Propene to 1-Iodopropane
To convert propene to 1-iodopropane, hydroiodic acid (HI) is used in an electrophilic addition reaction.
The double bond in propene reacts with HI. According to Markovnikov's rule, the hydrogen atom from HI attaches to the carbon with more hydrogen atoms, and the iodine attaches to the other carbon, forming the desired product.
\( \text{CH}_2=\text{CHCH}_3 + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{I} \)
Product: 1-iodopropane
Total number of nucleophiles from the following is: \(\text{NH}_3, PhSH, (H_3C_2S)_2, H_2C = CH_2, OH−, H_3O+, (CH_3)_2CO, NCH_3\)
Given below are two statements:
Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction.
Statement (II): In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the $ \beta $-carbon.
In the light of the above statements, choose the most appropriate answer from the options given below:
In the following substitution reaction: 