Molecule B will have a more acidic alpha-carbon because once the alpha proton becomes dissociated, the conjugate base will have relatively more stability than the conjugate base of molecule A.

When the alpha-carbon on molecule A loses it's proton, the conjugate base is not as stable. The reason for this is because the oxygen that is involved in the ester bond can contribute its electrons towards a resonance structure. Therefore, after the alpha-proton is lost, the alpha-carbon will have a negative charge that will be destabilized by the delocalized negative charge of the resonance structures.

From the start, we know we can eliminate answer choice  because it is the only answer choice that is not a strong acid. Now we have three strong acids, but we have to determine which is strongest. To do so, we take the conjugate base of each strong acid to see which conjugate base is the weakest acid. Remember: weaker conjugate base means a stronger acid.  is the largest ion of the bunch. Its large size allows it to better stabilize the negative charge and so it is the weakest (most stable) conjugate base. Because the weakest conjugate base leads to the strongest acid,  is our correct answer.

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.