Chapter 14 Work

Heckel's rule anti aromatic compounds

4n pi-electons 0,4,8,12,16,20

Heckels Rule Aromatic Compounds

4n+2 pi-electrons 2,6,10,14,18

What are the major resonance contributors for the formate anion HCO2

A and B

Three step synthesis of trans-2-pentene from acetylene

Acetylene contains 2 carbon atoms and 2-pentene contains 5 carbon atoms. To increase the starting material by 3 carbon atoms, we can use the alkylating reagents - MeBr and EtBr - as the reagents in the first and second steps. Acetylene and MeBr gives propyne as the step 1 product, whereas acetylene and EtBr gives 1-butyne as the step 1 product. Using the ethyl or methyl reagent first does not matter in this multistep synthesis route; subsequent alkylation with the other reagent gives the same step 2 product: 2-pentyne. Finally, 2-pentyne must be hydrogenated stereoselectively via Na in NH3 to give the trans-2-pentene product.

What are the minor resonance contributors for the formate anion HCO2

C

nonconjugated

Multiple bonds are seperated by more than one single bond

Conjugated

Double bond or triple bond separated by one single bond

What is the starting material and intermediate product?

Hydroboration-oxidation reactions are used to regioselectively hydrate alkynes. The first step of the reaction, hydroboration, involves addition of the bulky borane to the triple bond. (R2BH can be any bulky borane, such as 9-BBN, dicyclohexylborane or disiamylborane) Compared to Hg2 /acid-catalyzed hydration of an alkyne, hydroboration-oxidation displays a different regioselectivity and an apparent anti-Markovnikov reaction. However, the bulky borane is the electrophile of the reaction (and a nucleophilic H- , rather than an electrophilic H , is involved). The electrophilic boron adds to the unsubstituted carbon of the terminal alkyne, in accordance with Markovnikov\'s rule. Oxidation of the carbon-boron bond replaces the bulky borane with a hydroxyl group, yielding an enol. The enol quickly tautomerizes. Because the starting material was a terminal alkyne and the hydroboration reaction places the hydroxyl group on the terminal carbon, tautomerization yields an aldehyde.

Alkylation of an lakyne

In the first step, sodium amide deprotonates the terminal alkyne reactant. The resulting acetylide, RC≡ C- Na , reacts with the alkyl halide in the second step. Overall, the terminal alkyne is converted to an internal alkyne, and the number of carbon atoms in the product is the sum of the carbon atoms from both reactants.

Addition of one equivalent HCl to diene

It goes there because it is the most stable carbocation with the lowest transition energy state

Precursors of both products

Protonation occurs at one of the terminal carbons of the diene, forming an allylic carbocation, rather than at one of the middle carbons of the diene, which would form a less stable isolated carbocation. The products must come from the following allylic carbocation resonance structures, shown below. Each product represents the 1,2-addition product of one of the dienes and the 1,4-addition product of the other.

Two acid-base reaction are given below. Of the three reaction arrows presented in each reaction, select the arrow (via the multiple choice circle before it) that best indicates the acid base equilibrium position HC=-CH + NH2 ? HC=-C- + NH3 H2C=CH2 + HC=C- ? H2C=CH- + HC=-CH

Relative acidities of the compound classes are: Alkane<Alkene<Amine<Alkyne<Water SP carbond are more electronegative, thereby more acidic than SP3 carbon atoms.

Conjugated with c=o

enone (not diene)

Diels-Alder Reaction with stereochemistry

syn addition due to it not being trans