Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-12, Organic Chemistry — December 2014. 3 hours, closed-book
examination; any non-communicating (non-programmable) calculator permitted. Answer ALL FIVE
problems; each problem is of equal value (20 points), and the lettered sub-parts of a given
problem may be treated independently.
a) Cis/trans-3-heptene, C7H14. The C3=C4 double bond
carries an ethyl group on one side and a propyl group on the other; cis places both alkyl chains
on the same side of the double bond, trans places them on opposite sides.
Fig. Q3a — C7H14, both geometric isomers of 3-heptene
Both are C7H14 (one degree of unsaturation, the C=C);
they are configurational (E/Z) isomers of each other, not constitutional isomers — identical
connectivity, different 3-D arrangement about the fixed double bond.
Isomer
Formula
Geometry
trans-3-heptene (E)
C7H14
ethyl & propyl on opposite sides
cis-3-heptene (Z)
C7H14
ethyl & propyl on the same side
b) Friedel–Crafts acylation mechanism. AlCl3 is a strong Lewis
acid; it activates the acyl chloride into a resonance-stabilised acylium ion, which is
electrophilic enough to attack the electron-rich benzene ring. The mechanism runs in three
elementary steps.
Step 1 — Lewis-acid activation of the acyl chloride. AlCl3's
empty p-orbital accepts a lone pair from the chloride's Cl, polarising and then fully breaking the
C–Cl bond and generating a resonance-stabilised acylium ion,
CH3CH2C≡O+ (positive charge shared between the carbon and
the oxygen), plus the tetrahedral counter-ion AlCl4−:
$$\mathrm{CH_3CH_2C(=O)Cl + AlCl_3 \longrightarrow CH_3CH_2C{\equiv}O^+ + AlCl_4^-}$$
Step 2 — electrophilic attack on the aromatic ring. A pair of
π-electrons from the benzene ring attacks the electrophilic acylium carbon, forming a new
C–C bond. This breaks the ring's aromaticity and generates the resonance-stabilised
arenium (Wheland) intermediate: a cyclohexadienyl cation whose ipso carbon is now
sp3 (bonded to both the acyl group and its original H), with the positive charge
delocalised over the three remaining ring positions ortho/para to that carbon:
Fig. Q3b — sp3 ipso carbon, positive charge delocalised over 3 ring positions (only 2 of 3 resonance forms shown as double bonds here)
Step 3 — loss of H+ restores aromaticity. The
tetrahedral AlCl4− counter-ion (or a second equivalent of the leaving
Cl−) removes the ipso carbon's remaining H+; the C–H bonding pair
becomes the new ring π-bond, regenerating the aromatic sextet and releasing HCl, and AlCl3 is
regenerated (catalytic, not consumed overall):
$$\mathrm{[\text{arenium}]^+ + AlCl_4^- \longrightarrow C_6H_5C(=O)CH_2CH_3 + HCl + AlCl_3}$$
(Balanced with a leading coefficient of 2 on benzene, since benzene's six carbons and six
hydrogens both give an odd oxygen-atom count on a per-molecule basis; doubling everything clears
the fraction.)
Part
Result
(a)
cis- and trans-3-heptene, both C7H14
(b)
propiophenone + HCl, via acylium → arenium → deprotonation