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

If any word here feels unfamiliar, it is built from the ground up in the parent Isomerism note, and priorities lean on CIP Priority Rules.

The scenario matrix

Isomerism questions almost always fall into one of these cells. Read this table as "the enemy's whole playbook."

Cell	Case class	Degenerate / edge case it hides
A	Count all structural isomers of a formula	forgetting rings; miscounting metamers
B	Chain vs position vs functional — classify a pair	pairs that look chain but are functional
C	Tautomerism yes/no	zero α-H ⇒ no tautomerism (degenerate)
D	cis/trans possible or not	a carbon bearing two identical groups ⇒ no geometrical isomerism
E	Assign E/Z where cis ≠ Z	the "sign flip": priority on the cis side
F	Assign R/S at a stereocentre	lowest priority pointing toward you (must reverse)
G	Count stereoisomers, $2^{n}$	meso symmetry knocks the count down
H	Real-world / word problem	racemic mixture net rotation = zero (a "limiting" value)
I	Exam twist	"is every 4-bond carbon chiral?" — no

We now hit every cell.

Worked example Cell A — Count structural isomers of

$C_{4} H_{10}$ and of $C_{3} H_{6} O$ Forecast: guess how many for each before reading. (Most people undercount $C_{3} H_{6} O$ .)

1. $C_{4} H_{10}$ — a saturated alkane (formula $C_{n} H_{2 n + 2}$ , no ring, no double bond). Why this step? $2 (4) + 2 = 10$ H, so it is fully saturated; only the carbon skeleton can vary → this is pure chain isomerism. Skeletons: straight chain (n-butane) and one branch (2-methylpropane). Answer = 2.

2. $C_{3} H_{6} O$ — degree of unsaturation $= \frac{2 ( 3 ) + 2 - 6}{2} = 1$ . Why this step? One "degree of unsaturation" means one ring OR one double bond. That opens functional isomers, not just chain ones. Distinct constitutions: propanal ( $C H_{3} C H_{2} C H O$ ), propanone ( $C H_{3} C O C H_{3}$ ), prop-2-en-1-ol ( $C H_{2} = C H C H_{2} O H$ ), methyl vinyl ether ( $C H_{3} O C H = C H_{2}$ ), oxetane (4-membered ring), methyloxirane (epoxide ring). A solid exam answer = 6+.

Verify: every structure above has exactly 3 C, 6 H, 1 O and one degree of unsaturation. Aldehyde vs ketone = functional isomers; the ethers = another functional class; rings use the unsaturation as a ring instead of a π bond. ✔

Worked example Cell C — Tautomerism: the zero-α-H degenerate case

Forecast: which of these show keto–enol tautomerism — acetaldehyde $C H_{3} C H O$ , formaldehyde $H C H O$ , benzaldehyde $C_{6} H_{5} C H O$ ?

1. Identify the α-carbon = the carbon directly attached to the carbonyl carbon. Why this step? Tautomerism needs an H on the α-carbon to hop onto the oxygen (see parent's keto–enol derivation), so first we must locate α-C and check for H there.

2. Acetaldehyde: α-C is the $C H_{3}$ → 3 α-H's present ⇒ tautomerizes ✔.

3. Formaldehyde $H C H O$ : the only carbon is the carbonyl carbon; there is no carbon next to it → zero α-H ⇒ no tautomerism (degenerate case).

4. Benzaldehyde: the carbon next to C=O is an aromatic ring carbon carrying no removable sp³ H in the α-position → no α-H ⇒ no keto–enol tautomerism.

Verify: the rule "no α-H ⇒ no tautomerism" is the acidity idea from Acidity of Alpha-Hydrogens — no acidic H, nothing to shuttle. ✔

Worked example Cell D — Geometrical isomerism: the "two identical groups" degenerate case

Forecast: which of but-2-ene ( $C H_{3} C H = C H C H_{3}$ ) and prop-1-ene ( $C H_{2} = C H C H_{3}$ ) shows cis–trans?

1. Rule: each doubly-bonded carbon must carry two different groups, and rotation must be locked (the π bond does that). Why this step? If one alkene carbon holds two identical groups, swapping "sides" gives back the same molecule — no new isomer is created.

2. But-2-ene: left C has (CH₃, H), right C has (CH₃, H) — both carbons pass → cis and trans exist (look at the figure: same-side vs opposite-side methyls).

3. Prop-1-ene: the terminal $C H_{2} =$ carbon holds two H's (identical). Flipping them changes nothing → no geometrical isomerism.

Verify: the terminal carbon of prop-1-ene has H = H; the "same side / opposite side" pictures superimpose → confirms the degenerate case. ✔

Worked example Cell E — The sign flip: cis ≠ Z

Forecast: for 1-bromo-1-chloroprop-1-ene $C H C l = C (B r) C H_{3}$ ... actually take $(C H_{3}) (H) C = C (C l) (F)$ — is the "cis" arrangement E or Z? Guess before checking.

1. On the left carbon compare its two groups: $C H_{3}$ (carbon, Z=6) vs H (Z=1) → CH₃ wins (higher priority). Why this step? E/Z is decided by CIP Priority Rules: on each carbon rank the two attached groups by atomic number at the first point of difference.

2. On the right carbon compare Cl (Z=17) vs F (Z=9) → Cl wins.

3. Now ask: are the two winners (CH₃ and Cl) on the same side or opposite? If the drawing puts CH₃ and Cl on the same side → Z; opposite → E. Why this step? Z = zusammen = higher priorities together; this is the rigorous replacement for "cis by eye."

4. The trap: someone might call it "cis" because a big group (Cl) sits with a small group (H). But naming is by priority winners, not by size-similarity, so cis-by-eye and Z can disagree.

Verify: left winner = C (6) beats H (1); right winner = Cl (17) beats F (9). Same-side winners ⇒ Z, independent of what looks "cis." ✔

Worked example Cell F — R/S when lowest priority points toward you

Forecast: bromochlorofluoromethane $C H F C l B r$ drawn with H pointing toward the viewer and Br→Cl→F tracing clockwise — is it R or S?

1. Priorities: Br (35) > Cl (17) > F (9) > H (1). So H is priority 4 (lowest). Why this step? R/S is read only when the lowest priority points away from you; we must first confirm which way it points.

2. Here H points toward you (the wrong way). Why this step? The clock direction you see is the mirror of the true assignment when the low group faces you.

3. Trace 1→2→3 = Br→Cl→F: appears clockwise as drawn. Why this step? Clockwise normally = R, but the low group is toward us, so we flip the answer.

4. Flip clockwise → the true configuration is S.

Verify: rule — if lowest priority is toward the viewer, reverse the observed rotation. Observed clockwise ⇒ real answer S. ✔

Worked example Cell G —

$2^{n}$ vs the meso knock-down (tartaric acid) Forecast: tartaric acid $H O O C - C H O H - C H O H - C O O H$ has 2 stereocentres. Naive count $2^{2} = 4$ . How many real stereoisomers?

1. Count stereocentres: both central carbons have 4 different groups (–OH, –H, –COOH, and the rest of the chain) → $n = 2$ , naive max $= 2^{2} = 4$ . Why this step? $2^{n}$ is the ceiling for stereoisomers when there is no internal symmetry.

2. Because the two halves of the molecule are identical, one arrangement has an internal mirror plane — the (R,S) form is superimposable on its mirror image → this is a meso compound (achiral). Why this step? An internal mirror plane makes the molecule its own mirror image, so that "pair" collapses to one achiral compound.

3. Real count: (R,R) and (S,S) are an enantiomeric pair (2), plus one meso (R,S). Total = 3, not 4.

Verify: $2^{n} = 4$ but meso symmetry removes one ⇒ 3 distinct stereoisomers. The (R,S) and (S,R) labels describe the same meso molecule. ✔

Worked example Cell H — Word problem, and the "zero rotation" limiting value

Forecast: a chemist has pure (+)-2-butanol with specific rotation $[α] = + 13.5°$ . She mixes it 50:50 (by moles) with its enantiomer (−)-2-butanol. What is the observed rotation of the mixture?

1. Enantiomers rotate plane-polarized light equally in magnitude but opposite in sign (see Optical Activity and Polarimetry). Why this step? The (−) form contributes $- 13.5°$ per equal amount, exactly cancelling the $+ 13.5°$ .

2. Net rotation $= \frac{1}{2} (+ 13.5°) + \frac{1}{2} (- 13.5°) = 0°$ . Why this step? A 50:50 enantiomer mixture is a racemic mixture, and its net optical activity is the limiting value zero — optically inactive despite being made of two active molecules.

3. Conclusion: the polarimeter reads 0°; the mixture is a racemate ( $\pm$ ), sometimes written (RS).

Verify: $\frac{1}{2} (13.5) - \frac{1}{2} (13.5) = 0$ . Racemic ⇒ inactive. ✔

Recall Did every cell get covered?

A count ::: Example A ( $C_{4} H_{10}$ =2, $C_{3} H_{6} O$ ≈6) B classify ::: Example B (position / chain / functional) C zero-α-H ::: Example C (formaldehyde no tautomer) D two-identical-groups ::: Example D (prop-1-ene fails) E cis≠Z flip ::: Example E (priority winners decide) F low-priority-toward-you ::: Example F (reverse to S) G meso knock-down ::: Example G (tartaric acid = 3) H racemic zero ::: Example H (net rotation 0°) I 4 bonds ≠ chiral ::: Example I ( $C H_{2} C l B r$ achiral)