Carbohydrates — classification (mono - di - polysaccharides), Fischer - Haworth projections, mutarotation, glycosidic bo: Exercises

Intuition How to use this page — and what the coloured boxes mean

Work each problem before opening its solution. The problems climb from L1 (just recognise the word) to L5 (stitch several ideas together). Every idea here is built in the parent note; when a solution leans on a deeper concept it links out to Glucose — structure and reactions, Disaccharides — sucrose, maltose, lactose, Polysaccharides — starch, cellulose, glycogen, Optical isomerism and specific rotation, Hemiacetal and acetal formation, or Stereochemistry — anomers, epimers, enantiomers.

The coloured boxes on this page each have a job:

[!intuition] — the big-picture "why", read it first.
[!definition] — a term is pinned down precisely; memorise these.
[!recall]- — a collapsible box (click to expand). Problem solutions live here so you can self-test: try first, then reveal.
[!mistake] — the trap most students fall into, with why it feels right and how to fix it.

Definition Two terms this whole page leans on — pinned down once

Glycosidic bond — the C–O–C ether-type link made when the anomeric –OH of one sugar condenses with an –OH of another molecule, ejecting one H $_{2}$ O. Breaking it (hydrolysis) puts that water back. The schematic below shows the link and the lost water.
Specific rotation, $[α]_{D}$ — a number that says how much a solution of the sugar rotates the plane of polarised light. The subscript D = measured with the yellow sodium-D line ( $589 nm$ ); it is quoted for a $1 dm$ (10 cm) path length and $1 g mL^{- 1}$ concentration. Sign convention: a + value means the light is rotated clockwise (dextrorotatory), a − value anticlockwise. Units are conventionally written as degrees, $^{\circ}$ . (More depth: Optical isomerism and specific rotation.)

Level 1 — Recognition

Goal: name the class or group directly, no calculation.

L1.1

Problem. Classify each on hydrolysis: (a) fructose, (b) maltose, (c) cellulose.

Recall Solution

Hydrolysis breaks the glycosidic bond (see the schematic above): each break splits one link and adds one H $_{2}$ O. So the test is: how many simpler sugars come out?

(a) Fructose gives nothing simpler ⇒ monosaccharide (it is a ketohexose).
(b) Maltose gives 2 glucose ⇒ disaccharide.
(c) Cellulose gives many glucose ⇒ polysaccharide.

Why: the number of glycosidic bonds you can break equals (units − 1). Zero bonds → mono, one bond → di, many → poly. Compare the three chain lengths in the figure below.

Figure — Carbohydrates — classification (mono - di - polysaccharides), Fischer - Haworth projections, mutarotation, glycosidic bo

L1.2

Problem. Give the two-word code for a 5-carbon sugar bearing a terminal –CHO.

Recall Solution

Two pieces of information sit in the name:

Carbonyl type: a terminal –CHO means aldo-.
Carbon count: 5 carbons means -pentose.

Answer: aldopentose (e.g. ribose). The Fischer skeleton below shows exactly where the –CHO (top) and the counted carbons sit.

Level 2 — Application

Goal: apply one rule (Fischer↔Haworth, D/L, reducing test).

L2.1

Problem. In D-glucose the C2, C3, C4, C5 hydroxyls read (top→bottom) R, L, R, R in the Fischer projection. Using the "right→down, left→up" rule, state whether each of these –OH groups points up or down in the Haworth ring.

Recall Solution

The rule: a group on the right in Fischer goes down in Haworth; on the left goes up.

C2 = R → down
C3 = L → up
C4 = R → down
C5 = R → its –OH is consumed making the ring; the reference CH $_{2}$ OH (C6) on a D-sugar points up.

What the figure below shows: on the left is the vertical Fischer chain of D-glucose with each –OH drawn on its correct side (right = red arrow, left = black arrow); on the right is the flat Haworth ring. The middle arrow is the mapping — trace how a right –OH lands below the ring plane and a left –OH lands above it. This is the visual proof of "right→down, left→up".

L2.2

Problem. Is maltose a reducing sugar? Justify in one sentence using the anomeric-carbon idea.

Recall Solution

Maltose is glucose–α(1→4)–glucose. The (1→4) bond uses the anomeric C1 of the first glucose but the second glucose keeps its C1 anomeric carbon free. A free anomeric carbon can open to a –CHO ⇒ it reduces Fehling's/Tollens' ⇒ maltose is reducing. (Contrast: sucrose locks both anomeric carbons, so it is non-reducing.)

Level 3 — Analysis

Goal: reason across two concepts at once.

L3.1

Problem. Pure α-D-glucose has $[α]_{D} = + 11 2^{\circ}$ ; pure β-D-glucose has $[α]_{D} = + 1 9^{\circ}$ . A freshly dissolved sample is 50% α and 50% β by mole. Predict its observed specific rotation, ignoring the tiny open-chain fraction.

Recall Solution

Recall $[α]_{D}$ from the top box: it is the (signed) rotation of polarised light, additive across the components. Each anomer contributes in proportion to its mole fraction (see Optical isomerism and specific rotation). With $x_{α} = x_{β} = 0.5$ : $[α]_{obs} = 0.5 (112) + 0.5 (19) = 56 + 9.5 = 65. 5^{\circ}$ So the freshly made 50:50 mixture rotates $+ 65. 5^{\circ}$ . Over time it will drift down toward the equilibrium value $+ 52. 7^{\circ}$ (which is richer in β).

L3.2

Problem. An unknown D-hexose solution shows no mutarotation and gives a negative Fehling's test. What structural feature must it have, and name one sugar fitting this description.

Recall Solution

What is Fehling's test? Fehling's reagent is a deep-blue solution of Cu $^{2 +}$ ions. A free aldehyde (from an open anomeric carbon) reduces Cu $^{2 +}$ to Cu $^{+}$ , which drops out as a brick-red precipitate of Cu $_{2}$ O. So:

Positive = blue → brick-red precipitate (a reducing sugar is present).
Negative = stays blue, no precipitate (no free aldehyde available).

Now the two clues, one cause:

No mutarotation ⇒ the ring cannot open to the open-chain form ⇒ no free anomeric carbon.
Negative Fehling's (stays blue) ⇒ non-reducing ⇒ again no free –CHO / anomeric carbon.

Both point to every anomeric carbon being tied up in glycosidic bonds. A sugar fitting this: sucrose (glucose C1 ↔ fructose C2, both anomeric carbons locked). (Sucrose is technically a disaccharide, not a monosaccharide — the "no free anomeric carbon" deduction is what the data force.)

Level 4 — Synthesis

Goal: combine composition, additivity, and equilibrium.

L4.1

Problem. Using $[α]_{α} = + 11 2^{\circ}$ , $[α]_{β} = + 1 9^{\circ}$ , and equilibrium rotation $+ 52. 7^{\circ}$ , solve for the equilibrium mole fraction of the α anomer. Then state the β fraction.

Recall Solution

Let $x$ = mole fraction of α, so $(1 - x)$ = fraction of β. Additivity gives one linear equation: $112 x + 19 (1 - x) = 52.7$ Expand and collect: $112 x + 19 - 19 x = 52.7 \Rightarrow 93 x = 33.7 \Rightarrow x = 0.362$ So ≈ 36.2% α and ≈ 63.8% β at equilibrium. Why linear? Because rotation is a weighted average of contributions — a straight-line function of composition, so one equation pins one unknown.

L4.2

Problem. Maltose (C $_{12}$ H $_{22}$ O $_{11}$ , molar mass $342 g mol^{- 1}$ ) is fully hydrolysed to glucose (C $_{6}$ H $_{12}$ O $_{6}$ , $180 g mol^{- 1}$ ). Starting from $34.2 g$ of maltose, how many grams of glucose are produced?

Recall Solution

Balanced hydrolysis (water is added because the bond formed by losing water): $C_{12} H_{22} O_{11} + H_{2} O \to 2 C_{6} H_{12} O_{6}$ Moles of maltose $= 34.2/342 = 0.100 mol$ . Each maltose gives 2 glucose ⇒ $0.200 mol$ glucose. Mass $= 0.200 \times 180 = 36.0 g$ . Sanity check by mass: $34.2 g maltose + 0.100 \times 18 = 1.8 g water = 36.0 g$ — conservation of mass holds.

Level 5 — Mastery

Goal: chain classification, bonding, reducing behaviour and a numeric argument.

L5.1

Problem. Trehalose is a disaccharide made of two glucose units joined α,α(1↔1) (both anomeric carbons used). (a) Is it reducing? (b) Will it mutarotate? (c) On complete hydrolysis of $0.500 mol$ trehalose, how many moles of glucose form and how many moles of water are consumed?

Recall Solution

Recall a glycosidic bond ties up an anomeric –OH (top box). Here both anomeric carbons are used. (a) With both anomeric carbons locked, neither ring can open to a free –CHO ⇒ trehalose is non-reducing (same logic as sucrose). (b) No free anomeric carbon ⇒ no ring opening ⇒ no mutarotation. (c) One glycosidic bond per trehalose ⇒ one water added per molecule: $trehalose + H_{2} O \to 2 glucose$ $0.500 mol$ trehalose consumes $0.500 mol$ water and yields $2 \times 0.500 = 1.00 mol$ glucose.

L5.2

Problem. A student mixes pure α-D-glucose and pure β-D-glucose and measures an initial specific rotation of $+ 70. 0^{\circ}$ . Using $[α]_{α} = + 11 2^{\circ}$ and $[α]_{β} = + 1 9^{\circ}$ , find the initial mole fraction of α. Then, knowing the mixture will relax to the equilibrium value $+ 52. 7^{\circ}$ , state whether the α-fraction must increase or decrease over time.

Recall Solution

Step 1 — composition from rotation (additivity): $112 x + 19 (1 - x) = 70.0 \Rightarrow 93 x = 51.0 \Rightarrow x = 0.548$ So the mixture starts at ≈ 54.8% α. Step 2 — direction of change: equilibrium is ≈36% α (from L4.1). Since the sample starts above the equilibrium α-fraction, the α-fraction must decrease (α ⇌ β interconverts through the open chain until it reaches 36% α), and the rotation falls from $+ 70. 0^{\circ}$ toward $+ 52. 7^{\circ}$ .

L5.3

Problem (edge cases — ring size and branching). (a) Fructose in solution closes into a 5-membered furanose ring rather than a 6-membered pyranose. Does forming a furanose instead of a pyranose change whether fructose can mutarotate? (b) Glycogen and amylopectin are branched polysaccharides carrying α(1→6) links at branch points in addition to the main α(1→4) chain. On complete hydrolysis, does branching change the identity of the monosaccharide you recover?

Recall Solution

(a) Furanose vs pyranose. "Furanose" = 5-membered ring (named after furan); "pyranose" = 6-membered ring (named after pyran). What matters for mutarotation is not the ring size but whether the ring carries a hemiacetal at a free anomeric carbon that can open and reclose (see Hemiacetal and acetal formation). Fructofuranose still has a free anomeric carbon (its C2), so it still mutarotates and is still reducing — the smaller ring just means a different set of α/β rotation values, not a loss of ring-opening. Edge takeaway: ring size (furanose vs pyranose) changes the numbers, but ring-opening ability is what governs mutarotation and reducing behaviour. (b) Branching. The α(1→6) branch points are still glycosidic bonds between glucose units — they just join C1 of one glucose to C6 of another instead of C4. Hydrolysing every bond (both α(1→4) and α(1→6)) still releases only glucose. Branching changes the shape and the count of bond types, not the monomer identity: glycogen, amylopectin, amylose and cellulose all hydrolyse to glucose.

Active recall

Recall One-line self-check

Reducing needs what? → a free anomeric carbon.
Equilibrium α-fraction of glucose? → ≈36% (β-rich).
Why glucose from maltose weighs more than the maltose? → water added in hydrolysis.
Does a furanose ring mutarotate? → yes, if its anomeric carbon is free (ring size is irrelevant).
What do branched glycogen/amylopectin give on full hydrolysis? → only glucose.