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(a)
Interpretation:
The main product of the mononitration of benzoic acid should be predicted.
Concept introduction:
The electrophilic
Where,
The activating groups are the groups that have the ability to donate the electron density to the benzene ring.
The deactivating groups are the groups that have the ability to withdraw the electron density to the benzene ring. The ortho and para directing groups are the activating groups while meta direction groups are deactivating groups.
The nitration reaction takes place in the presence of nitric acid and sulphuric acid. In this reaction, the protonation of nitric acid occurs in order to produce the nitronium ion. The nitronium ion will attach on the benzene ring to form nitrobenzene. The general reaction is as follows:
(b)
Interpretation:
The main product of the monosulphonation of phenol should be predicted.
Concept introduction:
The electrophilic aromatic substitution is the type of reaction in which an electrophile substitutes the hydrogen atom of benzene. A general electrophilic aromatic substitution reaction of benzene can be written as:
Where,
The activating groups are the groups that have the ability to donate the electron density to the benzene ring.
The deactivating groups are the groups that have the ability to withdraw the electron density to the benzene ring. The ortho and para directing groups are the activating groups while meta direction groups are deactivating groups.
The sulfonation takes place in the presence of sulphuric acid. In this reaction, sulfur trioxide is formed that acts as an electrophile. Sulfur trioxide will attach on the benzene ring to form the final product. The general reaction is as follows:
(c)
Interpretation:
The main product of the monobromination of
Concept introduction:
The electrophilic aromatic substitution is the type of reaction in which an electrophile substitutes the hydrogen atom of benzene. A general electrophilic aromatic substitution reaction of benzene can be written as:
Where,
The activating groups are the groups that have the ability to donate the electron density to the benzene ring.
The deactivating groups are the groups that have the ability to withdraw the electron density to the benzene ring. The ortho and para directing groups are the activating groups while meta direction groups are deactivating groups.
Halogenation is the
The bromination takes place in the presence of Lewis acid and bromine molecule. In this reaction, the bromonium ion is produced that acts as an electrophile. Brominium will attach on the benzene ring to form the final product. The general reaction is as follows:
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Chapter 27 Solutions
GENERAL CHEMISTRY-MOD.MASTERINGCHEM.
- Nonearrow_forwardn Feb 3 A T + 4. (2 pts) Draw the structure of the major component of the Limonene isolated. Explain how you confirmed the structure. 5. (2 pts) Draw the fragment corresponding to the base peak in the Mass spectrum of Limonene. 6. (1 pts) Predict the 1H NMR spectral data of R-Limonene. Proton NMR: 5.3 pon multiplet (H Ringarrow_forwardPart VI. Ca H 10 O is the molecular formula of compound Tom and gives the in the table below. Give a possible structure for compound Tom. 13C Signals summarized C1 C2 C3 C4 C5 C6 C7 13C shift (ppm) 23.5 27.0 33.0 35.8 127 162 205 DEPT-90 + DEPT-135 + +arrow_forward
- 2. Using the following data to calculate the value of AvapH o of water at 298K. AvapH o of water at 373K is 40.7 kJ/mol; molar heat capacity of liquid water at constant pressure is 75.2J mol-1 K-1 and molar heat capacity of water vapor at constant pressure is 33.6 J mol-1 K-1.arrow_forwardPart VII. Below are the 'HNMR 13 3 C-NMR, COSY 2D- NMR, and HSQC 20-NMR (Similar with HETCOR but axes are reversed) spectra of an organic compound with molecular formula C6H13 O. Assign chemical shift values to the H and c atoms of the compound. Find the structure. Show complete solutions. Predicted 1H NMR Spectrum ли 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 f1 (ppm)arrow_forward3. Draw the expanded structural formula, the condensed structural formula, and the skeletal structural formula for 2-pentene. expanded structure: Condensed structure: Skeletal formula: 4. Draw the expanded structural formula, the condensed structural formula, and the skeletal structural formula for 2-methyl-3-heptene. expanded structure: Condensed structure: Skeletal formula: following structurearrow_forward
- Part IV. Propose a plausible Structure w/ the following descriptions: a) A 5-carbon hydrocarbon w/ a single peak in its proton decoupled the DEPT-135 Spectrum shows a negative peak C-NMR spectrum where b) what cyclohexane dione isomer gives the largest no. Of 13C NMR signals? c) C5H120 (5-carbon alcohol) w/ most deshielded carbon absent in any of its DEPT Spectivaarrow_forward13C NMR is good for: a) determining the molecular weight of the compound b) identifying certain functional groups. c) determining the carbon skeleton, for example methyl vs ethyl vs propyl groups d) determining how many different kinds of carbon are in the moleculearrow_forward6 D 2. (1 pt) Limonene can be isolated by performing steam distillation of orange peel. Could you have performed this experiment using hexane instead of water? Explain. 3. (2 pts) Using GCMS results, analyze and discuss the purity of the Limonene obtained from the steam distillation of orange peel.arrow_forward
- Part III. Arrange the following carbons (in blue) in order of increasing chemical shift. HO B NH 2 A CIarrow_forward6. Choose the compound that will produce the spectrum below and assign the signals as carbonyl, aryl, or alkyl. 100 ō (ppm) 50 0 7. 200 150 Assign all of the protons on the spectrum below. 8. A B 4 E C 3 ō (ppm) 2 1 0 Choose the compound that will produce the spectrum below and assign the signals to the corresponding protons. OH 6 OH 3 2 1 0 4 ō (ppm)arrow_forwardIn the Thermo Fisher application note about wine analysis (Lesson 3), the following chromatogram was collected of nine components of wine. If peak 3 has a retention time of 3.15 minutes and a peak width of 0.070 minutes, and peak 4 has a retention time of 3.24 minutes and a peak width of 0.075 minutes, what is the resolution factor between the two peaks? [Hint: it will help to review Lesson 2 for this question.] MAU 300 200 T 34 5 100- 1 2 CO 6 7 8 9 0 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 Minutes 3.22 0.62 1.04 O 1.24arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
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