please answer both questions if possible. thanks. epoxybutane (9.6g/100 ml. HO), ethoxyethene ( ounds with the molecular formula of Call O are shown in this list (with water solubility of each given i in parenthesis) are cis- .3- methyloxetane (mi accept 20), ethoxyethene (6g/100mL 1O), 1.2-epoxybutane inal (7.5 g/1 00ni L H2O ), 2-butanone ( 26g/ 100mL, H2O), and 2- sible). List these compounds in order of their hydrogen bonding ,weakest acceptor (I) to strongest acceptor (6). (2 points) 5. Pyrimidine is an electron poor aromatic ring and as such, it resists brominatio under extreme reaction conditions (Brz, 300°C). However, 5-hydroxypyrimidine1 ination except forms 2 (50°C, aqueous of 5- NazCO:). Draw resonance structures to show the electron rich ring positions c hydroxypyrimidine and draw the monobrominated products. (4 points) Solution Answer: 5- As each compounds have only one oxygen, so if any of these compounds are soluble it means that its oxygen is not free as much as compared to that which is not soluble becuase due to solubility its oxygen will be engaged with the water molecule. So the least soluble will be strongest hydrogen bond acceptor and the most soluble will be the least one.Therefore order will be (weakest to strongest): 1. 2-methyloxetane (Weakest) 2. 2-butanone 3. 1,2-epoxybutane 4. cis-2,3-epoxybutane 5. Butanal 6. Ethoxyethene (Strongest) .

1. please answer both questions if possible. thanks.
epoxybutane (9.6g/100 ml. HO), ethoxyethene ( ounds with the molecular formula of Call O are
shown in this list (with water solubility of each given i in parenthesis) are cis- .3- methyloxetane
(mi accept 20), ethoxyethene (6g/100mL 1O), 1.2-epoxybutane inal (7.5 g/1 00ni L H2O ), 2-
butanone ( 26g/ 100mL, H2O), and 2- sible). List these compounds in order of their hydrogen
bonding ,weakest acceptor (I) to strongest acceptor (6). (2 points) 5. Pyrimidine is an electron
poor aromatic ring and as such, it resists brominatio under extreme reaction conditions (Brz,
300°C). However, 5-hydroxypyrimidine1 ination except forms 2 (50°C, aqueous of 5-
NazCO:). Draw resonance structures to show the electron rich ring positions c
hydroxypyrimidine and draw the monobrominated products. (4 points)
Solution
Answer: 5- As each compounds have only one oxygen, so if any of these compounds are soluble
it means that its oxygen is not free as much as compared to that which is not soluble becuase due
to solubility its oxygen will be engaged with the water molecule. So the least soluble will be
strongest hydrogen bond acceptor and the most soluble will be the least one.Therefore order will
be (weakest to strongest):
1. 2-methyloxetane (Weakest)
2. 2-butanone
3. 1,2-epoxybutane
4. cis-2,3-epoxybutane
5. Butanal
6. Ethoxyethene (Strongest)