Concept explainers
(a)
Interpretation:
The electron-rich sites and electron-poor sites in the given elementary steps are to be identified.
Concept introduction:
An atom with a partial or full negative charge is an electron-rich site whereas an atom with a partial or full positive charge is an electron-poor site. In an elementary step, electrons tend to flow from an electron-rich site to an electron-poor site.

Answer to Problem 7.53P
The electron-rich and electron-poor sites for each elementary step are:
First elementary step:
Second elementary step:
Third elementary step:
Fourth elementary step:
Fifth elementary step:
Sixth elementary step:
Explanation of Solution
The first elementary step is:
In this step, the carbonyl oxygen of benzoic acid having lone pairs is the electron-rich site. The hydrogen atom carry partial positive charge due to adjacent positively charged oxygen atom and this is an electron-poor site. The electron-rich and electron-poor sites for this step are labelled below:
The second elementary step is:
In this step, the hydroxyl oxygen of isopropanol having lone pair is the electron-rich site. The carbonyl carbon carries a partial positive charge due to polar
The third elementary step is:
In this step, the hydroxyl oxygen of isopropannol having lone pairs is the electron-rich site. The hydrogen atom bonded to positively charged oxygen is the electron-poor site. The electron-rich and electron-poor sites for this step are labelled below:
The fourth elementary step is:
In this step, the hydroxyl groups bonded to the carbon having benzene ring attached are identical. One of the hydroxyl oxygen having lone pairs is the electron-rich site. The hydrogen atom carry partial positive charge due to adjacent positively charged oxygen atom and this is an electron-poor site. The electron-rich and electron-poor sites for this step are labelled below:
The fifth elementary step is:
In this step, the hydroxyl oxygen having lone pairs is the electron-rich site. The carbon bonded to three oxygen atoms has partial positive charge due to electronegativity of oxygen atoms is the electron-poor site. The electron-rich and electron-poor sites for this step are labelled below:
The sixth elementary step is:
In this step, the oxygen atom of a water molecule having lone pairs is an electron-rich site. The hydrogen atom bonded to positively charged oxygen is the electron-poor site. The electron-rich and electron-poor sites for this step are labelled below:
The electron-rich site and the electron-poor sites in each elementary step are identified on the basis of the negative and partial positive charge on respective atoms.
(b)
Interpretation:
In each of the given elementary steps, the appropriate curved arrows are to be drawn.
Concept introduction:
The curved arrow can be drawn from electron-rich site to an electron-poor site to show the flow of electron from electron-rich site to electron-poor site. The first curved arrow is drawn from the lone pair of negatively charged atom of electron-rich site to the less electronegative atom of electron-poor site. The second curved arrow is drawn from the region between the less electronegative atom and more electronegative atom towards the more electronegative atom indicating the breaking of bond.

Answer to Problem 7.53P
The curved arrow mechanism for each elementary steps are:
The first elementary step:
The second elementary step:
The third elementary step:
The fourth elementary step:
The fifth elementary step:
The sixth elementary step:
Explanation of Solution
The first elementary step is:
In this step, the carbonyl oxygen of benzoic acid having lone pairs is the electron-rich site. The hydrogen atom adjacent to the positively charged oxygen atom is an electron-poor site. The curved arrow mechanism for this step is shown below:
The first curved arrow is drawn from the lone pair of electron-rich oxygen to electron-poor hydrogen atom representing the formation of the
The second elementary step is:
In this step, the hydroxyl oxygen of isopropanol having lone pair is the electron-rich site. The carbonyl carbon is the electron-poor site. The curved arrow mechanism for this step is shown below:
The first curved arrow is drawn from the lone pair of electron-rich hydroxyl oxygen of isopropanol to the electron-poor carbonyl carbon representing the formation of
The third elementary step is:
In this step, the hydroxyl oxygen of isopropannol having lone pairs is the electron-rich site. The hydrogen atom bonded to positively charged oxygen is the electron-poor site. The curved arrow mechanism for this step is shown below:
The first curved arrow is drawn from the lone pair of electron-rich hydroxyl oxygen of isopropannol to the electron-poor hydrogen atom representing the formation of
The fourth elementary step is:
In this step, one of the hydroxyl groups bonded to the benzylic carbon is the electron-rich site. The hydrogen atom bonded to a positively charged oxygen atom is an electron-poor site. The curved arrow mechanism for this step is shown below:
The first curved arrow is drawn from the lone pair of electron-rich hydroxyl oxygen atom to the electron-poor hydrogen atom indicating the formation of the
The fifth elementary step is:
In this step, the hydroxyl oxygen having lone pairs is the electron-rich site. The carbon bonded to three oxygen atoms is the electron-poor site. The curved arrow mechanism for this step is shown below:
The first curved arrow is drawn from the lone pair of electron-rich hydroxyl oxygen atom to the electron-poor carbon atom indicating the formation of
The sixth elementary step is:
In this step, the oxygen atom of a water molecule having lone pairs is an electron-rich site. The hydrogen atom bonded to positively charged oxygen is the electron-poor site. The curved arrow mechanism for this step is shown below:
The first curved arrow is drawn from the lone pair of the electron-rich hydroxyl oxygen atom of a water molecule to the electron-poor hydrogen atom indicating the formation of the
The curved arrows for each of the given elementary steps are drawn from electron-rich site to electron-poor site and the less electronegative atom to more electronegative atom representing the formation and breaking of respective bonds.
(c)
Interpretation:
The names of each elementary step are to be identified.
Concept introduction:
In nucleophilic addition step, the nucleophile adds to the polar
In nucleophilic elimination step, the more electronegative atom bears full negative charge or partial negative charge. This electronegative atom forms a
An elementary step in which a proton is transferred from electron-poor site to electron-rich site and one bond is broken and another is formed simultaneously is called proton transfer step.

Answer to Problem 7.53P
The names for each of the given elementary steps are given below:
The first elementary step is proton transfer reaction.
The second elementary step is nucleophilic addition reaction.
The third elementary step is proton transfer reaction.
The fourth elementary step is proton transfer reaction.
The fifth elementary step is nucleophilic elimination reaction.
The sixth elementary step is proton transfer reaction.
Explanation of Solution
The first elementary step is:
This step involved the transfer of a proton from a positively charged oxygen atom to an electron-rich carbonyl oxygen atom. Thus this step is named as proton transfer reaction.
The second elementary step is:
In this step, the hydroxyl oxygen of isopropyl hydrogen having lone pair is acts as nucleophile. The carbonyl carbon is electron-poor site acts as electrophile due to adjacent electronegative oxygen atom. The lone pair on nucleophilic oxygen atom approached to the electrophilic carbonyl carbon by breaking
The third elementary step is:
This step involved the transfer of a proton from positively charged oxygen atom to an electron-rich oxygen atom of isopropanol. Thus this step is named as proton transfer reaction.
The fourth elementary step is:
This step involved the transfer of a proton from positively oxygen atom to an electron-rich oxygen atom. Thus this step is named as proton transfer reaction.
The fifth elementary step is:
In the above step, the hydroxyl oxygen having lone pair is electron-rich site and acts as a nucleophile and the carbon atom bonded to three oxygen atoms carry partial positive charge acts as an electrophile and the water molecule is leaving group. The lone pair of nucleophilic oxygen approached to the electrophilic carbon and polar
The sixth elementary step is:
This step involved the transfer of a proton from positively carbonyl oxygen atom to the electron-rich oxygen atom of the water molecule. Thus this step is named as proton transfer reaction.
The names for the given elementary steps are identified on the basis of the type of bond-forming and breaking.
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Chapter 7 Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
- Predict the major products of this organic reaction. If there aren't any products, because nothing will happen, check the box under the drawing area instead. No reaction. HO. O :☐ + G Na O.H Click and drag to start drawing a structure. XS xs H₂Oarrow_forwardWhat are the angles a and b in the actual molecule of which this is a Lewis structure? H H C H- a -H b H Note for advanced students: give the ideal angles, and don't worry about small differences from the ideal groups may have slightly different sizes. a = b = 0 °arrow_forwardWhat are the angles a and b in the actual molecule of which this is a Lewis structure? :0: HCOH a Note for advanced students: give the ideal angles, and don't worry about small differences from the ideal that might be caused by the fact that different electron groups may have slightly different sizes. a = 0 b=0° Sarrow_forward
- Determine the structures of the missing organic molecules in the following reaction: + H₂O +H OH O OH +H OH X Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structure of the missing organic molecule X. Click and drag to start drawing a structure.arrow_forwardIdentify the missing organic reactant in the following reaction: x + x O OH H* + ☑- X H+ O O Х Note: This chemical equation only focuses on the important organic molecules in the reaction. Additional inorganic or small-molecule reactants or products (like H₂O) are not shown. In the drawing area below, draw the skeletal ("line") structure of the missing organic reactant X. Click and drag to start drawing a structure. Carrow_forwardCH3O OH OH O hemiacetal O acetal O neither O 0 O hemiacetal acetal neither OH hemiacetal O acetal O neither CH2 O-CH2-CH3 CH3-C-OH O hemiacetal O acetal CH3-CH2-CH2-0-c-O-CH2-CH2-CH3 O neither HO-CH2 ? 000 Ar Barrow_forward
- What would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 2 2. n-BuLi 3 Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure.arrow_forwardPredict the products of this organic reaction: NaBH3CN + NH2 ? H+ Click and drag to start drawing a structure. ×arrow_forwardPredict the organic products that form in the reaction below: + OH +H H+ ➤ ☑ X - Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. Garrow_forward
- Predict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ marrow_forwardDetermine the structures of the missing organic molecules in the following reaction: + H₂O +H H+ Y Z ☑ ☑ Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X, Y, and Z. You may draw the structures in any arrangement that you like, so long as they aren't touching. Molecule X shows up in multiple steps, but you only have to draw its structure once. Click and drag to start drawing a structure. AP +arrow_forwardPlease help, this is all the calculations i got!!! I will rate!!!Approx mass of KMnO in vial: 3.464 4 Moss of beaker 3×~0. z Nax200: = 29.9219 Massof weacerv after remosimgain N2C2O4. Need to fill in all the missing blanks. ง ง Approx mass of KMnO4 in vials 3.464 Mass of beaker + 3x ~0-304: 29.9219 2~0.20 Miss of beaker + 2x- 29.7239 Mass of beaker + 1x~0.2g Naz (204 29-5249 Mass of beaver after removing as qa Na₂ C₂O T1 T2 T3 Final Buiet reading Initial butet reading (int)) Hass of NaOr used for Titration -reading (mL) calculation Results: 8.5ml 17mL 27.4mL Oml Om Oml T1 T2 T3 Moles of No CO Moles of KMO used LOF KM. O used Molenty of KMNO Averagem Of KMOWLarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
