Organic Chemistry - Functional group chemistry

Determine the products formed when propan-1-ol is heated under reflux with an excess of acidified $K_2Cr_2O_7$. Provide the balanced chemical equation.

$CH_3CH_2CH_2OH + 2[O] \xrightarrow{H^+/Cr_2O_7^{2-}, \text{reflux}} CH_3CH_2COOH + H_2O$

Propan-1-ol is a primary alcohol. Under reflux with excess oxidizing agent, it is fully oxidized from an aldehyde (propanal) to a carboxylic acid (propanoic acid). The color change observed is from orange ($Cr_2O_7^{2-}$) to green ($Cr^{3+}$).

Describe the mechanism of the reaction between chloromethane ($CH_3Cl$) and aqueous sodium hydroxide ($NaOH$).

$CH_3Cl + OH^- \rightarrow [HO\cdots CH_3 \cdots Cl]^{\ddagger} \rightarrow CH_3OH + Cl^-$

This follows an $S_N2$ mechanism because chloromethane is a primary halogenoalkane. The hydroxide ion ($OH^-$) acts as a nucleophile and attacks the electron-deficient carbon from the backside, leading to a transition state and inversion of configuration.

Predict the major product of the addition of $HBr$ to propene ($CH_3CH=CH_2$).

$CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3$

According to Markovnikov's rule, the hydrogen atom from $HBr$ attaches to the carbon atom with the greater number of hydrogen atoms already attached ($C_1$), while the $Br$ attaches to the more substituted carbon ($C_2$). This results in $2$-bromopropane as the major product because it proceeds via a more stable secondary carbocation.