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ADV-2026-P2-CHE-Q04

Building sucralose: three stereochemical decisions, one correct structure

JEE Advanced 2026 Paper 2 Chemistry Q.4 · Section 1 Biomolecules Carbohydrates
Answer
(A)
X = sucralose · C12H19Cl3O8

01Original question — verbatim

A known artificial sweetener X is composed of 4-chloro-4-deoxy-α-D-galactose and 1,6-dichloro-1,6-dideoxy-β-D-fructose joined by a glycosidic linkage. Structure of D-galactose is given below:
The correct structure of X is
  • (A)
    C4–Cl axial (galacto) · pyranose α · fructofuranose β
  • (B)
    C4–Cl equatorial (gluco) · pyranose α · fructofuranose β
  • (C)
    C4–Cl equatorial (gluco) · pyranose α · fructofuranose α
  • (D)
    C4–Cl axial (galacto) · pyranose α · fructofuranose α

Click an option to check it. The printed options are drawn as a chair plus a furanose; they are redrawn here as Haworth projections carrying exactly the same stereochemistry, with hydrogens omitted for clarity. All four printed options agree on the α-linkage — the two things they disagree about are the C4 chlorine and the fructose anomeric centre.

02Fischer → Haworth, one carbon at a time

The paper gives D-galactose as a Fischer projection, so the first job is to read it. The rule for a D-aldohexose drawn with the ring oxygen at the back-right and C1 on the right: a group on the right in Fischer points down in Haworth, a group on the left points up, and C6 always ends up. Click a carbon in the Fischer projection — or step through them — and watch where its OH lands.

D-galactose · Fischer

α-D-galactopyranose · Haworth

03Build the sweetener

Three independent stereochemical decisions turn "a galactose plus a fructose" into one specific molecule. Set each one and the builder draws the result, checks it against the target named in the question, and tells you which printed option you have just drawn.

Decisions

Your structure

Hydrogens omitted. The α/β labels mark the two anomeric carbons.

04Eliminating the options

Each printed option is one combination of the two contested decisions. Score all four against the two requirements the question states in words, and exactly one survives.

Note that no option can be eliminated by counting atoms — all four are C12H19Cl3O8 and all four are stereoisomers of one another. The question is purely about configuration.

05Sucrose → sucralose, and why anyone cares

Sucralose is sucrose with three specific hydroxyls swapped for chlorine and one centre inverted along the way. Those changes stop the body's enzymes from recognising it, so it passes through largely unmetabolised while binding the sweet receptor far more tightly than sucrose does.

What changes

PositionSucroseSucralose
Pyranose C4OH, gluco (equatorial) Cl, galacto (axial)
Furanose C1′CH2OHCH2Cl
Furanose C6′CH2OHCH2Cl
Glycosidic linkageα-1↔2′ βα-1↔2′ β (unchanged)
FormulaC12H22O11 C12H19Cl3O8

Replacing the C4 hydroxyl by chlorine with inversion is exactly what turns the gluco configuration of sucrose into the galacto configuration named in the question — which is why the paper says galactose, not glucose.

Sucrose

Sweetness relative to sucrose (= 1), on a logarithmic axis because the values span three orders of magnitude. Values are the commonly quoted mid-range figures; perceived sweetness depends on concentration and formulation.

06Step-by-step solution

  1. Recognise the compound

    4-chloro-4-deoxy-α-D-galactopyranose joined to 1,6-dichloro-1,6-dideoxy-β-D-fructofuranose is sucralose, C12H19Cl3O8 — the sweetener sold as Splenda. Nothing else needs to be recalled; the name in the question fully specifies the structure.

  2. Read the galactose Fischer projection

    The paper's Fischer projection has OH on the right at C2 and C5, and on the left at C3 and C4. Converting for a D-sugar (right → down, left → up, C6 up):

    C2–OH down  ·  C3–OH up  ·  C4–OH up (axial)  ·  C6 up

    That axial C4 is the single thing distinguishing galactose from glucose. Replacing it by chlorine keeps the position: the C4 chlorine must point up. This eliminates (B) and (C), which draw it equatorial.

  3. Fix the pyranose anomeric centre

    "α-D-galactose" means the C1 substituent — here the glycosidic oxygen — points down in the standard Haworth orientation, trans to the C6 branch. All four printed options draw this correctly, so it eliminates nothing, but getting it wrong would not give sucralose.

  4. Fix the fructose anomeric centre

    "β-D-fructofuranose" fixes the configuration at C2′. In the orientation the paper uses, that puts the C1′ CH2Cl up and the glycosidic oxygen down. Options (C) and (D) draw the CH2Cl down — the α anomer — so they go.

  5. Intersect the surviving sets

    C4 axial ⟹ {A, D}   ∩   fructose β ⟹ {A, B}   =   {A}

    Answer: (A).

✓ Independent verification

  • Structure matched against a reference, not against memory. Each drawn half was rebuilt as explicit 3D coordinates (Haworth ring in a plane, substituents at ±z) and its configuration assigned from that geometry by RDKit, then compared with the published sucralose structure (InChIKey BAQAVOSOZGMPRM-QBMZZYIRSA-N).
  • Galactose half: the axial-C4 build gives CO[C@H]1O[C@H](CO)[C@H](Cl)[C@H](O)[C@H]1O, identical to the galactose half of the reference; the equatorial build differs at exactly one stereocentre. So (B) and (C) are demonstrably the gluco isomer.
  • Fructose half: the CH2Cl-up build gives CO[C@]1(CCl)O[C@H](CCl)[C@@H](O)[C@@H]1O, identical to the reference; the CH2Cl-down build is its C2′ epimer. So (C) and (D) are the α-fructofuranoside.
  • Name cross-check: the IUPAC name of sucralose, "1,6-dichloro-1,6-dideoxy-β-D-fructofuranosyl 4-chloro-4-deoxy-α-D-galactopyranoside", reproduces the question's wording clause for clause.
  • Formula check: C12H19Cl3O8, M = 397.6 — consistent with three Cl replacing three OH on a C12H22O11 disaccharide (−3 OH, +3 Cl, −3 H… i.e. −3 O, +3 Cl, −3 H). All four options share this formula, confirming the question is purely stereochemical.
  • Redundant chemical argument: sucralose is made from sucrose, whose glucose unit is gluco; the C4 chlorination proceeds with inversion, which converts it to galacto. That independently predicts the axial chlorine without touching the Fischer projection.
  • Distractor audit: (B) fails the C4 test only; (D) fails the fructose test only; (C) fails both. Each wrong option is a single believable slip, which is why the elimination has to be done on both centres rather than one.

07Concepts and classification

Rules used

Fischer → Haworth (D-sugar): right → down, left → up, C6 → up
α-D-anomer: anomeric substituent down · β-D-anomer: up
galactose = glucose epimeric at C4 (C4–OH axial instead of equatorial)
glycoside = anomeric C–O–R; a 1↔2′ link joins both anomeric carbons, so a non-reducing disaccharide results
deoxy-halo sugar: –OH replaced by –Cl (here with inversion at C4)

Classification

Simulation IDADV-2026-P2-CHE-Q04
Exam / Year / PaperJEE Advanced · 2026 · Paper 2
SubjectChemistry (Organic / Biomolecules)
ChapterBiomolecules
TopicCarbohydrates — structure and glycosidic linkages
SubtopicsFischer and Haworth projections · Anomers and the α/β convention · Epimers (glucose vs galactose) · Pyranose and furanose rings · Glycosidic linkage in a non-reducing disaccharide · Deoxy-halo sugars · Artificial sweeteners
Question typeSingle correct MCQ (+3 / 0 / −1)
DifficultyModerate
Answer(A) sucralose