Question

4. Multiple Choice Place your answer in the space provided. (2 pts. each; 12 pts. total) Which is NOT true of the aldol condensation? A. It accomplishes the formation of a new carbon-carbon bond B. The enolate is favored at equilibrium (assume NaOH as the base) C. The key step is the mechanism is attack of the enolate ion on a carbonyl carbon atom D. Dehydration of the aldol product is often observed and is irreversible. _Which sets of hydrogen atoms in the compound below will undergo exchange with deuterium under the following conditions: CH-OD/CHONA? A. 2 and 3 only B. 1, 2 and 4 only C. 2, 3 and 4 only D. 1 and 4 only Which of the following compounds would not react in the iodoform test to produce a yellow precipitate? CHg-che-&-che-d-och, B. A CH₂-C-CH, B, H-C-CH, D. HE-CHapn Which of the following compounds could NOT undergo a self-Aldol condensation? cho-e-cich.) Which set of reagents would undergo a 1,4-addition to an Q.B-unsaturated ketone?? A. CH CH{MgBr B. (CH3) Culi CLIATH D. Hz, Pd/C What would you expect to be the major product of the reaction shown in the box below? Br Br Brz (excess) Br Br Br A. Br B. C D. NaOH 4. Predict the Product. Using appropriate structural formulas, indicate the major organic product for each of the following reactions. (39 pts.) Br cat. H 2.H20

Answer 1

First 4 sub-parts of first question are solved here. please post other questions separately.

Q4. i)

Option A) is correct as a new C-C bond is formed, as shown below.

Option B) is also correct, as NaOH is a strong base, and shifts the equilibrium to the right (see step 1) , favoring enolate formation.

Option C) is incorrect, as formation of enolate is the rate determining step.

Option D) is correct as dehydration is observed of the aldol product and is the only irreversible step of the reaction.

So only option C) is not correct about the aldol reaction

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ii) Hydrogens # 2 and 3 are acidic as they are on the carbons adjacent to C=O group, which is polar.

Hydrogen # 2 is most acidic as it has two adjacent C=O groups.

Hydrogen # 1 is not at all acidic, as it on a -CH3 carbon attached to Oxygen.

Similarly Hydrogen #4 is also not much acidic as it is far from C=O group.

In deuterium exchange, acidic hydrogens are exchanged.

Hence option A) is correct as hydrogen # 2 and 3 only can be exchanged

iii) For a compound to show positive iodoform test, it must have H₃C-(C=O) - group.

Option C) does not have this group, so cannot show iodoform test.

Please see below for attached solution.

8) in) A) 0 - c) D) 6) . X 0

iv) For aldol reaction, first step is formation of enolate ion, as discussed in part i) above. For this presence of alpha hydrogen is must.

options A) , B) and D) , all have alpha hydrogens, that is hydrogens attached to the carbon adjacent to the C=O group.

option C) does not any alpha hydrogen , hence can't form an enolate ions and cannot undergo self-aldol reaction.