Q7 :

In the hydroboration of alkenes, a 'syn' addition occurs so the H and BH2 added are on the same side of the bond.

When H2O2 and NaOH is added, the peroxide is deprotonated and you get the conjugate base -OOH , which now attacks the BH2.

Then, there is a migration of the C-B bond to the oxygen bound to boron.

Our -BH2 is being replaced by the -OH and there is a retention of stereochemistry.

Since your BH2 was 'down' at first (dashes mean pointing away from the paper), the OH will be down as well. The methyl was already given to be 'up' so this is important.

If there was no methyl group, there won't be a need to draw both dashes or wedges since the other C (C not bonded to the OH) has now 2 Hs bonded, and they are indistinguishable.

Good to read the following link :

masterorganicchemistry.com/reaction-guide/hydroboration of alkenes

For the oxymercuration in Qn 8, the end product basically has a plane of symmetry.

(Imagine a cutter/mirror plane slicing through the hexagon at the vertex where the methyl and OH is on, and the vertex that is directly opposite)

You easily ring flip the molecule and now the methyl and OH groups are in the opposite orientation.

Thank you very much for the detailed explanation! Will read up via the link provided

Welcome.

To clarify, same side of the bond means 'both up' or 'both down' , rather than 'to the same carbon'

Is it like cis or trans u mean?

Not exactly the same as cis/trans, but the idea is similar. Cis/trans is only for alkenes.

The terms used are anti/syn . For some cyclic structures, you will also see the terms 'exo' and 'endo'

https://en.m.wikipedia.org/wiki/Syn_and_anti_addition

https://www.masterorganicchemistry.com/2013/01/30/addition-reactions-stereochemistry/

https://goldbook.iupac.org/terms/view/E02094

Thanks for the links!