Organic Chemistry Practice Exam — Free

Classify the substrate and the conditions first, because that decides the mechanism. 2-Bromopentane is a secondary alkyl bromide; cyanide is a small, strong nucleophile but only a weak base (the $\text{p}K_a$ of HCN is about 9); and DMSO is a polar aprotic solvent that leaves the anion poorly solvated and highly reactive. Those three facts point to $\mathrm{S_N2}$, not $\mathrm{S_N1}$ (no carbocation forms from a secondary substrate in a polar aprotic solvent with a strong nucleophile present) and not E2 (cyanide is too weak a base).

An $\mathrm{S_N2}$ reaction is a single-step backside attack, so the stereocentre is inverted. To translate that inversion into an R/S label, rank the groups at C2. In the starting material the order is Br > propyl > methyl > H; in the product it is CN > propyl > methyl > H, because the nitrile carbon (N, N, N) outranks the propyl carbon (C, H, H). The incoming group therefore takes over the top priority exactly where the leaving group was, so inverting the geometry also flips the descriptor: (S) becomes (R).

The major product is (R)-2-methylpentanenitrile (option A).

Option B is the trap: it keeps the (S) label, which would require retention of configuration, something $\mathrm{S_N2}$ never gives. Option C would need a planar carbocation ($\mathrm{S_N1}$), which these conditions do not produce. Option D, the Zaitsev elimination product, would require a strong base such as sodium ethoxide; cyanide is not basic enough to drive E2.

Start with the degree of unsaturation, because it limits the structures worth considering. For $\mathrm{C_5H_{10}O}$,

so the molecule has exactly one ring or one double bond. The strong band at $1715\,\text{cm}^{-1}$ is a C=O stretch, which uses up that one degree of unsaturation, so there is no ring and no C=C. The absence of a broad band near $3300\,\text{cm}^{-1}$ rules out an alcohol, and the absence of the two weak aldehyde C–H stretches near $2720$ and $2820\,\text{cm}^{-1}$ rules out an aldehyde. The carbonyl must therefore be a ketone, and $1715\,\text{cm}^{-1}$ is the typical value for a saturated, acyclic ketone.

The ${}^{1}\mathrm{H}$ NMR now fixes the carbon skeleton. A 3H triplet next to a 2H quartet is the signature of an ethyl group ($\mathrm{CH_3CH_2}$, each set split by the neighbouring protons, $n + 1$ rule). The integrals 6H : 4H mean there are two ethyl groups, and because only two signals appear they must be equivalent. The quartet at $\delta\,2.45$ is deshielded, so the $\mathrm{CH_2}$ groups sit directly on the carbonyl carbon. Two equivalent ethyl groups on one C=O give $\mathrm{CH_3CH_2COCH_2CH_3}$, which has three distinct carbon environments ($\mathrm{CH_3}$, $\mathrm{CH_2}$ and C=O), matching the three ${}^{13}\mathrm{C}$ signals.

The compound is pentan-3-one.

The target has two substituents para to each other: a carboxylic acid and a nitro group. Neither can be placed directly in a way that gives the para relationship, so the plan is to install a methyl group first (an ortho/para director that can later be oxidized), nitrate it, and only then oxidize.

Step 1, Friedel–Crafts alkylation: benzene with $\mathrm{CH_3Cl}$ and $\mathrm{AlCl_3}$ gives toluene. This works because benzene is an unsubstituted, fully activated ring.

Step 2, nitration: toluene with $\mathrm{HNO_3}$ in $\mathrm{H_2SO_4}$. The methyl group is weakly activating and ortho/para directing, so the product is a mixture of 2-nitrotoluene (the larger fraction) and 4-nitrotoluene; the para isomer is separated (it is a solid at room temperature and crystallizes readily from the liquid ortho isomer).

Step 3, benzylic oxidation: hot $\mathrm{KMnO_4}$ followed by acidic workup ($\mathrm{H_3O^+}$) oxidizes the benzylic $\mathrm{CH_3}$ to $\mathrm{COOH}$. The nitro group survives, giving 4-nitrobenzoic acid.

Why order (a) fails: nitrobenzene is strongly deactivated, and Friedel–Crafts alkylation and acylation do not proceed on rings bearing $\mathrm{NO_2}$ (or other strongly withdrawing groups such as $\mathrm{SO_3H}$, acyl or $\mathrm{CN}$) because the ring is too electron-poor to attack the electrophile. Even if it did, $\mathrm{NO_2}$ is a meta director, so the methyl group would end up meta.

Why order (b) fails: if toluene is oxidized to benzoic acid before nitration, the $\mathrm{COOH}$ group is a deactivating meta director, so nitration gives 3-nitrobenzoic acid, the wrong isomer.

Reagents in order: (1) $\mathrm{CH_3Cl}$, $\mathrm{AlCl_3}$; (2) $\mathrm{HNO_3}$, $\mathrm{H_2SO_4}$, keep the para isomer; (3) $\mathrm{KMnO_4}$, heat, then $\mathrm{H_3O^+}$.