(\mathrm{2Fe^{2+}+ROOH+2H^+\longrightarrow2Fe^{3+}+ROH+H_2O}\). On adding potassium thiocyanate (KCNS) the \(\mathrm{Fe^{3+}}\) gives a blood-red \(\mathrm{Fe(CNS)_3}\). The other salts cannot be oxidised.
Question 15. What would be the product when sodium ethoxide is heated with chloroethane in the presence of dry ether?
- a) Ethyl ethanoate
- b) Methoxy ethane
- c) Ethoxyethane
- d) Butyl chloride
Correct Answer: c) Ethoxyethane
Reason: \(\mathrm{C_2H_5ONa+C_2H_5Cl\xrightarrow{dry\ ether,\ \Delta}C_2H_5OC_2H_5+NaCl}\). Both groups on oxygen are ethyl groups.
Question 16. What would be the product when iodoethane is heated with sodium ethoxide in the presence of dry ether?
- a) Ethanol
- b) Butanol
- c) Ethoxyethane
- d) Methoxyethane
Correct Answer: c) Ethoxyethane
Reason: \(\mathrm{C_2H_5ONa+C_2H_5I\xrightarrow{dry\ ether,\ \Delta}C_2H_5OC_2H_5+NaI}\). Iodide is a very good leaving group and the halide is primary, so the SN2 substitution takes place.
Question 17. 4-Nitroanisole is obtained as the major product when anisole reacts with which of the following?
Figure: Nitration of anisole with the nitrating mixture gives 4-nitroanisole as the major product
- a) Concentrated \(\mathrm{H_2SO_4}\)
- b) Concentrated \(\mathrm{HNO_3}\)
- c) Mixture of concentrated \(\mathrm{H_2SO_4}\) and \(\mathrm{HNO_3}\)
- d) Mixture of dilute \(\mathrm{H_2SO_4}\) and \(\mathrm{HNO_3}\)
Correct Answer: c) Mixture of concentrated \(\mathrm{H_2SO_4}\) and \(\mathrm{HNO_3}\)
Reason: The nitrating mixture produces the electrophile \(\mathrm{NO_2^+}\): \(\mathrm{HNO_3+2H_2SO_4\rightleftharpoons NO_2^++H_3O^++2HSO_4^-}\). The -OCH₃ group directs it to the para position (major) and ortho position (minor). Dilute acids do not form enough \(\mathrm{NO_2^+}\), and conc. \(\mathrm{H_2SO_4}\) alone only sulphonates or protonates.
3. Numerical and Subjective Questions
Short Answer Questions
i. Methoxypropane
Methoxypropane, \(\mathrm{CH_3OCH_2CH_2CH_3}\), is prepared by Williamson’s synthesis from sodium methoxide and a primary propyl halide (1-bromopropane) in dry ether:
Sodium propoxide with iodomethane also works. Isopropyl bromide is not used, because a secondary halide gives propene.
ii. Ethoxyethane by Williamson’s method
Sodium ethoxide is heated with chloroethane (or bromoethane or iodoethane) in dry ether. The ethoxide ion replaces the halide by an SN2 reaction:
Requirements: sodium alkoxide, a primary alkyl halide, dry ether and heat.
iii. Methoxybenzene
Phenol is first converted into sodium phenoxide with NaOH, and the phenoxide is then heated with iodomethane (or bromomethane) to give methoxybenzene (anisole), an aromatic ether:
Figure: Phenol is converted into sodium phenoxide, which reacts with iodomethane (an SN2 reaction) to give methoxybenzene, an aromatic ether
iv. Iodoethane from diethyl ether
Diethyl ether is heated with excess concentrated hydroiodic acid. Both C-O bonds break; the ethanol formed first is converted into the second molecule of iodoethane:
i. Ethanol is heated with concentrated sulphuric acid in the presence of about 140 °C?
Two molecules of ethanol lose one molecule of water (intermolecular dehydration) and ethoxyethane (diethyl ether) is formed. Ethanol must be in excess and the temperature must stay near 413 K; at 443 K ethene is formed.
ii. Ethoxyethane is treated with excess HI?
The ether is cleaved and only iodoethane is obtained (the alcohol formed first reacts with more HI):
iii. Ethoxyethane is exposed to air?
On standing in air and light, ethoxyethane slowly takes up oxygen and forms an ether peroxide, 1-ethoxyethyl hydroperoxide. Peroxides are explosive on heating or distilling. Their presence is shown by the blood-red colour of \(\mathrm{Fe(CNS)_3}\) with acidified \(\mathrm{FeSO_4}\) and KCNS, and they are prevented by storing the ether over iron or copper wire.
iv. Ethoxyethane is treated with chlorine in the presence of sunlight?
In bright sunlight, chlorine replaces all ten hydrogen atoms (free-radical substitution) and perchlorodiethyl ether is formed with hydrogen chloride:
In the dark only the hydrogens next to oxygen are replaced, giving alpha,alpha’-dichlorodiethyl ether: \(\mathrm{CH_3CH_2OCH_2CH_3+2Cl_2\xrightarrow{dark}CH_3CHClOCHClCH_3+2HCl}\).
v. Sodium ethoxide is heated with chloroethane in the presence of dry ether?
Williamson synthesis takes place; ethoxyethane and sodium chloride are formed:
Reasoning (working backwards). The only symmetrical ether with 4 carbon atoms is ethoxyethane, so S is ethoxyethane. It is made from sodium ethoxide and an ethyl halide, so Q, made with HI, is iodoethane. Iodoethane is obtained from an ether by excess HI, and the ether obtained from P at 140 °C with conc. \(\mathrm{H_2SO_4}\) is a symmetrical ether, so R is ethoxyethane and P is ethanol. (P → R is intermolecular dehydration of ethanol.)
| Name (IUPAC) | Structure | |
|---|---|---|
| P | Ethanol | \(\mathrm{CH_3CH_2OH}\) |
| R | Ethoxyethane | \(\mathrm{CH_3CH_2OCH_2CH_3}\) |
| Q | Iodoethane | \(\mathrm{CH_3CH_2I}\) |
| S | Ethoxyethane | \(\mathrm{CH_3CH_2OCH_2CH_3}\) |
R and S are the same compound, made in two different ways (dehydration of an alcohol and Williamson synthesis).
Reasoning. The second member of the alcohol series is ethanol, so P = \(\mathrm{CH_3CH_2OH}\). It fits every clue: it releases hydrogen with sodium (Q = sodium ethoxide), it contains the \(\mathrm{CH_3CH(OH)-}\) group and so gives yellow iodoform with iodine and NaOH (R), and at 140 °C it forms an ether with the formula \(\mathrm{C_4H_{10}O}\), which is ethoxyethane (S). Conc. HI converts S into iodoethane (T), and Q + T give S back by Williamson synthesis.
| Name | Structure | |
|---|---|---|
| P | Ethanol | \(\mathrm{CH_3CH_2OH}\) |
| Q | Sodium ethoxide | \(\mathrm{CH_3CH_2ONa}\) |
| R | Iodoform (triiodomethane) | \(\mathrm{CHI_3}\) |
| S | Ethoxyethane | \(\mathrm{CH_3CH_2OCH_2CH_3}\) |
| T | Iodoethane | \(\mathrm{CH_3CH_2I}\) |
i. Ethers should be stored in bottles with iron pieces.
Ethers slowly react with oxygen of the air to form explosive peroxides. Iron (or copper) has a greater tendency to combine with oxygen than the ether has, so the metal reacts with the peroxide and with traces of oxygen as they form, and the peroxide cannot build up in the bottle.
ii. It is dangerous to boil old samples of ether.
An old sample has formed ether peroxides. The peroxides are less volatile than the ether, so when the ether is boiled or distilled they stay behind and become concentrated in the flask. On further heating the concentrated peroxide decomposes violently and explodes. A sample is therefore tested with acidified \(\mathrm{FeSO_4}\) and KCNS (blood-red colour) before it is heated.
iii. Ethers are functional isomers of monohydric alcohols.
Both have the same general formula \(\mathrm{C}_n\mathrm{H}_{2n+2}\mathrm{O}\) but different functional groups: -OH in an alcohol and -O- in an ether.
| Molecular formula | Alcohol | Ether |
|---|---|---|
| \(\mathrm{C_2H_6O}\) | Ethanol, \(\mathrm{CH_3CH_2OH}\) | Methoxymethane, \(\mathrm{CH_3OCH_3}\) |
| \(\mathrm{C_4H_{10}O}\) | Butan-1-ol, \(\mathrm{CH_3CH_2CH_2CH_2OH}\) | Ethoxyethane, \(\mathrm{CH_3CH_2OCH_2CH_3}\) |
iv. Complete the following reactions.
a. Methoxybenzene + HI. The O-CH₃ bond breaks, giving phenol and iodomethane:
Iodobenzene and methanol are not formed, because the C₆H₅-O bond has partial double bond character by resonance and is too strong to break.
Figure: Cleavage of methoxybenzene with HI gives phenol and iodomethane; iodobenzene is never formed
b. Ethoxyethane + Cl₂ in sunlight.
The product is perchlorodiethyl ether.
| Reactant(s) | Condition(s) | Product(s) |
|---|---|---|
| (A) \(\mathrm{CH_3ONa+CH_3CH_2Cl}\) | Dry ether/heat | \(\mathrm{CH_3-CH_2-O-CH_3}\) |
| Ethanol | (B) conc. \(\mathrm{H_2SO_4}\), 413 K (140 °C), ethanol in excess | \(\mathrm{CH_3-CH_2-O-CH_2-CH_3}\) |
| \(\mathrm{CH_3-CH_2-O-CH_3}\) | Excess HI | (C) \(\mathrm{CH_3CH_2I}\) + (D) \(\mathrm{CH_3I}\) + \(\mathrm{H_2O}\) |
| (E) \(\mathrm{CH_3CH_2OCH_2CH_3}\) | Sunlight / chlorine | \(\mathrm{CCl_3-CCl_2-O-CCl_2-CCl_3}\) + HCl |
(A) Williamson synthesis of methoxyethane: \(\mathrm{CH_3ONa+CH_3CH_2Cl\longrightarrow CH_3OCH_2CH_3+NaCl}\). Sodium ethoxide with chloromethane (or iodomethane) is also correct, because both halides are primary.
(B) \(\mathrm{2CH_3CH_2OH\xrightarrow{conc.\ H_2SO_4,\ 413\ K}CH_3CH_2OCH_2CH_3+H_2O}\)
(C + D) Excess HI breaks both C-O bonds: \(\mathrm{CH_3OCH_2CH_3+2HI\longrightarrow CH_3I+CH_3CH_2I+H_2O}\).
(E) Ethoxyethane has ten hydrogens, all replaced by chlorine: \(\mathrm{CH_3CH_2OCH_2CH_3+10Cl_2\longrightarrow CCl_3CCl_2OCCl_2CCl_3+10HCl}\).
Long Answer Questions
Step 1: what each clue shows.
Step 2: identify. Methanol has one carbon and cannot give the iodoform test. Ethanol has two carbons and gives the test (the only primary alcohol that does), so P is ethanol. S is ethoxyethane, and excess HI changes it to iodoethane (T). Check: sodium ethoxide (Q) + iodoethane (T) gives ethoxyethane (S) by Williamson synthesis, as the question states.
| IUPAC name | Common name | Formula | |
|---|---|---|---|
| P | Ethanol | Ethyl alcohol | \(\mathrm{CH_3CH_2OH}\) |
| Q | Sodium ethoxide | Sodium ethoxide | \(\mathrm{CH_3CH_2ONa}\) |
| R | Triiodomethane | Iodoform | \(\mathrm{CHI_3}\) |
| S | Ethoxyethane | Diethyl ether | \(\mathrm{CH_3CH_2OCH_2CH_3}\) |
| T | Iodoethane | Ethyl iodide | \(\mathrm{CH_3CH_2I}\) |
Step 3: reactions.
Figure: Reaction scheme for P, Q, R, S and T. Ethanol is the second member of the alcohol series
1. Ethoxyethane with oxygen. In air and light, ethoxyethane is slowly oxidised at the carbon next to oxygen and forms an ether peroxide (1-ethoxyethyl hydroperoxide), which is explosive:
2. Methoxymethane with excess HI. Both C-O bonds are broken, so Y is iodomethane:
3. Y with sodium phenoxide. The phenoxide ion attacks the carbon of iodomethane (SN2, Williamson synthesis) and Z is methoxybenzene (anisole), an aromatic ether:
| Name | Formula | |
|---|---|---|
| Y | Iodomethane | \(\mathrm{CH_3I}\) |
| Z | Methoxybenzene (anisole) | \(\mathrm{C_6H_5OCH_3}\) |
i. Molecular formula and IUPAC name
| Element | Mass % | Atomic mass | Moles (% ÷ mass) | Ratio (÷ 1.375) |
|---|---|---|---|---|
| C | 65 | 12 | 5.417 | 3.94 |
| H | 13 | 1 | 13.00 | 9.45 |
| O | 22 | 16 | 1.375 | 1.00 |
The ratio is C : H : O = 3.94 : 9.45 : 1, so there are 4 carbon atoms and 1 oxygen atom. The percentages are rounded, so the hydrogen figure is not exactly a whole number. A saturated ether has the formula \(\mathrm{C}_n\mathrm{H}_{2n+2}\mathrm{O}\), so for n = 4 there are 10 hydrogen atoms, and the formula is \(\mathrm{C_4H_{10}O}\).
Check: M(\(\mathrm{C_4H_{10}O}\)) = 4(12) + 10(1) + 16 = 74. Percentages: C = 48/74 = 64.9% (65), H = 10/74 = 13.5% (about 13), O = 16/74 = 21.6% (22). The data agree.
The ether is symmetric, so the 4 carbons are divided into two ethyl groups: \(\mathrm{CH_3CH_2OCH_2CH_3}\), IUPAC name ethoxyethane (diethyl ether).
Figure: Mass percentage of carbon, hydrogen and oxygen in the symmetric ether
ii. Cold concentrated hydrochloric acid
The oxygen atom of ethoxyethane has lone pairs, so the ether acts as a base and accepts a proton from cold conc. HCl. An oxonium salt, diethyloxonium chloride, is formed and the ether is not cleaved:
Adding water decomposes the salt and gives back the ether.
iii. Storage of the ether for 80 years
Ether must be protected from oxygen, light and heat, because these form explosive peroxides in a bottle left for a long time:
Even with these precautions, no storage can make ether completely safe for such a long time, so in practice old samples are tested regularly and discarded safely if peroxide appears.
4. View PDF Notes
i. Handwritten Notes
ii. Typed PDF Notes
5. Frequently Asked Questions
What are ethers? What is the general formula of ethers? (Ethers class 12 notes)
How are ethers named? Give the common and IUPAC names of common ethers.
How are ethers classified? What types of isomerism do ethers show?
What is Williamson’s synthesis? What are its requirements?
How is ethoxyethane (diethyl ether) prepared?
What are the physical properties of ethers?
Why is the boiling point of ether lower than that of an alcohol of the same molecular mass?
What happens when ethoxyethane is treated with hydrogen iodide?
What are oxonium salts of ethers?
Why are ethers stored with iron pieces? Why is it dangerous to boil old ether?
What is the action of chlorine on ethoxyethane?
What happens when anisole undergoes electrophilic substitution?
What are the uses of ethers?
Are organometallic compounds, carboxylic acids and phenols part of Unit 12 Ethers? (Organometallic compounds, Carboxylic acid and Phenol Class 12 notes)
Where can I get Ethers class 12 notes PDF, handwritten notes and free download?
Where can I get Class 12 Chemistry notes of NEB (handwritten, PDF, Nepal)?
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