Hydroxide is the strongest base because (1) the charge is not delocalized (as in A and B) and (2) O is more electronegative than those elements further down Group 16 (S, Se).

Statements II and IV can be associated with the SN1 mechanism. Statement II is incorrect: the “1” in SN1 refers to the fact that the reaction operates under first-order kinetics (substitution, nucleophilic, unimolecular); that is, the slow (rate-determining) step is leaving group dissociation. Statement I describes the SN2 mechanism (backside attack, simultaneous nucleophile attack and leaving group dissociation).

E2 reactions occur quickest when the H-C bond and C-LG bonds involved are co-planar, or as close as possible to 180° with respect to each other. This is known as an antiperiplanar conformation. This conformation positions the sigma bonds being broken in the proper alignment to become the new pi bond.

First, determine the lowest energy chair conformation required for reaction, where the LG is axial and any other large groups remain equatorial. Once determined, two pathways must be considered (highlighted by the axial proton and deuterium below). Removal of the proton provides the more stable tri-substituted double bond (recall Zaitsev’s rule) and therefore is the preferred pathway.

This is a substitution reaction with inversion (SN2) of a methansulfonate (excellent leaving group) by halide, sodium iodide in this case (also known as the Finkelstein reaction. Impress your friends.). None of the other reaction conditions are suitable for this reaction.

The Grignard Reaction is the addition of an organomagnesium halide (Grignard reagent) to a ketone or aldehyde, to form a tertiary or secondary alcohol, respectively. The reaction with formaldehyde leads to a primary alcohol. The product is a tertiary alcohol; it follows that the starting compound must be a ketone. One strategy is to identify the Grignard fragment in the final product, and work backwards to identify the fragment derived from the starting material.

Aromatic compounds follow four rules: (1) They are conjugated—there needs to one “p” orbital from each atom in the ring, so each atom must be either sp2 or sp hybridized; 2) They are cyclic: linear systems are not aromatic; 3) They are planar: there is good overlap/interaction between p orbitals; 4) they follow the Huckel Rule — 4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.)

All three reagents are used for the esterification of carboxylic acids. Of these methods, III is the most general. Method I can be used if the alkyl halide is sufficiently electrophilic (methyl iodide). Method II (diazomethane) is used for making methyl esters on a small scale.

All of the answer choices accurately represent SN2 reactions. It is VITAL to know what SN1, SN2, E1, and E2 reactions are and their differences.

This is an organic lab question that deals with extraction. In order to extract the carboxylic acid ONLY, you can only use sodium bicarbonate. Although NaOH is capable of extracting the carboxylic acid, it will also extract the phenol, which is not what the question asks.

Methane is the prototypical tetrahedral molecule. The bond angle between the carbon atom and any pair of hydrogen atoms is 109.5 degrees. The molecule behaves according to VSPER (valence shell electron-pair repulsion) theory, which states that atoms surrounding a central atom will tend to move as far away from each other as possible. Four substituents around one carbon atom usually forms a tetrahedral molecule. The four electron pairs surrounding oxygen in water assort themselves in much the same way, but not quite. The two lone pairs actually exert a slightly stronger repulsive force on the hydrogen atoms making the bond angle about 5 degrees more acute.

A quick SN2 reminder: The SN2 reaction is a single-step displacement of a leaving group by a nucleophile. During the transition state, the bond to the nucleophile forms at the same time that the bond to the leaving group breaks. Thus, the nucleophile needs to approach from the back, and configuration at carbon is inverted. The best compound for the reaction, therefore, must be a primary alkyl halide for steric reasons. Assign the carbon centers containing the leaving group as 1°, 2°, 3° (drawing the molecules will assist you in this). C would react faster than A because of less steric hindrance around the carbon center.

A carbonyl has a distinct peak at ~1710 whereas a C=C bond as a peak at ~1650. Alcohols (-OH) have a broad peak around 3300, as do N-H bonds (not as broad as -OH however).

The bromine atom is heavier and larger so it belongs in the equatorial position. Generally, the larger atoms belong in the equatorial position to minimize the steric strain. Having the bromine point away from the ring is better than having it point in an axial direction, where it will meet resistance and strain from other hydrogen atoms.

Note that the problem says this is a cis isomer, and therefore answers A, B, and E are impossible when you draw out the chair conformation. If the atoms are on the same side of the molecule, then one must be equatorial and the other axially. For stabilization purposes, it is better for the larger bromine atom to be in the equatorial position and the chlorine in the axial position.

This is an example of a free-radical halogenation. Homolysis of the bromine bond by light initiates the reaction. Chain propagation occurs whereby hydrogen abstraction yields an alkyl radical plus HBr. The alkyl radical reacts by reaction with another molecule of bromine, regenerating a bromine radical to engage in further reactions. Chain termination occurs whereby two free radicals combine. Carbon-centered radicals follow the same rules of stability as carbocations: the greater number of alkyl groups directly connected to a carbon atom, the greater the stabilization.

Be sure to pay special attention to the ‘+’ symbol indicating a cation. If it lacked a positive or negative sign, it would be a sp³ hybridization.