The reaction uses a Grignard reagent's nucleophilic carbon to attack the carbon in carbon dioxide. Following treatment with water the resulting molecule (a carboxylate anion, of the form ) the carboxylate oxygen will be protonated, and the result is a carboxylic acid. Thus, the resulting molecule is a benzene ring with a carboxyl group attached (this is also known as benzoic acid).

There are two tetrahedral asymmetrical stereocenters in this molecule (the carbon atoms attached to each of the chlorine atoms). Thus, the combinations of R and S include RR, RS, SR, and SS. Note the plane of symmetry in the molecule; RS and SR are the same molecule (meso compounds). Thus, there are three distinct configurational stereoisomers of this compound.

Cyanide is a weak base and a good nucleophile, and the solvent is aprotic; therefore, the product is favored. This involves 100% inversion of stereochemistry; therefore II is favored.

The reaction given involves the direct alkylation of a ketone containing compound by the production of an enolate ion. The alpha hydrogen to the ketone group which is weakly acidic can be deprotonated in the presence of a base. LDA is a mild base that converts these compounds to the enolate ion. On the last step, through nucleophilic substitution by reaction with an alkyl halide with the enolate ion yields an alkyl group in the alpha position. LDA is sterically hindered therefore allowing for the production of an enolate ion without nucleophlic addition. Similarly, an ester or nitrile containing compound can also undergo this type of reaction.

This is a basic Claisen condensation reaction in which the functions to take off the acidic hydrogen (between the ketone and the ester) and then the same molecule can attack a carbonyl carbon to yield .

The attack at the ester carbonyl leads to the leaving of group.

The correct answer is "Bombardment by free radical electrons which causes the formation of a radical cation and spontaneous bond cleavage throughout the molecule, depending on instability of the bond and stability of the fragments." First the parent molecule (the compound being identified) is bombarded by free radicals, resulting in a molecular radical cation, and then the molecular ion spontaneously fragments at the weak bonds and where the resulting fragments will be thermodynamically stable. These fragments then reach the detector plate at different places depending on the mass of each fragment.

Incorrect answers:

"Bombardment by free radical electrons which radicalizes the molecule at different atoms, resulting in distinct fragments" is incorrect because it indicates that the desired fragmentation of the molecule is through direct abstraction of a radical by the free radical, rather than a spontaneous self-cleaving process. If this occurs, it is not a desirable result, because the molecular ion will no longer be structurally in tact for the next chamber of the mass spectrometry in which it is meant to fragment and be separated based on mass/charge ratio. It is also somewhat sterically unlikely for a free radical electron to directly cause the splitting of carbon-carbon bonds in a sterically bulky organic molecule -- the free radical electron is more likely to abstract a proton and leave a positive radical charge on the molecule.

"Spontaneous homolytic cleavage of unstable bonds with good leaving groups, unmediated by electron bombardment" is incorrect because mass spectrometry methods use electrons to induce radical cation formation.

"Ionization through proton bombardment, depending on instability of the bond and stability of the fragments" is false because electrons, rather than protons, are used to ionize the molecule.

"Putting the molecule through a strong magnetic field, resulting in various electron spins of the hydrogens in the molecule" is incorrect because it does not reflect the spectra produced by mass spectrometry, nor the method by which molecular fragments are produced.

The longest carbon chain is carbons long (thus ""), and the lack of double bonds makes it an alkane (thus ""). The highest priority functional group is the carboxylic acid group, and the suffix of the IUPAC name should reflect the highest priority functional group (thus ""). Because the carboxylic acid group is assumed to lie on carbon number there is no need to designate a locand for this functional group. The other substituents of this molecule are the methyl groups on carbons and when numbering the chain from left to right. Thus, the name is .

The reaction given involves the direct alkylation of a nitrile containing compound by the production of an enolate ion. The alpha hydrogen to the nitrile group which is weakly acidic can be deprotonated in the presence of a base. LDA is a mild base that converts these compounds to the enolate ion. On the last step, through nucleophilic substitution by reaction with an alkyl halide with the enolate ion yields an alkyl group in the alpha position. LDA is sterically hindered therefore allowing for the production of an enolate ion without nucleophlic addition. Similarly, an ester or ketone containing compound can also undergo this type of reaction.

Hydrogenation of a double bond involves the bond breaking and a hydrogen being added to each carbon of that double bond. You can tell the number of double bonds by taking the number of hydrogens added and dividing it by 2.

6 added hydrogen divided by 2 is 3 double bonds.

A hydrocarbon with zero degrees of unsaturation and carbons has hydrogens. For every two hydrogens less than , there is one degree of unsaturation. After hydrogenation, our final product has no double bonds. After calculation, we see that it has two degrees of unsaturation. This means that it has two rings.

The SN1 mechanism involves the formation of a carbocation intermediate in the rate-determining step. The most stable carbocation will produce the fastest reaction. We can immediately eliminate any answer choices that will produce primary or secondary carbocations, since a tertiary carbocation will be much more stable. When comparing tertiary carbocations, larger and more electronegative substituents will allow for more charge stabilization.

Since the tertiary carbocation formed by the dissociation of iodide from will the be most stable, this substrate will react the fastest.