The correct answer is "The charge of the molecular ion is most likely to be +1 because of the ionization process by bombardment of single electrons. This charge of +1 is maintained in all the reactions wherein the molecular ion spontaneously fragments, so the mass / charge ratio of each molecular ion or molecular ion fragment corresponds to mass / 1 which is equivalent to the mass itself."  Conservation of charge results in all fragments from the original molecular ion maintaining the +1 charge. A number divided by 1 is the same number, therefore m/z is representative of m when z = 1.

Incorrect answers:

The statement "Because the charge of the molecular fragments is always 0 (neutral)" is false because a) the fragments have a charge of +1 due to being ionized in the ionization chamber, and b) if z = 0, then the m/z quotient would not compute to m (it would be undefined).

The statement "The mass to charge ratio reflects the mass of each fragment since when the fragment is larger, the charge will be larger, and the ratio will come out to be the mass itself" is incorrect because the charge does not vary based on molecular size, but rather is a result of spontaneous homolytic cleavage of the parent cation and therefore always maintains the charge of +1 (it is thermodynamically unlikely for a cation to be ionized once further while still in the ionization chamber where electron bombardment occurs).

The statement "The mass to charge ratio reflects the mass because the mass of an electron is negligible" is not an accurate logical consequent -- the mass of an electron is very small compared to the mass of a proton, but that has nothing to do with the mass of the molecular ion itself or its charge (which is +1).

The statement "The mass to charge ratio reflects the mass because the molecular ion and fragments thereof are being passed through an oxidation potential which neutralizes the charge caused by the ionization chamber" is false because a) the ions are not being passed through an oxidation potential but an electric field, and b) if the charge were neutralized, z would be 0, which would result in m/z being undefined, which would not be experimentally useful.

The correct answer is "The McLafferty rearrangement -- a pericyclic reaction that occurs because of H-bonding between a carbonyl oxygen and a hydrogen on a carbon three carbons away from the carbonyl." This rearrangement readily occurs because of the radical cation nature of the molecular ion / fragments, which leads to stabilizing the oxygen on a carbonyl group through the hydrogen on the third carbon away from the carbonyl carbon, being held in a six-membered ring formation. The six-membered ring stabilizes the pericyclic transition state, which makes the rearrangement kinetically plausible. 

Incorrect answers:

"The Cope elimination -- a pericyclic reaction that results in the elimination of a nitro-group (formed through oxidation of an amine via mCPBA) and the formation of an alkene." -- the Cope elimination occurs specifically on molecules with nitro groups, and is made possible by high heat. It does not involve a radical cation.

The other answers are also incorrect because they do not rely on a radical cation being stabilized by a nearby hydrogen bond -- they occur on neutrally charged substrates.

The correct answer is "m/z 91. It corresponds to a benzylic radical cation, a fragment stabilized by the aromatic pi system." This fragment results from spontaneous cleaving of the benzylic and alpha-benzylic C-C bond, leading to a benzylic radical cation and an isopropyl radical cation.  The benzylic radical cation is more stable than the secondary isopropyl radical cation, so it is likely to be more abundant in the mass spec, and therefore correspond to the "base peak."

Incorrect answers:

"m/z 15. It corresponds to a methyl radical cation." -- This is not a very stabilized radical cation, it is unlikely to have the highest abundance.

"m/z 42. It corresponds to an isopropyl radical cation." -- This is a fairly stable fragment, and does result from the cleavage of the bond that results in the benzylic radical cation, but the benzylic radical cation itself is much more stable than a secondary isopropyl radical cation.

"m/z 57. It corresponds to an isobutyl radical cation." -- The isobutyl radical cation is a primary radical, which is not very stable, compared to the other fragments possible in this molecule.  It is not likely to be so abundant as to correspond to the base peak.

"m/z 77. It corresponds to an aryl radical cation."  -- An aryl radical cation is not nearly as stable as a benzylic radical cation, because the non-bonding electron is not in-plane with the pi-system of the benzene ring, but is rather orthogonal to the plane.

The correct answer is "This peak is usually accounted for by the naturally occurring isotopes of molecules and their fragments. The extra neutron results in a higher mass number by 1, while the charge is still +1 (as with all radical cations in the mass spec detector) The relative abundance of this higher-value peak depends on the natural abundance of the relevant isotope." All statements in this answer are true.

Incorrect answers:

"The smaller, close-by peak corresponds to the original molecular mass, and is the most common natural isotope." -- This statement is false because the most common natural isotope will be represented by a more abundant, higher-intensity peak.

"The smaller, close-by peak is a result of indeterminate "blips" on the mass spectra machine." -- This is not true.

"The smaller, close-by peak is a result of molecular fragments of impurities in the mass spec machine." -- The peaks correspond to isotopes, not to impurities in the machine.

"The smaller, close-by peak is a result of a molecular fragment which is ionized to a +2 radical cation state." -- This is not true, and such a radical cation would be highly unstable, and would result in a m/z peak of HALF the original m/z value (m/2 since z=2), not (m+1)/z = (m+1) when z = 1.

The correct answer is "an iodine radical cation" because the mass of such a cation would be 127. 

The correct answer is "It results in the formation of a chlorine radical" because a) this is something that does in fact happen, and b) it is true that the formation of a chlorine radical is more favorable than the formation of a hydrogen radical (which is the other option, in the case of ionizing aryl chloride). 

It probably not result in a chloride anion, since the carbon-chlorine bond is though to undergo spontaneous homolytic cleavage. If it does result in a chloride anion, it would not be detected by a mass spec, which only captures fragments with radical cation charge. The breaking of a carbon-chlorine bond is not thermodynamically favorable, because that is a strong bond with a high bond energy. And the formation of an aryl radical, though true, is not thermodynamically favorable because the lone electron is not stabilized by the aryl pi system (but rather is orthogonal to the pi plane).

The correct answer is  -- this can be found by corroborating the molecular weight of  with 100. It can also be found without looking at the multiple answer options, by knowing that in an alkane the ratio of carbon to hydrogen atoms will be . If  is the number of carbon atoms in an alkane, and  is the number of hydrogen atoms in an alkane, then the molecular weight of an alkane with  carbon atoms is:

Since in this question the molecular weight is 100, as per the parent peak on the mass spec, the equation is:

Therefore, there are 7 carbon atoms and  or 16 hydrogen atoms in this alkane. It's chemical formula is .

The correct answer is "it has an alcohol group." Primary alcohols are usually seen as strong, broad peaks in the  on the infrared spectrum.

The correct answer is "." This part of the spectrum is totally unique for each distinct molecule, depending on its overall vibrational frequencies and stretching possibilities.

In organic chemistry, chemists often exploit the acidic and basic properties of side products in order to remove them. Compounds I-IV are all organic-soluble, however, if they are converted to a sodium salt via deprotonation with sodium hydroxide, they will become fully soluble in the aqueous sodium hydroxide solution and be washed away with this layer. The question we must then address is which of these four compounds will be deprotonated by sodium hydroxide.

Sodium hydroxide is a relatively strong base. Given its conjugate acid is water, which has a pKa of 15.7, sodium hydroxide can deprotonate any compound with a pka lower than 15.7. Another way to say this is that sodium hydroxide has a pkaH (defined as the pKa of the conjugate acid) of 15.7. Thus, we see that I and III may be deprotonated and washed away, leaving II and IV behind in the reaction mixture.