Carbon Classification, Hydrogenation, Markovnikov Addition & Nitration: 04-BS-12, December 2014, Q4

Question 4: Carbon Classification, Hydrogenation, Markovnikov Addition & Nitration (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) CH₃CH₂C(CH₃)₂CH₂CH₃ is 3,3-dimethylpentane. A carbon's class is read directly off how many other carbons it is bonded to: 1→primary, 2→secondary, 3→tertiary — and this structure has a carbon bonded to four, which the p/s/t scheme alone does not label.

Fig. Q4a — C3 is bonded to four other carbons (quaternary, 0 H) — not tertiary

Main chain, C1–C5. C1 (CH₃, bonded only to C2) and C5 (CH₃, bonded only to C4) are each bonded to one carbon → primary. C2 (CH₂, bonded to C1 and C3) and C4 (CH₂, bonded to C3 and C5) are each bonded to two carbons → secondary.
C3 — the trap. C3 is bonded to C2, C4, and both methyl branches — four carbon neighbours in total, with zero hydrogens left on C3 itself. This is a quaternary carbon, a distinct category the p/s/t scheme does not cover; it is not tertiary, even though it superficially "has three things attached" if the branches are miscounted.
The two methyl branches on C3. Each branch CH₃ is bonded only to C3 — one carbon neighbour each → both primary.

Class	Count	Which carbons
Primary	4	C1, C5, both methyl branches on C3
Secondary	2	C2, C4
Tertiary	0	—
Quaternary	1	C3 (bonded to 4 carbons, 0 H)

b) Methylenecyclohexane + H₂. Pd hydrogenates the exocyclic C=C only (the ring itself is already fully saturated — there is no aromatic system here to reduce):

→

Fig. Q4b — hydrogenation of the exocyclic alkene only (ring stays saturated)

\mathrm{C_7H_{12} + H_2 \xrightarrow{Pd} C_7H_{14}}

Product: methylcyclohexane.

c) 3-Methylpent-2-ene + HCl. The alkene carbons are unevenly substituted: the left carbon (bearing the ethyl chain and a methyl branch) carries two alkyl substituents, the right carbon (bearing only a terminal methyl) carries one plus an H. Protonation therefore adds H⁺ to the less substituted right-hand carbon (it already has the H to begin with), generating the more stable tertiary carbocation on the left-hand carbon; chloride then captures that cation (Markovnikov's rule):

HCl

→

Fig. Q4c — Markovnikov addition: Cl and the new C–C bond land on the more substituted (tertiary) carbon

\mathrm{CH_3CH_2C(CH_3){=}CHCH_3 + HCl \longrightarrow CH_3CH_2C(CH_3)(Cl)CH_2CH_3}

Product: 3-chloro-3-methylpentane.

d) Benzene + HNO₃/H₂SO₄. H₂SO₄ protonates nitric acid to generate the nitronium electrophile NO₂⁺, which substitutes onto the ring by the same attack→arenium→deprotonation sequence used in Question 3(b):

HNO3

→

H2SO4

Fig. Q4d — electrophilic aromatic nitration

\mathrm{C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O}

Product: nitrobenzene.

Part	Product
(a)	4 primary, 2 secondary, 0 tertiary, 1 quaternary carbon
(b)	methylcyclohexane
(c)	3-chloro-3-methylpentane
(d)	nitrobenzene

Concept & Theory

Topic. Primary/secondary/tertiary/quaternary carbon classification; chemoselective hydrogenation; Markovnikov addition of HCl; nitration.

Key relations. Carbon class = number of carbon neighbours (1/2/3/4 → p/s/t/quaternary). Markovnikov: H to the alkene carbon with more H's already, forming the more substituted (more stable) carbocation. Nitronium generation: HNO₃+2H₂SO₄→NO₂⁺+H₃O⁺+2HSO₄⁻.

Why this works. A quaternary carbon has no C–H bonds left at all, so it cannot participate in radical halogenation or many oxidation reactions that require abstracting a C–H at that position — the classification has real chemical consequences, not just bookkeeping value. A catalyst hydrogenates whichever π-system is present without discriminating by position; if only one alkene is present (as here), only that one reacts.

Common pitfall. Mislabelling a quaternary carbon as tertiary by only counting "three other groups drawn nearby" instead of literally counting carbon-to-carbon bonds; or protonating the wrong alkene carbon in step (c) and reporting the anti-Markovnikov secondary chloride instead.

Source. McMurry, Organic Chemistry, 9th ed., Ch. 3 (carbon classification), Ch. 7 (hydrogenation, Markovnikov addition), Ch. 16 (nitration).

Question 4 of 5: Carbon Classification, Hydrogenation, Markovnikov Addition & Nitration

Question 4: Carbon Classification, Hydrogenation, Markovnikov Addition & Nitration (20 marks)