Organic Chemistry 1 final exam review: every reaction, plus a practice test

trans-1-Azido-4-methylcyclohexane. Ask the four questions: secondary substrate, strong nucleophile that is a weak base, polar aprotic solvent, no heat. That is SN2. Azide attacks the carbon from the side opposite bromine, and the carbon inverts. On a ring, inversion shows up as a cis/trans switch: bromine was on the same face as the methyl, so the azide ends up on the opposite face. (Neither product is chiral, so here you track cis and trans instead of R and S.)

Acetic acid (4.8) > phenol (about 10) > water (15.7) > ethanol (about 16). Compare the conjugate bases. Acetate shares the negative charge equally over two oxygens. Phenoxide spreads it into the ring, onto carbons, which is less helpful. Hydroxide and ethoxide have no resonance at all, so they are the least stable bases and their acids are the weakest. Water vs ethanol is a close call; it depends on the table your class uses.

trans-2-Methylcyclohexanol, as a racemic mixture. Hydroboration is anti-Markovnikov: boron bonds to the less substituted alkene carbon (C2), and oxidation swaps boron for OH with the same geometry. H and B add to the same face in one step (syn). So the H that lands on the methyl-bearing carbon is cis to the OH, and the methyl is trans. Borane attacks either face of the flat alkene equally, so both enantiomers form.

Three. List the combinations: (2R,3R), (2S,3S), (2R,3S), (2S,3R). The first two are mirror images that don't superimpose, so they are enantiomers. Both stereocenters of the last two carry the same set of groups (H, Br, CH₃, and the other half of the molecule), so (2R,3S) has an internal mirror plane that cuts the C2–C3 bond in half (easiest to see in the eclipsed conformation) and is identical to (2S,3R). That one compound is meso and achiral. 2ⁿ is a ceiling, not a guarantee.

SN1 and E1 compete. Bromide leaves on its own to give the tertiary carbocation (the slow step). Ethanol then either bonds to the carbon and loses a proton, giving tert-butyl ethyl ether (2-ethoxy-2-methylpropane), or removes an H from a neighboring carbon to give 2-methylpropene. Warming pushes the mix toward elimination. This is called solvolysis because the solvent is the nucleophile.

3-Methylpentane. Write out the carbons: CH₃–CH₂–CH–CH₂–CH₃ with a CH₃ hanging off the middle carbon. That's a five-carbon chain (pentane) with a methyl on C3. Numbering from either end gives 3, so there is no choice to make. It has six carbons total, the same as hexane, which is a nice check.

1-Bromo-1-methylcyclohexane. The pi bond grabs H⁺. Putting the H on the CH carbon leaves a tertiary carbocation on the carbon that holds the methyl; putting it the other way would give a less stable secondary cation. Bromide then bonds to the cation. Since that carbon has two identical ring paths plus a methyl and Br, the product has no stereocenter.

2,3-Dimethyl-1-butene. Tertiary substrate plus strong base means E2. The base can remove a hydrogen from one of the two CH₃ groups on C2 (giving 2,3-dimethyl-1-butene) or the one hydrogen on C3, a carbon crowded by two methyls (giving the tetrasubstituted 2,3-dimethyl-2-butene). Small ethoxide gives mostly the more substituted Zaitsev alkene. Bulky tert-butoxide is steered to the six exposed CH₃ hydrogens, so the less substituted Hofmann alkene becomes the major product.

sp, 180°. Count electron groups around the nitrile carbon: one single bond to CH₃ and one triple bond to N. Two groups means sp hybridization and a linear shape. The triple bond is one sigma bond plus two pi bonds made from the two leftover p orbitals. Quick rule: four groups sp³ (109.5°), three groups sp² (120°), two groups sp (180°).

trans-1,2-Dibromocyclohexane, racemic. The alkene attacks Br₂ to form a three-membered bromonium ion on one face of the ring. Bromide can only reach the carbons from the opposite face, like backside attack in SN2. The result is anti addition: one Br up, one Br down. Opening at either carbon gives the two enantiomers in equal amounts. (cis-1,2-Dibromocyclohexane would be meso, but it doesn't form here.)

Methyl equatorial. In the axial chair, the methyl sits right above the axial hydrogens two carbons away on both sides. Those two 1,3-diaxial interactions cost about 7.6 kJ/mol, so at room temperature about 95% of methylcyclohexane molecules are in the equatorial chair. The bigger the group, the stronger the preference.

Bromomethane > 1-bromopropane > 2-bromopropane > 2-bromo-2-methylpropane. Classify each carbon that holds Br: methyl, primary, secondary, tertiary. SN2 happens in one step with the nucleophile coming in from the back, so every alkyl group on that carbon slows it down. Tertiary halides essentially don't do SN2.

Oxymercuration–demercuration. The target has OH on C2, the more substituted alkene carbon, so you need Markovnikov hydration. Acid would form a secondary carbocation next to a carbon carrying three methyls; one methyl shifts over to make a tertiary cation, and the product is rearranged. Hg(OAc)₂ forms a bridged mercurinium ion instead, water opens it at the more substituted carbon, and NaBH₄ replaces the mercury with H. No free carbocation, no rearrangement.

From an oxygen lone pair to the carbon. SN2 needs two arrows drawn together: one from a lone pair on O to the carbon (forming C–O), and one from the C–Br bond to Br (breaking C–Br, so bromide leaves with both electrons). Check the result: oxygen goes from −1 to neutral, and bromine goes from neutral to −1.

Negative charge on oxygen. To rank resonance forms: first, more atoms with full octets wins (tie here). Next, fewer charges wins (tie, each form has one). Then, a negative charge belongs on the more electronegative atom. Oxygen beats carbon, so the oxygen-anion form contributes more. The real enolate is a single hybrid, with most of the charge on O and some on the alpha carbon, which is why it can react at either site.

Na in liquid NH₃. Sodium donates electrons one at a time, ammonia supplies the protons, and the vinyl intermediate takes the trans shape that keeps the two methyls apart. Pair it with H₂ and Lindlar's catalyst (cis-2-butene) and you can make either alkene from the same alkyne, a favorite synthesis question.