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Orgo 2 answer key

1. Answer = B

The dienophile was cis in the starting material, so it should stay cis in the end product

2. Answer = C

I has 8 pi e-II & III both have 6pi e-

Recall: only 1 lone pair on an atom can contribute (i.e. only 1 instead of both lone pair on the oxygen can contribute)

Recall: lone pairs on atoms connected to double bond inside the ring do not contribute (i.e. the lone pairs on the nitrogens don’t contribute)

3. Answer = A

They all obey Huckel’s rule which is just a difficult/fancy way of saying they have an odd # of participating pi e-

4. Answer = D because it gives off the least amount of heat, meaning it was the strongest bond.

Strong/stable bonds take a lot of energy to form, so when they form, they lose a lot of energy. Since they have already lost a lot of energy during formation, if you try to break them, they won’t give off much energy, since they don’t have as much left.

In contrast, a quickly formed unstable bond did not lose much energy to form. So when you break that bond, they give off a lot of energy.

Know this for your biochemistry class next semester : )

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5. Answer = AEthylene glycol is a protecting group for ketones and aldehydes

6. Answer = A

The H3O+ with heat turns the acetal into a ketone (it could of course also turn it into an aldehyde, it just depends on the structure)

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7. .Imine formation

8. Answer = A

NABH4 reduces all carbonyls except for esters, amides, and COOH

LiAlH4 reduces all carbonyls (more aggressive)

9. Answer = D

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10. Answer = D

11. Answer = BThe stronger the base 1,2 additionThe weaker the base 1,4 addition

A, C, & D are all strong bases so are 1,2

CN is a Nu- only reagent, so is 1,4 instead of 1,2

12. Answer = D because order of reactivity is acid chloride > anhydride > ester > amide

13. Answer = D

Recall: organolithium and Grignard would create an alcohol, whereas organocuprate makes carbonyl

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14. Answer = B

15. Answer = C because acidic conditions

Examples below had it been basic conditions or iodoform test

For the iodoform one in the pic above I made a mistake cuz it should actually be a carboxylate anion, not a carboxylic acid, since it’s basic conditions.

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16. Answer = A

The methyl is an ortho, para director

Recall: Friedel crafts alkylation can go through rearrangement. Don’t have to memorize that, just realize that things like to be more stable instead of less stable so a secondary carbocation is more stable than a primary carbocation, so the more stable is preferred.

17. Answer = C because the methyl and ethyl are ortho, para directors

(A) just garbage. (B) is ortho to both the methyl and ethyl, but it’s not favorable to be in between them

because it’s strained, so this is not the answer (C) is correct cause it’s ortho to one of the substituents and para to the other (D) this is meta, so no…

18. answer = A, it’s a radical reaction

19. answer = C because it was added to the benzylic position. Adding it there is more stable than adding it to the methyl, such as in A. Bromine with light is very regioselective, so prefers the more substituted area.

20. answer = D

it’s not a benzyl, so we don’t add it to the benzylic position. Adding it to the tertiary area is most stable since bromine with light is very regioselective

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21. answer = C

if you haven’t seen (C) as a reagent before, we can use process of elimination

a) turns alkynes trans alkenesb) garbage (reduces carbonyls, not alkynes or alkenes)

c) wut dis? Skip it and go to Dd) Fully reduces alkynes alkene alkane

If you don’t know regent C it’s ok, by process of elimination, a, b, and d don’t work, so go with C.

22. Answer = B

C and D are garbage. NaBH4 reduces carbonyls, not ethers. H2/Pd reduces carbonyls alcohols, it can’t turn a double bond into an alcohol

So now we choose between A and B.

Choice A would reduce the COOH alcohol, but it would be too many carbons

Choice B would reduce to ketone to an alcohol and yield the product shown below, which is what we want : ) yay

23. Answer = c because it would oxidize the secondary alcohol ketone but won’t oxidize the methyl

Gabriella Fiorino

Be careful, had it been a benzyl, the methyl could turn into a COOH, like option B. But because we have just a cyclohexane (not a benzyl), this cannot occur, so we only oxidize the OH and not the methyl

Recall: the reagent KMnO4 does the same thing as CrxOy stuffz

24. Answer = D

Have complements, comments, or concerns about the tutoring session? Click the link below to rate your tutor, Gabriella F.! :D It would make her very happy :Dhttps://ucf.qualtrics.com/jfe/form/SV_0jGMQ3pjtwHlzyRReferences:American Chemical Society (2002). Preparing for Your ACS Examination in Organic Chemistry: The Official Guide. American Chemical Society Division of Chemical Education Examinations Institute. Iowa state University: Ames, Iowa. Klien, D.R. (2015). Organic Chemistry: 2ndEd. Wiley: NJ. Print.

Smith, J.G.Organic Chemistry 5th edition. McGraw Hill.