Org 2 UE1 Answers - Tutorial 2015/16 Gary Midgley PDF

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Tutorial 2015/16 Gary Midgley...

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Organic Chemistry 2 (SIC2002) Unilearn Exercise No. 1 ANSWERS 1.

Name the following compounds.

Br

3 2

For a single ring with a single heteroatom, the heteroatom is numbered “1”. Keep lowest set for substituents 2,4- here (clockwise), not 3,5(anticlockwise).

5

1

Br

2,4-Dibromopyrrole

4

N H (i)

HO

H

7

1

N

6

2

HO

5 4

(ii)

Indole-5,6-diol (5,6-Indolediol is OK too.)

3

The compound without the OH groups is indole (see your notes). This has the N numbered “1”. The hetero ring (“pyrrole half”) is numbered first before transferring to the “benzene half” of the structure. This means that the diol OHs become “5,6-“ . OH is a principal functional group here, so is named as a prefix (diol) rather than a suffix (dihydroxy-). 5

4

6

O2N

7

7-Nitroisoquinoline

3

N 1

8

(iii) 5

2

For a fused ring, number in the same format as the parent hydrocarbon. The heteroatom then takes the lowest available number. The parent hydrocarbon in this case is naphthalene which is numbered as follows:

4

6

3

7

2

Naphthalene 8

1

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For quinoline, the N takes the number “1” (see notes), but for isoquinoline which we have in this question, the N is labelled “2” to correspond with naphthalene. As in the previous example, the hetero ring (the “pyridine half”) is numbered first before transferring to the “benzene half”. This means that the nitro group assumes position “7”. 1.

Imidazole is aromatic – it’s planar and has 6 x π-electrons (Hückel [4n+2], n = 1): sp 2 hybrid orbital

. .. N . ..

.

3

. 1

N H

Note that N1 is pyrrole-like (the lone pair is an integral part of the aromatic sextet) whereas the N3 heteroatom is pyridine-like (the lone pair is independent of the aromatic sextet). The lone pair on N3 is therefore more available for protonation since aromaticity isn’t destroyed when protonation of this N occurs; it’s therefore the most basic N. The N atoms are interchangeable by tautomerisation, but only the aromatic protonated ion of the two possibilities would be formed:

H N N H

N N

H

4: non-aromatic

H 6: still aromatic

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