Drawing of products for given chemical reactions

Question

Please assist in drawing the products of the following reactions (in the attached image), considering regiochemistry where applicable and noting that some reactions may yield multiple products:

https://ibb.co/jbvfRPz

Answer 1

Hi Chance, glad I can help:

(a) Hydrogenation (H₂/Pd):

1. The double bond in (R)-3-(tert-butyl)cyclohex-1-ene is hydrogenated, converting the alkene into an alkane. The product would be (R)-3-(tert-butyl)cyclohexane, with no more double bond.

(b) Bromination (Br₂/CH₂Cl₂):

1. Bromine will add across the double bond in an anti-addition manner, resulting in a dibromide product. For (R)-3-(tert-butyl)cyclohex-1-ene, bromine atoms would add at positions 1 and 2 of the cyclohexene ring.

(c) Hydrobromination (HBr/ether):

1. This is a Markovnikov addition of HBr across the double bond, where the bromine attaches to the more substituted carbon. The product would be (1-bromo-3-tert-butylcyclohexane).

(d) Acid-catalyzed hydration (H₂SO₄):

1. H₂SO₄ would lead to Markovnikov hydration of the double bond, producing an alcohol at the more substituted position. The product would be 3-(tert-butyl)-1-cyclohexanol.

(e) Dehydration (H₃PO₄, 250°C):

1. This reaction is an elimination reaction, and it will remove water from the alcohol formed in the previous reaction. The product would likely revert to 3-(tert-butyl)cyclohexene.

(f) Ozonolysis (1. O₃; 2. Zn, H₃O⁺):

1. Ozonolysis cleaves the double bond to form carbonyl-containing products. For 3-(tert-butyl)cyclohexene, this would split the ring at the double bond, leading to two ketones or aldehydes, depending on the substitution.

(g) Oxidation (KMnO₄, H₃O⁺, vigorous conditions):

1. Strong oxidation would cleave the double bond and oxidize the resulting fragments to carboxylic acids or ketones. In this case, you would likely get carboxylic acids where the double bond was located.

If you'd like further details or help with visualizing these products, I can assist in explaining further!