(a)
Interpretation:
The group number of the given electron-dot symbol should be stated.
Concept Introduction:
In an electron-dot symbol, dots are used to represent valence electrons around the elemental symbol.
When drawing or understanding an electron-dot symbol of an element, one should consider below facts;
- Each dot represents 1 electron.
- The dots are placed on the four sides of the elemental symbol.
- For 1 to 4 valence electrons, single dots are used.
- For more than 4 electrons, the dots are paired.
Valence electrons are the most loosely bound electrons of an element. They are in the outermost shell, which is also called "valence shell". Chemical properties of an element and its place in the periodic table depend on the number of valence electrons of that element.
For main group elements, the group number is also the number of valence electrons of the element.
(b)
Interpretation:
The charge of the ion formed by the given element should be determined.
Concept Introduction:
An atom of a main group element loses or gains electrons to obtain the electronic configuration of the noble gas closest to it in the periodic table to form ions.
Cations are formed by losing electrons, thus they have fewer electrons than protons and are positively charged.
Anions are formed by gaining electrons, thus they have more electrons than protons and are negatively charged.
For example, Sodium (Na) atom has 11 electrons (
(c)
Interpretation:
The formula of an ionic compound formed from Q and fluorine should be determined.
Concept Introduction:
Ionic compounds are composed of cations and anions, which are tightly attracted to each other.
The sum of the charges in an ionic compound must always be zero.
The formula for an ionic compound shows the ratio of ions that combine to give zero charge.
When cations and anions have different charges, the number of cations and anions differ so that the overall charge of the ionic compound be zero.
As an example, see the formulae of NaCl and MgCl2.
(d)
Interpretation:
The formula of an ionic compound formed from Q and oxygen should be determined.
Concept Introduction:
Ionic compounds are composed of cations and anions, which are tightly attracted to each other.
The sum of the charges in an ionic compound must always be zero.
The formula for an ionic compound shows the ratio of ions that combine to give zero charge.
When cations and anions have different charges, the number of cations and anions differ so that the overall charge of the ionic compound be zero.
As an example, see the formulae of NaCl and MgCl2.

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Chapter 3 Solutions
GENERAL,ORGANIC, & BIOLOGICAL CHEM-ACCES
- What is the reaction mechanism for this?arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. + Drawing Arrows CH3ONA, CH3OH heat : Br:O Na → H H Br Na + H H H H H :0: .H + Undo Reset Done Q CH3 Drag To Pan +arrow_forwardWhat is the reaction mechanism for this?arrow_forward
- 20.19 Predict the structure of the major 1,2-addition product formed by reaction of one mole of Cl₂ with 3-methylenecyclohexene. Also predict the structure of the 1,4-addition product formed under these conditions. 20.20 Which of the two molecules shown do you expect to be the major product formed by 1,2-addition of HCI to cyclopentadiene? Explain. Cyclopentadiene + HC 3-Chlorocyclopentene (racemic) or 4-Chlorocyclopentene (racemic)arrow_forward20.35 Propose structural formulas for compounds A and B and specify the configuration of compound B. EtO₂C 250°C C14H2004 CO₂Et 1. Oso, then NaHSO3 2. HIO4 C14H2006 A Barrow_forward20.21 Predict the major product formed by 1,4-addition of HCI to cyclopentadiene. 20.22 Draw structural formulas for the two constitutional isomers with the molecular for- mula C₂H,Br, formed by adding one mole of Br, to cyclopentadiene.arrow_forward
- Add substituents to draw the conformer below (sighting down the indicated bond), then rotate the back carbon to provide the conformation that will be capable of an E2 elimination. R/S stereochemistry is graded. + I I H CH3 Ph Досн Br OCH 3 Drawing Q H Atoms, Bonds and Rings Charges Tap a node to see suggestions. H H H H H Undo Reset Remove Done Rotatearrow_forward20.17 Predict the structure of the major product formed by 1,2-addition of HBr to 3-methylenecyclohexene. 3-Methylenecyclohexene 20.18 Predict the major product formed by 1,4-addition of HBr to 3-methylenecyclohexene.arrow_forward+ Draw a vicinal alkyl bromide that would produce the following alkene in an E2 elimination. Use a dash or wedge bond to indicate stereochemistry on asymmetric centers, where applicable. Ignore any inorganic byproducts. Br Drawing Strong Base H Q Atoms, Bonds Charges and Rings Draw or tap a new bond to see suggestions. Remove Done 語 Reset Undo + Drag To Panarrow_forward
- Draw a vicinal alkyl bromide that would produce the following alkene in an E2 elimination. Use a dash or wedge bond to indicate stereochemistry on asymmetric centers, where applicable. Ignore any inorganic byproducts. + Drawing Į Strong Base H Br Q Atoms, Bonds and Rings Charges Draw or tap a new bond to see suggestions. Undo Reset 謂 Remove Done Drag To Pan +arrow_forwardDraw the product of the E2 reaction shown below. Include the correct stereochemistry. Ignore any inorganic byproducts. + Br CH3 Q Strong Base Drawing Atoms, Bonds and Rings Charges Undo Reset H "Br H N Br. Remove Done .N. Drag To Panarrow_forwardCurved arrows are used to illustrate the flow of electrons. Use the reaction conditions provided and follow the curved arrows to draw the product of this elementary step in an elimination mechanism. Include all lone pairs and charges as appropriate. Ignore stereochemistry. Ignore byproducts. + Br: .. 8 0.01 M NaOH heat Drawing Q Atoms, Bonds and Rings Charges and Lone Pairs Draw or tap a new bond to see suggestions. Undo Reset Remove Done + Drag To Panarrow_forward
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