2. Write the appropriate resonance structure of the followings. $(2$ pts)
(a)
(b)
3. Place the necessary arrow(s) to convert the structures in the left to the ones in the right $(4 \mathrm{pts})$
(a)
(b)
Name $\qquad$
1. Answer the following miscellaneous questions:
I.D.
$\qquad$
$(4$ pts)
$$
\xrightarrow[\begin{array}{l}
\text { line-segment formula } \
\text { or stick-like diagram }
\end{array}]{\text { convert to }}
$$
b) Draw the most stable conformational structure for 2-methylpentane looking down at C2-C3
d) Indicate the numbers of mono-chlorinated products for:
e) What are the structures of $\mathrm{C}_{6} \mathrm{H}_{20}$ if its ozonolysis reaction in the presence of $\mathrm{Zn}$ and $\mathrm{HCl}$ produces $\mathrm{CH}_{3} \mathrm{C}(=\mathrm{O}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C}(=\mathrm{O}) \mathrm{CH}_{3}$

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Convert to line-segment formula**: To convert a skeletal structure into a line-segment or stick-like diagram, each vertex or end of a line represents a carbon atom, and the lines represent bonds between them. Hydrogen atoms are not shown but are implied to complete each carbon's four bonds. ‖ ‖ 2 ⋮ **Draw the most stable conformation for 2-methylpentane**: The most stable conformation of 2-methylpentane looking down the C2-C3 bond is the one that minimizes steric hindrance. This is typically a staggered conformation where the methyl group at C2 is as far away as possible from the other substituents. ‖ ‖ 3 ⋮ **Number of monochlorinated products**: To determine the number of possible monochlorinated products for a given alkane, identify the number of unique types of hydrogen atoms. Each unique type, when replaced by chlorine, will give a different monochlorinated product. ‖ ‖ 4 ⋮ **Ozonolysis products**: The ozonolysis of an alkene leads to the cleavage of the double bond and the formation of two carbonyl compounds. The given product suggests the original alkene had a double bond between the second and third carbon atoms in a six-carbon chain. ‖ ∻1 Answer∻ ⚹ The line-segment or stick-like diagram will show all the carbon atoms and the bonds between them, with implied hydrogens to satisfy the tetravalency of carbon. ⚹ ∻2 Answer∻ ⚹ The most stable conformation of 2-methylpentane is a staggered conformation with the methyl group at C2 anti to the largest group on C3. ⚹ ∻3 Answer∻ ⚹ The number of possible monochlorinated products depends on the number of unique types of hydrogen atoms in the alkane. ⚹ ∻4 Answer∻ ⚹ The structures resulting from the ozonolysis of $C_6H_{10}$ in the presence of Zn and HCl are two molecules of propanone, $CH_3C(=O)CH_3$. ⚹ ∻Key Concept∻ ⚹ Line-segment diagrams represent organic molecules with lines for bonds and vertices for carbon atoms, with hydrogen atoms being implicit. ⚹ ∻Explanation∻ ⚹ This representation simplifies the visualization of organic structures and is commonly used in organic chemistry. ⚹ ∻Key Concept∻ ⚹ The most stable conformation of an alkane minimizes steric hindrance between substituents. ⚹ ∻Explanation∻ ⚹ In 2-methylpentane, the staggered conformation with the methyl group anti to the largest substituent on the adjacent carbon is the most stable due to reduced steric clashes. ⚹ ∻Key Concept∻ ⚹ Unique types of hydrogen atoms in an alkane determine the number of monochlorinated products. ⚹ ∻Explanation∻ ⚹ Each type of hydrogen can be replaced by chlorine to give a distinct monochlorinated alkane, so identifying these types is key to determining the number of products. ⚹ ∻Key Concept∻ ⚹ Ozonolysis cleaves double bonds to form carbonyl compounds. ⚹ ∻Explanation∻ ⚹ The reaction splits the alkene into two parts, each becoming a carbonyl compound, which helps deduce the structure of the original alkene. ⚹◊What is the mechanism involved in the ozonolysis reaction of an alkene in the presence of Zn and HCl?⍭ Generate me a similar question◊

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