Draw the condensed structure of a compound that contains only carbon and hydrogen atoms and that has a. three sp 3 hybridized carbons. b. one sp 3 hybridized carbon and two sp 2 hybridized carbons. c. two sp 3 hybridized carbons and two sp hybridized carbons.

Hey, everyone, let's do this problem. It says draw the appropriate condensed structure of a hydrocarbon that contains part 12 carbons with SP three hybridization, part 22 carbons with SP three hybridization and two carbons with SP two hybridization, part 33 carbons with SP three hybridization and two carbons with SP hybridization. So we're looking at the different types of hybridization and we basically have to know how many bonds are going to be bonded to that type of carbon with a certain type of hybridization, right? Because if we want to determine the appropriate condensed structure, we need to know how many hydrogen each carbon should have. So looking at the different types of hybridization, we know that SP three hybridized atoms Have four charge centers. SP two hybridized atoms have three charge centers and S P Adams have to charge centers. So remember we're adding up the exponents to get the total number of charge centers, right? Three plus one is 42 plus one is 31 plus one is two. Now what is a charge center? Let's curly bracket. This a charge center is going to include a lone pair or any type of bond. So one lone pair on an atom, one charge center, one single bond is one charge center, one double bond is still one charge center. Okay. So it doesn't matter whether it's a single double or triple bond that still just makes it one charge center. So the steps for this problem, we, like we said, we need to know how many atoms are bound into each type of carbon. So it's going to be easiest if we first draw the Lewis structure for each compound and then we'll get the condensed formula from that. Okay. So let's draw our first compound Which is two carbons. Well, right, two carbons with SP three hybridization. So sp three hybridization means that those carbon should have four charge centers. Alright. Well, they're bonded to each other. So that means they each need three more bonds. So we can simply add three more hydrogen onto each carbon. So putting it in the condensed formula, we have two carbons each with three hydrogen. So we list CH three ch three And that is the condensed formula for compound one, not so bad, right? Alright. Compound to is two carbons with SP three hybridization and two carbons with SP two hybridization. So we have four carbons, let's connect our four carbons, 1234 and it said two with SP three hybridization and two carbons with SP two hybridization. So sp two hybridization means we're going to have a double bond right, two carbons with sp two hybridization means we'll have one double bond between two carbons. So let's make that the center two carbons, why not? And then the outside carbons on the ends will be sp three hybridized. So adding our hydrogen, the end carbons will have four single bonds, three hydrogen, three ch bonds and one cc bond. And then our inner two carbons will have one single C C bond, one double C C bond and one hydrogen C C H bond, right C H bond on each carbon. So for our condensed formula, we have four carbons From left to right, we have CH three CH double bonded to CH. So we write out that double bond and CH three. Alright, last compound, we have three carbons with sp three hybridization and two carbons with sp hybridization. So we have five carbons and two carbons have sp hybridization. So I know that's going to be a triple bond, right? Two carbons with sp hybridization will be one triple bond. So there are multiple structures we can do for this, right. Let's just make our triple bond the end between the two and between the 1st and 2nd carbon and then the rest will be single bonds. So with that information, we can add our hydrogen, our fifth carbon will get three hydrogen is fourth carbon will get to hydra Jen's third carbon will get to hydra Jen's second carbon already has full octet with these two charge centers. And our first carbon will get one hydrogen are condensed formula will be reading from left to right. We have our main carbon chain of five carbons. So ch three ch two ch two C triple bond C H. So those are the condensed formulas for these hydrocarbons, which makes a the correct answer for this problem. That's it for this one. I hope this video is helpful and thanks for watching.

Table of contents

Skip topic navigation

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

1. A Review of General Chemistry

Condensed Structural Formula

Organic Chemistry

1. A Review of General Chemistry

Condensed Structural Formula

Previous problemNext problem

6:53 minutes

Problem 52

Textbook Question

Draw the condensed structure of a compound that contains only carbon and hydrogen atoms and that has
a. three sp³ hybridized carbons.
b. one sp³ hybridized carbon and two sp² hybridized carbons.
c. two sp³ hybridized carbons and two sp hybridized carbons.

Verified step by step guidance

Step 1: Understand the hybridization of carbon atoms. Carbon atoms can be sp3, sp2, or sp hybridized depending on the number of sigma bonds and lone pairs around them. sp3 hybridized carbons form four sigma bonds, sp2 hybridized carbons form three sigma bonds and one pi bond, and sp hybridized carbons form two sigma bonds and two pi bonds.

Step 2: For part (a), identify a structure with three sp3 hybridized carbons. Each sp3 hybridized carbon must form four sigma bonds. A simple example is propane (C3H8), where all three carbons are sp3 hybridized. Draw the condensed structure by connecting the carbons with single bonds and adding enough hydrogens to satisfy the valency of each carbon.

Step 3: For part (b), identify a structure with one sp3 hybridized carbon and two sp2 hybridized carbons. The sp2 hybridized carbons must form a double bond between them, while the sp3 hybridized carbon forms single bonds. A simple example is propene (C3H6). Draw the condensed structure by placing a double bond between two carbons and adding hydrogens to satisfy the valency.

Step 4: For part (c), identify a structure with two sp3 hybridized carbons and two sp hybridized carbons. The sp hybridized carbons must form a triple bond between them, while the sp3 hybridized carbons form single bonds. A simple example is butyne (C4H6). Draw the condensed structure by placing a triple bond between two carbons and connecting the sp3 hybridized carbons with single bonds, adding hydrogens to satisfy the valency.

Step 5: Verify each structure by ensuring that the total number of bonds around each carbon matches its hybridization state and that the molecular formula matches the description provided in the problem.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

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

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate the bonding requirements of carbon atoms. In organic chemistry, sp3, sp2, and sp hybridizations correspond to different types of bonding: sp3 involves single bonds, sp2 involves one double bond, and sp involves triple bonds. Understanding hybridization is crucial for predicting the geometry and reactivity of organic compounds.

Condensed Structural Formula

A condensed structural formula is a way of representing a chemical compound that shows the arrangement of atoms without depicting all the bonds explicitly. It provides a simplified view of the molecule, indicating how atoms are connected. This format is particularly useful for visualizing the structure of organic compounds with multiple carbon atoms and functional groups.

Alkane, Alkene, and Alkyne Structures

Alkanes, alkenes, and alkynes are classes of hydrocarbons characterized by their bonding types. Alkanes contain only single bonds (sp3), alkenes have at least one double bond (sp2), and alkynes have at least one triple bond (sp). Recognizing these structures is essential for drawing the correct condensed structures based on the specified hybridization of carbon atoms in the question.

Watch next

Master How to interpret condensed structures. with a bite sized video explanation from Johnny

Start learning

Draw the condensed structure of a compound that contains only carbon and hydrogen atoms and that has a. three sp3 hybridized carbons. b. one sp3 hybridized carbon and two sp2 hybridized carbons. c. two sp3 hybridized carbons and two sp hybridized carbons.

Key Concepts

Hybridization

Condensed Structural Formula

Alkane, Alkene, and Alkyne Structures

Watch next

Draw the condensed structure of a compound that contains only carbon and hydrogen atoms and that has
a. three sp³ hybridized carbons.
b. one sp³ hybridized carbon and two sp² hybridized carbons.
c. two sp³ hybridized carbons and two sp hybridized carbons.