Phenol contains the hydroxide group, which is an electron donor, puts electron density into the benzene ring. Resonance structures are drawn as follows

Cis and trans-1,2-dibromocyclopropane (pictured below) share a diastereomeric relationship; that is, the molecules are stereoisomers that are not enantiomers. Diastereomerism occurs when two or more stereoisomers (i.e. same connectivities) of a compound have different configurations at one or more (but not all) of the equivalent stereocenters and are not mirror images of each other. Accordingly, diastereomers are distinct molecules with different chemical properties and can be separated by ordinary physical means. Recall that enantiomers are mirror images of each other and are identical in all physical aspects except for their direction of rotation of plane-polarized light.

2-butone is a carbonyl compound that can readily be reduced by NaBH4 into a secondary alcohol, 2-butanol. When 2-butanol is heated in acid, we get dehydration, which leads to 2-butene.

The electrons from one of the double bonds on the 1,3-dibutene create a new single bond. The other new single bond is created from the electrons in the double bond of the other reactant. These two new single bonds join the reactants to create a cyclic product.

The electrons from the other double bond in the 1,3-dibutene move between carbon 2 and 3. Thus, the final product is a 6-carbon cycloalkene with a halogen substituent.

An approximated 1H NMR spectrum of 1,1-dibromopropane (CH3CH2CHBr2) is shown below:

All carbonyl compounds absorb in the IR region of 1760–1665 cm^–1 owing to the stretching vibration of the C=O bond. This carbonyl band is distinct with a characteristic high intensity and therefore is quite useful for diagnostic purposes because few other functional groups absorb in this region. Conjugation with a double bond or benzene ring lowers the stretching frequency, usually by 30 to 40 cm^–1. The stretching frequency of the conjugated double bond is also lowered and may be enhanced in intensity. An example of this phenomenon is provided

Step 1 converts the carboxylic acid into an ester.

Step 2 adds two equivalents of Grignard reagent: one to turn the molecule into acetone, and a second one to turn it into tert-butoxide. There is no way to stop the reaction after the first addition of Grignard reagent.

Step 3 will neutralize the base, leaving only t-butyl alcohol (I).

PCC is the only compound listed that is not a reducing agent. Pyridinium chlorochromate is a weak oxidizing agent and is often used to oxidize alcohols into carbony compounds. All of the other compounds are similar in that they function as reducing agents.

Quinine has four centers of chirality (highlighted with asterisks). Thus, in principle, quinine has 2⁴ (2 x 2 x 2 x 2 = 16) stereoisomers.

NO2 is the only electron-withdrawing substituent because it contains two electronegative oxygen atoms which pull electrons from the benzene ring towards itself. This effect is electron-withdrawing and makes the ring slightly positive in charge. All the other substituents are electron-donating groups, which activate the ring for electrophilic addition.

The Wittig reaction involves the reaction of a phosphonium ylide (generated by treating a phosphonium salt with a strong base) with a ketone or aldehyde.

PPh3R1+ketone−R2→HR1C=CR2H+PPh3O

The reaction proceeds through a phosphaoxetane (4-membered ring containing both phosphorus and oxygen) intermediate to generate a new compound containing a carbon-carbon double bond, plus a phosphine oxide byproduct. It does not form trans double bonds exclusively; sometimes, a mixture of cis and trans isomers are obtained, and sometimes the cis isomer is the predominant product.

Diels-Alder reactions create cyclohexene rings (eliminate III, IV, and V), and starting dienophile is trans (E conformation), so the product is E (Eliminate I).

Reaction of an alkene with osmium tetroxide (OsO4) results in syn addition of to hydroxyl groups across the double bond, resulting in a 1,2-diol. The correct answer is the option in which hydroxide groups are added to the previously double-bonded carbons on the same side of the ring. The correct answer is compound II.

I− is a better leaving group than Cl− because it is a larger molecule and can distribute the negative charge over a larger area. SN2 works better with better leaving group and with less-substituted carbons (methyl > primary > secondary > tertiary)

An important criterion for diene reactivity is for the diene to maintain a cis conformation. Cyclopentadiene (1) has the double bonds permanently cis oriented due to its cyclical nature, thus is the most reactive of the two. Re-drawing diene (2) into a cis conformation shows the steric hindrance experienced by the terminal methyl groups; diene 2 is unable to maintain the geometry necessary for a Diels-Alder reaction.

SN1 reactions occur in two steps. First, the leaving group leaves, forming a carbocation. Next, the weak nucleophile attacks the carbocation (beware of rearrangements during this step). The protic solvent stabilizes the carbocation intermediate.

The Zaitsev product is the most stable alkene that can be formed. This is the major product formed in E1 elimination reactions because the carbocation can undergo hydride shifts to stabilize the positive charge. The most stable alkene is the most substituted alkene, and thus the correct answer.

The stability of each resulting conjugate base should be considered; the more stable this is, the stronger the parent acid. For enolates, the greater the degree of charge delocalization (i.e. the more resonance
structures that can be drawn) the stronger the acid. Proton Hb is most acidic due to the charge (and lone pair of electrons) on the conjugate base being delocalized over both carbonyl groups and the benzene ring.