answers to chapter 2 - · pdf fileintroduction to bioorganic chemistry and chemical biology 1...

Introduction to Bioorganic Chemistry and Chemical Biology 1

Answers to Chapter 2(in-text & asterisked problems)

Answer 2.1

Answer 2.2

Answer 2.3

Answer 2.4

Introduction to Bioorganic Chemistry and Chemical Biology | A2084Van Vranken & Weiss | 978-0-8153-4214-4 © www.garlandscience.com design by www.blink.biz

NHN

....

SN CN

R

R..

+

-+

- ..

Basicity:

Basicity:

Basicity:

OOH..: .. ..

>

sp2

sp3 >

sp2sp3

sp2sp

>


F

*σC–FNu:

O

Cl

�*C–O

Nu:

NN

�*N–N

Nu:

pB

F BCH3

CH3

Nu:

NO H

*H–O

:Nu(:B)

σ

A B C ED


O....

nO nN

N.. N

NH2..

..sp2sp3

nNHB

H HH

-σB-H

A B C D


O

OEtCl

O

OEtNn-BuNH2

.. O:

OEtCl

Nn-Bu

HH

+ -n-Bu

H H

+

B:

O

OEtCl

O

OEtNn-BuNH2

.. O

OEtCl

Nn-Bu

HH

+n-Bu

H H

+

B:..

also acceptable...

Ph

OEt:

HA

OEt

Ph

+

H2O:PhHO

OEt

H

+

-A:

PhHO

OEt.. H A

PhHO

OEtH+

.. PhOH

+

-A:

A

B

2 Introduction to Bioorganic Chemistry and Chemical Biology: Answers to ChApter 2

Answer 2.5

note: amide bond resonance precludes the lone pair of the amide nh from acting as a hydrogen-bond acceptor.

Answer 2.6

Answer 2.7

A Keq = 10[pKa (phoh) – pKa (h20)]

Keq = 10[10 – –1.7)

k1 ≈ 1011 M–1 s–1 (diffusion limit for h3o+ in water) = 1011.7 ≈ 1012

Keq =

k–1 ≈ 1011 M–1 s–1 / 1012 = 0.1 M–1 s–1

B (h2n– anions are unlikely intermediates in water.)

Keq = 10[pKa (h3n) – pKa (h20)]

Keq = 10[38 ––1.7)

k1 ≈ 1011 M–1 s–1 (diffusion limit for h3o+ in water) = 1039.7 ≈ 1040

Keq =

k–1 ≈ 1011 M–1 s–1 / 1040 = 10–29 M–1 s–1

the deprotonation of nh3 by h2o would be inconceivably slow.

Answer 2.8


O+H3N

HN

O -

O

O

HO

NH

OH

CH3

OHO

HOHO

OH

OH

NH3+

H-bond acceptors = boxedH-bond donors = circled


N

O

N

O

N

O

O

N

N

O

N

O

N

O

O

N

H H

H

H

H

H

H

H

Kevlar


NH

OH

NH2

O

AH

:

N

OHH

H

+

A-:

NH2OH

NH3+O-

B:

NH2O-..

HB+

A

B

Introduction to Bioorganic Chemistry and Chemical Biology: Answers to ChApter 2 3

Answer 2.9

A 10(9.2–7.2) = 100; [hCn]/[Cn–] = 100:1

B 10(9.2–7.2) = 100; [nh4+]/[nh3] = 100:1

C 10(4.2–7.2) = 0.001; [phCo2h]/[phCo2–] = 1:1000

Answer 2.10

note that the actual equilibrium will involve all possible diastereomers.

Answer 2.11

Answer 2.12

A Use the following equation.

only the Coulombic, first term is relevant for the charge–charge repulsion. substi-tute the distances given as r into the Coulombic equation as follows.

In other words, the repulsive interaction potential decreases by about 33%.

B here, only the third term for the attractive interaction is relevant, because the first two terms can be approximated as zero. we can compare the attractive energies at those two differences by taking the ratio.

thus, the attractive interaction potential decreases by more than 90%.


O OH

OHR

..(HO)2Ca

O O

OHR

CaHO OH

H

-

+ :B O O

OHR

CaHO OH-

O O

OHR

CaOH..

O OCa+

-OH

..

OHR

O OCa +-

..

..

OH

OHR

O OCa

+-

OH

..

OHR

O OH

OHR


Henderson–Hasselbalch: log[A–]

[HA]= pKa – pH

Rearranged:[A–]

[HA]= 10(pKa – pH)


E ε r

4� q1 × q2

1.987 cal mol–1 K–1 × 298.15 K

–9800 cal mol–1

r12

x

r6

y+ –∝


ε 34� q1 × q2

= 2/3

ε 24� q1 × q2


36

y

= (2/3)6 = 0.088

26

y


O

H3CNH3:

O -

NH2

+H3C

H

A: O

NH2H3C

HA

- : OH

NH2H3C

NH2+H3C

NC-: CN

NH2H3C

:HA

OH2

NH2H3C ..

+


Answer 2.13

*Answer 2.14

*Answer 2.16

*Answer 2.17

*Answer 2.18


N

OH2NO OH

O

O

areneinteraction

net dipole net dipole

net dipole

net dipole

H-bond donors = circled H-bond acceptors = boxed

Every part of these molecules is capable of engaging in dispersion interactions (not shown)

areneinteraction

areneinteraction


O

OH3CH

: :....

O

HH3C

: : O

H

: :

H H

H OO

HH

..

......

C O:- +

O

OHCH3

: :....

O

H

: :HH

HH

H

N C+ -

H3C: :

A B C D

E F G H


O

O

O O

CH2H3C..

- H

C CNN

NH2

:nO

:nO

nC-

O

CH2H3C

:nO-

actually, these "lone pairs" arein orbitals with � symmetry

..nN

..

or

A B C D


NH

NH2N

O.. H

CH

-

H

NN

..

+

-+

NH

NHN

O+

H

CC

HH

CN

CCN

..:

:

-

N N

..

+

-A B C


OC

OCH3Ph

:: H A OC

OCH3Ph

:

H+

H2O..

OHC

OCH3

PhHOH+

-A

OHC

OCHPhHO ..

HA

OHC

OCH3

PhHO

H

+

: OC

OHPh

H+ : A-OC

OHPh

O OH

:B - O O-

..O O-..

O O-

.. H B O OH


OC

OCH3Ph

:: H A OC

OCH3Ph

:

H+

H2O..

OHC

OCH3

PhHOH+

-A

OHC

OCHPhHO ..

HA

OHC

OCH3

PhHO

H

+

: OC

OHPh

H+ : A-OC

OHPh

O OH

:B - O O-

..O O-..

O O-

.. H B O OH

Introduction to Bioorganic Chemistry and Chemical Biology: Answers to ChApter 2 5

*Answer 2.20

*Answer 2.21

*Answer 2.24


ROH

O

R NR'

Me

HO R"

R'

R

pH

2

phenol amine carboxylic acid

RO -

O

R NR'

Me

HO R"

R'

R

7

RO -

O

R NR'

Me

- O R"

R'

R

12

H+

H+


HN

O

HO

OHCl

OH

N

N

O

OH

N

OHN

OMe

O

pKa 5 pKa' 10

pKa 5

pKa 16

pKa' 10

pKa' 10

H2N

O

- O

OHCl

OH

pH 7+

pH 7N

N

O

OO

pH 7

N

OHN

OMe

+

H+ -

H


+H3N

HN

O -

O

O

HO

PO

NH3+

O-O

O

HO

O

O

CO2-

OHHO

HOOH

NH2

NH

A B

C


*Answer 2.28

*Answer 2.30

A 23 = 8 stereoisomers

B

*Answer 2.33


O

H

OH

OH

OH

OH

O

HOOH

HOO -

..

H

HA .. O

HOOH

HOOH

H+

O

HOOH

HOO -

H

HA ..

O

HOOH

HOOH

H+

riboseα-ribofuranose

-A :

-A :

β-ribofuranose

+

+


O

H

OH

OH

OH

OH

D-ribose L-ribose

O

H

OH

OH

OH

OH


N C NH2

NH3N

NH2NH

H H+

-

B N

NH2NH

-

H

H NH3+ NH

NH2H2N

::

::

aminonitrile guanidine

answers to chapter 2 - · pdf fileintroduction to bioorganic chemistry and chemical biology 1...

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