There are two compounds of the formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}:\) The compound on the right, cisplatin, is used in cancer therapy. The compound on the left, transplatin, is ineffective for cancer therapy. Both compounds have a square- planar geometry. (a) Which compound has a nonzero dipole moment? (b) The reason cisplatin is a good anticancer drug is that it binds tightly to DNA. Cancer cells are rapidly dividing, producing a lot of DNA. Consequently, cisplatin kills cancer cells at a faster rate than normal cells. However, since normal cells also are making DNA, cisplatin also attacks healthy cells, which leads to unpleasant side effects. The way both molecules bind to DNA involves the \(\mathrm{Cl}^{-}\) ions leaving the Pt ion, to be replaced by two nitrogens in DNA. Draw a picture in which a long vertical line represents a piece of DNA. Draw the \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\) fragments of cisplatin and transplatin with the proper shape. Also draw them attaching to your DNA line. Can you explain from your drawing why the shape of the cisplatin causes it to bind to DNA more effectively than transplatin?

Short Answer

Expert verified
Cisplatin has a nonzero dipole moment due to the adjacent location of its Cl atoms, causing an uneven distribution of electron charges. It binds more effectively to DNA than transplatin because its adjacent Cl atoms are replaced by two nitrogens in the DNA, forming a stronger bond with the DNA strand. In contrast, transplatin's opposite Cl atoms lead to a weaker bond, making it less effective in cancer therapy.

Step by step solution

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1. Determine the molecular geometry of the compounds

Both cisplatin and transplatin have a square-planar geometry, as mentioned in the exercise. This means their atoms are arranged in a square with the central atom, Platinum (Pt), at the center. In cisplatin, the two Chloride (Cl) atoms are located next to each other, while in transplatin, they are located opposite each other.
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2. Identify the compound with a nonzero dipole moment

A compound has a nonzero dipole moment if it has an uneven distribution of electron charges. In cisplatin, the Cl atoms are next to each other, creating an area of higher electron density on one side of the molecule. This creates a net dipole moment in the molecule. On the other hand, in transplatin, the Cl atoms are located opposite each other, causing the electron density to be balanced, resulting in a net dipole moment of zero. Therefore, cisplatin has a nonzero dipole moment.
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3. Draw both compounds and their interaction with DNA

To understand why cisplatin binds more effectively to DNA than transplatin, first draw a vertical line to represent a piece of DNA. Then, draw the \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\) fragments of both compounds interacting with the DNA line: - Cisplatin: two adjacent Cl atoms are replaced by two nitrogens in DNA, allowing the compound to form a strong bond with the DNA strand. - Transplatin: Cl atoms are opposite to each other, so when they are replaced by nitrogens in DNA, the resulting bond is weaker and less efficient. This difference in binding can be visualized in the following manner: ``` DNA | cisplatin: Pt(NH3)2-N-N | DNA DNA | transplatin: N-Pt(NH3)2-N | DNA ```
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4. Explain the difference in binding efficiency between cisplatin and transplatin

The shape of cisplatin allows it to form a strong bond with DNA, as its two adjacent Cl atoms can effectively be replaced by two nitrogens from the DNA strand. This strong bond allows the compound to bind tightly to DNA, which is why it is a more effective anticancer drug than transplatin. In contrast, the shape of transplatin prevents it from forming strong bonds with DNA, making it less effective in cancer therapy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square-Planar Geometry
Square-planar geometry is a specific molecular shape that is crucial for understanding the behavior of certain compounds, including cisplatin. In this geometry, the central atom, platinum (Pt), is surrounded by four ligands (atoms or molecules that are bonded to the central atom) at the corners of a square. This arrangement places the ligands in the same plane as the central atom, giving the molecule a flat, or 'planar', appearance.

In the context of cisplatin, two of the ligands are ammonia (NH3) molecules, and the other two are chloride (Cl-) ions. With cisplatin, the chloride ions are adjacent to each other, while in transplatin, they are across from one another. This subtle difference in structure between the isomers leads to significantly different chemical behaviors and biological activities, highlighting the importance of molecular geometry in drug design and function.
Dipole Moment
A dipole moment is a vector quantity that represents the separation of positive and negative charges within a molecule. A molecule with a nonzero dipole moment has an asymmetric distribution of charges, leading to one side being slightly more negative and the other side slightly more positive.

Cisplatin possesses such an asymmetry because its chloride ions are next to each other, creating an area with a higher concentration of negative charge, resulting in a nonzero dipole moment. The transplatin molecule, on the other hand, has its chloride ions at opposite ends, balancing the electron density and resulting in no net dipole moment. This property not only affects the molecule's physical characteristics, like solubility and boiling point, but also its biological interactions. In the case of cisplatin and transplatin, the nonzero dipole moment of cisplatin may contribute to its stronger interaction with the negatively charged DNA backbone.
Anticancer Drug Mechanism
Understanding the mechanism of anticancer drugs is vital for appreciating how these compounds can selectively kill cancer cells while minimizing damage to normal cells. Cisplatin works by binding to DNA and interfering with the cell's replication process. As cancer cells divide rapidly, they are more vulnerable to the effects of DNA-binding drugs like cisplatin.

The two adjacent chloride ions in cisplatin are replaced by nitrogen atoms found in the DNA bases, forming cross-links. This cross-linking distorts the DNA helix and prevents it from being 'read' correctly by the cell's machinery. Consequently, the cell cannot duplicate its DNA or divide, which leads to programmed cell death, or apoptosis. While cisplatin is effective at targeting rapidly dividing cells, it also affects healthy cells, which produces the side effects associated with chemotherapy. Tailoring the shape and reactivity of such compounds is an ongoing research area aimed at developing more selective and less harmful cancer treatments.

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Most popular questions from this chapter

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{6},\) is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g)$$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4 ) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4},\) and write two possible molecular geometries for the molecule based on this electron- domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on \(\mathrm{F}\) interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

What hybridization do you expect for the atom indicated in red in each of the following species? (a) \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-} ;\) (b) \(\mathrm{PH}_{4}^{+}\) (c) \(\mathrm{AlF}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}^{+}\)

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \AA\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D} .\) (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

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