Question

With explanation please!

10.(30 pts) Provide the major product(s) for the following reactions. (Remember: Stereochemistry and regiochemistry!). ( = H HBr (1 equiv) ROOR Bra (1 equiv) HgSO4, H20, H2SO4 1) RzBH 2) H202, NaOH 1) 03 2) H30 مجلس

Answer 1

Here there are alkynes present on the reactant side. So, the reaction is said to be addition reaction. Depending upon the different reagents used here we will get the different final products as discussed below.

a) In the first reaction, there is a alkyne (ethynylcyclohexane) present. This will undergo bromination reaction in the presence of a peroxide. During the reaction, the addition of the bromide ion will takes place at the less substituted carbon atom (terminal carbon atom of alkyne) and the addition of hydrogen takes place to the more substituted carbon atom. This condition is called as Anti Markovnikov rule. So, we will get the major product which is according to the anti Markovnikov's rule.

Here we got two products, both are anti Markovnikov's product. But its configuration is different. One is trans and another one is cis. But because of the steric hindrance present in the cis configuration ( [(Z)-2-bromoethenyl]cyclohexane ) it is not that stable. So, the trans product ( [(E)-2-bromoethenyl]cyclohexane ) is the major one.

HBr ( 1 equivalent) SCH Ozorion ROOR (peroxide) ethynylcyclohexane [(E)-2-bromoethenyl] cyclohexane [(Z)-2-bromo ethenyl] cyclohexane (minor) (major)

b) This reaction is bromination of alkyne (3,6-diethyloct-4-yne). In the reactant there is a triple bond and the bromide ions attack to the both carbon atom of C triple bond C bond. Here only one product forms. That is trans product ((4E)-4,5-dibromo-3,6-diethyloct-4-ene). Because of the highest steric hindrance there is 'no' posibility of forming a cis-product.

CH3 CH3 на сна Brz (1 equivalent) H₃C H2C CH3 CH3 3,6-diethyloct-4-yne (4E)-4,5-dibromo-3,6-diethyloct-4-ene

c) This reaction is a hydration of alkynes ((but-3-yn-2-yl)cyclobutane) using oxymercuration and acid. Here we will get a ketone as a product.

The -OH group from the reactant attacks on the more substituted alkyne. And first we will get a enol (3-cyclobutylbut-1-en-2-ol) as a product. But the keto form is more stable than the enol form. So, because of the tautomerisation, we will get a ketone (3-cyclobutylbutan-2-one) as the final product.

H3C HC HgSO4, H,0 H3C OH tautomerisation CH HSO4 CH2 O CHE CH4 (but-3-yn-2-yl)cyclobutane 3-cyclobutylbut-1-en-2-ol 3-cyclobutylbutan-2-one

d) This reaction performs through the hydroboration of terminal alkyes (ethynylcyclopentane). And then in the presence of base and peroxide, the attack of the -OH group takes place at the less substituted terminal alkyne. So, we will get an enol as a product ((E)-2-cyclopentylethen-1-ol). But this form is not that stable. So, because of the tautomerisation, we will get a aldehyde (cyclopentylacetaldehyde) as a product.

1) R,BH tautomerisation =CH HOH 2) H2O2, NaOH ethynylcyclopentane (E)-2-cyclopentylethen-1-ol cyclopentylacetaldehyde

e) This reaction is a oxidative cleavage of alkyne in the presence of ozone (O3) and water. Here the cleavage of the triple bond takes place and we will get two molecules of carboxylic acid as the final product. Here hence the alkyne (1,1'-(ethyne-1,2-diyl)dicyclohexane) has two same groups attached to the triple bond , we will get two molecules of cyclohexanecarboxylic acid as final product.

1,1'-(ethyne-1,2-diyl)dicyclohexane cyclohexanecarboxylic acid