Question 1 of 5: Carbon Classification & Alkene Stability Ranking
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-12, Organic Chemistry — May 2014. 3 hours, closed-book
examination; no calculator allowed. Answer ALL FIVE problems; each problem is of equal value
(20 points), and the lettered/numbered sub-parts of a given problem may be treated
independently.
a meta (1,3) relationship, matching the drawing
convention used two questions later for Question 3(a)(i)'s explicitly-labelled meta compound. Solved below as meta; this changes the product from phthalic acid (ortho) to isophthalic acid
(meta).
a) The compound is 1-ethyl-2-(2-methylpropyl)cyclohexane — an ethyl
group and an isobutyl (2-methylpropyl) group on two adjacent ring carbons.
Fig. Q1a — 1-ethyl-2-(2-methylpropyl)cyclohexane, C12H24 (substituents on adjacent ring carbons)
(i)/(ii) Classification. A carbon's class depends only on how many
other carbons it is directly bonded to: primary (p) = 1 carbon neighbour, secondary (s)
= 2, tertiary (t) = 3. Working around the 12-carbon skeleton:
Ring carbons. The two substituent-bearing ring carbons (C1: ethyl; C2:
isobutyl) are each bonded to two ring neighbours plus the substituent's first carbon
— three carbon neighbours each → both tertiary (t). The remaining four
ring carbons (C3–C6) are each bonded only to their two ring neighbours → all four
secondary (s).
Ethyl substituent (–CH2–CH3). The
–CH2– is bonded to the ring carbon and the terminal methyl — two
neighbours → secondary (s). The terminal CH3 has one neighbour
→ primary (p).
Isobutyl substituent (–CH2–CH(CH3)–CH3).
The first –CH2– (bonded to the ring and the next carbon) →
secondary (s). The branch CH (bonded to the CH2, and both methyl
groups) — three neighbours → tertiary (t). Both terminal methyls (one
neighbour each) → primary (p) × 2.
Class
Count
Which carbons
Primary (p)
3
ethyl CH3; both isobutyl branch CH3's
Secondary (s)
6
ring C3–C6 (4); ethyl CH2; isobutyl CH2
Tertiary (t)
3
ring C1, ring C2; isobutyl branch CH
b) Alkene stability. All three alkenes share the same degree of substitution
pattern at the double bond itself (each alkene carbon bears one alkyl chain and one H, i.e. a
1,2-dialkyl/disubstituted alkene) — the differences are (1) E vs. Z geometry and (2) how
bulky the two flanking alkyl groups are right at the double bond.
Fig. Q1b — the three alkenes (decreasing stability left→right)
Trans (E) vs. cis (Z): trans is always more stable. In the trans isomer the
two ethyl groups point to opposite sides of the double bond, minimizing steric (van der Waals)
crowding; in the cis isomer they are forced onto the same side, creating steric strain that
raises the molecule's energy (lower stability). This ranks trans-3-hexene above cis-3-hexene.
Bulkier flanking groups increase cis strain further. Cis-2,5-dimethyl-3-hexene
replaces each plain ethyl flank with an isopropyl-type flank (the branch methyl sits directly on
the allylic carbon, right next to the double bond), so the two bulkier groups clash even
more severely on the same side than cis-3-hexene's two ethyls do. This makes it the least stable
of the three.
