Sebastien B. answered 10/05/24
Math, Physics, Chemistry - Experienced Credentialed Tutor/Teacher.
a) (R)-(tert-butyl)cyclohex-1-ene with H2/Pd:
We have: (R)-(tert-butyl)cyclohexane with H- and H- on the same side and the opposite side of (R)-(tert-butyl) which is less sterically encumbered (but it change nothing because we can't have enantiomer or diastereoisomer because because there is already an H- on each carbon in the double bond.
b) 1-Buten-2-ylcyclopropane Br2 in CH2Cl2:
We have an anti-addition of -Br so the products are: 1,2(R)-dibromobutan-2-ylcyclopropane and 1,2(S)-dibromobutan-2-ylcyclopropane in same quantity (racemic solution).
c) 1-Buten-2-ylcyclopropane HBr in ether:
The reaction proceeds as a typical electrophilic addition, with the bromine atom -Br attaching to the more substituted carbon of the alkene, following Markovnikov's rule, resulting in the formation of an alkyl bromide product; the ether solvent does not significantly affect the reaction mechanism itself. So we have:
2(R)-bromobutan-2-ylcyclopropane and 2(S)-bromobutan-2-ylcyclopropane in same quantity (racemic solution)...