Hi Keira,

Q8: Chromatography in general is a set of techniques that take advantage of different properties of molecules to aid separation or extraction.

We see this in more detail in some of the biochem and ochem lectures, if you haven't had those yet! 😀

Gas-liquid chromatography or GC takes advantage of the different volatilities or boiling points of organic molecules to help separate them. I'm not a big fan of the explanation that the AAMC gives here in regards to molecular weight.

Instead, the intermolecular forces explanation makes more sense. Molecules that have LOW intermolecular forces are EASIER to boil, and will therefore have LOWER boiling points.

As an example, compare water (high intermolecular forces due to polarity and hydrogen bonding) to methane (low intermolecular forces). Water (100°C) has a MUCH higher boiling point than methane (-162°C).

So, we now need to identify which molecule has the weakest or lowest intermolecular forces.

A: has a hydroxyl group (alcohol group), allowing it to H-bond
B: nothing obvious stands out...
C: has an electron-withdrawing group (chlorine), which induces bond polarity
D: has an electron-withdrawing group (bromine), which induces bond polarity

It's B! 😁

Q9: D is "heavy hydrogen" or deuterium -- hydrogen with an extra neutron. You'll see it used for a couple of reasons: (1) as a tracer, (2) to regulate the speed of nuclear reactions.

Because it is not usually found in nature, you can artifically add deuterium to a molecule and trace its presence through a reaction mechanism. This allows us to be able to figure out how reactions work! This is similar to how we use radioactive dye tracers in imaging processes. Follow the dye... 😮

We also use D2O (heavy water) instead of H2O ('normal water') in nuclear reactors to help regulate the speed of nuclear reactions.

Back to the question!

HBr is an acid. It will protonate the -OH group on the reactant to create an -OH2+ group.

This reaction will then proceed by an SN2 mechanism. SN2 reactions ALWAYS OCCUR WITH AN INVERSION OF CONFIGURATION. In other words, if your reactant is (R), it the product will be (S), and vice versa.

The reactant given to us is (R), so the product must be (S). The answer is (B) or (C).

Since there are 5 carbons, we have a pentane. The answer is (C).

Q23: Compound 1 is described in great detail in the passage for us, starting from left and moving to the right.

(1) Hydrophobic region: long carbon chain on far left

(2) Residues that could be cross-linked: cysteines, which contain thiol groups

(3) Spacer residues: glycines

(4) Phosphorylated residues: serine (used in a previous question)

(5) Residues with an affinity for cells (AHA!!!!!): so what's over heeeeeeere 🤔🤔🤔

The group on the far right with two COOH is aspartic acid (Asp). So, we are down to answers C and D. Is the rest cysteine (Cys) or arginine (Arg)? Well, there's no sulfur in this location, so it's not Cys!

Therefore, D is the answer. Click and run away. 🤭