Concept explainers
Interpretation:
The steps how the two products formed from reaction of methylenecyclohehane with NBS has to be given. Stereoisomers has to be disregarded.
Concept introduction:
Bromination of Allylic Carbons:
N-bromosuccinimide (NBS) is used for the allylic bromination through radical reaction. Bromination of allylicc carbon requires low concentration of bromine and low concentration of hydrobromic acid. If high concentration of bromine and high concentration of hydrobromic acid which leads to the formation of bromonation in the double bond.
Bromination reaction starts with the homolytic cleavage of N-Br bond in N-bromosuccinimide (NBS) which creates bromine radical to initiate the radical bromination reaction.
NBS bromine radical removes the allylic hydrogen which forms hydrogen bromide and allylic radical in the first propagation step, the allylic radical is stabilized by the double bond in ring. This allylic radical reaction with bromine molecule and forms allylic bromide in the second propagation step which are shown above.
Want to see the full answer?
Check out a sample textbook solutionChapter 13 Solutions
Organic Chemistry; Modified MasteringChemistry with Pearson eText -- ValuePack Access Card; Study Guide and Student Solutions Manual for Organic Chemistry, Books a la Carte Edition (7th Edition)
- Draw all resonance forms of the molecules. Include curved arrow notation. Label major resonance contributor.arrow_forward: Resonance Forms a) Draw all resonance forms of the molecules. Include curved arrow notation. Label major resonance contributor. SO₂ NO3arrow_forward1d. Use Le Chatelier's principle to describe the effect of the following changes on the position of the Haber-Bosch equilibrium: N2(g) + 3H2(g)= 2NH3(9) AH = -92kJ Choose one of the following answers: shift to reactant side, shift to product side or no change and draw the resulting graph. I. Increase the [N2(g)] Effect: H₂ N₂ NH3 II. Decrease the volume of the container. Effect: H₂ N₂2 NH3arrow_forward
- f) The unusual molecule [2.2.2] propellane is pictured. 1) Given the bond length and bond angles in the image, what hybridization scheme best describes the carbons marked by the askerisks? 2) What types of orbitals are used in the bond between the two carbons marked by the askerisks? 3) How does this bond compare to an ordinary carbon-carbon bond (which is usually 1.54 Å long)? CH2 1.60Å H₂C * H₂C CH2 C H2C * C Of H₂ 120°arrow_forwarde) Determine the hybridization and geometry around the indicated carbon atoms. H3C CH3 B HC CH2 A C C C CH3arrow_forwardDon't used Ai solution and hand raitingarrow_forward
- Don't used Ai solutionarrow_forwardDon't used Ai solution and hand raitingarrow_forward75.0 grams of an unknown metal was heated to 95.0°C, it was then placed into 150.0 grams of water at23.1°C, when the metal and water reached thermal equilibrium, the temperature was 27.8°C. Calculatethe specific heat of the metal. (Assume that the specific heat of water is 4.18 J/g °C)arrow_forward