Question

Please help explain these questions. thank you.

10) Circle there has the compounds listed below. To receive il credit, bewelly explain your selection H 1) Identity by name the cos of compounds cancional group that each of the following molecules belongs to OH NHE 12) Sepce problem: Fill in the banks A - D for each compound or reagent that is missing OH 4.HNO, B 1] NaNH 2) D

Answer 1

10. Pyridine is more basic than pyrrole.

Basicity means the ability to donate electrons. More easily the electrons are donated, more basic the species is.

The lone pair of electrons on pyrrole is not available for donation. The electrons takes part in the delocalisation (to impart aromaticity of the ring). Hence, it is less basic.

On the other hand, the lone pair of electrons on nitrogen in pyridine are readily available for donation (they do no take part in delocalisation). Hence, it is more basic.

11. From left to right:

a. Ketone (CO group with both side alkyl group).

b. Amide (CONH2 group).

c. Alcohol (OH group).

d. Ester (COO group is present).

e. Acid chloride (COCl group).

f. Cyclic ether (oxygen attached to alkyl groups on both side, present in a cyclic ring).

g. Cyanide (-CN group)

h. Cyclic ketal.

12.

A = Iodobenzene, C₆H₅I

B = NaOH, high temperature (~ 130 - 140℃)

C = Ethoxybenzene, C₆H₅OCH₂CH₃ .

D = Birch reduction, Na ( or Li) + NH₃ + EtOH.

13. At low temperature, 1,2-adduct will be formed. Because the reaction, at low temperature, is irreversible and proceeds through lowest energy transition state. In other words, you can say that the reaction proceeds via most stable carbocation (secondary, resonance-stabilised carbocation).