Hydrocarbons

The chlorination of methane proceeds via a Free Radical Mechanism.
During the chain termination step, two methyl free radicals ($\dot{C}H_3$) can combine to form an ethane molecule.

(a) $CH_3CH=C(CH_3)_2$: 2-Methylbut-2-ene.

(b) $CH_2=CH-C \equiv C-CH_3$: Pent-1-en-3-yne (Numbering from double bond end priority).

(c) (Image implied): 1,3-Butadiene or similar cyclic structure (Assuming standard text question 1,3-Butadiene: Buta-1,3-diene).

(d) $–CH_2–CH_2–CH=CH_2$ attached to Phenyl: 4-Phenylbut-1-ene.

(e) (Image implied): Methylphenol/Cresol? (Assuming standard text question: 2-Methylphenol).

(f) $CH_3(CH_2)_4CH(CH_2)_3CH_3$ with side chain $-CH_2-CH(CH_3)_2$:
Longest chain: 10 carbons (Decane).
Substituent at C5: 2-methylpropyl (isobutyl).
Name: 5-(2-Methylpropyl)decane.

(g) $CH_3-CH=CH-CH_2-CH=CH-CH(C_2H_5)-CH_2-CH=CH_2$:
Longest chain containing double bonds: 10 carbons (Deca…triene).
Numbering from right gives double bonds at 1, 4, 7.
Substituent: Ethyl at C4.
Name: 4-Ethyldeca-1,5,8-triene.

(a) $C_4H_8$ (One double bond):

• $CH_2=CH-CH_2-CH_3$: But-1-ene
• $CH_3-CH=CH-CH_3$: But-2-ene (cis/trans)
• $(CH_3)_2C=CH_2$: 2-Methylprop-1-ene

(b) $C_5H_8$ (One triple bond):

• $HC \equiv C-CH_2-CH_2-CH_3$: Pent-1-yne
• $CH_3-C \equiv C-CH_2-CH_3$: Pent-2-yne
• $HC \equiv C-CH(CH_3)_2$: 3-Methylbut-1-yne

(i) Pent-2-ene: $CH_3-CH=CH-CH_2CH_3 \xrightarrow{O_3} CH_3CHO + CH_3CH_2CHO$ (Ethanal + Propanal).

(ii) 3,4-Dimethylhept-3-ene: Cleavage at C3-C4 double bond.
Product 1: Butan-2-one ($CH_3COCH_2CH_3$).
Product 2: Pentan-2-one ($CH_3COCH_2CH_2CH_3$).

(iii) 2-Ethylbut-1-ene: Cleavage gives Methanal ($HCHO$) + Pentan-3-one.

(iv) 1-Phenylbut-1-ene: $Ph-CH=CH-CH_2CH_3$. Gives Benzaldehyde + Propanal.

Products: $CH_3CHO$ (Ethanal) and $CH_3CH_2COCH_2CH_3$ (Pentan-3-one).
Remove oxygen and join carbons with double bond.
Structure: $CH_3-CH=C(CH_2CH_3)_2$.
IUPAC Name: 3-Ethylpent-2-ene.

Aldehyde with molar mass 44 u is Ethanal ($CH_3CHO$).
Since ozonolysis gives two moles of ethanal, the alkene is symmetric.
Join two ethanal molecules: $CH_3-CH=O + O=CH-CH_3 \rightarrow CH_3-CH=CH-CH_3$.
Structure: But-2-ene.

Propanal: $CH_3CH_2CH=O$.
Pentan-3-one: $O=C(CH_2CH_3)_2$.
Join double bond: $CH_3CH_2-CH=C(CH_2CH_3)_2$.
Structure: 3-Ethylhex-3-ene.

(i) Butane ($C_4H_{10}$):
$2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O$

(ii) Pentene ($C_5H_{10}$):
$2C_5H_{10} + 15O_2 \rightarrow 10CO_2 + 10H_2O$

(iii) Hexyne ($C_6H_{10}$):
$2C_6H_{10} + 17O_2 \rightarrow 12CO_2 + 10H_2O$

(iv) Toluene ($C_7H_8$):
$C_7H_8 + 9O_2 \rightarrow 7CO_2 + 4H_2O$

Cis-isomer has a higher boiling point.
Reason: The cis-isomer has a net dipole moment because the alkyl groups are on the same side, leading to stronger dipole-dipole interactions. The trans-isomer is more symmetric, has near-zero dipole moment, and weaker intermolecular forces.

Benzene is extraordinarily stable due to Resonance. The 6 $\pi$-electrons are delocalized over the entire carbon ring, lowering the potential energy of the molecule. This delocalization (aromaticity) makes it much more stable than a hypothetical cyclic triene with localized bonds.

1. Planarity: The molecule must be flat to allow orbital overlap.
2. Complete Delocalization: Continuous ring of overlapping p-orbitals.
3. Hückel’s Rule: The ring must contain $(4n + 2)$ $\pi$-electrons, where $n$ is an integer (0, 1, 2…).

(i) Cyclooctatetraene (8 $\pi$ e-): It has $4n$ $\pi$-electrons ($n=2$), not $4n+2$. It is anti-aromatic (though actually non-aromatic as it adopts a tub shape to avoid planarity).

(ii) Cyclopentadiene ($sp^3$ carbon): One carbon is $sp^3$ hybridised, breaking the cyclic conjugation/planarity.

(iii) Cycloheptatriene ($sp^3$ carbon): One carbon is $sp^3$, breaking conjugation.

(i) p-Nitrobromobenzene: $C_6H_6 \xrightarrow{Br_2, FeBr_3} Bromobenzene \xrightarrow{HNO_3, H_2SO_4} \text{p-Nitrobromobenzene}$ (Br is o,p-directing).

(ii) m-Nitrochlorobenzene: $C_6H_6 \xrightarrow{HNO_3, H_2SO_4} Nitrobenzene \xrightarrow{Cl_2, AlCl_3} \text{m-Nitrochlorobenzene}$ ($NO_2$ is m-directing).

(iii) p-Nitrotoluene: $C_6H_6 \xrightarrow{CH_3Cl, AlCl_3} Toluene \xrightarrow{HNO_3, H_2SO_4} \text{p-Nitrotoluene}$.

(iv) Acetophenone: $C_6H_6 \xrightarrow{CH_3COCl, Anhyd. AlCl_3} Acetophenone$ (Friedel-Crafts Acylation).

Structure: $H_3C^a – C^bH_2 – C^c(CH_3^d)_2 – C^eH_2 – C^fH(CH_3^g)_2$

• $1^\circ$ (Primary): a, d(x2), g(x2). 5 carbons (Bonded to 3 H each).
• $2^\circ$ (Secondary): b, e. 2 carbons (Bonded to 2 H each).
• $3^\circ$ (Tertiary): f. 1 carbon (Bonded to 1 H).
• $4^\circ$ (Quaternary): c. 1 carbon (Bonded to 0 H).

Branching decreases the boiling point of alkanes. Branching makes the molecule more spherical, reducing the surface area for contact. This weakens the Van der Waals forces between molecules, lowering the boiling point.

Without Peroxide (Markovnikov): Electrophilic addition. $H^+$ adds first to form stable $2^\circ$ carbocation. $Br^-$ attacks $2^\circ$ C. Product: 2-Bromopropane.

With Peroxide (Anti-Markovnikov/Kharasch): Free radical mechanism. Peroxide breaks to radicals, generates $\dot{Br}$. $\dot{Br}$ adds to propene to form more stable $2^\circ$ free radical (Br adds to terminal C). Then H abstraction yields 1-Bromopropane.

o-Xylene is a resonance hybrid of two Kekulé structures (one with double bond between methyl carbons, one single).
Products: Methylglyoxal ($CH_3COCHO$), Glyoxal ($OHC-CHO$), Dimethylglyoxal ($CH_3COCOCH_3$).
The formation of all three products proves that the double bonds are delocalized and exist in both positions, supporting the Kekulé resonance model.

Order: Ethyne > Benzene > n-Hexane.
Reason: Acidic character depends on the % s-character of the carbon attached to hydrogen.
Ethyne (sp, 50% s) > Benzene ($sp^2$, 33% s) > Hexane ($sp^3$, 25% s).
Higher s-character means electrons are held closer to the nucleus (more electronegative carbon), making $H^+$ release easier.

Benzene is electron-rich due to its $\pi$-electron cloud. It attracts electrophiles ($E^+$) easily.
It repels nucleophiles (electron-rich species). Also, nucleophilic substitution requires disrupting the aromaticity or attacking a repelling cloud, which is energetically difficult under normal conditions.

(i) Ethyne: Pass through Red Hot Iron Tube at 873K (Cyclic Polymerization) $\rightarrow$ Benzene.

(ii) Ethene: $Br_2/CCl_4 \rightarrow$ 1,2-Dibromoethane $\xrightarrow{alc. KOH}$ Ethyne $\xrightarrow{\text{Red Hot Fe}}$ Benzene.

(iii) Hexane: Aromatization ($Cr_2O_3/Al_2O_3$, 773K, 10-20 atm) $\rightarrow$ Benzene.

Hydrogenation adds $H_2$ to double bonds.
Possible Alkenes:
1. 2-Methylbut-1-ene ($CH_2=C(CH_3)CH_2CH_3$)
2. 2-Methylbut-2-ene ($CH_3C(CH_3)=CHCH_3$)
3. 3-Methylbut-1-ene ($CH_2=CHCH(CH_3)_2$)

(a) Chlorobenzene > p-Nitrochlorobenzene > 2,4-Dinitrochlorobenzene.
(Nitro group is electron withdrawing, deactivating the ring).

(b) Toluene > p-H3C-C6H4-NO2 > p-O2N-C6H4-NO2.
(Methyl activates, Nitro deactivates).

Order: Toluene > Benzene > m-Dinitrobenzene.
Reason: Methyl group in Toluene is electron-donating (+I, Hyperconjugation), activating the ring for electrophilic attack ($NO_2^+$). Nitro groups are electron-withdrawing (-I, -R), strongly deactivating the ring.

Other than Anhydrous $AlCl_3$, we can use Anhydrous $FeCl_3$, $SnCl_4$, or $BF_3$ as Lewis acids for Friedel-Crafts alkylation.

Wurtz reaction involves coupling of alkyl halides ($2RX + 2Na \to R-R$).
If we try to prepare an odd carbon alkane (e.g., Propane) using two different halides ($CH_3Cl + C_2H_5Cl$), we get a mixture of three products:
1. Ethane ($CH_3-CH_3$)
2. Propane ($CH_3-C_2H_5$)
3. Butane ($C_2H_5-C_2H_5$)
Separating these closely boiling alkanes is difficult, making the yield of the desired odd alkane low.