CHEM1611 Answers to Problem Sheet 10

1.

(a) pyranose

6-membered ring:

OOH

H

H

OH

OH

H

H

OH

H

CH2OH

D-glucopyranose

furanose

5-membered ring:

O

OH

CH2OH

H

OH

OH

H

CH2OH

H

D-fructofuranose

Fischer projection

Change the stereochemistry

at any stereogenic centre

(but not at all). For

example:

CHO

OHH

CH2OH

OH

OH

OH

H

H

H

2. If the pH < pKa of an ionisable group, then it will be predominantly in the

conjugate acid form.

If the pH > pKa, then it will be predominantly in the conjugate base form.

The pI (i.e. the pH at which the predominant species has a net charge of zero) is

the mean of the two pKa values associated with the formation of the +1 and –1

species. In this case, the mean is ½ (2.10 + 4.07) = 3.09.

If the pH = pKa, then 50% will be in the conjugate acid form and 50% in the

conjugate base form.

(a) pH = 1.0:

pKa (sidechain) > pH so it will exist in the protonated (-COOH) form.

pKa (α-COOH) > pH so it will also exist in the protonated (-COOH) form.

pKa (NH3+) > pH so it will also exist in the protonated (-NH3

+ form).

(b) pH = 4.0:

pKa (sidechain) > pH so it will exist in the protonated (-COOH) form.

pKa (α-COOH) < pH so it will exist in the deprotonated (-COO) form.

pKa (NH3+) > pH so it will exist in the protonated (-NH3

+ form).

(b) pH = 12.0:

pKa (sidechain) < pH so it will exist in the deprotonated (-COO) form.

pKa (α-COOH) < pH so it will exist in the deprotonated (-COO) form.

pKa (NH3+) > pH so it will exist in the deprotonated (-NH2 form).

pH 1.0 pH 4.0 pH 12.0

HOOCCOOH

NH3

H

+1 charge

Major (over 50%)

HOOCCOO

NH3

H

0 charge

Minor (under 50%)

OOCCOO

NH3

H

-1 charge

OOCCOO

NH2

H

-1 charge

3. (a)

H3N CH

CH2OH

C

O

NH C

CH3

CO2

hydroxyl

ammonium carboxylateamide

(b)

H3N CH

CH2OH

C

O

NH C

CH3

CO2HCl

(c)

H2N CH

CH2OH

C

O

NH C

CH3

CO2 Na

(d)

H3N CH

CH2OH

CO2H NH3 CH

CH3

CO2HCl Cl (e)

H3N CH

CH3

C

O

NH CH

CH2OH

CO2

4. By convention, the N-terminal is on the left:

H3N CH

O

NH CH

CH2

O

NH CH

CH2

O

NH CH

CH

CO2

CH3 CH3

CH

CH3CH3

CH

CH3CH3

CH2

N

NH

5. Note that this tetrapeptide has a modified C-terminus, which has been

converted to an amide with ammonia. On acid hydrolysis this will be converted

to the acid + ammonium ion.

C

CO2H

H

CH2SH

H3N

N H

CO2HH

H

C

CO2H

H

CH2

H3N

CH

CH3H3C

C

CO2H

H

H

H3N

absolute

configuration – L

absolute

configuration - L

absolute

configuration - L

absolute

configuration –

not applicable

6. (a) U is

CHO

CH2OH

H

OH

OHH

H

H

(b) The Haworth stereoformula for V is:

OH

OH

H

H

H

OH

CH2OH

H

(c) T is -2-deoxy-D-ribofuranose. V is -2-deoxy-D-ribofuranose

7. It is a fragment of RNA: the sugar component is ribose (rather than 2’-deoxy

ribose) and it contains the base uracil (rather than thymine).

H

N

H

O OH

H H

OOP

O

O

O

P OO

O

HH

O OH

H H

O

P OO

O

HH

OH OH

H H

O

NH

O

O

N

NN

N

NH2

N

N

NH2

O

phosphate

sugar: ribose

nucleic base: uracil (U)

nucleic base: adenine (A)

H

N

H

O OH

H H

OOP

O

O

O

P OO

O

HH

O OH

H H

O

P OO

O

HH

OH OH

H H

O

NH

O

O

N

NN

N

NH2

N

N

NH2

O

nucleotide:phosphate-sugar-base

nucleoside: sugar-base