Question

Please help me!!

1. (8) Fill in the blank with the best word or words. What is the % of the R enantiomer in a sample of carvone that has a specific rotation of 30°, given that the specific rotation of R- carvone is -61°? What is the rate law for the following reaction? HBI How many stereoisomers have the condensed structure CHCHBICH:CHBTCH?. Write the IUPAC name for the isomer with the molecular formula CsH11Br that undergoes El elimination at the fastest rate.

Answer 1

• The formula of enantiomeric excess is:

$e.e =\frac{observed \, specific\, rotation}{specific \, rotation \, of\, pure\, enantiomer}\times100^o/_o$

Obviously here, excess enantiomer is 'S', having specific roation just opposite of that of 'R', i.e, +61

$So,e.e =\frac{30^o}{61^o}\times100^o/_o=49.2^o/_o$

thus, rest of the solution is recimic having 1:1 composition of 'R' and 'S'.

So, % of R enantiomer is 1/2 x (100 - 49.2) % = 25.4 %.

• Two possibilities can be there, S_N1 or S_N2. Now, if OH leaves as OH₂ then the system will have a 2^o carbocation, which is not much stable and of high energy. So, the reaction follows mainly the S_N2 pathway. Firstly protonation of alcohol which is very fast step and second is nucleophilic attack by Br^-, which is rate determining step. Thus, rate law is: RATE = K [ 2-butanol ] [ Br^- ]
• If two different stereocenters are there in a molecule, then possible stereoisomer number is 2² = 4. But here both stereocenters have same bonding connectivity so two meso isomers will be homomers. So, total number of stereoisomer is 3.
• If the elimination is to be done fastest then, the Br containing carbn must be 3^o, so that a 3^o carbocation is generated. So, we have to accomodate 4 carbons on a carbon attached to Br.

4 carbons -C- Br

So, two methyl and one ethyl can be the option. So the molecule will be:

3 4 H2C—CH3 H3cz6—Br 1 CH3

So, the name is: 2-bromo-2-methylbutane.