Paper II — Q8
(a) (i) Estimate the expected splitting (coupling constant J in Hz) for the lettered protons in the ¹H NMR spectrum of the…
Estimate the expected splitting (coupling constant J in Hz) for the lettered protons in the ¹H NMR spectrum of the following compounds: (1) A, (2) B, (3) C 5 marks
Compare the chemical shifts of the labelled protons Hᵃ and Hᵇ in the ¹H NMR spectrum of the following compounds and justify your answer. 10 marks
Count the number of peaks observed in the ¹H NMR spectrum of the following compounds. Justify your answer: (1), (2) 5 marks
A halogenated ester shows M⁺ peak at m/z 166 (10%) and M+2 peak at m/z 168 (9·8%) in mass spectrum. ¹H NMR spectrum of this compound shows two triplets and a singlet at δ 2·9, 3·6 and 3·8 ppm, respectively in the intensity ratio 1:1:1·5. Deduce the structure of the compound. Justify your answer. 10 marks
Two isomeric alkenes with same molecular formula C₆H₁₂ show strong peaks at m/z 42 and 56 in the mass spectrum. Propose fragmentation pattern for both the peaks. 5 marks
(1) Phthalic acid diethyl ester shows a characteristic peak at m/z 149 in the mass spectrum. Account for the observance of this peak by fragmentation pattern. (2) The mass spectrum of ethylbenzene shows a characteristic peak at m/z 91 while n-propylbenzene shows strong peak at m/z 92. Explain with the help of fragmentation pattern. 10 marks
An unknown organic compound with molecular formula C₄H₅NO₂ displays a band at 2250 cm⁻¹ and a strong band at 1740 cm⁻¹ in the IR spectrum. The compound shows only two signals in 3:2 ratio in the ¹H NMR spectrum. Find out the structure of the compound. Justify your answer. 10 marks
हिंदी में प्रश्न पढ़ें
(क) (i) निम्नलिखित यौगिकों के ¹H NMR स्पेक्ट्रम में अक्षरित प्रोटोनों के लिए अपेक्षित विपाटन (युग्मन स्थिरांक J, Hz में) का अनुमान लगाइए: (1) A, (2) B, (3) C (5 अंक)
निम्नलिखित यौगिकों के ¹H NMR स्पेक्ट्रम में Hᵃ एवं Hᵇ अंकित प्रोटोनों की रासायनिक सृति की तुलना कीजिए तथा अपने उत्तर का औचित्य सिद्ध कीजिए। (10 अंक)
निम्नलिखित यौगिकों के ¹H NMR स्पेक्ट्रम में दिखने वाले शिखरों की संख्या की गणना कीजिए। अपने उत्तर का औचित्य सिद्ध कीजिए: (1), (2) (5 अंक)
(ख) (i) एक हैलोजीनेटेड एस्टर द्रव्यमान स्पेक्ट्रम में M⁺ शिखर m/z 166 (10%) पर और M+2 शिखर m/z 168 (9·8%) पर दर्शाता है। ¹H NMR स्पेक्ट्रम में यह यौगिक दो त्रिक व एक एकल क्रमशः δ 2·9, 3·6 और 3·8 ppm पर तीव्रता अनुपात 1:1:1·5 में दर्शाता है। यौगिक की संरचना का निगमन कीजिए। अपने उत्तर का औचित्य सिद्ध कीजिए। (10 अंक)
दो समावयवी एल्कीन, जिनका समान आणविक सूत्र C₆H₁₂ है, द्रव्यमान स्पेक्ट्रम में प्रबल शिखर m/z 42 तथा 56 पर दर्शाते हैं। दोनों शिखरों के खण्ड प्रतिरूप को प्रस्तावित कीजिए। (5 अंक)
(ग) (i) (1) थैलिक एसिड डाइएथिल एस्टर द्रव्यमान स्पेक्ट्रम में एक अभिलाक्षणिक शिखर m/z 149 पर दर्शाता है। खंडन प्रतिरूप दर्शाते हुए इस शिखर के प्रेक्षण का कारण बताइए। (2) एथिलबेंज़ीन द्रव्यमान स्पेक्ट्रम में एक अभिलाक्षणिक शिखर m/z 91 पर दर्शाता है, जबकि n-प्रोपिलबेंज़ीन m/z 92 पर प्रबल शिखर दर्शाता है। खंडन प्रतिरूप की सहायता से इसे समझाइए। (10 अंक)
एक अज्ञात कार्बनिक यौगिक, जिसका आण्विक सूत्र C₄H₅NO₂ है, IR स्पेक्ट्रम में 2250 cm⁻¹ पर एक बैंड तथा 1740 cm⁻¹ पर एक प्रबल बैंड दर्शाता है। यह यौगिक ¹H NMR स्पेक्ट्रम में केवल दो सिग्नल 3:2 के अनुपात में दर्शाता है। यौगिक की संरचना का पता लगाइए। अपने उत्तर का औचित्य सिद्ध कीजिए। (10 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(a) Two chemical structures labeled (1) and (2). Structure (1) is a skeletal formula of a chiral alkyl chloride. It shows a central carbon atom bonded to a chlorine atom (Cl) via a solid wedge bond, a methyl group, and an ethyl group. Structure (2) is a benzene ring substituted with two methyl groups (Me) at adjacent positions (ortho). The bond to the lower methyl group is drawn as a solid wedge.
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a)(i)
- A, CH₂=C(Br)Cl. Hᵃ and Hᵇ sit on the same sp² carbon, so this is geminal (²J) coupling of a terminal =CH₂: J ≈ 0–3 Hz. The two protons are non-equivalent (one is cis to Br, the other cis to Cl), so each appears as a narrow doublet.
- B, (E)-1,2-dichloroethene. Hᵃ and Hᵇ are on adjacent carbons and trans (dihedral angle 180°). ³J(trans) ≈ 12–18 Hz, about 12 Hz for this compound. Because the molecule is centrosymmetric, Hᵃ and Hᵇ are chemically and magnetically equivalent, so the spectrum shows one singlet; the coupling is present but not observed as splitting.
- C, HC≡C–CH₂Cl. Hᵃ and Hᵇ are four bonds apart. The cylindrical π system of the triple bond transmits a long-range ⁴J ≈ 2–3 Hz. Hᵃ appears as a triplet and CH₂ᵇ as a doublet.
(a)(ii)
In both D and E the labelled proton is on the terminal (β) carbon, cis to the substituent. Hᵃ in D (methyl vinyl ether) is strongly shielded, δ ≈ 4.2 ppm. Hᵇ in E (methyl vinyl ketone) is deshielded, δ ≈ 6.2 ppm. For comparison, ethene resonates at δ 5.28.
- D, CH₂=CH–OCH₃: the OCH₃ group donates by resonance (+M, from the oxygen lone pair), which outweighs its −I effect; the resonance form ⁻CH₂–CH=O⁺–CH₃ puts extra electron density on the β-carbon, so Hᵃ appears at about δ 4.2, upfield of ethene.
- E, CH₂=CH–COCH₃: the carbonyl withdraws by −M and −I; the resonance form ⁺CH₂–CH=C(CH₃)–O⁻ lowers the electron density on the β-carbon, so Hᵇ appears at about δ 6.2, downfield of ethene.
Donation from the oxygen lone pair shields Hᵃ; withdrawal by the carbonyl, reinforced by its anisotropy, deshields Hᵇ. Hence δ(Hᵇ, E) > δ(Hᵃ, D) by about 2 ppm.
(a)(iii)
(1) (S)-2-Chlorobutane, CH₃–CH(Cl)–CH₂–CH₃: 5 signals.
- C-1 CH₃: doublet, δ about 1.5.
- C-2 CH: multiplet, δ about 4.0.
- C-3 CH₂: the carbon is adjacent to a stereocentre, so its two protons are diastereotopic (no symmetry operation interchanges them). They give two separate multiplets near δ 1.7–1.8, which often overlap in practice.
- C-4 CH₃: triplet, δ about 1.0.
(2) As printed, the ring has non-conjugated double bonds, C1=C2 and C4=C5. The ring carries a vinylic Me on C-1 and a wedged Me on sp³ C-6, with C-3 a CH₂. This is 1,6-dimethylcyclohexa-1,4-diene, which has no symmetry element (C-6 is a stereocentre). It gives 8 signals:
- 2 CH₃ signals: vinylic about 1.7, allylic doublet about 1.0.
- 1 C-6 methine.
- 3 vinylic H (C-2, C-4, C-5).
- 2 for the diastereotopic C-3 protons.
If accidental overlaps occur, fewer peaks are resolved.
How this answer will be evaluated
What earns full marks: Accurate mechanisms, correct structures, and clear justifications for all parts.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a)(i)-(ii) Estimate coupling constants for A, B, C and compare chemical shifts for D, E. 10 marks
compare— paired headings or point-wise contrast → key differences → significance → conclusion
Must cover
- A: CH₂=C(Br)Cl — geminal ²J between the non-equivalent terminal vinyl H's (≈0–3 Hz; each a doublet)
- B: (E)-1,2-dichloroethene — trans ³J ≈ 12–18 Hz (not seen as splitting: the two H are equivalent — one singlet)
- C: HC≡C–CH₂Cl — long-range propargylic ⁴J ≈ 2–3 Hz (Hᵃ triplet, CH₂ᵇ doublet)
- D vs E shifts: Hᵃ (vinyl ether) ≈4.2 ppm shielded by +M OMe; Hᵇ (vinyl ketone) ≈6.2 ppm deshielded by the −M/−I carbonyl (cf. ethene 5.28)
Loses marks
- Confusing cis and trans coupling magnitudes
- Ignoring the effect of the substituent on shift
Earns more
- Typical J ranges quoted (cis ≈ 6–12 Hz, trans ≈ 12–18 Hz, geminal 0–3 Hz)
- Reference anisotropic effects of C=O or OMe
Extra mark
- Draw Newman projections for coupling analysis
- (a)(iii) Count the number of peaks in the 1H NMR spectra of the two compounds. 5 marks
enumerate— list the items in order → one line each → no commentary
Must cover
- (1) 2-chlorobutane: 5 signals — the C3 CH₂ protons are diastereotopic
- (2) 1,6-dimethylcyclohexa-1,4-diene: no symmetry element (C6 is a stereocentre) → 8 signals
- Count: 2 CH₃ + C6-H + 3 vinylic H + 2 diastereotopic C3-H
- Note that accidental overlap can reduce the observed number of peaks
Loses marks
- Failing to recognize symmetry in the ring
- Counting non-equivalent protons as equivalent
Earns more
- Label equivalent protons on the structures
Extra mark
- Mention splitting patterns for the peaks
- (b)(i) Deduce the structure of the halogenated ester from MS and NMR data. 10 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Halogen from the ~1:1 M⁺/M+2 pair (10:9.8) → one Br
- Formula C₄H₇BrO₂ (166 = 79 + CO₂CH₃ + C₂H₄)
- NMR: 3.8 (s, 3H) OCH₃; 3.6 (t, 2H) BrCH₂; 2.9 (t, 2H) CH₂C=O — two triplets prove an isolated –CH₂–CH₂– unit
- Structure: methyl 3-bromopropanoate, BrCH₂CH₂CO₂CH₃; isomers excluded (ethyl bromoacetate, methyl 2-bromopropanoate)
Loses marks
- Ignoring the M+2 peak for halogen identification
- Incorrectly assigning the NMR signals
Earns more
- Calculate the molecular weight from the M+ peak
- Assign specific chemical shifts to specific protons
Extra mark
- Draw the structure clearly
- (b)(ii) Propose fragmentation patterns for m/z 42 and 56 in C6H12 isomers. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- Alkene radical cations undergo the McLafferty rearrangement (γ-H transfer; six-membered TS; C3–C4 cleavage)
- Hex-1-ene → propene radical cation C₃H₆⁺• m/z 42
- 2-Methylpent-1-ene (and hex-2-ene) → isobutene radical cation C₄H₈⁺• m/z 56
- Even-mass radical-cation products are the hallmark of a rearrangement
Loses marks
- Incorrect fragmentation pathways
- Failing to show the charge on the fragments
Earns more
- Label the fragments with their chemical formulas
Extra mark
- Mention the stability of the carbocations
- (c)(i) Explain the m/z 149 peak in phthalate and m/z 91/92 in alkylbenzenes. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Diethyl phthalate: m/z 149 = cyclic protonated phthalic anhydride ion C₈H₅O₃⁺ (177 − 28: •OC₂H₅ loss, then C₂H₄ loss or vice versa)
- m/z 149 is diagnostic of ortho-diesters — both CO₂Et groups are required
- Ethylbenzene: benzylic cleavage → benzyl cation → tropylium C₇H₇⁺ m/z 91
- n-Propylbenzene: γ-H McLafferty (H transfer to an ortho ring carbon) + ethene loss → C₇H₈⁺• m/z 92 (ethylbenzene has no γ-H)
Loses marks
- Incorrect fragmentation mechanism
- Failing to show the electron movement
Earns more
- Draw the resonance structures of the tropylium ion
Extra mark
- Mention the specific mechanism name (McLafferty)
- (c)(ii) Deduce the structure of C4H5NO2 from IR and NMR data. 10 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- DBE = 3 for C₄H₅NO₂
- IR: 2250 cm⁻¹ C≡N; 1740 cm⁻¹ ester C=O
- NMR: two singlets 3:2 → OCH₃ (≈3.8) and the active CH₂ (≈3.5), both without neighbouring H
- Structure: methyl cyanoacetate NC–CH₂–CO₂CH₃; the CH₃CO–O–CH₂CN isomer excluded
Loses marks
- Ignoring the IR data for functional groups
- Incorrectly assigning the NMR signals
Earns more
- Calculate the degree of unsaturation
- Assign specific chemical shifts to specific protons
Extra mark
- Draw the structure clearly
What "Justify" is asking you to do
Defend a position with reasons that carry evidence, and show why the contrary view does not hold. Where the stem runs as a question and asks you to justify your answer, the position is yours to choose and the marks lie wholly in the defence.
Structure that answers it
Position stated plainly → reason 1 with evidence → reason 2 with evidence → strongest objection, met → position restated as qualified
Where marks are lost
Reasons stated and none of them evidenced. The other standard loss is fence-sitting — an answer that finds merit on both sides and commits to neither has justified nothing.
Practice this exact question
Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.
Evaluate my answer →More from Chemistry 2025 Paper II
- Q5 (a) Write the structure of the product(s) and the intermediate formed in the following re…
- Q6 (a) (i) Elucidate the structure of the product and the intermediate (if any) in the follo…
- Q7 (a) (i) Acetone shows a weak absorption at 280 nm and a strong absorption at 190 nm in th…
- Q8 (a) (i) Estimate the expected splitting (coupling constant J in Hz) for the lettered prot…