Question

N-bomo soccinamide NBS)-usel speifilly for ally lic bonnction 1. (16 points) The molecule shown below, 3-ethyl-4-methylenehexane, is subjected to free-radical mono-bromination. a) How many products could be obtained? b) Draw the bond-line structure for the major product c) In the space below, provide the complete mechanism using the arrow notation, for th formation of the major product shown is part b above. HINT A steady and low concentration of bromine is provided to avoid poly-bromination. NBS Methyleyclohexane is subjected to the conditions shown below. Br2, hv Draw all possible monobromination products that could take place resulting in substitution for a) secondary hydrogens: b) Draw all possible monobromination products that could take place resulting in substitution for tertiary hydrogens:

can you plz explain how to solve question 1&2?

Answer 1

Sol :-

1.(a) (d) () resa ) Here, (a), (b), (c), (d), (e) and (t) are 6 different types of hydrogens foresente in this structure. Therefore, No. of monochlorinated products are : 6 | (b). + Allyl boomide Br 3-Bromo-3-ethyl-4famethylenehexane As, terrary folee Radical is more stable than Parimary and secondary free radicals. Therefore, Tertiary ally! bromide will be the major product. Also, Torhary allyl free radical is resonating stabalized.

(C) ANylic bromination catalyst → Mechanism . N-891: H-BY- + :B.1 BI NBS ON TEN-H+ : B- BA Succinimide ja :B3-89; 28 Hunan + H + B - + H-B91 + Ba-B91 -> fB + Bon (Allylbromide product) Major Birhe V 824 IT B1 in 3 Secondary products Br, ho Tertiary Product .