7.5 Nitric Acid Nitrogen forms oxoacids such as H2N2O2 (hyponitrous acid), HNO2 (nitrous acid) and HNO3 (nitric acid). Amongst them HNO3 is the most important. Preparation In the laboratory, nitric acid is prepared by heating KNO3 or NaNO3 and concentrated H2SO4 in a glass retort. NaNO3 + H2SO4 → NaHSO4 + HNO3 On a large scale it is prepared mainly by Ostwald’s process. This method is based upon catalytic oxidation of NH3 by atmospheric oxygen. 4NH3(g) + 5O2(g) (from air) →(Pt/Rh gauge catalyst, 500K, 9 bar) 4NO(g) + 6H2O(g) Nitric oxide thus formed combines with oxygen giving NO2. 2NO(g) + O2(g) ⇌ 2NO2(g) Nitrogen dioxide so formed, dissolves in water to give HNO3. 3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) NO thus formed is recycled and the aqueous HNO3 can be concentrated by distillation upto ~ 68% by mass. Further concentration to 98% can be achieved by dehydration with concentrated H2SO4. Properties It is a colourless liquid (f.p. 231.4 K and b.p. 355.6 K). Laboratory grade nitric acid contains ~ 68% of the HNO3 by mass and has a specific gravity of 1.504. In the gaseous state, HNO3 exists as a planar molecule with the structure as shown. In aqueous solution, nitric acid behaves as a strong acid giving hydronium and nitrate ions. HNO3(aq) + H2O(l) → H3O+(aq) + NO3–(aq) Concentrated nitric acid is a strong oxidising agent and attacks most metals except noble metals such as gold and platinum. The products of oxidation depend upon the concentration of the acid, temperature and the nature of the material undergoing oxidation. 3Cu + 8 HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2O Zinc reacts with dilute nitric acid to give N2O and with concentrated acid to give NO2. 4Zn + 10HNO3(dilute) → 4 Zn (NO3)2 + 5H2O + N2O Zn + 4HNO3(conc.) → Zn (NO3)2 + 2H2O + 2NO2 Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid because of the formation of a passive film of oxide on the surface.
-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 179Ethers Structure
Ethers Structure, explained for NEET
Structure of Ethers — What NEET Expects You to Know
Ethers are organic compounds with the general formula R–O–R′, where R and R′ are alkyl or aryl groups bonded to an oxygen atom. The oxygen in an ether carries two lone pairs and adopts an approximately sp³ hybridisation geometry, similar to water. The C–O–C bond angle in dimethyl ether is about 111.7° — slightly larger than the tetrahedral angle of 109.5° — due to repulsion between the two bulky alkyl groups (NCERT Class 12 Chemistry, Chapter 7, Part 2, page 199).
Nomenclature — two systems you must know:
- Common naming: name the two alkyl/aryl groups alphabetically, followed by "ether." CH₃–O–C₂H₅ is ethyl methyl ether. When both groups are identical, use the prefix "di-": CH₃–O–CH₃ is dimethyl ether.
- IUPAC naming: ethers are named as alkoxyalkanes. The smaller alkyl group + oxygen becomes the alkoxy prefix; the larger chain is the parent alkane. CH₃–O–C₂H₅ is methoxyethane.
Classification:
- Simple (symmetrical) ethers: both R groups identical (e.g., diethyl ether, C₂H₅–O–C₂H₅).
- Mixed (unsymmetrical) ethers: R groups differ (e.g., ethyl methyl ether).
Physical properties relevant to NEET:
- Ethers have much lower boiling points than isomeric alcohols because ethers cannot form hydrogen bonds with themselves (no O–H bond). However, ethers CAN accept hydrogen bonds from water, making lower ethers (like dimethyl ether and diethyl ether) slightly soluble in water.
- Ethers are relatively unreactive — the C–O bond is difficult to cleave under ordinary conditions, which is why diethyl ether is widely used as a solvent.
Common confusion: students sometimes conflate the bond angle and hybridisation of ethers with that of epoxides (cyclic ethers with ~60° angles and significant ring strain). The C–O–C angle in open-chain ethers is near tetrahedral; in epoxides it is forced to ~60° by the three-membered ring.
Can you answer these Ethers Structure MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The general formula of an ether is:
Show answer and why every option is right or wrong
Answer: B. Ethers have the general structure R–O–R′, where R and R′ are alkyl or aryl groups bonded to an oxygen atom (NCERT Class 12 Chemistry, Chapter 7, Part 2, page 199).
Why A is wrong: A represents the general formula of an alcohol (hydroxyl group bonded to R), not an ether.
Why C is wrong: C represents the general formula of an aldehyde (–CHO functional group), not an ether.
Why D is wrong: D represents the general formula of a carboxylic acid (–COOH functional group), not an ether.
The hybridisation of oxygen in dimethyl ether is:
Show answer and why every option is right or wrong
Answer: A. Oxygen in ethers is sp³ hybridised — it forms two σ bonds (to two C atoms) and holds two lone pairs, giving four electron-pair domains and approximately tetrahedral geometry (NCERT Class 12 Chemistry, Chapter 7, Part 2, page 199).
Why B is wrong: B: sp² hybridisation gives a trigonal planar geometry (120°). Oxygen in ethers has four electron-pair domains (two bonds + two lone pairs), not three.
Why C is wrong: C: sp hybridisation gives a linear geometry (180°), which would require only two electron-pair domains. Oxygen in ethers has four.
Why D is wrong: D: sp³d hybridisation gives a trigonal bipyramidal geometry and requires d-orbitals, which is not applicable to second-period elements like oxygen.
The IUPAC name of CH₃–O–C₂H₅ is:
Show answer and why every option is right or wrong
Answer: C. In IUPAC nomenclature, the smaller alkyl group + oxygen forms the alkoxy prefix (methoxy from CH₃O–), and the larger alkyl chain is the parent alkane (ethane from C₂H₅). Hence methoxyethane (NCERT Class 12 Chemistry, Chapter 7, Part 2).
Why A is wrong: A is the common name, not the IUPAC name. The question specifically asks for the IUPAC name.
Why B is wrong: B reverses the naming convention — the smaller group should be the alkoxy prefix (methoxy), and the larger chain should be the parent (ethane), not the other way around.
Why D is wrong: D is a common name written in non-alphabetical order. In common naming, groups are listed alphabetically (ethyl methyl ether), and this is still not the IUPAC name.
Which of the following is a symmetrical ether?
Show answer and why every option is right or wrong
Answer: A. A symmetrical (simple) ether has identical groups on both sides of oxygen. C₂H₅–O–C₂H₅ (diethyl ether) has two ethyl groups, making it symmetrical.
Why B is wrong: B has a methyl group on one side and an ethyl group on the other — this is a mixed (unsymmetrical) ether.
Why C is wrong: C has a methyl group and a phenyl group — different groups make this a mixed ether (methyl phenyl ether / anisole).
Why D is wrong: D has an ethyl group and a phenyl group — different groups make this a mixed ether.
Diethyl ether has a much lower boiling point than 1-butanol, despite having similar molecular masses. The primary reason is:
Show answer and why every option is right or wrong
Answer: D. Ethers lack an O–H group and therefore cannot form hydrogen bonds with other ether molecules. 1-Butanol has an –OH group enabling strong intermolecular H-bonding, which raises its boiling point significantly (NCERT Class 12 Chemistry, Chapter 7, Part 2).
Why A is wrong: A is factually wrong — diethyl ether (MW ~74) and 1-butanol (MW ~74) have very similar molecular masses. The boiling point difference is not due to mass.
Why B is wrong: B is incorrect — 1-butanol is polar (it has an –OH group), which is precisely why it has the higher boiling point.
Why C is wrong: C is incorrect — London dispersion forces depend on molecular surface area and are similar for these two isomeric compounds. The key difference is the presence of H-bonding in the alcohol.
Lower ethers like diethyl ether are slightly soluble in water because:
Show answer and why every option is right or wrong
Answer: C. The oxygen atom in ethers carries two lone pairs that can accept hydrogen bonds from water molecules (water acts as the H-bond donor). This interaction provides enough stabilisation for lower ethers to be slightly water-soluble (NCERT Class 12 Chemistry, Chapter 7, Part 2).
Why A is wrong: A is incorrect — ethers have no O–H or N–H bond, so they cannot donate hydrogen bonds. They can only accept them via the lone pairs on oxygen.
Why B is wrong: B is incorrect — ethers are covalent compounds, not ionic. Their slight water solubility comes from hydrogen-bond acceptance, not ion-dipole interactions.
Why D is wrong: D is factually wrong — diethyl ether boils at 34.6°C, far below water's 100°C. Boiling point comparison does not explain water solubility.
The C–O–C bond angle in dimethyl ether (~111.7°) is slightly larger than the ideal tetrahedral angle (109.5°). The most likely reason is:
Show answer and why every option is right or wrong
Answer: D. Although oxygen is sp³ hybridised, the two bulky methyl groups attached to it create steric repulsion that pushes the C–O–C angle slightly beyond the ideal tetrahedral angle of 109.5° (NCERT Class 12 Chemistry, Chapter 7, Part 2, page 199).
Why A is wrong: A is incorrect — oxygen in ethers retains two lone pairs. The lone pairs are present and contribute to the overall electron geometry.
Why B is wrong: B is incorrect — the C–O bond in ethers is a polar covalent bond, not ionic. Ionicity would imply complete electron transfer, which does not occur here.
Why C is wrong: C is incorrect — if oxygen were sp² hybridised, the bond angle would be closer to 120°, not ~112°. The oxygen is sp³ hybridised with a slight deviation due to steric effects.
A student claims that the C–O–C bond angle in ethylene oxide (an epoxide) is similar to that in dimethyl ether. Which statement correctly identifies the error?
Show answer and why every option is right or wrong
Answer: B. Ethylene oxide is a three-membered ring (two C + one O), forcing the C–O–C angle to approximately 60°. This is dramatically different from the ~111.7° in open-chain dimethyl ether and introduces significant ring strain. The student's claim is wrong because ring geometry overrides the preference for tetrahedral angles.
Why A is wrong: A incorrectly validates the student's claim. While oxygen does tend toward sp³ hybridisation in both, the three-membered ring in ethylene oxide forces the bond angle to ~60°, making the angles very different.
Why C is wrong: C reverses the structures — dimethyl ether is the open-chain compound and ethylene oxide is the cyclic (epoxide) compound.
Why D is wrong: D is incorrect — oxygen in ethylene oxide retains two lone pairs. The constrained bond angle is due to ring geometry, not absence of lone pairs.
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How do you solve a Ethers Structure question? A worked example
- 1
Given
• Molecular formula: C₄H₁₀O• The compound is an ether (confirmed by absence of reaction with Na)• Both alkyl groups on oxygen are identical (symmetrical ether)
- 2
Required
• Structural formula• Common name• IUPAC name• Approximate C–O–C bond angle
- 3
Concept
An ether with formula R–O–R (symmetrical) must partition the non-oxygen atoms equally between the two R groups. The oxygen in an open-chain ether is sp³ hybridised, giving a bond angle near the tetrahedral value.
- 4
Formula / Principle
• General formula of ether: R–O–R′• For C₄H₁₀O with identical groups: C₄H₁₀O = R–O–R, so 2R + O = C₄H₁₀O• Each R must account for C₂H₅ (ethyl group)
- 5
Substitution
• C₄H₁₀O → C₂H₅–O–C₂H₅• Common name: the two alkyl groups are both ethyl → diethyl ether• IUPAC name: smaller group = ethyl → ethoxy prefix; larger chain = ethane → ethoxyethane
- 6
Calculation
No arithmetic calculation is needed here. The structural deduction is:• Total carbons = 4, split equally → 2 carbons per alkyl group → C₂H₅ each• Hydrogen count check: 2 × (C₂H₅) + O = C₄H₁₀O ✓
- 7
Final answer
• Structural formula: C₂H₅–O–C₂H₅• Common name: Diethyl ether• IUPAC name: Ethoxyethane• C–O–C bond angle: approximately 112° (slightly greater than tetrahedral 109.5° due to steric repulsion between the two ethyl groups; oxygen is sp³ hybridised)
- 8
Common trap
A common confusion is writing "ethoxymethane" or choosing the larger group as the alkoxy prefix. IUPAC convention: the smaller group becomes the alkoxy prefix, the larger chain is the parent alkane. When both groups are identical, either can be the prefix — but the systematic name is ethoxyethane, not "diethyl ether" (which is the common name only).
- 9
Similar NEET-style question
An ether with molecular formula C₅H₁₂O gives two different alkyl groups on cleavage with HI. One fragment is iodomethane and the other is butan-2-ol. Write the structural formula of the ether and give its IUPAC name. (Answer: CH₃–O–CH(CH₃)CH₂CH₃, 2-methoxybutane — check C₅H₁₂O against the structure before naming. Three things fix the fragments: HI gives IODIDES, never bromides; in a methyl/secondary ether I⁻ attacks the LESS hindered methyl carbon by SN2, so the methyl leaves as CH₃I; and the secondary side keeps the oxygen and emerges as the alcohol, butan-2-ol.)
*(Answer: CH₃–O–CH(CH₃)₂, 2-methoxypropane)*
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What to remember before solving Ethers Structure questions
More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.
Ethers Structure questions from past NEET papers
1 question from NEET 2020. Answers verified against NTA official keys.
Anisole on cleavage with HI gives :
All 23 past-paper questions from Alcohols, Phenols, Ethers and Carbonyl Compounds →
Sources
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