Reactions of Alcohols

Answer: The suffix -ol is added to the name of the molecule when an -OH functional group is the highest priority group.

Answer: A secondary alcohol is defined as an alcohol with the -OH group attached to a carbon atom that has two alkyl groups connected to it.

Ethanol and Water Hydrogen Bonding
(Hydrogen bonding representation; actual diagram to be inserted)

Alcohols can undergo various reactions. There are two main types of reactions:
- Substitution Reactions: In these reactions, the hydroxyl group (-OH) is replaced by a different functional group.
- Elimination Reactions: In these reactions, the alcohol loses a molecule of water, resulting in the formation of an alkene.

Alcohols react with compounds that contain halide ions (e.g., sodium bromide, NaBr) during substitution reactions. This results in the hydroxyl (-OH) group being replaced by a halide, thus converting the alcohol into a haloalkane.
An acid catalyst, such as concentrated sulfuric acid (H₂SO₄), is required for the reaction to proceed.
General Equation:
$R-OH + H-X <br /> ightarrow R-X + H₂O$
(where R is the rest of the molecule and X is a halide ion)

To synthesize 2-bromo-2-methylpropane: Shake 2-methylpropan-2-ol (a tertiary alcohol) with sodium bromide and concentrated sulfuric acid at room temperature.

Alkene compounds are valuable chemicals used as precursors for various organic products, including polymers.
Elimination reactions can be employed to convert alcohols into alkenes by removing water.
In this case, the alcohol is combined with an acid catalyst (H₂SO₄ or H₃PO₄) and heated.
This process often referred to as a dehydration reaction, leads to the elimination of water from the alcohol.
General Reaction for Dehydration of Alcohols:
$R-OH <br /> ightarrow R=C + H₂O$

Synthesis of Ethene: Ethanol can be converted into ethene through dehydration by mixing it with an acid catalyst and heating it to 170°C.

The eliminated water molecule in an elimination reaction comprises the hydroxyl (-OH) group and a hydrogen atom that was bonded to a carbon atom next to the hydroxyl carbon.
Consequently, elimination can lead to the formation of two different alkene products, depending on which side of the hydroxyl group the hydrogen is eliminated from.
Additionally, E/Z isomerism is possible if the formed alkenes are capable of representing geometrical isomers.

When butan-2-ol is heated with concentrated phosphoric acid, elimination can take place between the hydroxyl group and a hydrogen atom on either carbon-1 or carbon-3. This results in a mixture of alkene products: but-1-ene, E-but-2-ene, and Z-but-2-ene.

Elimination between the Hydroxyl Group and Hydrogen on Carbon-1:
- Results in the formation of but-1-ene:
  
  H
  H
  I
  H-C-C-C-C-H \
  H-OH\ \
  acid \
  catalyst \
  heat \
  H-H-H\ \
  H-C-C=C-C \
Elimination between the Hydroxyl Group and Hydrogen on Carbon-3:
- Results in the formation of but-2-ene, which can form E/Z isomers. A mixture of E-but-2-ene and Z-but-2-ene will be produced:
  
  H
  H
  H-C-C-C-C-H \
  H-OH \ \
  acid \
  catalyst \
  heat \
  H=C \
  H-H\ \

Answer: Use 2-methylpropan-2-ol and sodium chloride (NaCl) as reagents under appropriate conditions.

Products:
- E-4-methyl-hex-2-ene:
  
  H\
  H{12}C \ C=C \ CH(CH)CH2CH_3 \
- Z-4-methyl-hex-2-ene:
  
  H\
  H{12}C \ C=C \ H{12}C \
  CH(CH{2})CH{2}CH_{3} \
- E-3-methyl-hex-3-ene:
  
  H{2}C & \ H \ C=C & \ H{2}C \
  CH{2}CH{3} \
- Z-3-methyl-hex-3-ene:
  
  H{2}C & \ H \ C=C & \ CH{2}CH{3} \ amespaceH{12}C
To determine the products of the elimination reaction of an alcohol, first identify what alkenes form when the hydrogen atoms on either side of the hydroxyl group are excluded. Additionally, check if either produced alkene can give rise to E/Z isomers.