Question

can someone help me understand these two problems?

(b) Rank in order of increasing number of unique monobrominated products resulting from radical halogenation with N-bromosuccinimide (NBS) and UV light (1 fewest, 4most). (c) Rank in order of increasing pKa for the indicated H (1 = lowest pKa, 4-highest pKa). 2 ㄐ

Answer 1

b. Stability of free radical: The stability of the free radical is like carbocation which is as follows; Stability order of free radical: CH3 CH3 -CH, > HC=C-CH, > Học-c: > “CH > H,C-CH, > CH3 CH3 CH3 Benzylic Allylic Saturated aliphatic H The benzylic free radial is the most stable due to resonance effect once the radical formation tasks place it could participate in resonance of benzene ring that's why the order of allylic bromination is as follows: 2° Benzylylic allylic No allylic position Answer: Do@ c. Acidity: The ability of a compound to donate hydrogen ion or accept hydroxide ions, is called acidity. It is also explained as, the ability of compound to donate protons or accept electrons is called acidity. Factor affecting acidity of organic compounds: (i) Electronegativity: More the electronegativity of atom, attached from a compound more will be the acidic if the electronegative atom not belongs from second period. Due to smaller size of first and second period, as electronegativity decreases down the group acidity increases. (i) Percentage of s character: More the percentage of s character on hybridization of carbon atom of organic compound more will be the acidity. (v) pK, value: More the pK, value of compound less will be the acidity and higher will be basicity. More acidic alkyl gives least conjugated base and vice versa, due to % s characters follows: Alkane (25%) = sp Alkene (23%)=sp? Alkyne ( 50%) = sp hydroxyl H Ammonia alkyne H alkene IV. N H spk H : H C: oH H3C— LH H3C spsp pka = 16 pka = 26 H LH: sp? pka = 42 pka=37 Answer: 0- H H2C-E H N - H 13 H3C H pka = 16 pka = 26 pka=37 pka = 42