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.

In an oxidation-reduction (redox) reaction, reduction and oxidation both occur. Thus, not all of the elements can be oxidized, and not all of them can be reduced. In this equation, the oxidation number of oxygen is . Multiplying that by the number of oxygen atoms (), the overall charge on oxygen is .  has an overall  charge, so the oxidation number on  must combine with  to form . Thus, the oxidation number of  at the beginning of the reaction is . The iodine at the beginning of the reaction has an oxidation number of , as seen by the negative superscript.

At the end of the reaction,  has an oxidation number of , as seen by the positive superscript. Iodine has an oxidation number of  (there is no charge on the ).

Thus,  went from having an oxidation number of  to one of . Iodine went from having an oxidation number of  to one of . Oxidation occurs when electrons are lost (the number becomes more positive), and reduction occurs when electrons are gained (the number becomes more negative). Because the oxidation number of iodine became more positive, iodine was oxidized. Because the oxidation number of manganese became more negative (less positive), manganese was reduced.