CHM125 Workshop 3

Interactive Learning Guide — Ionic Compounds, Balancing, Functional Groups & Organic Reactions

Part 1 — Ionic Compound Formulas (Q1–4)

The Big Idea

Ions combine until their charges cancel to zero (electrical neutrality). Think of it like balancing a scale.

The Cross Method: Take each ion's charge number and make it the subscript of the OTHER ion.
If both charges have the same magnitude, reduce to 1:1 — don't write subscripts.
GroupChargeExamples
Group 1+1Na⁺, K⁺, Li⁺
Group 2+2Ca²⁺, Mg²⁺, Ba²⁺
Group 13+3Al³⁺
Group 16−2O²⁻, S²⁻
Group 17−1F⁻, Cl⁻, Br⁻, I⁻

Question 1

a. Ca and Cl

1Ca is Group 2 → Ca²⁺
2Cl is Group 17 → Cl⁻
3Cross: Ca's charge (2) becomes Cl's subscript; Cl's charge (1) becomes Ca's subscript

CaCl₂

(+2) + 2(−1) = 0 ✓

Question 2

b. Mg and O

1Mg → Mg²⁺  |  O → O²⁻
2Same magnitude (both 2) → 1:1 ratio

MgO

(+2) + (−2) = 0 ✓ — Equal charges always simplify to 1:1.

Question 3

c. K and O

1K → K⁺  |  O → O²⁻
2Cross: O's charge (2) → K's subscript. K's charge (1) → O's subscript.

K₂O

2(+1) + (−2) = 0 ✓

Question 4

d. Pb⁴⁺ and F

1Pb given as Pb⁴⁺  |  F → F⁻
2Cross: Pb's charge (4) → F's subscript

PbF₄

(+4) + 4(−1) = 0 ✓

Part 2 — Compound (Polyatomic) Ions (Q5–8)

The Big Idea

Polyatomic ions are groups of atoms that travel together as one charged unit — like a team that stays together.

Parenthesis Rule: If you need MORE THAN ONE polyatomic ion, wrap it in parentheses first, then add the subscript.
✗ CaOH₂ (wrong)    ✓ Ca(OH)₂ (correct)
NameFormulaCharge
HydroxideOH⁻−1
AmmoniumNH₄⁺+1
CarbonateCO₃²⁻−2
NitrateNO₃⁻−1
SulfateSO₄²⁻−2
SulfiteSO₃²⁻−2
PhosphatePO₄³⁻−3

Question 5

a. Ca and hydroxide

1Ca²⁺ and OH⁻ — need 2 hydroxides to cancel +2
2More than one polyatomic ion → use parentheses

Ca(OH)₂

Question 6

b. Ammonium and Cl

1NH₄⁺ and Cl⁻ — same magnitude → 1:1

NH₄Cl

Question 7

c. Cu²⁺ and sulfate

1Cu²⁺ and SO₄²⁻ — equal magnitude → 1:1

CuSO₄

Question 8

d. K and sulfite

1K⁺ and SO₃²⁻ — need 2 K⁺ to balance −2

K₂SO₃

No parentheses because K is a simple ion, not polyatomic.

Part 3 — Balancing Equations (Q9–12)

The Big Idea

Atoms don't appear or disappear — they just rearrange. Every atom on the left must appear on the right. You only change coefficients (big numbers in front), never subscripts.

Balancing Order: 1) Metals → 2) Non-metals (skip O & H) → 3) Hydrogen → 4) Oxygen last
If you get fractions, multiply ALL coefficients to clear them.

Question 9

a. NaCl(aq) + Pb(NO₃)₂(aq) → NaNO₃(aq) + PbCl₂(s)

1Pb: 1 = 1 ✓
2Cl: 1 on left, 2 on right → put 2 in front of NaCl
3Na: now 2 on left → put 2 in front of NaNO₃
4Check: N 2=2 ✓ O 6=6 ✓

2NaCl(aq) + Pb(NO₃)₂(aq) → 2NaNO₃(aq) + PbCl₂(s)

Question 10

b. CaBr₂(aq) + Na₂CO₃(aq) → NaBr(aq) + CaCO₃(s)

1Ca: 1=1 ✓ | CO₃: 1=1 ✓
2Na: 2 on left, 1 on right → put 2 in front of NaBr
3Br: 2=2 ✓

CaBr₂(aq) + Na₂CO₃(aq) → 2NaBr(aq) + CaCO₃(s)

Question 11

c. HCl(aq) + Ba(OH)₂(aq) → H₂O(l) + BaCl₂(aq)

1Ba: 1=1 ✓
2Cl: 1 left, 2 right → put 2 in front of HCl
3H: 4 left (2 from HCl + 2 from Ba(OH)₂) → put 2 in front of H₂O
4O: 2=2 ✓

2HCl(aq) + Ba(OH)₂(aq) → 2H₂O(l) + BaCl₂(aq)

Question 12

d. C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

1Balance C: 3 on left → 3CO₂
2Balance H: 8 on left → 4H₂O (4×2=8)
3Balance O: right has 3(2)+4(1) = 10 O → 5O₂

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Part 4 — Identifying Functional Groups (Q13–20)

The Big Idea

The functional group is the specific atom arrangement that controls how a molecule behaves. The carbon chain is just a handle — the functional group is where the chemistry happens.

Identification shortcut: Find the "special" atoms (O, N, double bonds) and check how they're connected.
What You SeeFunctional Group
C=C in the chainAlkene
−OH on a carbon (no C=O next to it)Alcohol
C=O between two carbonsKetone
C=O with −OH at end of chain (−COOH)Carboxylic Acid
C=O − O − C (ester linkage)Ester
C=O bonded to NAmide
−NH₂ with no C=O nearbyAmine

Question 13

a) Chain with internal C=C double bond

Alkene

The C=C double bond is the defining feature. No oxygen, no nitrogen — just a carbon-carbon double bond.

Question 14

b) Chain with −OH group

Alcohol

An −OH bonded to a carbon chain. No carbonyl (C=O) nearby, so it's an alcohol, not a carboxylic acid.

Question 15

c) Chain with C=C near the end

Alkene

Same rule — a C=C double bond makes it an alkene regardless of position in the chain.

Question 16

d) Chain with C=O flanked by two carbon chains

Ketone

C=O flanked by carbons on BOTH sides. No −OH on the carbonyl carbon, no N → ketone.

Question 17

e) Chain ending in −C(=O)OH

Carboxylic Acid

C=O PLUS an −OH on the same carbon = −COOH = carboxylic acid. The most oxidized single-carbon functional group.

Question 18

f) Chain with C=O bonded to −NH₂

Amide

C=O bonded to nitrogen = amide. Key distinction: amides have C=O next to N. Amines do NOT have a C=O.

Question 19

g) Cyclopentane ring with −OH

Alcohol

−OH on a ring carbon = cyclic alcohol (cyclopentanol). Ring or chain doesn't matter — −OH on carbon = alcohol.

Question 20

h) Cyclohexane ring with −NH₂

Amine

−NH₂ on a carbon with NO carbonyl nearby = amine. If there were a C=O next to the N, it would be an amide.

Part 5 — Hydrogenation of Alkenes (Q21–23)

The Big Idea

Hydrogenation adds H₂ across a C=C double bond using a Pt catalyst. The double bond breaks and each carbon grabs one H.

Alkene + H₂ ⟶Pt Alkane
The carbon skeleton stays EXACTLY the same. The only change: the double bond becomes a single bond. Always produces a fully saturated alkane.

Question 21

a) Hex-2-ene + H₂ / Pt →

Product: Hexane

The double bond disappears. The 6-carbon chain stays intact, now fully saturated.

Question 22

b) Cyclohexene + H₂ / Pt →

Product: Cyclohexane

The double bond in the ring disappears. Cyclohexene → cyclohexane.

Question 23

c) 3-Methyl-2-pentene + H₂ / Pt →

Product: 3-Methylpentane

The double bond saturates. The methyl branch stays put. Skeleton unchanged — just no more double bond.

Part 5 continued — Hydration of Alkenes (Q24–26)

The Big Idea

Hydration adds water (H₂O) across a C=C double bond using an acid catalyst (H⁺). The −OH attaches to one carbon, H to the other.

Markovnikov's Rule: OH goes to the carbon with MORE carbon substituents (the more substituted carbon). "The rich get richer."
Result: Always an alcohol.

Question 24

d) Hex-2-ene + H₂O / H⁺ →

Products: 2-Hexanol or 3-Hexanol

2-Hexanol
or
3-Hexanol

C2 and C3 are similarly substituted, so OH can attach to either, giving a mixture.

Question 25

e) Cyclohexene + H₂O / H⁺ →

Product: Cyclohexanol

Water adds across the ring's double bond. Both carbons equivalent by symmetry.

Question 26

f) 3-Methyl-1-pentene + H₂O / H⁺ →

Product: 3-Methyl-2-pentanol

Markovnikov: OH goes to C2 (the more substituted carbon of the C1=C2 double bond).

Part 6 — Carboxylic Acids + Alcohols & Amines (Q27–30)

The Big Idea

Carboxylic acids react with alcohols to form esters, and with amines to form amides. Both release water — condensation reactions.

Ester: Acid (−COOH) + Alcohol (HO−R) → Ester (−COO−R) + H₂O
Amide: Acid (−COOH) + Amine (H₂N−R) → Amide (−CONH−R) + H₂O

Where does the water come from? The −OH leaves the acid, the H leaves the alcohol/amine. Together → H₂O.
Naming Esters: [alcohol]-yl + [acid]-ate  (propanol + butanoic acid → propyl butanoate)
Naming Amides: N-[amine] + [acid]-amide  (propylamine + butanoic acid → N-propylbutanamide)

Question 27

a) Butanoic acid + Propanol →

+

Propyl Butanoate + H₂O

+H₂O

The −OH from the acid and H from the alcohol leave as water. The fragments join through a new C−O ester bond.

Question 28

b) Acetic acid + 1-Butanol →

+

Butyl Acetate + H₂O

+H₂O

Acetic acid (2C) + butanol (4C) → butyl acetate. The alcohol-derived part (butyl) goes first in the name.

Question 29

c) 3-Methylbutanoic acid + Cyclopentanol →

+

Cyclopentyl 3-Methylbutanoate + H₂O

+H₂O

Cyclopentyl (from the alcohol) + 3-methylbutanoate (from the acid).

Question 30

d) Butanoic acid + Propylamine →

+

N-Propylbutanamide + H₂O

+H₂O

Acid + amine (not alcohol) → amide (−CONH−). The N- prefix in the name indicates the propyl group is on nitrogen.

Part 7 — Hydrolysis of Esters & Amides (Q31–34)

The Big Idea

Hydrolysis is the REVERSE of ester/amide formation. Water BREAKS the bond apart, regenerating the original acid and alcohol (or amine).

Ester + H₂O → Carboxylic Acid + Alcohol
Amide + H₂O → Carboxylic Acid + Amine

Where to break: Find the C−O (ester) or C−N (amide) bond next to the C=O. Break there. Carbonyl side gets −OH (acid). Other fragment gets H (alcohol/amine).

Question 31

a) Ethyl butanoate + H₂O →

Butanoic Acid + Ethanol

Butanoic acid
+
Ethanol

Break the ester C−O. The 4C side gets −OH → butanoic acid. The 2C side gets H → ethanol.

Question 32

b) Isobutyl acetate + H₂O →

Acetic Acid + 2-Methyl-1-propanol (Isobutanol)

Acetic acid
+
Isobutanol

2C acid side → acetic acid. Branched 4C alcohol side → isobutanol (2-methyl-1-propanol).

Question 33

c) N-Ethylbutanamide + H₂O →

Butanoic Acid + Ethylamine

Butanoic acid
+
Ethylamine

Break the amide C−N. Carbonyl side gets −OH → butanoic acid. N side gets H → ethylamine.

Question 34

d) N-(2-Methylpropyl)acetamide + H₂O →

Acetic Acid + Isobutylamine (2-Methylpropanamine)

Acetic acid
+
Isobutylamine

2C acid → acetic acid. Branched amine → isobutylamine (2-methylpropanamine).