Question

Write equations for the reaction of I-hatanol with each reagent. Where you predict no reaction, write \(N R\). (a) Na metal (b) \(\mathrm{HBr}\), heat (c) HI, heat (d) \(\mathrm{PBr}_{\mathrm{a}}\) (e) \(\mathrm{SOCl}_{2}\), proridine (f) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{H}_{2} \mathrm{O}\), heat (g) \(\mathrm{HIO}_{4}\) (h) PCC (i) \(\mathrm{CH}_{4} \mathrm{SO} \mathrm{C}\), pyridine

Short answer

Question: List the products formed by the reaction of I-hatanol with the following reagents: (a) Na metal, (b) HBr and heat, (c) HI and heat, (d) PBr3, (e) SOCl2 and pyridine, (f) K2Cr2O7, H2SO4, H2O, and heat, (g) HIO4, (h) PCC, and (i) CH3SOCl and pyridine. Answer: (a) I-Hatanol-O^{-}Na^{+} and NaI (b) I-Hatanol-Br and H2O (c) I-Hatanol-I and H2O (d) I-Hatanol-Br, HBr, and waste products (e) I-Hatanol-Cl, SO2, and HCl (f) I-Hatanol-COOH, K2SO4, and Cr2(SO4)3 (g) No reaction (h) I-Hatanone and PCC waste products (i) I-Hatanol-OCH3, SO2, and HCl

Step-by-step solution

(a) Reaction with Na metal

Sodium metal is a strong reducing agent. When it reacts with I-hatanol, it will act like a reducing agent and give electrons to the hydroxyl group of I-hatanol. This will result in the formation of a sodium alkoxide ion and a sodium iodide salt. The reaction is as follows: $$\ce{I-Hatanol + Na -> I-Hatanol-O^{-}Na^{+} + NaI}$$

(b) Reaction with HBr and heat

HBr is a strong acid, and when it is heated, it can react with the hydroxyl group of I-hatanol, resulting in the formation of an alkyl halide and water. The reaction equation is: $$\ce{I-Hatanol + HBr -> I-Hatanol-Br + H2O}$$

(c) Reaction with HI and heat

HI, similar to HBr, is a strong acid that can react with the hydroxyl group of I-hatanol under heat. This reaction results in the formation of an alkyl iodide and water. The equation for this reaction is: $$\ce{I-Hatanol + HI -> I-Hatanol-I + H2O}$$

(d) Reaction with PBr3

PBr3 is a halogenation reagent and reacts with the –OH group in I-hatanol, forming an alkyl bromide. The equation for this reaction is: $$\ce{I-Hatanol + PBr3 -> I-Hatanol-Br + HBr+Waste}$$

(e) Reaction with SOCl2 and pyridine

SOCl2 is a chlorination reagent and reacts with the hydroxyl group of I-hatanol in the presence of a base like pyridine, forming an alkyl chloride. The equation for this reaction is: $$\ce{I-Hatanol + SOCl2 -> I-Hatanol-Cl + SO2 + HCl}$$

(f) Reaction with K2Cr2O7, H2SO4, H2O, and heat

This set of reagents is a strong oxidizing agent. In the presence of heat, it will oxidize primary I-hatanol to a carboxylic acid. The equation for this reaction is: $$\ce{I-Hatanol + K2Cr2O7 + H2SO4 + H2O -> I-Hatanol-COOH + K2SO4 + Cr2(SO4)3}$$

(g) Reaction with HIO4

HIO4 is a strong oxidizing agent that cleaves diols (molecules with adjacent alcohol functional groups). As there is no diol in the I-hatanol structure, we predict no reaction. Thus: $$\ce{I-Hatanol + HIO4 -> NR}$$

(h) Reaction with PCC

Pyridinium chlorochromate (PCC) is a mild oxidant that selectively oxidizes primary alcohols to aldehydes. The reaction equation is: $$\ce{I-Hatanol + PCC -> I-Hatanone + PCC-waste}$$

(i) Reaction with CH3SOCl and pyridine

CH3SOCl is a methylation reagent that reacts with hydroxyl groups in the presence of pyridine. It will react with I-hatanol and give a methyl ether. The reaction equation is: $$\ce{I-Hatanol + CH3SOCl -> I-Hatanol-OCH3 + SO2 + HCl}$$

Key concepts

Reduction reaction

Reduction reactions are fundamental in organic chemistry, involving the gain of electrons or hydrogen atoms by a molecule. In the case of I-hatanol, reduction occurs when it reacts with sodium metal. This reaction generates a sodium alkoxide ion as the alcohol accepts an electron from sodium.
This forms a new bond, reducing the original molecule by converting the hydroxyl group ( abla) into one charged with a negative ion.

• Involve transfer of electrons
• Change functional groups
• Produce ions or simplified molecules

Reduction is paired with oxidation, where one substance loses electrons, and another gains them. Understanding this helps predict reactions, balancing electron shifts during organic transformations.

Oxidation reaction

Oxidation reactions are characterized by the loss of electrons or an increase in oxygen bonds in a molecule. I-hatanol undergoes oxidation with reagents like chromium trioxide derivatives and PCC. These oxidizing agents convert primary alcohols like I-hatanol into different carbonyl-containing compounds such as aldehydes or carboxylic acids, depending on the strength of the oxidant used.

• Involve addition of oxygen or removal of hydrogen
• Lead to a change in molecular structure
• Essential for transforming alcohols

Using pyridinium chlorochromate (PCC) allows for controlled oxidation. Understanding oxidation is crucial for synthesizing desired organic compounds efficiently.

Halogenation reaction

Halogenation reactions involve replacing a hydrogen atom with a halogen atom (F, Cl, Br, I) in organic compounds. I-hatanol reacts with reagents like HBr, HI, and PBr3 during halogenation. These reactions typically involve converting alcohols into alkyl halides through the substitution of the hydroxyl group with a halogen.

• Reactions with halogens
• Substitute heteroatoms
• Create reactive sites on organic molecules

Understanding how halogenation alters molecular composition provides insight into forming more reactive or stable compounds, foundational in synthetic organic pathways.

Esterification

Esterification is an important organic reaction where an alcohol and an acid form an ester and water. Although not directly part of the I-hatanol reactions discussed, understanding esterification is key in organic synthesis. Esters are valuable for their pleasant odors and flavors, often used in food and fragrance industries.

• Produces esters from acids and alcohols
• Often involves acid catalysis
• Releases water as a byproduct

Mastering esterification aids in synthesizing various organic esters, broadening the applications and functionality of simple organic molecules.

Alcohol reactions

Alcohol reactions are fundamental in organic chemistry, involving changes to the hydroxyl (-OH) group. I-hatanol, as an alcohol, provides an excellent example through its diverse reactions. Whether engaging in oxidation, reduction, or halogenation, its reactivity illustrates how alcohols can undergo transformation.

• Transform alcohols through various reactions
• Include oxidation, esterification, and substitution
• Provide pathways to different functional groups

Through identifying the type of reaction, predictability of alcohol transformation supports the synthesis of complex organic molecules. Mastering alcohol reactions is crucial for students aiming to develop versatile organic chemistry knowledge.