there's a few questions I have...actually it's a bit more than a few...

1. Identify the carbon atom(s) in the structure shown that has (have) each of the following hybridizations: a. sp3, b. sp, c. sp2
N*triplebond*C-CH2-CH=CH-CHOH
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CH=O
|
H
2. Draw all the possible noncyclic structural isomers of C5H10. Name each compound. (I don't understand the cyclic part at all...)

3. What are the approximate bond angles a. about a double bonded carbon atom in an alkene, b. about a triply bonded carbon atom in an alkyne?

4. What are the characteristic hybrd orbitals employed by a. carbon in an alkane, b. carbon in a double bond in an alkene, c. carbon in the benzene ring, d. carbon in a triple bond in an alkyne? --I know what the hybrid orbitals are but I don't know how to figure out how to know what the orbitals are...

5. Write a balanced chemical equation for the substitution reaction of CL2 with para-dichlorobenzene in the presence of FeCl3 as a catalyst.

6. Write a series of reactions leading to para-bromoethylbenzene, beginning with benzene ad using other reagents as needed. What isomeric side products might also be formed?

7. Draw all the distinct geometric isomers of 2,4-hexadeine.

8. Describe the intermediate that is thought to form in the addition of a hydrogen halide to an alkene, using cyclo-hexene as the alkene in your description.

9. The rate law for addition of Br2 to an alkene is first order in Br2 and first order in the alkene. Does this fact prve that the mechanism of addition of Br2 to an alkene proceeds in the same manner as for addition of HBr? Explain.

10. Write a balanced chemical equation using condensed structural formulas for the saponification(base hydrolysis) of a. methyl propionate, b. phenyl acetate

11. Write the condensed structural formula for each o the following compounds: a. 2-butanol, b. 1,2-ethanediol, c. methyl formate, d. diethyl ketone, e. diethyl ether

12. (same question as above): a. 3,3-dichlorobutyraldehyde, b. methyl phenyl ketone, c. para-bromobenzoic acd, d. methyl-trans-2-butenyl ether, e. N, N-diamethyl-benzamide

13. Does 3-chloro-3-methylhexane have optical isomers? Why or why not?

that's sooo many questions...but...this stuff is just HARD and without a teacher it's really difficult....:(

The topics of the questions you are asking are not covered adequately even in a college General Chemistry textbook. An Organic Chemistry textbook would be more helpful for most of the 13 questions.

A good starting point for organic chemistry is the online textbook, General Organic and Biochemistry by James K. Hardy here:
http://ull.chemistry.uakron.edu/genobc/
Start with the chapter on Alkanes.
Below are comments or partial answers with no explanations. Even that took far too long. We merely assist here. We do not do in depth tutoring, teaching, or providing answers. We assume you already know the basics in the topics of your questions.
The answers to #1, #3, and #4 you should be able to find on your own in almost any chemistry textbook.
#2. The open chain isomers of C5H10 are
1-pentene, cis-2-pentene, trans-2-pentene, 2-methyl-1-butene, 2-methyl-2-butene. We can't draw these here.
#5 and #6 involve substitution reactions on the benzene ring. That is a topic you need to study after studying substitution reactions of alkanes.
#7. The answer is cis-cis-, cis-trans-, and trans-trans- isomers of 2,4-haxadiene. It would take too long to explain. You can start by studying cis- and trans- isomerism.
#8. Possibly a carbonium ion with the X- ion linking to the positive carbon atom. See my next comment.
#9. You need to study reaction mechanisms for different types of addition reactions with an alkene. This is a fairly advanced topic even for an Organic Chemistry class.
#10. This has to do with the reaction of an ester with a strong base, such as NaOH. It is also called hydrolysis of an ester.
#11. CH3CH(OH)CH2CH3, CH2(OH)CH2(OH), HCOOCH3, CH3C(CO)CH3, CH3CH2-O-CH2CH3. These can be drawn better with the functional group off the carbon chain.
#12. (Similar to #11)
#13. Yes (four different atoms or groups attached to carbon #3)

12

I understand that you have multiple questions regarding organic chemistry. I will do my best to help you understand and answer each question one by one. Let's start with question 1.

1. To identify the carbon atom(s) with specific hybridizations in the given structure, we need to determine the number of sigma bonds and lone pairs around each carbon atom.

a. sp3 hybridization: In the given structure, the carbon atom bonded to four other atoms (hydrogens or carbon) is sp3 hybridized. In this case, the carbon atom labeled with an asterisk (*) would have sp3 hybridization.

b. sp hybridization: In the structure, the carbon atom triple-bonded to nitrogen is sp hybridized. So, the carbon atom in the triple bond would have sp hybridization.

c. sp2 hybridization: In the structure, the carbon atom in the double bond with oxygen is sp2 hybridized. Therefore, the carbon atom in the double bond would have sp2 hybridization.

Now, let's move on to question 2.

2. Drawing noncyclic structural isomers of C5H10: To generate structural isomers, you need to rearrange the atoms while keeping the same molecular formula.

In this case, the molecular formula is C5H10. It means there are five carbon atoms and ten hydrogen atoms in the molecule.

Possible noncyclic structural isomers of C5H10:
a. Pentane (n-pentane)
b. Isopentane (Methylbutane)
c. Neopentane (Dimethylpropane)

These are the three possible noncyclic structural isomers of C5H10. Naming conventions can be found in organic chemistry textbooks or online resources.

Now, let's move on to question 3.

3. Approximate bond angles:
a. Double bonded carbon atom in an alkene: The approximate bond angle around a double bonded carbon atom in an alkene is 120 degrees. The bond angle is slightly larger than the ideal sp2 hybridized angle of 120 degrees due to slight repulsion between the pi bonds.

b. Triply bonded carbon atom in an alkyne: The approximate bond angle around a triply bonded carbon atom in an alkyne is 180 degrees. Alkynes have linear geometry because the triple bond consists of one sigma bond and two pi bonds, resulting in a straight line shape.

Now, let's move on to question 4.

4. Characteristic hybrid orbitals:
a. Carbon in an alkane: In an alkane, carbon atoms are sp3 hybridized. Each carbon is bonded to four other atoms or groups with approximately tetrahedral geometry.

b. Carbon in a double bond in an alkene: In an alkene, the carbon atoms involved in the double bond are sp2 hybridized. Each carbon is bonded to three other atoms or groups with approximately trigonal planar geometry.

c. Carbon in the benzene ring: Each carbon atom in a benzene ring is sp2 hybridized. The delocalized pi bonds in benzene result in a planar hexagonal shape.

d. Carbon in a triple bond in an alkyne: In an alkyne, the carbon atoms involved in the triple bond are sp hybridized. Each carbon is bonded to two other atoms or groups with a linear geometry.

These hybridizations are determined based on the number of sigma bonds and lone pairs around each carbon atom in the molecule.

Now, let's move on to question 5.

5. Balanced chemical equation for the substitution reaction of Cl2 with para-dichlorobenzene:
The reaction is described as the substitution of chlorine (Cl2) with para-dichlorobenzene (C6H4Cl2). The presence of FeCl3 acts as a catalyst.

C6H4Cl2 + Cl2 → C6H4Cl3 + HCl

This balanced equation represents the substitution reaction taking place, where one chlorine atom is substituted by another chlorine atom, forming para-trichlorobenzene (C6H4Cl3) and hydrochloric acid (HCl).

If you have any further questions or need more explanations, please let me know.