Part 6: MECHANISMS
Note that no other reagents are needed in order to complete any of these sequences, you should only be using what is there.

A
HBr in the presence of peroxides will be a radical addition type process with a radical chain type process. This means it needs to be drawn with single headed arrows indicating the motion of individual electrons. Note that the question indicates that you should be considering an excess of HBr. The major product turns out to be the 1,2-dibromide for reasons explained below.

mechanism for the addition of HBr to an alkyne under radical conditions
The weak O-O bond is broken homolytically to give two alkoxy radicals that react with H-Br generating the electrophilic Br radical. This sequence provides the initiation steps. Then the Br radical adds to the alkyne giving a bromovinyl radical which then reacts with more H-Br forming the stronger C-H bond to give the bromoalkene and a new Br radical that can keep the process going.  When the Br radical reacts with the bromoalkene, the selectivity favours the formation of the radical I where the electrons of the Br already in place can stabilise the new radical (in the same way as Br stabilises the C+ formed in the addition of HBr to alkynes under heterolytic conditions).  The alternative regioselectivity gives radical II which is less stable. This new radical I reacts with another molecule of H-Br to give the vicinal dibromide and another Br radical that can keep the process going. Note that the selectivity of the reaction is different to that of the addition of H+ and Br- to the alkyne because in the radical process the Br radical is the electrophile (and not H+) and so it is Br that adds first to the C=C.

B
Hydration of an alkyne gives the Markovnikov product, which is the more highly substituted enol via the more stable secondary vinyl cation which then undergoes acid catalysed tautomerism to the ketone. Notice that the proton donor is the acid catalyst and the nucleophile is water (no hydroxide under acidic conditions).
alkyne hydration

C
Alcohols resemble water in many ways. The reaction here is like the hydration of an alkene except that Nu is now the ROH rather than H2O.
reaction of an alcohol with an enol ether
First protonate the C=C with the acid catalyst (the ROH is not acidic enought to do this on its own) to give the more stable C+.  Note the O atom can donate its electrons to stabilise the C+ in a resonance interaction. Then the alcohol reacts as the Nu via the O lone pairs (just like water would) to give an oxonium ion that would be then deprotonate to give the desired product and regenerating the H+ catalyst.

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