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):
Step 1: The label S_N2 tells you three facts at once: a group is substituted, the attacking species is a nucleophile, and two species take part in the slow (rate-deciding) step. Hence the rate law is first order in the halide and first order in the nucleophile, second order overall: \(Rate = k[RX][Nu^-]\).
Step 2: Picture the halide carbon as the centre. The nucleophile comes in along the line that is \(180^\circ\) away from the C–halogen bond. Bond making and bond breaking happen together, so there is only one energy hump (transition state) and no free carbocation is formed. The three groups that stay on carbon flip from one side to the other, like an umbrella turning inside out in the wind, giving inversion of configuration.
Step 3: Illustration: \(CH_3I + OH^- \rightarrow CH_3OH + I^-\).
Step 4: Since a crowded carbon is hard to approach from behind, adding alkyl groups slows S_N2. Reactivity therefore falls in the order primary methyl \(>\) 1° \(>\) 2° \(>\) 3°.

Part (ii): A weak/controlled oxidant halts oxidation at the aldehyde, so \(RCH_2OH\) to \(RCHO\) needs PCC. A vigorous oxidant pushes it to the acid, so \(RCH_2OH\) to \(RCOOH\) needs acidified KMnO₄ (or hot \(K_2Cr_2O_7/H_2SO_4\)).

Option 2 (OR):
Step 1: Naming rules: pick the longest chain, give the senior functional group the lowest number, then add substituents alphabetically.
Step 2 (i): tert-butyl chloride, a propane skeleton with Cl and CH₃ on C2: 2-chloro-2-methylpropane.
Step 3 (ii): A four-carbon chain with Br on the second carbon: 2-bromobutane.
Step 4 (iii): Parent is the aryl methyl ketone (ethanone attached to phenyl); the ring fluorine is para, i.e. position 4: 1-(4-fluorophenyl)ethan-1-one.
Step 5 (iv): With OH the top-priority group, it becomes phenol (C1). Two ring methyls give a xylenol: 3,5-dimethylphenol.
Step 6 (v): Six-carbon parent (hexanol); numbering from the tert-butyl side makes OH C3 and places methyls at 2,2,4,5: 2,2,4,5-tetramethylhexan-3-ol.
\[\boxed{\text{2-chloro-2-methylpropane, 2-bromobutane, 1-(4-fluorophenyl)ethan-1-one, 3,5-dimethylphenol, 2,2,4,5-tetramethylhexan-3-ol}}\]