The correct ranking from most basic to least basic is:

The best bases are negatively charged, and the worst bases are positively charged (acidic). The stronger the base, the weaker (more stable) it's conjugate acid. An alkane is a very stable conjugate acid, which tells us that is the most basic of the set.

We know based on charge alone, that is more basic than but less basic than or .

We know that is more basic because the electronegativity of  is less than that of . This means the lone pair of electrons on are held less tightly and more likely to pick up a proton.

Remember that the strongest acids have the weakest conjugate bases.  is more acidic than  because iodine has a larger atomic radius than bromine. , , and  are strong acids and should be at the beginning of the list. Alkanes are not acidic. Acetic acid is a weak acid (pKa =4-5).

Recall that the stronger an acid, the lower the pKa.

II (two fluorine atoms really close to  has largest inductive effect, so bond is most weakened, and pKa is lowest)

V (one fluorine really close to  has strong inductive effect)

I (one fluorine a little further away from  has weaker inductive effect)

IV (no inductive effect)

III (alcohols are much less acidic than carboxylic acids, and it has the highest pKa of all)

The side of the reaction that is favored will have the acid with the higher , because the reaction goes (strong acid + strong base  weak acid + weak base).

The double bond in 1-hexene is across carbons one and two. Just adding HBr would put the bromine on carbon 2 because the bromine is more stable on the more highly subtituted carbon. If you want to put the bromine on the less substituted carbon (generating 1-bromohexane), you must carry out the reaction in the presence of peroxides. The mechanism involves a free radical on the secondary carbon and a bromine addition on the primary carbon.

The most substituted atom is most likely to be attacked, as a radical intermediate would be created. The radical intermediate would be most stable on the most substituted carbon.

The radical reaction would create the anti-Markovnikov product shown and both _, hv and , AIBN (azobisisobutyronitrile) effectively cause radical reactions. The bromine would then be substituted with the  group.

Termination is the only step that does not generate a radical product. The termination step ends the reaction by bringing together two radical molecules to produce a stable non-radical product.

Homolytic bond breaking occurs when a molecule breaks up to form two or more new products. In the reaction given, molecular chlorine forms two radicals in which one electron stays with each fragment formed.

Lithium aluminum hydride is correct because it is a reducing agent, and is therefore not capable of oxidizing secondary alcohols. Instead, LAH could be used to perform the reverse reaction, reducing a ketone to an alcohol. The other answer choices are oxidizing agents.