Draw all the isomers that have molecular formula C 5 H 11 Br. ( Hint: There are eight.) a. Give the systematic name for each of the isomers.

Hello everyone. Let's do this problem. It says draw all the cyclic isomers of C 4h seven B R and give the systematic names for each of them. So when we are given the molecular formula, and we want to draw all of the isomers, we first should calculate the degrees of uns saturation to determine if there are any characteristics such as rings, double bonds or triple bonds. And that will help us know what isomers we should be drawing. So this formula is two times the number of carbons plus two plus the number of nitrogens minus the number of hydrogens minus the number of halogens all over two. So, plugging in the values from our molecular formula, we have two times four, the number of carbons plus two plus zero, we have no nitrogens, right, minus seven hydrogens minus one halogen, right? The bromine is a halogen all over two. And once we crunch those numbers, we get a degree of uns saturation of one and one degree of uns saturation tells us that we have either one ring or one double bond. But our problem tells us that we are drawing cyclic isomers. So all of our isomers need to have one ring. So that is the only degree of uns saturation. We will have no double bonds or triple bonds or anything else like that. All right. So our molecular formula has four carbons. Let's start by drawing the smallest ring, we can, which would be three carbons. And then we can add the other carbon as a substituent group. So, starting with a three carbon ring, I can add a methyl group and I can add my bromine to a few different spots right? On the first isomer, let's put the bromine on the same carbon as that methyl group. I can also draw a three carbon ring and I'll have that methyl group on one carbon and then I can have a bromine, one carbon away from that methyl group. So on the carbon next door, I can also do one more three carbon ring where the bromine is attached to the end of a carbon substituent group. OK? And now let's go on to a four carbon ring, 1, 2, 3, 4, which is basically a square and I can only add the bromine to one position, right? Because no matter which carbon I add it to, it will be the same compound. Let me draw this a little more neatly on that third isomer. So now let's assign systematic names for all of these. So in our first three membered ring, the base is going to be cyclopropane, right? Prop means three carbons and listing our substituent groups in alphabetical order that will be bromine and then methyl and they are both on carbon number one. So one bromo, one methyl oops cyclo propane. All right. So one bromo, one methyl cyclo propane is our first cyclic isomer. Moving on to our next cyclic isomer, which is the three carbon ring with a methyl group and then a bromine next door. So which substituent group gets priority? A bromine or methyl, the bromine. So we'll have one bromo and that's listed first because we list our substituent in alphabetical order. Then we have two methyl and we'll have the same base name cyclopropane. All right, moving on to our last three membered carbon ring, we have one substituent group and it's a methyl group with a bromine on the end of the methyl group. So this is going to be carbon number one. It's the only substituent group on our ring. So we have one and then in parentheses, bromo, methyl and our base is still cyclopropane. OK. Our last cyclic isomer is the four carbon ring. So the base will be cyclo butane or the root name, I should say cyclo butane and we have one substituent which is the bromine. So we'll number that carbon one. So one, bromo, cyclo butane. So these are the four cyclic isomers, four C 4h seven B R. That's it for this problem. I hope this video was helpful and thank you 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

4. Alkanes and Cycloalkanes

Naming Alkyl Halides

Organic Chemistry

4. Alkanes and Cycloalkanes

Naming Alkyl Halides

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5:46 minutes

Problem 72a

Textbook Question

Draw all the isomers that have molecular formula C₅H₁₁Br. (Hint: There are eight.)
a. Give the systematic name for each of the isomers.

Verified step by step guidance

Step 1: Understand the molecular formula C5H11Br. This represents a compound with 5 carbon atoms, 11 hydrogen atoms, and 1 bromine atom. The presence of bromine indicates that these are alkyl bromides (haloalkanes). The task is to draw all possible structural isomers and name them systematically.

Step 2: Begin by considering the possible carbon chain arrangements. The parent chain can be a straight chain (pentane) or branched chains (butane or propane). For each chain, place the bromine atom at different positions to generate unique isomers.

Step 3: For the straight-chain isomer (pentane), place the bromine atom at each unique carbon position. This will result in 1-bromopentane and 2-bromopentane. Note that placing bromine on the third carbon is equivalent to the second position due to symmetry.

Step 4: For the branched-chain isomers, consider the butane backbone (a four-carbon chain with one methyl group as a branch). Place the bromine atom at each unique position on the chain. This will result in isomers such as 1-bromo-2-methylbutane, 2-bromo-2-methylbutane, and 2-bromo-3-methylbutane.

Step 5: Finally, consider the propane backbone (a three-carbon chain with two methyl groups as branches). Place the bromine atom at each unique position. This will result in isomers such as 1-bromo-2,2-dimethylpropane and 2-bromo-2-methylpropane. Verify that all eight isomers are unique and name them systematically following IUPAC nomenclature rules.

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Key Concepts

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

Isomerism

Isomerism refers to the phenomenon where compounds have the same molecular formula but different structural arrangements or spatial orientations. In organic chemistry, isomers can be classified into structural isomers, which differ in the connectivity of atoms, and stereoisomers, which have the same connectivity but differ in the arrangement of atoms in space. Understanding isomerism is crucial for identifying and naming the various forms of a compound.

Nomenclature

Nomenclature in organic chemistry is the system of naming chemical compounds based on specific rules set by the International Union of Pure and Applied Chemistry (IUPAC). Each isomer of a compound must be systematically named to reflect its structure, including the longest carbon chain, branching, and functional groups. Familiarity with nomenclature rules is essential for accurately identifying and communicating the identity of isomers.

Branched vs. Unbranched Alkanes

Branched and unbranched alkanes are types of hydrocarbons that differ in their carbon chain structure. Unbranched alkanes have a straight-chain structure, while branched alkanes contain one or more carbon branches off the main chain. The molecular formula C5H11Br can yield multiple isomers, including both branched and unbranched forms, which must be considered when drawing and naming the isomers.

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