Hi

Question 54:

This is a reduction reaction with LiAlH4, which reduces a carboxylic acid to an alcohol.

We can eliminate D as it shows an oxidation of the alcohol on lactic acid, which doesn't match the reagent.

We can eliminate C as you cannot reduce a carboxylic acid to an aldehyde using MCAT only reagents

We can eliminate A as you have a hydrate which is unstable and would form an aldehyde (which would be the same structure as C)

Therefore, you are left with B which has a primary alcohol where the carboyxlic acid was.

Question 58

This one is pretty tricky - what I would do first is figure out what step 2 does, and the main consequences of this.

Step 2 overall:

-takes an aldehyde to a carboxylic acid
-makes this oxidized carboxylic acid leave
-generates a sugar that is one carbon shorter, with all other stereocentres being the same
-following this reaction, glucose is optically active (not meso)

Therefore, we are looking for molecules that are eliminated by this step, which are going to be meso and therefore have a mirror plane.

The next thing I would do is try to test structures to see if they pass the test.

I would start with structure 5 as an example because it would eliminate all other answers.

Here is how you do this step 2 for structure 5: https://ibb.co/6vwCsvJ.

Because this renders structure 5 meso (see blue dotted mirror plane line), structure 5 is eliminated as a possible structure of glucose and therefore C must be correct (as no other answer includes 5). If I had time to double check this question on the MCAT, I would check all other structures in this same way - here I will leave it for your practice to verify.

For questions 64 and 66:

We look at double bonds similarly to carbocations - something with more carbons attached directly to it and less hydrogens is considered more substituted. You can see this kind of thing with primary (1 C), secondary (2 C), and tertiary (3C) carbocations.

In this example, the alkene I. has 2 Hs, 1 O and 1 C attached to it, making it less substituted than alkene II. which has 1 H, 2 Cs, and one O attached to it.

For question 66, try to practice applying this above definition to tell me if compound 1 or compound 2 contains the more substituted alkene.

As for the alpha carbon, recall that alpha just means one carbon away from the C=O. In this particular question, you can also see where the alpha carbon was as the alkene will form there.


Try to to draw and label the more substituted alkenes and the alpha carbon, and reply with an image on here to practice, and I can let you know if the image is correct.

Hope this helps, please let me know if this is unclear!