Here we see an  reaction with rearrangement. The bromine, an excellent leaving group, leaves the carbon chain and a carbo-cation (positively charged carbon) is formed on that carbon. A positive charge is more stable on a more substituted carbon, and so the positive charge rearranges itself onto the branched carbon. Essentially, the positive charge and a hydrogen on the branched carbon switched positions. The methanol was then free to attack the branched carbon to form the major product shown.

This reaction adds  and  (eliminate II). The reaction is Markovnikov (Eliminate I). A hydride shift occurs putting the carbocation on the more substituted carbon before addition of (eliminate III and V).

After carbocation is formed, a rearrangement reaction stabilizes positive charge by putting it on a tertiary carbon. This is done by a methyl shift. Recall that tertiary carbocations are the most stable due to the inductive effect of alkyl groups on the electron-deficient carbocation.

The first step of this reaction will be protonation of the hydroxyl oxygen to create a good leaving group. When the leaving group leaves, what's left is a secondary carbocation that is vicinal to (next to) a quaternary carbon. A methyl shift is thermodynamically favored in this case, as the rearrangement will leave a tertiary carbocation. Following the rearrangement the nucleophile (bromide) will attack the tertiary carbocation, forming a sigma bond with the carbon. The answer is thus .

The pK_a value indicates how strong an acid is, and acid strength increases as pK_a decreases. The side of a reaction with a lower pK_a is going to dissociate more, pushing the equilibrium over to the other side. The equilibrium will thus lie on the side with the HIGHER pK_a.

Since the pK_a of acetic acid (4.76) is higher than the pK_a of trifluoroacetic acid (0), the reaction will shift to the left to reach equilibrium.

Use the Henderson Hasselbalch equation:

An easy way to consider relative base strengths is to consider the strength of the compounds' conjugate acids.The stronger the conjugate acid, the weaker the base. Water (compound IV) is the least basic of the compounds because its conjugate acid,  is the strongest of the given compounds' conjugate acids . Carboxylate ions (compound III) are highly stabilized by resonance and predominate at neutral pH. The conjugate carboxylic acids readily donate protons (acetic acid: ). Ammonia (compound I) has considerable basicity; binding a fourth hydrogen produces ammonium ion , which predominates at neutral pH  Sodium propoxide (compound II) is a strong base, bearing a full negative charge on its oxygen. Its conjugate acid, 1-propanol, is a rather weak acid . Since its conjugate acid is the weakest (highest ), sodium propoxide is the strongest base. Based on the previous observations the correct ordering of the compounds is: II, I, III, IV.

The governing principle regarding the prediction of  values (relative to other compounds) is to assess the stability of the product formed by the release of a proton. The release of the alkyne hydrogen in compound III results in a carbanion, a highly unstable species, so it is expected that this compound is the least acidic and has the highest . Intuition serves well in this instance and we see that hydrogens bound to a triple bond have a  value of around 25. The relative stabilities of the remaining compounds may be assessed in the same manner. Compound IV is the second weakest acid  because the three methyl groups donate electron density such that if the oxygen is deprotonated, the resulting negative charge is destabilized. Methanol and water have a unique, non-intuitive relationship regarding their relative acidities. One would assume that water should be a stronger acid than other acids bound to alkyl groups (by the reasoning expressed for compound IV). This is the case for all alcohols except methanol, in which the delocalization of charge allowed by the increased molecular size outweighs the destabilization caused by electron donation. Thus methanol is a slightly stronger acid than water. This is evidenced in their  values: 15.7 for water and 15.5 for methanol. The correct ordering of the given compounds is: III, IV, II, I.

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.