Question

(i) Explain the mechanism of the Bimolecular Nucleophilic Substitution reaction (S_N2) in a haloalkane and give the order of reactivity of the alkyl halides based on S_N2. (3)
(ii) Name the reagent used in the following reactions: (x) Oxidation of a primary alcohol to an aldehyde. (y) Oxidation of a primary alcohol to a carboxylic acid. (2)
OR
Write the IUPAC names of the following compounds (1+1+1+1+1):
(i) (CH₃)₃C–Cl
(ii) CH₃–CH₂–CH(Br)–CH₃
(iii) 4-F–C₆H₄–COCH₃
(iv) a benzene ring bearing two CH₃ groups and one OH group
(v) CH₃–CH(CH₃)–CH(CH₃)–CH(OH)–C(CH₃)₃

Hint:

S_N2 is a one-step back-side attack with inversion; reactivity is CH₃X > 1° > 2° > 3°. For oxidation use PCC (to aldehyde) and KMnO₄ (to acid). For the OR part, pick the longest chain and give OH/Cl/Br the lowest locant.

Answer 1

Option 1

Part (i): S_N2 mechanism and reactivity order
Step 1 (What S_N2 means): S_N2 stands for Substitution, Nucleophilic, Bimolecular. It is a one-step (concerted) reaction whose rate depends on the concentration of BOTH the alkyl halide and the nucleophile, so it follows second order kinetics: Rate = \( k[RX][Nu^-] \).
Step 2 (How it happens): The nucleophile attacks the carbon atom carrying the halogen from the side exactly opposite (back side) to the leaving halogen. In the transition state this carbon is partly bonded to both the incoming nucleophile and the outgoing halogen; the three other groups lie in a plane (trigonal bipyramidal transition state). As the new bond forms, the C–X bond breaks, and the halide leaves.
Step 3 (Result): The reaction goes through a single transition state with no intermediate, and the configuration at carbon is turned inside-out (inversion of configuration, called Walden inversion). Example: \( CH_3Br + OH^- \rightarrow CH_3OH + Br^- \).
Step 4 (Reactivity order): Because the nucleophile must reach the carbon from the back, bulky alkyl groups block the attack (steric hindrance). Fewer alkyl groups mean an easier attack, so the S_N2 reactivity order is: CH₃X > primary (1°) > secondary (2°) > tertiary (3°) halide. Methyl halide reacts fastest and the tertiary halide is essentially unreactive by S_N2.

Part (ii): Reagents
(x) To stop at the aldehyde and avoid over-oxidation, a mild oxidising agent is used: PCC (Pyridinium Chlorochromate) in dichloromethane (Collins reagent also works). \( RCH_2OH \xrightarrow{PCC} RCHO \).
(y) To go all the way to the carboxylic acid, a strong oxidising agent is used: acidified KMnO₄ (or K₂Cr₂O₇/H₂SO₄). \( RCH_2OH \xrightarrow{KMnO_4/H^+} RCOOH \).

Option 2 (OR): IUPAC names
Step 1 (i): (CH₃)₃C–Cl has a central carbon bonded to three methyl groups and one Cl. The longest chain is propane; the central C2 carries a methyl and a chlorine. Name: 2-chloro-2-methylpropane.
Step 2 (ii): CH₃CH₂CH(Br)CH₃ is a four-carbon chain (butane) with Br on C2 (lowest locant). Name: 2-bromobutane.
Step 3 (iii): A benzene ring carries a –COCH₃ (acetyl, the parent ketone is acetophenone = 1-phenylethan-1-one) and an F at the para (4) position. Name: 1-(4-fluorophenyl)ethan-1-one (para-fluoroacetophenone).
Step 4 (iv): The –OH on a benzene ring makes it a phenol, so OH gets C1. Two methyl groups on the ring give a dimethylphenol (xylenol); for the shown structure the name is 3,5-dimethylphenol (the two methyls occupy positions 3 and 5).
Step 5 (v): CH₃–CH(CH₃)–CH(CH₃)–CH(OH)–C(CH₃)₃. The longest chain that includes the –OH carbon is six carbons (hexane). Numbering from the tert-butyl end gives OH the lowest locant (C3): C2 carries two methyls, C4 and C5 each carry one methyl. Name: 2,2,4,5-tetramethylhexan-3-ol.
\[\boxed{\text{2-chloro-2-methylpropane; 2-bromobutane; 1-(4-fluorophenyl)ethan-1-one; 3,5-dimethylphenol; 2,2,4,5-tetramethylhexan-3-ol}}\]