
(a)
Interpretation:
The major product of given reaction should be given.
Concept introduction:
In the nucleophilic substitution reaction, the
In
The rate determination step is formation of carbocation.
The stability order of carbocation is,
Tertiary > Secondary > Primary
Therefore, tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation. Primary alcohol is less stable therefore it won’t undergoes
(b)
Interpretation:
The major product of given reaction should be given.
Concept introduction:
In the nucleophilic substitution reaction, the rate of reaction depends on reactant as well as nucleophile, which are involved in reaction is called bimolecular nucleophilic substitution reaction.
In
Reactant and nucleophile are present at the rate determination step.
The order of species involving in
Tertiary < Secondary < Primary
(c)
Interpretation:
The major product of given reaction should be given.
Concept introduction:
Chromic Acid:
Chromic Acid (
(d)
Interpretation:
The major product of given reaction should be given.
Concept introduction:
Dehydration reaction:
Removal of water molecule from the reaction when the alcohol is treated with strong acid like sulfuric acid.
The stability of carbocation is given below,
Tertiary carbocation is more stable than the secondary and primary.
(e)
Interpretation:
The major product of given reaction should be given.
Concept introduction:
In the presence of acid catalyst, this reaction takes place through partial SN1 and partial SN2 pathway.
Epoxides are reactive, methoxide ion attacks the Epoxides in a less sterically hindered position which forms the alkoxide ion, and then it gets proton from alcohol which form the product.
(f)
Interpretation:
The major product of given reaction should be given.
Concept introduction:
In the presence of acid catalyst, this reaction takes place through partial SN1 and partial SN2 pathway. Epoxides are reactive, Epoxides get protonated followed by alcohol attacks to the stable carbocation and form the product.
Epoxides are reactive, methoxide ion attacks the Epoxides in a less sterically hindered position which forms the alkoxide ion, and then it gets proton from alcohol which form the product.
(g)
Interpretation:
The major product should be identified.
Concept introduction:
Tosylation reaction:
The alcohol is treated with any tosyl chloride (methane sulfonyl chloride) which yields tosylated product this reaction is called as alkyl tosylate and which is shown below,
The alcohols is reaction with acids like hydrochloric acid or hydrobromic acid, the bromine atom attacks back side of the carbon atoms which is bearing alcohol group which yield the corresponding inversion product.
(h)
Interpretation:
The major product of given reaction should be given.
Concept introduction:
Oxidation of alcohol:
Alcohols reaction with hypochlorous (oxidizing agent) in the presence of acetic acid which yields the corresponding
Primary alcohols gives aldehyde, secondary alcohols gives ketone.
(i)
Interpretation:
The major product should be identified.
Concept introduction:
Generally
For elimination reaction quartnary ammonium halide can be converted in to quartnary ammonium hydroxide by using aqueous silver oxide.
Hofmann elimination:
Quartnary ammonium ion undergoes elimination when using strong base like hydroxide ion this reaction is called as hofmann elimination. Proton abstraction is takes place in β- carbon atom which is having more number of hydrogen.
(j)
Interpretation:
The major product should be identified.
Concept introduction:
The alcohols is reaction with acids like hydrochloric acid or hydrobromic which yield the corresponding carbocation intermediate, this carbocation intermediate undergoes substitution reaction which yields the corresponding substitution product.
Tertiary alcohols undergo substitution very fast than the secondary alcohols because tertiary carbocation is more stable than the secondary carbocation than the primary carbocation.
Primary alcohol is less stable therefore it won’t undergoes

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Chapter 10 Solutions
Organic Chemistry Study Guide and Solutions Manual, Books a la Carte Edition (8th Edition)
- PROBLEMS Q1) Label the following salts as either acidic, basic, or neutral a) Fe(NOx) c) AlBr b) NH.CH COO d) HCOON (1/2 mark each) e) Fes f) NaBr Q2) What is the pH of a 0.0750 M solution of sulphuric acid?arrow_forward8. Draw all the resonance forms for each of the fling molecules or ions, and indicate the major contributor in each case, or if they are equivalent (45) (2) -PH2 سمة مدarrow_forwardA J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3arrow_forward
- 1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.arrow_forward3. For the reactions below, draw the arrows corresponding to the transformations and draw in the boxes the reactants or products as indicated. Note: Part A should have arrows drawn going from the reactants to the middle structure and the arrows on the middle structure that would yield the final structure. For part B, you will need to draw in the reactant before being able to draw the arrows corresponding to product formation. A. B. Rearrangement ΘΗarrow_forward2. Draw the arrows required to make the following reactions occur. Please ensure your arrows point from exactly where you want to exactly where you want. If it is unclear from where arrows start or where they end, only partial credit will be given. Note: You may need to draw in lone pairs before drawing the arrows. A. B. H-Br 人 C Θ CI H Cl Θ + Br Oarrow_forward
- 4. For the reactions below, draw the expected product. Be sure to indicate relevant stereochemistry or formal charges in the product structure. a) CI, H e b) H lux ligh Br 'Harrow_forwardArrange the solutions in order of increasing acidity. (Note that K (HF) = 6.8 x 10 and K (NH3) = 1.8 × 10-5) Rank solutions from least acidity to greatest acidity. To rank items as equivalent, overlap them. ▸ View Available Hint(s) Least acidity NH&F NaBr NaOH NH,Br NaCIO Reset Greatest acidityarrow_forward1. Consider the following molecular-level diagrams of a titration. O-HA molecule -Aion °° о ° (a) о (b) (c) (d) a. Which diagram best illustrates the microscopic representation for the EQUIVALENCE POINT in a titration of a weak acid (HA) with sodium. hydroxide? (e)arrow_forward
- Answers to the remaining 6 questions will be hand-drawn on paper and submitted as a single file upload below: Review of this week's reaction: H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O ---> H₂NC(=NH)N(CH3)CH2COOH (creatine) Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts) Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts) Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3 pts) NH2(C=NH)-N(CH)CH2COOH This bond Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem Q9 is valid). (4 pts) Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should be ready to understand the first half of one of the Grignard reactions presented in a past…arrow_forwardPropose a synthesis pathway for the following transformations. b) c) d)arrow_forwardThe rate coefficient of the gas-phase reaction 2 NO2 + O3 → N2O5 + O2 is 2.0x104 mol–1 dm3 s–1 at 300 K. Indicate whether the order of the reaction is 0, 1, or 2.arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning

