Give the structures of the products you would expect when each alcohol reacts with (1) HCl, ZnCl 2 ; (2) HBr; (3) PBr 3 ; (4) P/I 2 ; and (5) SOCl 2 . (a) butan-1-ol (b) 2-methylbutan-2-ol

Hey everyone. Let's do this problem and says, predict the expected products when the following two alcohols are treated with each of the given re agents and the Regents are one hcl and zinc chloride to H B R three P B r 34 phosphorus iodide and five S O C 02. And the alcohols that we are reacting them with our propane one all and two methyl propane to all. So let's go through our re agents and discuss how they usually work with alcohols. And then we'll show the predicted products. So first we have hcl and zinc chloride, so hydrochloric acid reacts with alcohols to give Hey lights. However, the chloride ion is a weak nuclear file. Thus an additional lewis acids such as zinc chloride is sometimes necessary to promote this reaction with primary and secondary alcohols. Zinc chloride coordinates with the oxygen of the alcohol. In the same way that a proton does accept zinc chloride corresponds more strongly. This re agent of hcl and zinc chloride is called the Lucas re agent. And when primary alcohols react with Lucas reagents, ionization is not really possible because the primary carbon katan is too unstable. Primary substrates substrates would react by an SN two mechanism. So, primary via s. n. two. But we said it's not really common or really possible and secondary and tertiary alcohols React via and sn one to give a Hey lied. So substituting the alcohol with a chlorine atom. So I drew our little template here with all of our reagents or our alcohols rather so we can add our re agents and add our predicted product. So first hcl and zinc chloride. We are simply replacing the alcohol group with our chlorine atom. Right? So these would be our predicted products. And we said that would be via SN two for our primary alcohol and sn one for our tertiary alcohol. Alright, moving on to H. B. R. So alcohol is a weak electro file because the hydroxy group is a poor leaving group. Right? But it can become a good leaving group when we make it water when we protein ate it to make water. So HBR reacts with a primary alcohol by sN two. Attack of bromide on the protein ated alcohol. So we convert Oh h. two H 20. To make a good leaving group. And then we have A primary alcohol reacting via SN two. Okay. And a second or tertiary and secondary alcohol would react by SN one. And that is it for that reaction. So let's predict the product, we're going to substitute our alcohol with me. And again that reactant is HBR. Alright, moving on to the next one, we have PBR three phosphorous tri bromide. So this is often the best regent for converting primary or secondary alcohols to alcohol bromide. So better than HBR especially if the alcohol might rearrange into a strong acid. So best for primary secondary alcohol to alcohol bromine. Mhm. So again just converting that O. H group into br so PBR three PBR three, primary alcohol. Hey line tertiary alcohol. Hey line alcohol bromine. Alright. Our next re agent is phosphorus iodine. And this one reacts similarly to phosphorus tribe Roma. Except instead of adding a bromine atom, it's going to produce an alkali iodide. So best for all of this same thing except alcohol I died. So the same product except adding an iodine adam instead of me. So P. I. Two, P. I. Two primary alcohol iodine, tertiary alkali died. And lastly We have the final chloride s. o. c. l. two. So this is often the best re agent for converting an alcohol to an alcohol chloride. And in this mechanism the nuclear file is actually delivered to the carbo catan by the leaving group which usually gives a retention of the configuration under different conditions. The retention of configuration is not always observed. So in this case we have an alcohol to an alcohol chloride and we have retention of configuration. So again converting that alcohol group to a chlorine group. And it may be worth noting that in the last three reactions. So the case of phosphorus hey lied and vinyl chloride, the reaction follows the sn two mechanism. Since the compound is since compound to is a tertiary alcohol, the sn two reaction is not feasible due to hysteric hindrance during the backside attack of the nuclear file. So because they are tertiary they are not good with the backside attack. Alright. So that is it for this problem. These are all our possible reactions, some of them more feasible than others. But that makes c the correct answer for this problem. That's it for this one. I hope this video is helpful, and thanks for watching.

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1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

12. Alcohols, Ethers, Epoxides and Thiols

Leaving Group Conversions Summary

Organic Chemistry

12. Alcohols, Ethers, Epoxides and Thiols

Leaving Group Conversions Summary

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7:37 minutes

Problem 21a,b

Textbook Question

Give the structures of the products you would expect when each alcohol reacts with (1) HCl, ZnCl₂; (2) HBr; (3) PBr₃; (4) P/I₂; and (5) SOCl₂.
(a) butan-1-ol
(b) 2-methylbutan-2-ol

Verified step by step guidance

Step 1: Understand the reactivity of alcohols with the given reagents. Alcohols can undergo substitution reactions where the hydroxyl group (-OH) is replaced by a halide (Cl, Br, or I). The reactivity depends on the type of alcohol (primary, secondary, or tertiary) and the reagent used.

Step 2: For (a) butan-1-ol (a primary alcohol): (1) With HCl and ZnCl₂ (Lucas reagent), the reaction proceeds via an SN2 mechanism, forming 1-chlorobutane. (2) With HBr, the reaction also proceeds via an SN2 mechanism, forming 1-bromobutane. (3) With PBr₃, the hydroxyl group is replaced by bromine, forming 1-bromobutane. (4) With P/I₂, iodine replaces the hydroxyl group, forming 1-iodobutane. (5) With SOCl₂, the hydroxyl group is replaced by chlorine, forming 1-chlorobutane.

Step 3: For (b) 2-methylbutan-2-ol (a tertiary alcohol): (1) With HCl and ZnCl₂, the reaction proceeds via an SN1 mechanism due to the stability of the tertiary carbocation, forming 2-chloro-2-methylbutane. (2) With HBr, the reaction also proceeds via an SN1 mechanism, forming 2-bromo-2-methylbutane. (3) With PBr₃, tertiary alcohols generally do not react efficiently with PBr₃ due to steric hindrance. (4) With P/I₂, tertiary alcohols generally do not react efficiently for the same reason. (5) With SOCl₂, the reaction forms 2-chloro-2-methylbutane.

Step 4: Write the chemical equations for each reaction to visualize the transformations. For example, for butan-1-ol with HCl and ZnCl₂: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} \). Similarly, write equations for all other reactions.

Step 5: Summarize the products for each reaction and note the differences in reactivity between primary and tertiary alcohols. Primary alcohols typically undergo SN2 reactions, while tertiary alcohols favor SN1 mechanisms due to carbocation stability. This understanding helps predict the products for each reagent.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Alcohol Reactivity

Alcohols are organic compounds containing a hydroxyl (-OH) group. Their reactivity is influenced by the structure of the alcohol (primary, secondary, or tertiary) and the type of reagents used. Different reagents can lead to various substitution or elimination reactions, resulting in different products. Understanding the reactivity of alcohols is crucial for predicting the outcomes of their reactions with acids and halogenating agents.

Nucleophilic Substitution Mechanisms

Nucleophilic substitution is a fundamental reaction mechanism in organic chemistry where a nucleophile replaces a leaving group in a molecule. The two main types are SN1 and SN2 mechanisms. SN1 involves a two-step process with a carbocation intermediate, while SN2 is a one-step process where the nucleophile attacks the substrate simultaneously as the leaving group departs. The choice of mechanism depends on the structure of the alcohol and the nature of the nucleophile.

Halogenation of Alcohols

Halogenation refers to the introduction of halogen atoms into organic compounds, often replacing the hydroxyl group in alcohols. Different reagents, such as HCl, HBr, PBr3, and SOCl2, can be used for this purpose, each yielding different halides. The choice of reagent affects the reaction pathway and the type of halide produced, which is essential for predicting the products when alcohols react with these halogenating agents.

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