Isomerism — structural (chain, position, functional, metamerism, tautomerism) and stereo (geometrical - cis-trans - E-Z,: Exercises

Level 1 — Recognition

You are only asked to name the category. No 3D thinking yet.

Problem 1.1

$C_{4} H_{10}$ can be n-butane ( $C H_{3} C H_{2} C H_{2} C H_{3}$ ) or isobutane ( $(C H_{3})_{3} C H$ ). What type of isomerism relates them?

Recall Solution

What we check first: do the atoms connect differently, or only sit differently in 3D? Here n-butane is a straight 4-carbon chain; isobutane is a 3-carbon chain with a branch. The skeleton itself changed. Answer: Chain isomerism (a structural/constitutional isomerism). Both are $C_{4} H_{10}$ (check: n-butane $= 4 C + 10 H$ ; isobutane $= 4 C + 10 H$ ✓), so the formula is shared — it is the carbon backbone that differs.

Problem 1.2

Classify each pair: (a) $C H_{3} C H_{2} O H$ vs $C H_{3} O C H_{3}$ (b) propan-1-ol vs propan-2-ol (c) $C H_{3} O C_{3} H_{7}$ vs $C_{2} H_{5} O C_{2} H_{5}$

Recall Solution

The decision tree: same functional group? → if NO, functional isomerism. If YES, same skeleton but group moved? → position isomerism. If it's an ether/amine and only the alkyl split differs → metamerism. (a) Alcohol vs ether — the functional group changed → functional isomerism. (b) Both alcohols, same 3-carbon chain, $- O H$ just slides from C1 to C2 → position isomerism. (c) Both ethers ( $C_{4} H_{10} O$ ), same $- O -$ centre; the carbons split $1 + 3$ vs $2 + 2$ around it → metamerism.

Level 2 — Application

Now you apply a rule — count, or check a condition.

Problem 2.1

Draw/list all structural isomers of $C_{5} H_{12}$ (pentanes). How many?

Recall Solution

Strategy: fix the longest carbon chain, then shorten it by one and attach the freed carbon as a branch, avoiding duplicates.

Longest chain 5: n-pentane $C H_{3} (C H_{2})_{3} C H_{3}$ .
Longest chain 4, one methyl branch (only unique spot is C2): 2-methylbutane (isopentane). Putting the methyl on C3 gives the same molecule flipped, so no new isomer.
Longest chain 3, two methyls on the central C: 2,2-dimethylpropane (neopentane). Answer: 3 structural isomers of $C_{5} H_{12}$ . (Each still has $5 C + 12 H$ ✓.)

Problem 2.2

Does but-2-ene ( $C H_{3} C H = C H C H_{3}$ ) show geometrical isomerism? What about propene ( $C H_{2} = C H C H_{3}$ )?

Recall Solution

The two conditions (from parent note): (1) restricted rotation — a $C = C$ or a ring; (2) each doubly-bonded carbon carries two different groups.

But-2-ene: each alkene carbon holds one $C H_{3}$ and one $H$ — two different groups on each → both conditions met → YES, cis and trans exist.
Propene: the $C H_{2} =$ carbon holds $H$ and $H$ — two identical groups. Flipping them gives nothing new → NO geometrical isomerism. See figure below for why identical groups kill the isomerism.

Figure — Isomerism — structural (chain, position, functional, metamerism, tautomerism) and stereo (geometrical - cis-trans - E-Z,

Level 3 — Analysis

Here you must compare, reason, or distinguish look-alikes.

Problem 3.1

For 2-butenoic acid $C H_{3} - C H = C H - C O O H$ , assign E or Z to the isomer that has the two $C H_{3}$ and $C O O H$ groups on the same side.

Recall Solution

Tool: CIP priority (see CIP Priority Rules) — on each alkene carbon, rank its two substituents by atomic number of the first atom; the higher-priority one is the "flag."

Left carbon: substituents are $C H_{3}$ (first atom C) and $H$ (first atom H). $C > H$ → $C H_{3}$ is higher priority.
Right carbon: substituents are $C O O H$ (first atom C) and $H$ . → $C O O H$ is higher priority. Where do the flags sit? We're told $C H_{3}$ and $C O O H$ are on the same side. Those are exactly the two higher-priority groups. Higher-priority groups on the same side ⇒ Z (zusammen = together).

Problem 3.2

In 1-chloropropene $C l C H = C H - C H_{3}$ , the isomer with $C l$ and $C H_{3}$ on opposite sides — is it cis or trans, and is it E or Z? Do the two labels agree?

Recall Solution

cis/trans (by-eye similarity): the "reference" groups are usually the carbon chain / the like groups. With $C l$ and $C H_{3}$ on opposite sides, by the everyday "main groups opposite" reading many call this trans. E/Z (rigorous CIP):

Left carbon: $C l$ (Z=17) vs $H$ → $C l$ higher.
Right carbon: $C H_{3}$ (C, Z=6) vs $H$ → $C H_{3}$ higher.
The two higher-priority flags ( $C l$ , $C H_{3}$ ) are on opposite sides ⇒ E. Here trans and E happen to agree, but only because the "obvious" groups also won the priority contest. This is not guaranteed.

Level 4 — Synthesis

Combine two or more ideas: chirality + counting + symmetry.

Problem 4.1

2-chlorobutane $C H_{3} - C H C l - C H_{2} - C H_{3}$ : is there a chiral centre? If yes, how many stereoisomers, and what are they called?

Recall Solution

Find candidate carbons: we need a carbon with four different groups (a stereocentre). Look at C2. C2's four attachments: $H$ , $C l$ , $- C H_{3}$ (methyl), $- C H_{2} C H_{3}$ (ethyl). All four different ⇒ C2 is a chiral centre. Count: one stereocentre, no internal symmetry ⇒ $2^{n} = 2^{1} = 2$ stereoisomers. These two are enantiomers (non-superimposable mirror images), labelled $R$ and $S$ . Related content: Optical Activity and Polarimetry. Answer: 2 stereoisomers (the $R$ and $S$ enantiomers).

Problem 4.2

Assign R or S to bromochlorofluoromethane $C H F C l B r$ when, with $H$ pointing away from you, the trace $B r \to C l \to F$ runs clockwise.

Recall Solution

Step 1 — rank by atomic number: $B r (35) > C l (17) > F (9) > H (1)$ . So priorities are $B r = 1, C l = 2, F = 3, H = 4$ . Step 2 — lowest ( $H$ ) already points away (given). Good — no re-orientation needed. Step 3 — trace $1 \to 2 \to 3$ , i.e. $B r \to C l \to F$ . We're told this is clockwise. Clockwise with lowest away ⇒ R (rectus).

Level 5 — Mastery

Full multi-concept problems: counting with symmetry, tautomer feasibility, and a stereocentre census.

Problem 5.1

Tartaric acid $H O O C - C H O H - C H O H - C O O H$ has two stereocentres. Naïvely $2^{2} = 4$ . How many distinct stereoisomers actually exist, and why the discrepancy?

Recall Solution

Census of the 4 combinations of (C2, C3): $(R, R)$ , $(S, S)$ , $(R, S)$ , $(S, R)$ .

$(R, R)$ and $(S, S)$ are non-superimposable mirror images → a genuine enantiomer pair (2 compounds).
$(R, S)$ and $(S, R)$ : the molecule has an internal mirror plane (top half mirrors bottom half). So $(R, S)$ is $(S, R)$ — the same single molecule, which is achiral despite two stereocentres. This is a meso compound. Count: $2^{n} = 4$ minus the $(R, S) / (S, R)$ overlap $=$ 3 distinct stereoisomers ((+), (−), and meso).

Problem 5.2

Which of these can undergo keto–enol tautomerism? (a) acetone $C H_{3} C O C H_{3}$ (b) formaldehyde $H C H O$ (c) benzaldehyde $C_{6} H_{5} C H O$ . State how many tautomerize.

Recall Solution

The one requirement: an α-hydrogen — a hydrogen on the carbon directly next to the carbonyl $C = O$ . Without it, there is no proton to shuttle and no way to form the enol $C = C$ . See Acidity of Alpha-Hydrogens.

(a) Acetone: the $C H_{3}$ groups flanking $C = O$ carry α-H's → tautomerizes ✓.
(b) Formaldehyde: the carbonyl carbon holds only two $H$ 's and no neighbouring carbon — no α-carbon at all → no α-H → cannot tautomerize ✗.
(c) Benzaldehyde: the group next to $C = O$ is the aromatic ring carbon, which bears no H on an sp³ α-carbon available to shuttle → cannot classically tautomerize ✗. Answer: only 1 (acetone) tautomerizes.

Problem 5.3

For a molecule with 3 independent stereocentres and no symmetry, and separately a molecule with 4 stereocentres one pair of which creates a meso relationship — give the maximum stereoisomer counts.

Recall Solution

Rule: max $= 2^{n}$ for $n$ stereocentres, reduced whenever an internal mirror plane makes a mirror-pair identical.

3 centres, no symmetry: $2^{3} =$ 8 stereoisomers.
4 centres, naïve $2^{4} = 16$ ; but a meso relationship makes some mirror-pairs coincide. For the symmetric $2, 3, 4$ -type case where the molecule can fold on a central plane, the count drops. The clean textbook case (2 centres) gives $2^{2} - 1 = 3$ ; scaling the same fold to 4 centres with a central plane gives $2^{n - 1} + 2^{(n /2) - 1} = 2^{3} + 2^{1} =$ 10 stereoisomers. Answer: 8 and 10 respectively.

Recall Rapid re-test (all answers)

1.1 chain · 1.2 functional / position / metamerism · 2.1 3 · 2.2 but-2-ene yes, propene no · 3.1 Z · 3.2 trans = E (they agree) · 4.1 2 enantiomers · 4.2 R · 5.1 3 (meso) · 5.2 1 (acetone) · 5.3 8 and 10.

Active Recall

Structural isomers of pentane (

C_{5} H_{12}

)

3 — n-pentane, isopentane (2-methylbutane), neopentane (2,2-dimethylpropane).

Stereoisomers of 2-chlorobutane

2 — the R and S enantiomers.

Distinct stereoisomers of tartaric acid

3 — (+), (−), and one meso form.

C H F C l B r

with H back and Br→Cl→F clockwise

R configuration.

Which of acetone / formaldehyde / benzaldehyde tautomerizes

only acetone (it alone has an α-hydrogen).