Paper II — Q7
(a) Discuss the different types of secondary structures of proteins and compare these structures with tertiary structure of…
Discuss the different types of secondary structures of proteins and compare these structures with tertiary structure of proteins. 15 marks
Complete the following reactions and give suitable mechanisms for the formation of products: CH₃—CH₂—CH₂—C(=O)—Cl (1) 2NaBH₄ → ? (2) H₃O⁺
CH₃—C≡C—CH₃ Na in liq. NH₃ → ? 10+5=15 marks
A compound (A) containing C, H and O has molecular weight 102 and displays two signals in the ¹H NMR spectrum at δ 1·1 (d) and 3·55 (septet) in the integral ratio of 6:1. Treatment of A with 1 mole of HI gives rise to B and C. In the IR spectrum, B gives a strong absorption band at 3330 cm⁻¹, whereas its ¹H NMR spectrum shows signals at δ 1·05(d, 6H), 3·6 (septet, 1H) and 4·4 (s, 1H, disappeared with D₂O). The ¹H NMR spectrum of C gives signals at δ 1·9(d, 6H) and 4·25 (septet, 1H). Reaction of A with excess of HI gives only C. Identify A, B and C. Write all the reactions involved. 10 marks
Three isomeric compounds having MF C₅H₁₀O give positive 2,4-DNP test and display the following NMR spectral characteristics. Identify the compounds. Among them, one isomeric compound on treatment with KOH (concentrated) gives two products. Write their structures also: (1) A triplet at δ1.05 and a quartet at δ2.47 (2) Two singlets (3) A doublet at δ1.0, a singlet at δ2.1 and a septet at δ2.2 10 marks
हिंदी में प्रश्न पढ़ें
प्रोटीन के भिन्न प्रकारों की द्वितीयक संरचनाओं की विवेचना कीजिए तथा इन संरचनाओं की तुलना प्रोटीन के तृतीयक संरचना से कीजिए। 15 अंक
निम्नलिखित अभिक्रियाओं को पूर्ण कीजिए तथा उत्पादों के बनने की उपयुक्त क्रियाविधि दीजिए : CH₃—CH₂—CH₂—C(=O)—Cl (1) 2NaBH₄ → ? (2) H₃O⁺
CH₃—C≡C—CH₃ तरल NH₃ में Na → ? 10+5=15 अंक
C, H तथा O के एक यौगिक (A) का आणविक भार 102 है और ¹H NMR स्पेक्ट्रम में दो सिग्नल δ 1·1 (d) तथा 3·55 (सेप्टेट) पर अभिन्न अनुपात 6:1 में देता है। A का विवेचन जब 1 मोल HI के साथ किया जाता है, तो B तथा C उत्पाद बनते हैं। B के IR स्पेक्ट्रम में 3330 cm⁻¹ पर एक प्रबल अवशोषण बैंड है, जबकि ¹H NMR स्पेक्ट्रम में δ 1·05(d, 6H), 3·6 (सेप्टेट, 1H) और 4·4 (s, 1H, D₂O में लुप्त) पर सिग्नल दिखाता है। C के ¹H NMR स्पेक्ट्रम में δ 1·9(d, 6H) तथा 4·25 (सेप्टेट, 1H) पर सिग्नल देता है। यदि A की अभिक्रिया अतिरिक्त HI से की जाती है, तो केवल C बनता है। A, B तथा C को पहचानिए। सभी संबंधित अभिक्रियाओं को लिखिए। 10 अंक
तीन समावयवी यौगिकों, जिनका आणविक सूत्र (MF) C₅H₁₀O है, 2,4-DNP परीक्षण सकारात्मक देते हैं तथा निम्नलिखित NMR स्पेक्ट्रमी अभिलक्षण देते हैं। यौगिकों को पहचानिए। इनमें एक समावयवी यौगिक KOH (संद्रित) के साथ विवेचन करने पर दो उत्पाद देता है। उत्पादों की संरचना भी बताइए : (1) δ1.05 पर एक ट्रिप्लेट और δ2.47 पर एक क्वार्टेट (2) दो सिंगलेट (3) δ1.0 पर एक डबलेट, δ2.1 पर एक सिंगलेट और δ2.2 पर एक सेप्टेट 10 अंक
Model answer
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(a) Secondary structures of proteins compared with tertiary structure
Protein secondary structures are local, regularly repeating backbone conformations stabilized mainly by C=O...H-N hydrogen bonds. The alpha-helix is a right-handed helix with 3.6 residues per turn, a rise of 1.5 Å and a pitch of 5.4 Å; each carbonyl oxygen accepts a hydrogen bond from the N-H group four residues ahead. The beta-pleated sheet is formed by extended strands linked by interstrand or interchain hydrogen bonds; antiparallel sheets have nearly linear hydrogen bonds, while parallel sheets have slightly angled bonds and a different spacing. Beta-turns are short four-residue reversals that change the direction of the chain, often involving proline or glycine, and may contain a hydrogen bond between residues i and i+3. Random coils or loops are irregular segments without a fixed hydrogen-bond pattern; they connect helices, sheets and turns. In globular proteins, alpha-helices and beta-sheets are often separated by turns and loops, while in fibrous proteins beta-sheets may form long, stable fibres. Tertiary structure is the overall three-dimensional folding of the entire polypeptide chain, in which these secondary elements pack into a compact native shape. It is controlled by side-chain interactions: hydrophobic residues cluster in the interior, disulfide bridges form between cysteine residues, ionic or salt bridges link charged side chains, and hydrogen bonds and van der Waals contacts further stabilize the fold. Thus secondary structure is local, regular and backbone-driven, whereas tertiary structure is global, side-chain-driven and determines the final biological shape; secondary elements are the building blocks organized by tertiary interactions.
(b)(i) Butanoyl chloride with NaBH₄
Butanoyl chloride, CH₃CH₂CH₂COCl, is reduced by NaBH4. Borohydride delivers hydride to the carbonyl carbon; the tetrahedral intermediate collapses, expelling chloride and giving butanal. The aldehyde formed is still readily reduced, so a second hydride adds, and H3O+ protonates the alkoxide to give butan-1-ol, CH3CH2CH2CH2OH.
(b)(ii) But-2-yne with Na in liquid NH₃
But-2-yne undergoes dissolving-metal reduction with Na in liquid NH₃. Sodium transfers an electron to the alkyne to form a radical anion; protonation by NH₃ gives a vinyl radical, which takes up the less crowded trans geometry. A second electron gives the trans vinyl anion, and a second protonation gives trans-but-2-ene, (E)-but-2-ene. The anti relationship of the two added hydrogens explains the trans alkene.
(c)(i) Identification of A, B and C
A contains C, H and O, has molecular weight 102, and shows 1H NMR signals at delta 1.1 (doublet, 6H) and 3.55 (septet, 1H). The formula C6H14O is consistent with M=102. The 6:1 doublet-septet pattern indicates two equivalent isopropyl groups, so A is diisopropyl ether, (CH3)2CHOCH(CH3)2. The septet at 3.55 is the methine proton coupled to six equivalent methyl protons; the doublet at 1.1 is the six methyl protons coupled to that methine. With 1 mol HI, the ether oxygen is protonated and iodide attacks the more accessible carbon, cleaving the C-O bond to give isopropanol B, (CH3)2CHOH, and isopropyl iodide C, (CH3)2CHI. C shows the corresponding isopropyl iodide pattern at delta 1.9 and 4.25. B has a strong IR band at 3330 cm-1 for O-H, 1H NMR signals at delta 1.05 (d, 6H) and 3.6 (septet, 1H), and a D2O-exchangeable singlet at delta 4.4. With excess HI, the alcohol is converted further to iodide, so only C remains: (CH3)2CHOCH(CH3)2 + HI gives (CH3)2CHOH + (CH3)2CHI, and (CH3)2CHOH + HI gives (CH3)2CHI + H2O.
(c)(ii) The three C₅H₁₀O carbonyl isomers
The C₅H₁₀O compounds giving a positive 2,4-DNP test are carbonyl compounds. The NMR patterns distinguish ethyl, tert-butyl and isopropyl environments attached to the carbonyl carbon. (1) Pentan-3-one, CH₃CH₂COCH₂CH₃: two equivalent ethyl groups give only a triplet (CH₃, δ 1.05) and a quartet (CH₂ next to C=O, δ 2.47). (2) 2,2-Dimethylpropanal, (CH₃)₃CCHO: a 9H singlet near δ 1.1 for the tert-butyl group and a 1H singlet near δ 9.5 for CHO, since neither has neighbours to couple with. (3) 3-Methylbutan-2-one, CH₃COCH(CH₃)₂: a 6H doublet (δ 1.0), a 3H singlet for the acetyl CH₃ (δ 2.1) and a 1H septet for the methine (δ 2.2). Among these, 2,2-dimethylpropanal, which has no α-hydrogen, is the isomer that, on treatment with concentrated KOH, undergoes the Cannizzaro reaction to give 2,2-dimethylpropan-1-ol, (CH3)3CCH2OH, and potassium 2,2-dimethylpropanoate, (CH3)3CCOOK. In the mechanism, OH⁻ adds to one aldehyde, and the tetrahedral anion transfers a hydride to a second aldehyde, which is reduced while the first is oxidised. Pentan-3-one and 3-methylbutan-2-one have α-hydrogens and enolise instead.
How this answer will be evaluated
Framework: Concept > Structure or mechanism > Reasoning > Result
What earns full marks: Accurate structures, mechanisms, and spectral analysis with clear reasoning.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Define secondary structures and compare them with tertiary structure. 15 marks
discuss— intro → 3-4 dimensions → example → balanced close
Must cover
- Define alpha-helix and beta-pleated sheet structures
- Identify hydrogen bonding as the stabilizing force
- Distinguish secondary from tertiary structure (R-group interactions)
- Mention specific interactions in tertiary (disulfide, ionic, hydrophobic)
Loses marks
- Confusing primary structure with secondary
- Omitting the role of hydrogen bonding
Earns more
- Mention random coil structure
- Reference specific amino acid residues (e.g., Gly, Pro) affecting structure
Extra mark
- Include a labelled diagram of alpha-helix or beta-sheet
- (b)(i) Complete the reaction of butanoyl chloride with NaBH4 and show mechanism. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Identify product as butan-1-ol
- Show nucleophilic attack of hydride on carbonyl carbon
- Show elimination of chloride ion
- Show protonation of the alkoxide intermediate
Loses marks
- Product without mechanism
- Incorrect nucleophile identification
Earns more
- Mention the reactivity of acid chlorides vs esters
- Correct arrow pushing for electron movement
Extra mark
- Mention the specific role of the second equivalent of NaBH4
- (b)(ii) Complete the reaction of 2-butyne with Na in liq. NH3 and show mechanism. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- Identify product as trans-2-butene
- Show radical anion formation
- Explain anti-addition of protons
- Justify trans stereochemistry via steric repulsion
Loses marks
- Product without mechanism
- Incorrect stereochemistry (cis instead of trans)
Earns more
- Names it a dissolving-metal reduction (Na/NH₃)
Extra mark
- Comparison with cis-addition using Lindlar's catalyst
- (c)(i) Identify compounds A, B, and C based on spectral data and reactions. 10 marks
analyse— intro → causes → effects → stakeholders/linkages → way forward
Must cover
- Identify A as diisopropyl ether
- Identify B as isopropanol (2-propanol)
- Identify C as isopropyl iodide (2-iodopropane)
- Write the reaction of A with HI to form B and C
Loses marks
- Incorrect identification of A, B, or C
- Failure to write the reaction equations
Earns more
- Justify A's structure using the 6:1 doublet-septet NMR ratio
- Justify B's structure using the IR band at 3330 cm-1
Extra mark
- Explanation of why excess HI yields only C
- (c)(ii) Identify three isomeric compounds and the products of one with KOH. 10 marks
analyse— intro → causes → effects → stakeholders/linkages → way forward
Must cover
- Identify (1) as pentan-3-one (triplet + quartet only)
- Identify (2) as 2,2-dimethylpropanal (two singlets: 9H t-Bu, 1H CHO)
- Identify (3) as 3-methylbutan-2-one
- 2,2-Dimethylpropanal (no α-H) undergoes the Cannizzaro reaction with conc. KOH: 2,2-dimethylpropan-1-ol + potassium 2,2-dimethylpropanoate
Loses marks
- Incorrect identification of any of the three isomers
- Failure to identify the products of the KOH reaction
Earns more
- Justify (1) using the triplet/quartet NMR pattern
- Justify (2) using the two singlet NMR signals
- Justify (3) using the doublet/singlet/septet NMR pattern
Extra mark
- Names the Cannizzaro reaction and shows the hydride transfer
What "Discuss" is asking you to do
Lay the issue out from more than one side — how it arose, what is claimed for it, what is held against it, and where it now stands. UPSC attaches discuss to broad topics with several live dimensions, so coverage of the dimensions earns more than the strength of your opinion.
Structure that answers it
Set the issue up → the case as it is made → the case against → the dimension both sides leave out → where the balance now lies
Where marks are lost
Listing facts with no thread between them, or arguing one side throughout and calling it a discussion.
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