Arrange the following alkyl radicals in terms of their relative stabilities: Methyl radical, Secondary…

Question

Arrange the following alkyl radicals in terms of their relative stabilities: Methyl radical, Secondary radical, Tertiary radical, Primary radical

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Here are the solutions to the questions from the image:

Question 1:
A carbocation is a species that contains a carbon atom bearing a positive charge.

\boxed{\text{Carbocation}}

Question 2:
The product of a Friedel-Crafts acylation reaction is an acylbenzene (or aryl ketone).

\boxed{\text{Acylbenzene}}

Question 3:
The Lewis structure for the tert-butyl radical (CH₃)₃C• is:

\begin{tikzpicture}[scale=0.8]
    \node (C) at (0,0) {C};
    \node (CH3_top) at (0,1) {CH$_3$};
    \node (CH3_left) at (-0.8,-0.5) {CH$_3$};
    \node (CH3_right) at (0.8,-0.5) {CH$_3$};
    \node (radical) at (0.3,0.3) {$\cdot$};

    \draw (C) -- (CH3_top);
    \draw (C) -- (CH3_left);
    \draw (C) -- (CH3_right);
\end{tikzpicture}

\boxed{\text{See diagram above}}

Question 4:

Heterolysis (or heterolytic cleavage) is the unsymmetrical breaking of a covalent bond where one atom retains both electrons from the bond, forming a cation and an anion.
Homolysis (or homolytic cleavage) is the symmetrical breaking of a covalent bond where each atom retains one electron from the bond, forming two radicals.

Question 5:
A trivalent carbon atom that bears a positive charge is a carbocation. (Assuming the question refers to a carbocation based on common organic chemistry contexts).

\boxed{\text{Positive charge}}

Question 6:
Markovnikov's Rule states that in the electrophilic addition of a protic acid (HX) to an alkene, the hydrogen atom adds to the carbon atom of the double bond that already has more hydrogen atoms, while the halogen (X) adds to the carbon atom with fewer hydrogen atoms.

Question 7:

a) CH₃CH₂CH(CH₃)₂ (2-methylbutane)

\begin{tikzpicture}[scale=0.8]
    \node (C1) at (0,0) {CH$_3$};
    \node (C2) at (1,0) {CH$_2$};
    \node (C3) at (2,0) {CH};
    \node (C4) at (3,0) {CH$_3$};
    \node (C5) at (2,-1) {CH$_3$};

    \draw (C1) -- (C2) -- (C3) -- (C4);
    \draw (C3) -- (C5);

    \node at (0.5,0.3) {1$^\circ$};
    \node at (1.5,0.3) {2$^\circ$};
    \node at (2.5,0.3) {3$^\circ$};
    \node at (2.5,-0.7) {1$^\circ$};
\end{tikzpicture}

CH₃ (C1) hydrogens: Primary (1°)
CH₂ (C2) hydrogens: Secondary (2°)
CH (C3) hydrogen: Tertiary (3°)
CH₃ (C4) hydrogens: Primary (1°)
CH₃ (C5) hydrogens: Primary (1°)

b) (CH₃)₃CCH₂CH₃ (2,2-dimethylbutane)

\begin{tikzpicture}[scale=0.8]
    \node (C1) at (0,0) {CH$_3$};
    \node (C2) at (1,0) {C};
    \node (C3) at (2,0) {CH$_2$};
    \node (C4) at (3,0) {CH$_3$};
    \node (C5) at (1,1) {CH$_3$};
    \node (C6) at (1,-1) {CH$_3$};

    \draw (C1) -- (C2) -- (C3) -- (C4);
    \draw (C2) -- (C5);
    \draw (C2) -- (C6);

    \node at (0.5,0.3) {1$^\circ$};
    \node at (1.5,0.3) {4$^\circ$}; % Quaternary carbon, no hydrogens
    \node at (2.5,0.3) {2$^\circ$};
    \node at (3.5,0.3) {1$^\circ$};
    \node at (1.5,1.3) {1$^\circ$};
    \node at (1.5,-1.3) {1$^\circ$};
\end{tikzpicture}

CH₃ (C1) hydrogens: Primary (1°)
C (C2) has no hydrogens (quaternary carbon)
CH₂ (C3) hydrogens: Secondary (2°)
CH₃ (C4) hydrogens: Primary (1°)
CH₃ (C5) hydrogens: Primary (1°)
CH₃ (C6) hydrogens: Primary (1°)

c) (CH₃)₂CHCH(CH₃)CH₂CH₃ (2,3,4-trimethylpentane)

\begin{tikzpicture}[scale=0.8]
    \node (C1) at (0,0) {CH$_3$};
    \node (C2) at (1,0) {CH};
    \node (C3) at (2,0) {CH};
    \node (C4) at (3,0) {CH$_2$};
    \node (C5) at (4,0) {CH$_3$};
    \node (C6) at (1,-1) {CH$_3$};
    \node (C7) at (2,1) {CH$_3$};

    \draw (C1) -- (C2) -- (C3) -- (C4) -- (C5);
    \draw (C2) -- (C6);
    \draw (C3) -- (C7);

    \node at (0.5,0.3) {1$^\circ$};
    \node at (1.5,0.3) {2$^\circ$}; % This is a secondary carbon, but the hydrogen is tertiary. Let's clarify.
    \node at (2.5,0.3) {3$^\circ$};
    \node at (3.5,0.3) {2$^\circ$};
    \node at (4.5,0.3) {1$^\circ$};
    \node at (1.5,-0.7) {1$^\circ$};
    \node at (2.5,1.3) {1$^\circ$};
\end{tikzpicture}

Let's re-evaluate the structure for (c) (CH₃)₂CHCH(CH₃)CH₂CH₃. This is 2,3-dimethylpentane.

\begin{tikzpicture}[scale=0.8]
    \node (C1) at (0,0) {CH$_3$};
    \node (C2) at (1,0) {CH};
    \node (C3) at (2,0) {CH};
    \node (C4) at (3,0) {CH$_2$};
    \node (C5) at (4,0) {CH$_3$};
    \node (C6) at (1,1) {CH$_3$}; % Methyl on C2
    \node (C7) at (2,-1) {CH$_3$}; % Methyl on C3

    \draw (C1) -- (C2) -- (C3) -- (C4) -- (C5);
    \draw (C2) -- (C6);
    \draw (C3) -- (C7);

    \node at (0.5,0.3) {1$^\circ$};
    \node at (1.5,0.3) {3$^\circ$}; % H on C2 is tertiary
    \node at (2.5,0.3) {3$^\circ$}; % H on C3 is tertiary
    \node at (3.5,0.3) {2$^\circ$};
    \node at (4.5,0.3) {1$^\circ$};
    \node at (1.5,1.3) {1$^\circ$};
    \node at (2.5,-0.7) {1$^\circ$};
\end{tikzpicture}

CH₃ (C1) hydrogens: Primary (1°)
CH (C2) hydrogen: Tertiary (3°)
CH₃ (C6) hydrogens: Primary (1°)
CH (C3) hydrogen: Tertiary (3°)
CH₃ (C7) hydrogens: Primary (1°)
CH₂ (C4) hydrogens: Secondary (2°)
CH₃ (C5) hydrogens: Primary (1°)

Question 8:
Reaction of methylcyclopentane with bromine under irradiation ( $h\nu$ ) is a free radical halogenation. Bromine is selective for the most substituted carbon. The methyl group is on a tertiary carbon in methylcyclopentane.
The product is 1-bromo-1-methylcyclopentane.

\begin{tikzpicture}[scale=0.8]
    % Methylcyclopentane
    \draw (0,0) -- (1,0) -- (1.5,0.866) -- (0.5,1.732) -- (-0.5,0.866) -- (0,0);
    \draw (0.5,1.732) -- (0.5,2.232) node[above] {CH$_3$};
    \node at (2.5,1) {+ Br$_2$ $\xrightarrow{h\nu}$};

    % 1-bromo-1-methylcyclopentane
    \begin{scope}[xshift=5cm]
        \draw (0,0) -- (1,0) -- (1.5,0.866) -- (0.5,1.732) -- (-0.5,0.866) -- (0,0);
        \draw (0.5,1.732) -- (0.5,2.232) node[above] {CH$_3$};
        \draw (0.5,1.732) -- (1,1.732) node[right] {Br};
    \end{scope}
\end{tikzpicture}

\boxed{\text{1-bromo-1-methylcyclopentane}}

Question 9:
To prepare cyclohexene from cyclohexane:

Free radical halogenation* of cyclohexane with bromine and light ( $h\nu$ ) to form bromocyclohexane.
\begin{tikzpicture}[scale=0.8] % Cyclohexane \draw (0,0) -- (1,0) -- (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); \node at (2,0.866) {+ Br$_2$ $\xrightarrow{h\nu}$}; % Bromocyclohexane \begin{scope}[xshift=4cm] \draw (0,0) -- (1,0) -- (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); \draw (0.5,1.732) -- (0.5,2.232) node[above] {Br}; \end{scope} \end{tikzpicture}
Dehydrohalogenation* (E2 elimination) of bromocyclohexane using a strong base (e.g., potassium hydroxide in ethanol) to form cyclohexene.
\begin{tikzpicture}[scale=0.8] % Bromocyclohexane \draw (0,0) -- (1,0) -- (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); \draw (0.5,1.732) -- (0.5,2.232) node[above] {Br}; \node at (2.5,0.866) {$\xrightarrow{\text{KOH, EtOH, heat}}$}; % Cyclohexene \begin{scope}[xshift=5cm] \draw (0,0) -- (1,0); \draw[double] (1,0) -- (1.5,0.866); \draw (1.5,0.866) -- (1,1.732) -- (0,1.732) -- (-0.5,0.866) -- (0,0); \end{scope} \end{tikzpicture}
Question 10:
A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond. Nucleophiles are typically electron-rich and can be neutral or negatively charged.

Question 11:
The relative stabilities of alkyl radicals from least stable to most stable are:
Methyl radical < Primary radical < Secondary radical < Tertiary radical
$\boxed{\text{Methyl radical < Primary radical < Secondary radical < Tertiary radical}}$
Question 12:
Assuming the question refers to (2S,3S)-2,3-dibromobutane, as "1,2-dibromobutane" with an (S) configuration is ambiguous for a compound with two chiral centers.
The Fischer projection for (2S,3S)-2,3-dibromobutane is:
\begin{tikzpicture}[scale=0.8] \node at (0,1.5) {CH$_3$}; \node at (0,0.5) {C}; \node at (0,-0.5) {C}; \node at (0,-1.5) {CH$_3$}; \draw (0,0.5) -- (0,-0.5); % Vertical bond \draw (-0.5,0.5) -- (0.5,0.5); % Horizontal bond for C2 \draw (-0.5,-0.5) -- (0.5,-0.5); % Horizontal bond for C3 \node at (-0.5,0.5) {H}; \node at (0.5,0.5) {Br}; % For (2S) \node at (-0.5,-0.5) {H}; \node at (0.5,-0.5) {Br}; % For (3S) \end{tikzpicture} $\boxed{\text{See diagram above for (2S,3S)-2,3-dibromobutane}}$
Question 13:
An asymmetric carbon (or chiral carbon) is a carbon atom that is bonded to four different groups.

Question 14:
The compound is CH₃-CH(Cl)-CH₂-CH₃, which is 2-chlorobutane.
Identify asymmetric centers.
The carbon at position 2 (CH(Cl)) is bonded to four different groups: -CH₃, -H, -Cl, and -CH₂CH₃. Therefore, it is an asymmetric (chiral) center.
The other carbons are not asymmetric.
Number of asymmetric centers = 1.
Calculate the number of stereoisomers.
For a molecule with $n$ chiral centers and no meso compounds, the number of stereoisomers is $2^n$ .
Number of stereoisomers = $2^1 = 2$ .
$\boxed{\text{1 asymmetric center, 2 stereoisomers}}$
Question 15:
The compound is 1,3-dimethylcyclohexane.
$\boxed{\text{1,3-dimethylcyclohexane}}$
Section B: Answer All Questions

Question 16 (a):
Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. They determine the chemical properties and reactivity of organic compounds.

Question 16 (b):
Grignard reagents (RMgX) react with carbonyl compounds (aldehydes and ketones) to form alcohols. The type of alcohol (primary, secondary, or tertiary) depends on the starting carbonyl compound:
- To synthesize a primary alcohol: React a Grignard reagent with formaldehyde (methanal, HCHO).
\begin{tikzpicture}[scale=0.8] \node (RMgX) at (0,0) {RMgX}; \node (HCHO) at (2,0) {HCHO}; \node at (1,0) {+}; \node at (3.5,0) {$\xrightarrow{\text{1. Ether, 2. H}_3\text{O}^+}$}; \node (RCH2OH) at (6,0) {RCH$_2$OH}; \end{tikzpicture}
Example: CH₃MgBr + HCHO $\rightarrow$ CH₃CH₂OH (ethanol)
- To synthesize a secondary alcohol: React a Grignard reagent with any aldehyde other than formaldehyde (R'CHO).
\begin{tikzpicture}[scale=0.8] \node (RMgX) at (0,0) {RMgX}; \node (R'CHO) at (2,0) {R'CHO}; \node at (1,0) {+}; \node at (3.5,0) {$\xrightarrow{\text{1. Ether, 2. H}_3\text{O}^+}$}; \node (RCHOHR') at (6,0) {RCH(OH)R'}; \end{tikzpicture}
Example: CH₃MgBr + CH₃CHO $\rightarrow$ CH₃CH(OH)CH₃ (2-propanol)
- To synthesize a tertiary alcohol: React a Grignard reagent with a ketone (R'COR'').
\begin{tikzpicture}[scale=0.8] \node (RMgX) at (0,0) {RMgX}; \node (R'COR'') at (2,0) {R'COR''}; \node at (1,0) {+}; \node at (3.5,0) {$\xrightarrow{\text{1. Ether, 2. H}_3\text{O}^+}$}; \node (RC(OH)R'R'') at (6,0) {RC(OH)R'R''}; \end{tikzpicture}
Example: CH₃MgBr + (CH₃)₂CO $\rightarrow$ (CH₃)₃COH (2-methyl-2-propanol)

Arrange the following alkyl radicals in terms of their relative stabilities: Methyl radical, Secondary radical, Tertiary radical, Primary radical

Step-by-step solution

Free radical halogenation* of cyclohexane with bromine and light ( $h\nu$ ) to form bromocyclohexane.

Dehydrohalogenation* (E2 elimination) of bromocyclohexane using a strong base (e.g., potassium hydroxide in ethanol) to form cyclohexene.

Identify asymmetric centers.

Calculate the number of stereoisomers.

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More Chemistry Questions

Arrange the following alkyl radicals in terms of their relative stabilities: Methyl radical, Secondary radical, Tertiary radical, Primary radical

Step-by-step solution

Free radical halogenation* of cyclohexane with bromine and light (hνh\nuhν) to form bromocyclohexane.

Dehydrohalogenation* (E2 elimination) of bromocyclohexane using a strong base (e.g., potassium hydroxide in ethanol) to form cyclohexene.

Identify asymmetric centers.

Calculate the number of stereoisomers.

Still stuck on this one?

More Chemistry Questions

Free radical halogenation* of cyclohexane with bromine and light ( $h\nu$ ) to form bromocyclohexane.